Klein-Gordon equation from harmonic oscillator: Hamiltonian, creation and annihilation operators

Michael E. Peskin & Daniel V. Schroeder, An Introduction to Quantum Field Theory, (Perseus Books, 1995) – Chapter 2.

Continuing our discussion of the derivation of the Klein-Gordon field by analogy with the harmonic oscillator, we arrived at the field and its conjugate momentum density:

 $\displaystyle \phi\left(\mathbf{x},t\right)$ $\displaystyle =$ $\displaystyle \frac{1}{\left(2\pi\right)^{3}}\int d^{3}p\;e^{i\mathbf{p\cdot x}}\frac{1}{\sqrt{2\omega_{\mathbf{p}}}}\left(a_{\mathbf{p}}+a_{-\mathbf{p}}^{\dagger}\right)\ \ \ \ \ (1)$ $\displaystyle \pi\left(\mathbf{x},t\right)$ $\displaystyle =$ $\displaystyle \frac{-i}{\left(2\pi\right)^{3}}\int d^{3}p\;e^{i\mathbf{p\cdot x}}\sqrt{\frac{\omega_{\mathbf{p}}}{2}}\left(a_{\mathbf{p}}-a_{-\mathbf{p}}^{\dagger}\right) \ \ \ \ \ (2)$

The operators ${a_{\mathbf{p}}}$ and ${a_{\mathbf{p}}^{\dagger}}$ are analogues of the raising and lowering operators ${a}$ and ${a^{\dagger}}$ in the harmonic oscillator, with the extra condition that we have one pair of operators for each momentum ${\mathbf{p}}$. In the harmonic oscillator, the operators satisfied the commutation relation

$\displaystyle \left[a,a^{\dagger}\right]=1 \ \ \ \ \ (3)$

When applied to the Klein-Gordon field, we assume that operators with different momenta commute, but those with the same momenta do not. The assumption is that

 $\displaystyle \left[a_{\mathbf{p}},a_{\mathbf{p}^{\prime}}^{\dagger}\right]$ $\displaystyle =$ $\displaystyle \left(2\pi\right)^{3}\delta^{\left(3\right)}\left(\mathbf{p}-\mathbf{p}^{\prime}\right)\ \ \ \ \ (4)$ $\displaystyle \left[a_{\mathbf{p}},a_{\mathbf{p}^{\prime}}\right]$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (5)$ $\displaystyle \left[a_{\mathbf{p}}^{\dagger},a_{\mathbf{p}^{\prime}}^{\dagger}\right]$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (6)$

[These equations differ from those in Klauber’s book as discussed earlier in that the factors of ${2\pi}$ turn up in different places. However, the final result for the commutator ${\left[\phi\left(\mathbf{x},t\right),\pi\left(\mathbf{x}^{\prime},t\right)\right]}$ is the same, which is what matters.]

From here, we can work out the commutator ${\left[\phi\left(\mathbf{x},t\right),\pi\left(\mathbf{x}^{\prime},t\right)\right]}$ by plugging the operator commutators into the integrals for ${\phi}$ and ${\pi}$. This is similar to the derivation we did earlier, except here we’re assuming the commutators for ${a_{\mathbf{p}}}$ and ${a_{\mathbf{p}}^{\dagger}}$ are as given above, rather than deriving them from the assumed commutator ${\left[\phi\left(\mathbf{x},t\right),\pi\left(\mathbf{x}^{\prime},t\right)\right]}$. This calculation proves to be somewhat easier than the previous one, in that we can throw away all integrals containing ${\left[a_{\mathbf{p}},a_{\mathbf{p}^{\prime}}\right]}$ and ${\left[a_{\mathbf{p}}^{\dagger},a_{\mathbf{p}^{\prime}}^{\dagger}\right]}$. We get

 $\displaystyle \left[\phi\left(\mathbf{x},t\right),\pi\left(\mathbf{x}^{\prime},t\right)\right]$ $\displaystyle =$ $\displaystyle \frac{-i}{2\left(2\pi\right)^{6}}\int d^{3}pd^{3}p^{\prime}\sqrt{\frac{\omega_{\mathbf{p}^{\prime}}}{\omega_{\mathbf{p}}}}\left(\left[a_{-\mathbf{p}}^{\dagger},a_{\mathbf{p}^{\prime}}\right]-\left[a_{\mathbf{p}},a_{-\mathbf{p}^{\prime}}^{\dagger}\right]\right)e^{i\left(\mathbf{p}\cdot\mathbf{x}+\mathbf{p}^{\prime}\cdot\mathbf{x}^{\prime}\right)}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{-i\left(2\pi\right)^{3}}{2\left(2\pi\right)^{6}}\int d^{3}pd^{3}p^{\prime}\sqrt{\frac{\omega_{\mathbf{p}^{\prime}}}{\omega_{\mathbf{p}}}}\left(-\delta^{\left(3\right)}\left(\mathbf{p}^{\prime}+\mathbf{p}\right)-\delta^{\left(3\right)}\left(\mathbf{p}^{\prime}+\mathbf{p}\right)\right)e^{i\left(\mathbf{p}\cdot\mathbf{x}+\mathbf{p}^{\prime}\cdot\mathbf{x}^{\prime}\right)}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{i}{\left(2\pi\right)^{3}}\int d^{3}pe^{i\mathbf{p}\cdot\left(\mathbf{x}-\mathbf{x}^{\prime}\right)}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\delta^{\left(3\right)}\left(\mathbf{x}-\mathbf{x}^{\prime}\right) \ \ \ \ \ (10)$

This is the same result we got earlier using Klauber’s method. Remember that we’re still taking the field ${\phi}$ to be a real field.

P&S now derive the total Hamiltonian in their equation 2.31. The technique is very similar to that used by Klauber. The differences are (i) we take the momentum ${\mathbf{p}}$ to be continuous rather than the discrete ${\mathbf{k}}$ used by Klauber; (ii) the Klein-Gordon field is real, rather than the two complex fields used by Klauber; (iii) the Hamiltonian density has a factor of ${\frac{1}{2}}$ not found in Klabuer; and (iv) the factors of ${2\pi}$ show up in different places.

P&S’s Hamiltonian density is

$\displaystyle \mathcal{H}=\frac{1}{2}\left[\pi^{2}+\left(\nabla\phi\right)^{2}+m^{2}\phi^{2}\right] \ \ \ \ \ (11)$

From 1 we have

$\displaystyle \nabla\phi=\frac{i}{\left(2\pi\right)^{3}}\int d^{3}p\;\mathbf{p}e^{i\mathbf{p\cdot x}}\frac{1}{\sqrt{2\omega_{\mathbf{p}}}}\left(a_{\mathbf{p}}+a_{-\mathbf{p}}^{\dagger}\right) \ \ \ \ \ (12)$

Plugging this and 2 into 11 and integrating over ${d^{3}x}$ gives the total Hamiltonian

$\displaystyle H=\int d^{3}x\int\frac{d^{3}pd^{3}p^{\prime}}{4\left(2\pi\right)^{6}}e^{i\mathbf{x}\cdot\left(\mathbf{p}+\mathbf{p}^{\prime}\right)}\left(A+B\right) \ \ \ \ \ (13)$

where ${A}$ comes from the ${\pi^{2}}$ term in 11 and ${B}$ comes from ${\left(\nabla\phi\right)^{2}+m^{2}\phi^{2}}$:

 $\displaystyle A$ $\displaystyle =$ $\displaystyle -\sqrt{\omega_{\mathbf{p}}\omega_{\mathbf{p}^{\prime}}}\left(a_{\mathbf{p}}-a_{-\mathbf{p}}^{\dagger}\right)\left(a_{\mathbf{p}^{\prime}}-a_{-\mathbf{p}^{\prime}}^{\dagger}\right)\ \ \ \ \ (14)$ $\displaystyle B$ $\displaystyle =$ $\displaystyle \frac{-\mathbf{p}\cdot\mathbf{p}^{\prime}+m^{2}}{\sqrt{\omega_{\mathbf{p}}\omega_{\mathbf{p}^{\prime}}}}\left(a_{\mathbf{p}}+a_{-\mathbf{p}}^{\dagger}\right)\left(a_{\mathbf{p}^{\prime}}+a_{-\mathbf{p}^{\prime}}^{\dagger}\right) \ \ \ \ \ (15)$

Doing the ${x}$ integral first, we have

$\displaystyle \int d^{3}x\frac{e^{i\mathbf{x}\cdot\left(\mathbf{p}+\mathbf{p}^{\prime}\right)}}{\left(2\pi\right)^{3}}=\delta^{\left(3\right)}\left(\mathbf{p}+\mathbf{p}^{\prime}\right) \ \ \ \ \ (16)$

so we can set ${\mathbf{p}^{\prime}=-\mathbf{p}}$ and eliminate the integral over ${\mathbf{p}^{\prime}}$. Further, using ${\omega_{\mathbf{p}}^{2}=p^{2}+m^{2}}$ converts ${A}$ and ${B}$ to

 $\displaystyle A$ $\displaystyle =$ $\displaystyle -\omega_{\mathbf{p}}\left(a_{\mathbf{p}}-a_{-\mathbf{p}}^{\dagger}\right)\left(a_{-\mathbf{p}}-a_{\mathbf{p}}^{\dagger}\right)\ \ \ \ \ (17)$ $\displaystyle B$ $\displaystyle =$ $\displaystyle \omega_{\mathbf{p}}\left(a_{\mathbf{p}}+a_{-\mathbf{p}}^{\dagger}\right)\left(a_{-\mathbf{p}}+a_{\mathbf{p}}^{\dagger}\right)\ \ \ \ \ (18)$ $\displaystyle A+B$ $\displaystyle =$ $\displaystyle 2\omega_{\mathbf{p}}\left(a_{\mathbf{p}}a_{\mathbf{p}}^{\dagger}+a_{-\mathbf{p}}^{\dagger}a_{-\mathbf{p}}\right) \ \ \ \ \ (19)$

Since we’re integrating over all ${\mathbf{p}}$ we can replace ${-\mathbf{p}}$ by ${\mathbf{p}}$ in the last term, and use 4 on the first term:

 $\displaystyle H$ $\displaystyle =$ $\displaystyle \int\frac{d^{3}p}{4\left(2\pi\right)^{3}}\left(A+B\right)=2\int\frac{d^{3}p}{4\left(2\pi\right)^{3}}\omega_{\mathbf{p}}\left(2a_{\mathbf{p}}^{\dagger}a_{\mathbf{p}}+\left[a_{\mathbf{p}},a_{\mathbf{p}}^{\dagger}\right]\right)\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int\frac{d^{3}p}{\left(2\pi\right)^{3}}\omega_{\mathbf{p}}\left(a_{\mathbf{p}}^{\dagger}a_{\mathbf{p}}+\frac{1}{2}\left[a_{\mathbf{p}},a_{\mathbf{p}}^{\dagger}\right]\right) \ \ \ \ \ (21)$

Since the commutator in the last term has two operators both with suffix ${\mathbf{p}}$, it is an infinite quantity, so its integral is infinite. This is swept under the carpet by saying that since this energy is present in all states and it’s only the difference between a given state and the ground state that can be measured, we can ignore it. In the harmonic oscillator, the ground state ${\left|0\right\rangle }$ has energy ${\frac{1}{2}\omega}$ so this infinite term can be thought of as the sum of this zero-point energy over all momentem states.

In field theory, a state ${\left|0\right\rangle }$ is postulated such that ${a_{\mathbf{p}}\left|0\right\rangle =0}$ for all ${\mathbf{p}}$. This is the vacuum state, which has the infinite zero-point energy. If we operate on ${\left|0\right\rangle }$ with ${a_{\mathbf{p}}^{\dagger}}$ this produces an eigenstate of ${H}$ as we can show using the commutator 4 and ignoring the infinite energy term:

 $\displaystyle Ha_{\mathbf{p}}^{\dagger}\left|0\right\rangle$ $\displaystyle =$ $\displaystyle \int\frac{d^{3}p^{\prime}}{\left(2\pi\right)^{3}}\omega_{\mathbf{p}^{\prime}}a_{\mathbf{\mathbf{p}^{\prime}}}^{\dagger}a_{\mathbf{p}^{\prime}}a_{\mathbf{p}}^{\dagger}\left|0\right\rangle \ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int\frac{d^{3}p^{\prime}}{\left(2\pi\right)^{3}}\omega_{\mathbf{p}^{\prime}}a_{\mathbf{\mathbf{p}^{\prime}}}^{\dagger}\left(a_{\mathbf{p}}^{\dagger}a_{\mathbf{p}^{\prime}}+\left(2\pi\right)^{3}\delta\left(\mathbf{p}-\mathbf{p}^{\prime}\right)\right)\left|0\right\rangle \ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\int\frac{d^{3}p^{\prime}}{\left(2\pi\right)^{3}}\omega_{\mathbf{p}^{\prime}}a_{\mathbf{\mathbf{p}^{\prime}}}^{\dagger}a_{\mathbf{p}}^{\dagger}a_{\mathbf{p}^{\prime}}\left|0\right\rangle \right]+\omega_{\mathbf{p}}a_{\mathbf{p}}^{\dagger}\left|0\right\rangle \ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0+\omega_{\mathbf{p}}a_{\mathbf{p}}^{\dagger}\left|0\right\rangle \ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \omega_{\mathbf{p}}a_{\mathbf{p}}^{\dagger}\left|0\right\rangle \ \ \ \ \ (26)$

The operator ${a_{\mathbf{p}}^{\dagger}}$ acts on the vacuum to create an excited state with energy ${\omega_{\mathbf{p}}}$ in the same way that the ${a^{\dagger}}$ operator in the harmonic oscillator operates on the ground state to produce an oscillator in the next highest energy state. In field theory, this excitation is called a particle, so ${a_{\mathbf{p}}^{\dagger}}$ becomes a creation operator that creates a particle with energy ${\omega_{\mathbf{p}}}$. A similar calculation shows that ${a_{\mathbf{p}}}$, acting on a state containing a particle of energy ${\omega_{\mathbf{p}}}$, removes this particle from the state and produces an eigenstate with energy lowered by ${\omega_{\mathbf{p}}}$, so ${a_{\mathbf{p}}}$ is called an annihilation operator. These operators can produce and destroy multiple particles within a single state.

Anticommutators, creation and annihilation operators in the Dirac equation

Reference: References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 4, Problem 4.25.

In the case of the scalar field, the second quantization postulate is that the Poisson brackets of classical field theory translate into commutators in quantum field theory. For the Dirac equation, however, the four solution vectors contain 4-d spinors, for which there are no analogies in classical theory. As a result, there are no Poisson brackets for spinor fields, so we can’t apply second quantization in the same way as for the Klein-Gordon field. In fact, it turns out that the coefficients in the general solution of the Dirac equation in quantum field theory obey anticommutation relations. The general solutions are

 $\displaystyle \psi$ $\displaystyle =$ $\displaystyle \sum_{r=1}^{2}\sum_{\mathbf{p}}\sqrt{\frac{m}{VE_{\mathbf{p}}}}\left[c_{r}\left(\mathbf{p}\right)u_{r}\left(\mathbf{p}\right)e^{-ipx}+d_{r}^{\dagger}\left(\mathbf{p}\right)v_{r}\left(\mathbf{p}\right)e^{ipx}\right]\ \ \ \ \ (1)$ $\displaystyle$ $\displaystyle \equiv$ $\displaystyle \psi^{+}+\psi^{-}\ \ \ \ \ (2)$ $\displaystyle \bar{\psi}$ $\displaystyle =$ $\displaystyle \sum_{r=1}^{2}\sum_{\mathbf{p}}\sqrt{\frac{m}{VE_{\mathbf{p}}}}\left[d_{r}\left(\mathbf{p}\right)\bar{v}_{r}\left(\mathbf{p}\right)e^{-ipx}+c_{r}^{\dagger}\left(\mathbf{p}\right)\bar{u}_{r}\left(\mathbf{p}\right)e^{ipx}\right]\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle \equiv$ $\displaystyle \bar{\psi}^{+}+\bar{\psi}^{-} \ \ \ \ \ (4)$

where the coefficients ${c_{r}\left(\mathbf{p}\right)}$ and ${d_{r}\left(\mathbf{p}\right)}$ (and their Hermitian conjugates) are operators rather than numbers. It is these operators that obey the anticommutation relations, which as far as I can tell, are just postulates of the theory. The anticommutation relations are

$\displaystyle \left[c_{r}\left(\mathbf{p}\right),c_{s}^{\dagger}\left(\mathbf{p}^{\prime}\right)\right]_{+}=\left[d_{r}\left(\mathbf{p}\right),d_{s}^{\dagger}\left(\mathbf{p}^{\prime}\right)\right]_{+}=\delta_{rs}\delta_{\mathbf{pp}^{\prime}} \ \ \ \ \ (5)$

[Note that some books use braces ${\left\{ \right\} }$ to indicate anticommutators rather than the ${\left[\right]_{+}}$ notation used by Klauber.] All other anticommutators are taken to be zero. One consequence of this latter property is that all the operators anticommute with themselves, that is

 $\displaystyle \left[c_{r}\left(\mathbf{p}\right),c_{r}\left(\mathbf{p}\right)\right]_{+}$ $\displaystyle =$ $\displaystyle \left[c_{r}^{\dagger}\left(\mathbf{p}\right),c_{r}^{\dagger}\left(\mathbf{p}\right)\right]_{+}=0\ \ \ \ \ (6)$ $\displaystyle \left[d_{r}\left(\mathbf{p}\right),d_{r}\left(\mathbf{p}\right)\right]_{+}$ $\displaystyle =$ $\displaystyle \left[d_{r}^{\dagger}\left(\mathbf{p}\right),d_{r}^{\dagger}\left(\mathbf{p}\right)\right]_{+}=0 \ \ \ \ \ (7)$

Or, in other words

$\displaystyle \left[c_{r}\left(\mathbf{p}\right)\right]^{2}=\left[c_{r}^{\dagger}\left(\mathbf{p}\right)\right]^{2}=\left[d_{r}\left(\mathbf{p}\right)\right]^{2}=\left[d_{r}^{\dagger}\left(\mathbf{p}\right)\right]^{2}=0 \ \ \ \ \ (8)$

Just as in the case of the scalar field, these operators prove to be creation and annihilation operators. As shown by Klauber in his section 4.6.1, if we operate on a state containing a single particle in the state ${\left|\psi_{r,\mathbf{p}}\right\rangle }$ with the operator ${c_{r}\left(\mathbf{p}\right)}$ we get an eigenstate of the number operator ${N_{r}\left(\mathbf{p}\right)}$ with eigenvalue 0:

$\displaystyle N_{r}\left(\mathbf{p}\right)c_{r}\left(\mathbf{p}\right)\left|\psi_{r,\mathbf{p}}\right\rangle =0\times c_{r}\left(\mathbf{p}\right)\left|\psi_{r,\mathbf{p}}\right\rangle \ \ \ \ \ (9)$

That is, the state ${c_{r}\left(\mathbf{p}\right)\left|\psi_{r,\mathbf{p}}\right\rangle }$ is the vacuum state ${\left|0\right\rangle }$. Similarly, the operator ${c_{r}^{\dagger}\left(\mathbf{p}\right)}$ creates a particle with spin ${r}$ and momentum ${\mathbf{p}}$ when it operates on the vacuum state (or on any state that does not already contain a particle with spin ${r}$ and momentum ${\mathbf{p}}$):

$\displaystyle N_{r}\left(\mathbf{p}\right)c_{r}^{\dagger}\left(\mathbf{p}\right)\left|0\right\rangle =1\times c_{r}^{\dagger}\left(\mathbf{p}\right)\left|0\right\rangle \ \ \ \ \ (10)$

so that

$\displaystyle c_{r}^{\dagger}\left(\mathbf{p}\right)\left|0\right\rangle =\left|\psi_{r,\mathbf{p}}\right\rangle \ \ \ \ \ (11)$

However, if we try to create another particle of the same spin and momentum in the same state, we get zero because of 8:

$\displaystyle \left[c_{r}^{\dagger}\left(\mathbf{p}\right)\right]^{2}\left|0\right\rangle =0 \ \ \ \ \ (12)$

Note the subtle distinction between 9 and 12. In the former, we destroy an existing particle thus producing the vacuum state, which is real physical state containing no particles. In the latter, we actually annihilate the vacuum state by attempting to create two identical particles in it, producing zero, that is, nothing at all (not even the vacuum). This is the theory’s way of telling us that it is physically impossible to create two identical fermions, an effect we encountered previously in non-relativistic quantum theory.

Creation and annihilation operators: normalization

Reference: References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 3, Problem 3.11.

The number operators are defined in terms of the creation and annihilation operators for the free scalar Hamiltonian as

 $\displaystyle N_{a}\left(\mathbf{k}\right)$ $\displaystyle =$ $\displaystyle a^{\dagger}\left(\mathbf{k}\right)a\left(\mathbf{k}\right)\ \ \ \ \ (1)$ $\displaystyle N_{b}\left(\mathbf{k}\right)$ $\displaystyle =$ $\displaystyle b^{\dagger}\left(\mathbf{k}\right)b\left(\mathbf{k}\right) \ \ \ \ \ (2)$

We’ve seen that ${a^{\dagger}\left(\mathbf{k}\right)}$ creates a particle of energy ${\omega_{\mathbf{k}}}$ when it operates on a state, and ${a\left(\mathbf{k}\right)}$ destroys a particle with energy ${\omega_{\mathbf{k}}}$ when it operates on a state (if that state contains such a particle). That is, the state ${a^{\dagger}\left(\mathbf{k}\right)\left|n_{\mathbf{k}}\right\rangle }$ is an eigenstate of ${N_{a}\left(\mathbf{k}\right)}$ with eigenvalue ${n_{\mathbf{k}}+1}$ and ${a\left(\mathbf{k}\right)\left|n_{\mathbf{k}}\right\rangle }$ is an eigenstate of ${N_{a}\left(\mathbf{k}\right)}$ with eigenvalue ${n_{\mathbf{k}}-1}$. However, if we require all multiparticle states to be normalized, so that ${\left\langle n_{\mathbf{k}}\left|n_{\mathbf{k}}\right.\right\rangle =1}$, the states ${a^{\dagger}\left(\mathbf{k}\right)\left|n_{\mathbf{k}}\right\rangle }$ and ${a\left(\mathbf{k}\right)\left|n_{\mathbf{k}}\right\rangle }$ do not produce normalized states. Rather, we have

 $\displaystyle a^{\dagger}\left(\mathbf{k}\right)\left|n_{\mathbf{k}}\right\rangle$ $\displaystyle =$ $\displaystyle A\left|n_{\mathbf{k}}+1\right\rangle \ \ \ \ \ (3)$ $\displaystyle a\left(\mathbf{k}\right)\left|n_{\mathbf{k}}\right\rangle$ $\displaystyle =$ $\displaystyle B\left|n_{\mathbf{k}}-1\right\rangle \ \ \ \ \ (4)$

for some constants ${A}$ and ${B}$ that are determined by requiring normalization.

To find ${A}$ and ${B}$, we can take the modulus of the states above. We get (we’ll leave off the ${\left(\mathbf{k}\right)}$ dependence of the ${a^{\dagger}}$ and ${a}$ operators to save typing; everything in what follows occurs at wave number ${\mathbf{k}}$; we’ll also assume ${A}$ and ${B}$ are real, although they could also be multiplied by some phase factor ${e^{i\alpha}}$, but this just complicates things unnecessarily). By using the commutation relaton

$\displaystyle \left[a,a^{\dagger}\right]=1 \ \ \ \ \ (5)$

we get, from 3

 $\displaystyle \left\langle n_{\mathbf{k}}\left|aa^{\dagger}\right|n_{\mathbf{k}}\right\rangle$ $\displaystyle =$ $\displaystyle A^{2}\left\langle n_{\mathbf{k}}+1\left|n_{\mathbf{k}}+1\right.\right\rangle =A^{2}\ \ \ \ \ (6)$ $\displaystyle \left\langle n_{\mathbf{k}}\left|aa^{\dagger}\right|n_{\mathbf{k}}\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle n_{\mathbf{k}}\left|a^{\dagger}a+1\right|n_{\mathbf{k}}\right\rangle \ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle n_{\mathbf{k}}\left|N_{a}+1\right|n_{\mathbf{k}}\right\rangle \ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(n_{\mathbf{k}}+1\right)\left\langle n_{\mathbf{k}}\left|n_{\mathbf{k}}\right.\right\rangle \ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(n_{\mathbf{k}}+1\right)\ \ \ \ \ (10)$ $\displaystyle A$ $\displaystyle =$ $\displaystyle \sqrt{n_{\mathbf{k}}+1} \ \ \ \ \ (11)$

Therefore

$\displaystyle a^{\dagger}\left(\mathbf{k}\right)\left|n_{\mathbf{k}}\right\rangle =\sqrt{n_{\mathbf{k}}+1}\left|n_{\mathbf{k}}+1\right\rangle \ \ \ \ \ (12)$

For the annihilation operator, we have from 4:

 $\displaystyle \left\langle n_{\mathbf{k}}\left|a^{\dagger}a\right|n_{\mathbf{k}}\right\rangle$ $\displaystyle =$ $\displaystyle B^{2}\left\langle n_{\mathbf{k}}-1\left|n_{\mathbf{k}}-1\right.\right\rangle =B^{2}\ \ \ \ \ (13)$ $\displaystyle \left\langle n_{\mathbf{k}}\left|a^{\dagger}a\right|n_{\mathbf{k}}\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle n_{\mathbf{k}}\left|N_{a}\right|n_{\mathbf{k}}\right\rangle \ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle n_{\mathbf{k}}\left\langle n_{\mathbf{k}}\left|n_{\mathbf{k}}\right.\right\rangle \ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle n_{\mathbf{k}}\ \ \ \ \ (16)$ $\displaystyle B$ $\displaystyle =$ $\displaystyle \sqrt{n_{\mathbf{k}}} \ \ \ \ \ (17)$

Therefore

$\displaystyle a\left(\mathbf{k}\right)\left|n_{\mathbf{k}}\right\rangle =\sqrt{n_{\mathbf{k}}}\left|n_{\mathbf{k}}-1\right\rangle \ \ \ \ \ (18)$

This relation implies that applying ${a\left(\mathbf{k}\right)}$ to a state that contains no particles with energy ${\omega_{\mathbf{k}}}$ (that is, where ${n_{\mathbf{k}}=0}$) produces 0. In particular, if we apply ${a\left(\mathbf{k}\right)}$ to the vacuum state, we end up with no state at all:

$\displaystyle a\left(\mathbf{k}\right)\left|0\right\rangle =0 \ \ \ \ \ (19)$

Note that ${\left|0\right\rangle }$ and 0 aren’t the same thing: ${\left|0\right\rangle }$ is a quantum state with no particles in it, while 0 is mathematically zero, that is, nothing. As an analogy, ${\left|0\right\rangle }$ is like having a bucket with nothing in it, while 0 corresponds to removing the bucket as well.

We can repeat exactly the same calculations for the antiparticle operators ${b^{\dagger}}$ and ${b}$ and get the results

 $\displaystyle b^{\dagger}\left(\mathbf{k}\right)\left|\bar{n}_{\mathbf{k}}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\bar{n}_{\mathbf{k}}+1}\left|\bar{n}_{\mathbf{k}}+1\right\rangle \ \ \ \ \ (20)$ $\displaystyle b\left(\mathbf{k}\right)\left|\bar{n}_{\mathbf{k}}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\bar{n}_{\mathbf{k}}}\left|\bar{n}_{\mathbf{k}}-1\right\rangle \ \ \ \ \ (21)$

Creation and annihilation operators: commutators and anticommutators

References: Amitabha Lahiri & P. B. Pal, A First Book of Quantum Field Theory, Second Edition (Alpha Science International, 2004) – Chapter 1, Problems 1.1 – 1.2.

As a bit of background to the quantum field theoretic use of creation and annihilation operators we’ll look again at the harmonic oscillator. The creation and annihilation operators (called raising and lowering operators by Griffiths) are defined in terms of the position and momentum operators as

 $\displaystyle a^{\dagger}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2\hbar m\omega}}\left[-ip+m\omega x\right]\ \ \ \ \ (1)$ $\displaystyle a$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2\hbar m\omega}}\left[ip+m\omega x\right] \ \ \ \ \ (2)$

From the commutator ${\left[x,p\right]=i\hbar}$ we can work out

 $\displaystyle \left[a,a^{\dagger}\right]$ $\displaystyle =$ $\displaystyle \frac{1}{2\hbar m\omega}\left(-im\omega\left[x,p\right]\right)\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1 \ \ \ \ \ (4)$

The annihilation operator ${a}$ acting on the vacuum or ground state ${\left|0\right\rangle }$ gives 0, and the creation operator ${a^{\dagger}}$ produces a state ${a^{\dagger}\left|0\right\rangle =\left|1\right\rangle }$ with energy eigenvalue ${\frac{3}{2}\hbar\omega}$. Successive applications of ${a^{\dagger}}$ produce states with higher energy, where each quantum of energy is ${\hbar\omega}$.

Normalization

Given that the ground state is normalized so that ${\left\langle \left.0\right|0\right\rangle =1}$, we can find the factor required to normalize higher states so that ${\left\langle \left.n\right|n\right\rangle =1}$. Consider ${n=2}$. We have

$\displaystyle a^{\dagger}a^{\dagger}\left|0\right\rangle =A\left|2\right\rangle \ \ \ \ \ (5)$

where ${A}$ is to be determined. We have

 $\displaystyle \left\langle 0\left|aaa^{\dagger}a^{\dagger}\right|0\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle 0\left|a\left(1+a^{\dagger}a\right)a^{\dagger}\right|0\right\rangle \ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle 0\left|aa^{\dagger}\right|0\right\rangle +\left\langle 0\left|aa^{\dagger}aa^{\dagger}\right|0\right\rangle \ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle 0\left|\left(1+a^{\dagger}a\right)\right|0\right\rangle +\left\langle 0\left|aa^{\dagger}\left(1+a^{\dagger}a\right)\right|0\right\rangle \ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle \left.0\right|0\right\rangle +\left\langle 0\left|aa^{\dagger}\right|0\right\rangle \ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle \left.0\right|0\right\rangle +\left\langle 0\left|\left(1+a^{\dagger}a\right)\right|0\right\rangle \ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle \left.0\right|0\right\rangle +\left\langle \left.0\right|0\right\rangle \ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{A^{2}}\ \ \ \ \ (13)$ $\displaystyle A$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}} \ \ \ \ \ (14)$

For ${n=3}$ we get ${\left\langle 0\left|aaaa^{\dagger}a^{\dagger}a^{\dagger}\right|0\right\rangle }$. We need to commute each ${a}$ through the ${a^{\dagger}}$ operators to its right. The first ${a}$ will generate the factor ${\left(1+a^{\dagger}a\right)}$ 3 times as it commutes with each ${a^{\dagger}}$ operator. Each of these terms will be ${\left\langle 0\left|aaa^{\dagger}a^{\dagger}\right|0\right\rangle }$ and we already know that this term produces a factor of 2. Therefore

$\displaystyle \left\langle 0\left|aaaa^{\dagger}a^{\dagger}a^{\dagger}\right|0\right\rangle =3\times2=6 \ \ \ \ \ (15)$

We can extend this result to the general case:

$\displaystyle \left\langle 0\left|a^{n}\left(a^{\dagger}\right)^{n}\right|0\right\rangle =n! \ \ \ \ \ (16)$

The normalization must then be

$\displaystyle \left|n\right\rangle =\frac{1}{\sqrt{n!}}\left(a^{\dagger}\right)^{n}\left|0\right\rangle \ \ \ \ \ (17)$

Number operator

We’ve met the number operator ${N}$ in the field case, but there is an analogous operator for the harmonic oscillator. We have

$\displaystyle N\equiv a^{\dagger}a \ \ \ \ \ (18)$

As with the field case, we can work out its commutators:

 $\displaystyle \left[N,a^{\dagger}\right]$ $\displaystyle =$ $\displaystyle a^{\dagger}aa^{\dagger}-a^{\dagger}a^{\dagger}a\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle a^{\dagger}a^{\dagger}a+a^{\dagger}-a^{\dagger}a^{\dagger}a\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle a^{\dagger}\ \ \ \ \ (21)$ $\displaystyle \left[N,a\right]$ $\displaystyle =$ $\displaystyle a^{\dagger}aa-aa^{\dagger}a\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle a^{\dagger}aa-a+a^{\dagger}aa\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -a \ \ \ \ \ (24)$

Applying this to ${\left|n\right\rangle }$ we get

$\displaystyle N\left|n\right\rangle =\frac{1}{\sqrt{n!}}N\left(a^{\dagger}\right)^{n}\left|0\right\rangle \ \ \ \ \ (25)$

We get

 $\displaystyle N\left(a^{\dagger}\right)^{n}$ $\displaystyle =$ $\displaystyle \left[a^{\dagger}+a^{\dagger}N\right]\left(a^{\dagger}\right)^{n-1}\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(a^{\dagger}\right)^{n}+\left(a^{\dagger}\right)^{2}\left(1+N\right)\left(a^{\dagger}\right)^{n-2}\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \ldots\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle n\left(a^{\dagger}\right)^{n}+\left(a^{\dagger}\right)^{n}N\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle n\left(a^{\dagger}\right)^{n}+\left(a^{\dagger}\right)^{n}a^{\dagger}a \ \ \ \ \ (30)$

When operating on ${\left|0\right\rangle }$, the last term gives 0, so

$\displaystyle N\left|n\right\rangle =\frac{n}{\sqrt{n!}}\left(a^{\dagger}\right)^{n}\left|0\right\rangle \ \ \ \ \ (31)$

Multiple oscillators

If we now have a system of ${N}$ non-interacting harmonic oscillators with equal masses and frequencies ${\omega_{i}}$, ${i=1,\ldots,N}$, the Hamiltonian is

$\displaystyle H=\frac{1}{2m}\sum_{i}\left(p_{i}^{2}+m^{2}\omega_{i}^{2}x_{i}^{2}\right) \ \ \ \ \ (32)$

Since the oscillators are not coupled, the creation and annihilation operators for different operators all commute, so that

$\displaystyle \left[a_{i},a_{j}^{\dagger}\right]=\delta_{ij} \ \ \ \ \ (33)$

so the normalized state where oscillator ${i}$ is in the ${n_{i}}$th excited state is

$\displaystyle \left|n_{1}n_{2}\ldots n_{N}\right\rangle =\prod_{i=1}^{N}\frac{\left(a_{i}^{\dagger}\right)^{n_{i}}}{\sqrt{n_{i}!}}\left|0\right\rangle \ \ \ \ \ (34)$

The number operator in this case is

$\displaystyle \mathcal{N}=\sum_{i=1}^{N}\left(a_{i}^{\dagger}a_{i}\right) \ \ \ \ \ (35)$

This works because the commutation relation 33 allows each term ${a_{i}^{\dagger}a_{i}}$ in the sum to pick out the number of quanta of oscillator ${i}$.

Anticommutators

Now suppose that instead of the commutation relations 33 we have anticommutation relations as follows:

 $\displaystyle \left\{ a_{i},a_{j}^{\dagger}\right\}$ $\displaystyle \equiv$ $\displaystyle a_{i}a_{j}+a_{j}a_{i}=\delta_{ij}\ \ \ \ \ (36)$ $\displaystyle \left\{ a_{i}^{\dagger},a_{j}^{\dagger}\right\}$ $\displaystyle =$ $\displaystyle \left\{ a_{i},a_{j}\right\} =0 \ \ \ \ \ (37)$

If we start with the vacuum state ${\left|0\right\rangle }$ and require ${a_{i}^{\dagger}\left|0\right\rangle =\left|0\ldots1_{i}\ldots0\right\rangle }$ (that is, ${a_{i}^{\dagger}}$ creates one quantum in category ${i}$), then if we try to create another quantum in the same state, we get

 $\displaystyle \left\langle 0\left|a_{i}a_{i}a_{i}^{\dagger}a_{i}^{\dagger}\right|0\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle 0\left|a_{i}\left(1-a_{i}^{\dagger}a_{i}\right)a_{i}^{\dagger}\right|0\right\rangle \ \ \ \ \ (38)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle 0\left|a_{i}a_{i}^{\dagger}\right|0\right\rangle -\left\langle 0\left|a_{i}a_{i}^{\dagger}a_{i}a_{i}^{\dagger}\right|0\right\rangle \ \ \ \ \ (39)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle 0\left|a_{i}a_{i}^{\dagger}\right|0\right\rangle -\left\langle 0\left|a_{i}a_{i}^{\dagger}\left(1-a_{i}^{\dagger}a_{i}\right)\right|0\right\rangle \ \ \ \ \ (40)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle 0\left|a_{i}a_{i}^{\dagger}\right|0\right\rangle -\left\langle 0\left|a_{i}a_{i}^{\dagger}\right|0\right\rangle +\left\langle 0\left|a_{i}a_{i}^{\dagger}a_{i}^{\dagger}a_{i}\right|0\right\rangle \ \ \ \ \ (41)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (42)$

Thus, attempting to create two quanta in the same state produces zero, so at most one quantum can occupy each state. The commutator case 33 thus behaves like bosons and the anticommutator case like fermions.

Quantum field theory representation of non-relativistic quantum mechanics

References: Mark Srednicki, Quantum Field Theory, (Cambridge University Press, 2007) – Chapter 1, Problem 1.2.

One of the main problems faced in developing a relativistic quantum theory is that in non-relativistic quantum mechanics, position and time are not on an equal footing. Position is treated as an operator, while time is just a parameter that labels a particular instance of a state or wave function. In special relativity, space and time are treated equivalently, in the sense that they form equal components of four-dimensional spacetime. [Time and space aren’t the same, of course, since although we can travel to any point in space whenever we want, we can move only forward in time, even in relativity. However, space and time components are treated equally in the sense that they transform into each other in the Lorentz transformations.]

Attempts to develop a relativistic quantum theory therefore can take one of two paths in an attempt to solve this disparity. One way is to promote time to an operator, but this leads to complex theories (although they do work). The other way is to demote position from an operator to just a label, so its status is the same as that of time. This idea leads to quantum field theory.

The idea is that the position ${\mathbf{x}}$ becomes, like time, a label on an operator. We can define a set of operators ${\phi\left(\mathbf{x}\right)}$ such that at each point ${\mathbf{x}}$ in space, there is a separate operator. The position ${\mathbf{x}}$ becomes a label telling us which operator we’re dealing with. The set of all such operators (that is, the set of operators defined over all space) is called a quantum field, and hence we get quantum field theory by studying such sets of operators. In the general case, each operator is also a function of time so that a quantum field is actually made up of a set of operators ${\phi\left(\mathbf{x},t\right)}$.

To get an idea of how this works, we can rewrite non-relativistic quantum mechanics using a quantum field. Non-relativistic quantum mechanics is governed by the Schrödinger equation, which in its most general form is

$\displaystyle i\hbar\frac{\partial}{\partial t}\left|\psi,t\right\rangle =H\left|\psi,t\right\rangle \ \ \ \ \ (1)$

where ${H}$ is the hamiltonian. Now suppose we define a quantum field ${a\left(\mathbf{x}\right)}$ and its hermitian conjugate ${a^{\dagger}\left(\mathbf{x}\right)}$ that satisfy the commutation relations

 $\displaystyle \left[a\left(\mathbf{x}\right),a\left(\mathbf{x}^{\prime}\right)\right]$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (2)$ $\displaystyle \left[a^{\dagger}\left(\mathbf{x}\right),a^{\dagger}\left(\mathbf{x}^{\prime}\right)\right]$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (3)$ $\displaystyle \left[a\left(\mathbf{x}\right),a^{\dagger}\left(\mathbf{x}^{\prime}\right)\right]$ $\displaystyle =$ $\displaystyle \delta^{3}\left(\mathbf{x}-\mathbf{x}^{\prime}\right) \ \ \ \ \ (4)$

where ${\delta^{3}\left(\mathbf{x}\right)}$ is the 3-d Dirac delta function. These operators are similar to the raising and lowering operators we used to solve the harmonic oscillator, although in this case the operators are labelled by positions in space rather than energy states in an oscillator.

We require one additional property for this field: if ${\left|0\right\rangle }$ represents the vacuum state, that is, a state with no particles in it, then

$\displaystyle a\left(\mathbf{x}\right)\left|0\right\rangle =0 \ \ \ \ \ (5)$

That is, ${a\left(\mathbf{x}\right)}$ eliminates the vacuum state for all values of ${\mathbf{x}}$.

Now suppose we define a hamiltonian using this field, as follows:

 $\displaystyle H$ $\displaystyle =$ $\displaystyle \int d^{3}x\;a^{\dagger}\left(\mathbf{x}\right)\left(-\frac{\hbar^{2}}{2m}\nabla^{2}+U\left(\mathbf{x}\right)\right)a\left(\mathbf{x}\right)+\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{1}{2}\int d^{3}x\;d^{3}y\;V\left(\mathbf{x}-\mathbf{y}\right)a^{\dagger}\left(\mathbf{x}\right)a^{\dagger}\left(\mathbf{y}\right)a\left(\mathbf{y}\right)a\left(\mathbf{x}\right) \ \ \ \ \ (6)$

Here, ${U}$ is an external potential energy and ${V}$ is an interaction energy between two particles at locations ${\mathbf{x}}$ and ${\mathbf{y}}$.

Also, suppose we have a time-dependent quantum state defined by

$\displaystyle \left|\psi,t\right\rangle =\int d^{3}x_{1}\ldots d^{3}x_{n}\psi\left(\mathbf{x}_{1},\ldots,\mathbf{x}_{n};t\right)a^{\dagger}\left(\mathbf{x}_{1}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)\left|0\right\rangle \ \ \ \ \ (7)$

If we require 1 to be true, what condition does this place on the function ${\psi\left(\mathbf{x}_{1},\ldots,\mathbf{x}_{n};t\right)}$ inside the integral? To determine this, we need to apply ${H}$ to ${\left|\psi,t\right\rangle }$ by using the commutation relations. Consider the first integral in 6. We can propagate the operator ${a\left(\mathbf{x}\right)}$ through the list of operators ${a^{\dagger}\left(\mathbf{x}_{1}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)}$ in 7 by applying the commutator 4. We get

 $\displaystyle a\left(\mathbf{x}\right)a^{\dagger}\left(\mathbf{x}_{1}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)$ $\displaystyle =$ $\displaystyle \left[\delta^{3}\left(\mathbf{x}-\mathbf{x}_{1}\right)+a^{\dagger}\left(\mathbf{x}_{1}\right)a\left(\mathbf{x}\right)\right]a^{\dagger}\left(\mathbf{x}_{2}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \delta^{3}\left(\mathbf{x}-\mathbf{x}_{1}\right)a^{\dagger}\left(\mathbf{x}_{2}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)+\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle a^{\dagger}\left(\mathbf{x}_{1}\right)\left[\delta^{3}\left(\mathbf{x}-\mathbf{x}_{2}\right)+a^{\dagger}\left(\mathbf{x}_{2}\right)a\left(\mathbf{x}\right)\right]a^{\dagger}\left(\mathbf{x}_{3}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \delta^{3}\left(\mathbf{x}-\mathbf{x}_{1}\right)a^{\dagger}\left(\mathbf{x}_{2}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)+\ldots+\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle a^{\dagger}\left(\mathbf{x}_{1}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n-1}\right)\left[\delta^{3}\left(\mathbf{x}-\mathbf{x}_{n}\right)+a^{\dagger}\left(\mathbf{x}_{n}\right)a\left(\mathbf{x}\right)\right] \ \ \ \ \ (10)$

When we apply this expansion to the vacuum state ${\left|0\right\rangle }$, the last term in the final bracket vanishes because of 5. Doing the integral over ${x}$ in 6 results in

 $\displaystyle \int d^{3}x\;a^{\dagger}\left(\mathbf{x}\right)\left(-\frac{\hbar^{2}}{2m}\nabla^{2}+U\left(\mathbf{x}\right)\right)a\left(\mathbf{x}\right)\left|\psi,t\right\rangle$ $\displaystyle =\nonumber$ $\displaystyle \sum_{i}\int d^{3}x_{1}\ldots d^{3}x_{n}a^{\dagger}\left(\mathbf{x}_{i}\right)\left(-\frac{\hbar^{2}}{2m}\nabla_{i}^{2}+U\left(\mathbf{x}_{i}\right)\right)\psi a^{\dagger}\left(\mathbf{x}_{1}\right)\ldots a^{\dagger}\left(\mathbf{x}_{i-1}\right)a^{\dagger}\left(\mathbf{x}_{i+1}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)\left|0\right\rangle \ \ \ \ \ (11)$ $\displaystyle =$ $\displaystyle \sum_{i}\int d^{3}x_{1}\ldots d^{3}x_{n}\left(-\frac{\hbar^{2}}{2m}\nabla_{i}^{2}+U\left(\mathbf{x}_{i}\right)\right)\psi a^{\dagger}\left(\mathbf{x}_{1}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)$

where the last line follows from 3.

We can do a similar calculation for the second integral in 6, although it’s a bit more complicated because we have to integrate over both ${x}$ and ${y}$. Applying ${a\left(\mathbf{x}\right)}$ to ${a^{\dagger}\left(\mathbf{x}_{1}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)}$ gives the result 10. Applying ${a\left(\mathbf{y}\right)}$ to the first term of this result gives

 $\displaystyle a\left(\mathbf{y}\right)\delta^{3}\left(\mathbf{x}-\mathbf{x}_{1}\right)a^{\dagger}\left(\mathbf{x}_{2}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)\left|0\right\rangle$ $\displaystyle =$ $\displaystyle \delta^{3}\left(\mathbf{x}-\mathbf{x}_{1}\right)\times\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle \left[\delta^{3}\left(\mathbf{y}-\mathbf{x}_{2}\right)a^{\dagger}\left(\mathbf{x}_{3}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)+\delta^{3}\left(\mathbf{y}-\mathbf{x}_{n}\right)a^{\dagger}\left(\mathbf{x}_{2}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n-1}\right)\right]\left|0\right\rangle \ \ \ \ \ (12)$

with similar terms arising from the other terms in 10. When we integrate over ${x}$ and ${y}$ this first term gives us

$\displaystyle \sum_{j=2}^{n}\int d^{3}x_{1}\ldots d^{3}x_{n}V\left(\mathbf{x}_{1}-\mathbf{x}_{j}\right)\psi\left(\mathbf{x}_{1},\ldots,\mathbf{x}_{n};t\right)a^{\dagger}\left(\mathbf{x}_{1}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)\left|0\right\rangle \ \ \ \ \ (13)$

The other terms in the expansion 10 each contribute a sum

$\displaystyle \sum_{\begin{array}{c} j=1\\ j\ne i \end{array}}^{n}\int d^{3}x_{1}\ldots d^{3}x_{n}V\left(\mathbf{x}_{i}-\mathbf{x}_{j}\right)\psi\left(\mathbf{x}_{1},\ldots,\mathbf{x}_{n};t\right)a^{\dagger}\left(\mathbf{x}_{1}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)\left|0\right\rangle \ \ \ \ \ (14)$

so the overall result for the second integral is

 $\displaystyle \frac{1}{2}\int d^{3}x\;d^{3}y\;V\left(\mathbf{x}-\mathbf{y}\right)a^{\dagger}\left(\mathbf{x}\right)a^{\dagger}\left(\mathbf{y}\right)a\left(\mathbf{y}\right)a\left(\mathbf{x}\right)\left|\psi,t\right\rangle$ $\displaystyle =\nonumber$ $\displaystyle \sum_{j=1}^{n}\sum_{i=1}^{j-1}\int d^{3}x_{1}\ldots d^{3}x_{n}V\left(\mathbf{x}_{i}-\mathbf{x}_{j}\right)\psi\left(\mathbf{x}_{1},\ldots,\mathbf{x}_{n};t\right)a^{\dagger}\left(\mathbf{x}_{1}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)\left|0\right\rangle \ \ \ \ \ (15)$

Requiring 11 plus 15 to satisfy 1 gives us (from equating the integrands on both sides):

$\displaystyle i\hbar\frac{\partial\psi}{\partial t}=\sum_{i=1}^{n}\left(-\frac{\hbar^{2}}{2m}\nabla_{i}^{2}+U\left(\mathbf{x}_{i}\right)\right)\psi+\sum_{j=1}^{n}\sum_{i=1}^{j-1}V\left(\mathbf{x}_{i}-\mathbf{x}_{j}\right)\psi \ \ \ \ \ (16)$

This is just the Schrödinger equation in its more traditional form, for a hamiltonian containing kinetic energy, overall potential energy ${U}$ and particle-particle interaction energy ${V}$ for a collection of ${n}$ particles. Thus the state ${a^{\dagger}\left(\mathbf{x}_{1}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)\left|0\right\rangle }$ corresponds to a state containing ${n}$ particles at locations ${\mathbf{x}_{i}}$, so the hermitian conjugate field operators ${a^{\dagger}\left(\mathbf{x}_{i}\right)}$ act as creation operators, with each operator creating its particle at the location ${\mathbf{x}_{i}}$ used to label the operator.

The operator

$\displaystyle N\equiv\int d^{3}x\;a^{\dagger}\left(\mathbf{x}\right)a\left(\mathbf{x}\right) \ \ \ \ \ (17)$

counts the particles present, in effect by annihilating a particle at location ${\mathbf{x}}$ (if one exists there), then creating it again in the same location. More formally, if we apply ${N}$ to the state ${a^{\dagger}\left(\mathbf{x}_{1}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)\left|0\right\rangle }$ then the annihilation operator ${a\left(\mathbf{x}\right)}$ acting on the state produces the result 10, so we get

 $\displaystyle Na^{\dagger}\left(\mathbf{x}_{1}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)\left|0\right\rangle$ $\displaystyle =$ $\displaystyle \int d^{3}x\;a^{\dagger}\left(\mathbf{x}\right)\delta^{3}\left(\mathbf{x}-\mathbf{x}_{1}\right)a^{\dagger}\left(\mathbf{x}_{2}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)\left|0\right\rangle +\ldots+\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle \int d^{3}x\;a^{\dagger}\left(\mathbf{x}\right)a^{\dagger}\left(\mathbf{x}_{1}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n-1}\right)\delta^{3}\left(\mathbf{x}-\mathbf{x}_{n}\right)\left|0\right\rangle \ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{i=1}^{n}a^{\dagger}\left(\mathbf{x}_{1}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)\left|0\right\rangle \ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle n\;a^{\dagger}\left(\mathbf{x}_{1}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)\left|0\right\rangle \ \ \ \ \ (20)$

Second quantizing operators – examples

Reference: Tom Lancaster and Stephen J. Blundell, Quantum Field Theory for the Gifted Amateur, (Oxford University Press, 2014), Section 4.2.

We’ve seen that we can second quantize a single-particle operator ${\hat{\mathcal{A}}}$ using creation and annihilation operators to get the multi-particle version:

$\displaystyle \hat{A}=\sum_{\alpha,\beta}\mathcal{A}_{\alpha\beta}a_{\alpha}^{\dagger}a_{\beta} \ \ \ \ \ (1)$

Using this result, we can get second quantized versions of some common operators. The unit operator is

$\displaystyle \hat{1}=\sum_{\gamma}\left|\gamma\right\rangle \left\langle \gamma\right| \ \ \ \ \ (2)$

so

 $\displaystyle \left\langle \alpha\left|\hat{1}\right|\beta\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle \alpha\left|\sum_{\gamma}\left|\gamma\right\rangle \left\langle \gamma\right|\right|\beta\right\rangle \ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{\gamma}\delta_{\alpha\gamma}\delta_{\gamma\beta}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \delta_{\alpha\beta} \ \ \ \ \ (5)$

so the multi-particle version is

$\displaystyle \hat{n}=\sum_{\alpha}a_{\alpha}^{\dagger}a_{\alpha} \ \ \ \ \ (6)$

Since ${a_{\alpha}^{\dagger}a_{\alpha}}$ is the number operator, it counts the number of particles in state ${\alpha}$ so ${\hat{n}}$ gives the total number of particles in the multi-particle state. [I’m still not clear as to whether this result is supposed to apply to states where there are more than one particle in a given momentum state. The derivation of 1 appears to assume that each particle is in a different single-particle state, so it seems safer to assume that ${a_{\alpha}^{\dagger}a_{\alpha}}$ can return only 0 or 1.]

For the momentum operator (we’re still looking at the particle in a box, so momentum states are still discrete) we have

 $\displaystyle \hat{\mathbf{p}}\left|\mathbf{p}\right\rangle$ $\displaystyle =$ $\displaystyle \mathbf{p}\left|\mathbf{p}\right\rangle \ \ \ \ \ (7)$ $\displaystyle \left\langle \mathbf{q}\left|\hat{\mathbf{p}}\right|\mathbf{p}\right\rangle$ $\displaystyle =$ $\displaystyle \mathbf{p}\left\langle \mathbf{q}\left|\mathbf{p}\right.\right\rangle \ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \mathbf{p}\delta_{\mathbf{qp}} \ \ \ \ \ (9)$

The multi-particle version is therefore

 $\displaystyle \hat{p}$ $\displaystyle =$ $\displaystyle \sum_{\mathbf{q},\mathbf{p}}\mathbf{p}\delta_{\mathbf{qp}}a_{\mathbf{q}}^{\dagger}a_{\mathbf{p}}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{\mathbf{p}}\mathbf{p}a_{\mathbf{p}}^{\dagger}a_{\mathbf{p}} \ \ \ \ \ (11)$

We can extend this result to functions of momentum ${f\left(\mathbf{p}\right)}$. First, we look at powers of the momentum operator, where we can use induction to prove that ${\left(\hat{\mathbf{p}}\right)^{n}\left|\mathbf{p}\right\rangle =p^{n}\left|\mathbf{p}\right\rangle }$. We know this is true for ${n=1}$ so assume it’s true for ${n-1}$. Then

 $\displaystyle \left(\hat{\mathbf{p}}\right)^{n}\left|\mathbf{p}\right\rangle$ $\displaystyle =$ $\displaystyle \hat{\mathbf{p}}\left(\hat{\mathbf{p}}\right)^{n-1}\left|\mathbf{p}\right\rangle \ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle p^{n-1}\hat{\mathbf{p}}\left|\mathbf{p}\right\rangle \ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle p^{n}\left|\mathbf{p}\right\rangle \ \ \ \ \ (14)$

QED. That is, ${\left|\mathbf{p}\right\rangle }$ is an eigenvector of ${\left(\hat{\mathbf{p}}\right)^{n}}$ with eigenvalue ${p^{n}}$.

Now if the function ${f\left(\hat{\mathbf{p}}\right)}$ can be expanded in powers of ${\hat{\mathbf{p}}}$ then

$\displaystyle f\left(\hat{\mathbf{p}}\right)=f_{0}+f_{1}\hat{\mathbf{p}}+f_{2}\left(\hat{\mathbf{p}}\right)^{2}+\dots \ \ \ \ \ (15)$

where the ${f_{i}}$ are constants. Now${\left|\mathbf{p}\right\rangle }$ is an eigenvector of the term ${f_{i}\left(\hat{\mathbf{p}}\right)^{i}}$ in the series with eigenvalue ${p^{i}}$. In other words, we’re replacing a series in the operator ${\hat{\mathbf{p}}}$ with an identical series in its eigenvalue, so

 $\displaystyle f\left(\hat{\mathbf{p}}\right)\left|\mathbf{p}\right\rangle$ $\displaystyle =$ $\displaystyle f\left(\mathbf{p}\right)\left|\mathbf{p}\right\rangle \ \ \ \ \ (16)$ $\displaystyle \left\langle \mathbf{q}\left|f\left(\hat{\mathbf{p}}\right)\right|\mathbf{p}\right\rangle$ $\displaystyle =$ $\displaystyle f\left(\mathbf{p}\right)\left\langle \mathbf{q}\left|\mathbf{p}\right.\right\rangle \ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle f\left(\mathbf{p}\right)\delta_{\mathbf{qp}} \ \ \ \ \ (18)$

Therefore the second-quantized version of ${f\left(\hat{\mathbf{p}}\right)}$ is

 $\displaystyle \hat{A}$ $\displaystyle =$ $\displaystyle \sum_{\mathbf{p}}f\left(\mathbf{p}\right)a_{\mathbf{p}}^{\dagger}a_{\mathbf{p}}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{\mathbf{p}}f\left(\mathbf{p}\right)\hat{n}_{\mathbf{p}} \ \ \ \ \ (20)$

The interpretation is that the operator ${f}$ acts separately on each particle with the total result being the sum of its values for all particles.

For example, the hamiltonian for a single free particle is ${\hat{H}=\hat{\mathbf{p}}^{2}/2m}$ so the hamiltonian for a collection of free particles is

$\displaystyle \hat{H}=\sum_{\mathbf{p}}\frac{\mathbf{p}^{2}}{2m}\hat{n}_{\mathbf{p}} \ \ \ \ \ (21)$

The potential energy is usually given as a function of position, so using the momentum eigenfunction ${\left|\mathbf{p}\right\rangle =\frac{1}{\sqrt{\mathcal{V}}}e^{-i\mathbf{p}\cdot\mathbf{x}}}$ (where ${\mathcal{V}}$ is the volume of the box) we have from 1

 $\displaystyle \left\langle \mathbf{q}\left|\hat{V}\right|\mathbf{p}\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{\mathcal{V}}\int d^{3}x\; e^{i\mathbf{q}\cdot\mathbf{x}}V\left(\mathbf{x}\right)e^{-i\mathbf{p}\cdot\mathbf{x}}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\mathcal{V}}\int d^{3}x\; e^{-i\left(\mathbf{p}-\mathbf{q}\right)\cdot\mathbf{x}}V\left(\mathbf{x}\right)\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle \equiv$ $\displaystyle \tilde{V}_{\mathbf{p}-\mathbf{q}} \ \ \ \ \ (24)$

The potential can then be second quantized as

$\displaystyle \hat{V}=\sum_{\mathbf{p},\mathbf{q}}\tilde{V}_{\mathbf{p}-\mathbf{q}}a_{\mathbf{p}}^{\dagger}a_{\mathbf{q}} \ \ \ \ \ (25)$

Example Suppose we have a 3 state system with a hamiltonian

 $\displaystyle \hat{H}$ $\displaystyle =$ $\displaystyle E_{0}\sum_{i=1}^{3}a_{i}^{\dagger}a_{i}+W\left[a_{1}^{\dagger}a_{2}-a_{1}^{\dagger}a_{3}+a_{2}^{\dagger}a_{1}+a_{2}^{\dagger}a_{3}-a_{3}^{\dagger}a_{1}+a_{3}^{\dagger}a_{2}\right]\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle \equiv$ $\displaystyle T+V \ \ \ \ \ (27)$

where ${T}$ is the kinetic energy (the first term) and ${V}$ is the potential energy (the second term). ${T}$ is diagonal but ${V}$ is not; we can see the effect of ${V}$ on the basis states ${\left|100\right\rangle }$, ${\left|010\right\rangle }$ and ${\left|001\right\rangle }$ by observing that ${a_{1}^{\dagger}a_{2}\left|010\right\rangle =\left|100\right\rangle }$ (annihilate state 2 and create state 1), ${a_{1}^{\dagger}a_{2}\left|100\right\rangle =0}$ (no particle in state 2 so annihilation of state 2 produces 0) and so on.

 $\displaystyle V\left|100\right\rangle$ $\displaystyle =$ $\displaystyle W\left(\left|010\right\rangle -\left|001\right\rangle \right)\ \ \ \ \ (28)$ $\displaystyle V\left|010\right\rangle$ $\displaystyle =$ $\displaystyle W\left(\left|100\right\rangle +\left|001\right\rangle \right)\ \ \ \ \ (29)$ $\displaystyle V\left|001\right\rangle$ $\displaystyle =$ $\displaystyle W\left(-\left|100\right\rangle +\left|010\right\rangle \right) \ \ \ \ \ (30)$

We can write the hamiltonian as a matrix

 $\displaystyle \hat{H}$ $\displaystyle =$ $\displaystyle T+V\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle E_{0}\left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\right]+W\left[\begin{array}{ccc} 0 & 1 & -1\\ 1 & 0 & 1\\ -1 & 1 & 0 \end{array}\right] \ \ \ \ \ (32)$

In this form, for example, 28 would be written as

$\displaystyle V\left|100\right\rangle =W\left[\begin{array}{ccc} 0 & 1 & -1\\ 1 & 0 & 1\\ -1 & 1 & 0 \end{array}\right]\left[\begin{array}{c} 1\\ 0\\ 0 \end{array}\right]=W\left[\begin{array}{c} 0\\ 1\\ -1 \end{array}\right] \ \ \ \ \ (33)$

Finding the energies and eigenstates of this hamiltonian means we need to find the eigenvalues and eigenvectors of ${\hat{H}}$, which turn out to be

$\displaystyle E=E_{0}+W,\; E_{0}+W,\; E_{0}-2W \ \ \ \ \ (34)$

The ground state ${\left|\Omega\right\rangle }$ (assuming ${W>0}$) has energy ${E_{0}-2W}$ and its eigenvector is

$\displaystyle \left|\Omega\right\rangle =\frac{1}{\sqrt{3}}\left(\left|100\right\rangle -\left|010\right\rangle +\left|001\right\rangle \right) \ \ \ \ \ (35)$

The other energy level ${E_{0}+W}$ is doubly degenerate and its 2-d space of eigenvectors is spanned by

$\displaystyle \frac{1}{\sqrt{2}}\left(-\left|100\right\rangle +\left|001\right\rangle \right),\;\frac{1}{\sqrt{2}}\left(\left|100\right\rangle +\left|010\right\rangle \right) \ \ \ \ \ (36)$

Second quantizing a single-particle operator

Reference: Tom Lancaster and Stephen J. Blundell, Quantum Field Theory for the Gifted Amateur, (Oxford University Press, 2014), Section 4.2.

Second quantization is the expression of quantum mechanical states using particles rather than waves. We use creation and annihilation operators acting on particle states to make transitions between states. If we have a single-particle operator ${\hat{\mathcal{A}}}$ we can expand it using two unit operators as follows:

 $\displaystyle \hat{\mathcal{A}}$ $\displaystyle =$ $\displaystyle \sum_{\alpha}\left|\alpha\right\rangle \left\langle \alpha\right|\hat{\mathcal{A}}\sum_{\beta}\left|\beta\right\rangle \left\langle \beta\right|\ \ \ \ \ (1)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{\alpha,\beta}\left|\alpha\right\rangle \mathcal{A}_{\alpha\beta}\left\langle \beta\right| \ \ \ \ \ (2)$

where

$\displaystyle \mathcal{A}_{\alpha\beta}\equiv\left\langle \alpha\left|\hat{\mathcal{A}}\right|\beta\right\rangle \ \ \ \ \ (3)$

Each of the sets ${\left|\alpha\right\rangle }$ and ${\left|\beta\right\rangle }$ are complete, orthonormal sets of states for a single particle. For multi-particle systems, we use the creation and annihilation operators ${a_{\mathbf{p}}^{\dagger}}$ and ${a_{\mathbf{p}}}$ to add or subtract particles from a state, so we’d like to know how to define a multi-particle version ${\hat{A}}$ of the single-particle operator ${\hat{\mathcal{A}}}$.

The route to this end is a bit complex, so bear with me. Suppose we have a system of ${N}$ particles. To satisfy the symmetry rules for bosons and fermions, we can write a state of these ${N}$ particles as

$\displaystyle \left|\psi_{1},\dots,\psi_{N}\right\rangle =\frac{1}{\sqrt{N!}}\sum_{P}\xi^{P}\prod_{i=1}^{N}\left|\psi_{P\left(i\right)}\right\rangle \ \ \ \ \ (4)$

This notation requires a bit of explanation. First, we’re assuming that each of the ${N}$ particles is in a different state and that these states are orthonormal. Only in this case is the normalization factor ${\frac{1}{\sqrt{N!}}}$. (For example, if we put 2 bosons into the same state ${\left|\psi_{1}\right\rangle }$, the wave function is ${\frac{1}{2}\left|\psi_{1}\right\rangle \left|\psi_{1}\right\rangle }$, not ${\frac{1}{\sqrt{2}}\left|\psi_{1}\right\rangle \left|\psi_{1}\right\rangle }$.) The ${P}$ refers to a permutation of the integers ${1,\dots,N}$, and ${P\left(i\right)}$ is the ${i}$th integer in the permutation ${P}$. The product term ${\prod_{i=1}^{N}\left|\psi_{P\left(i\right)}\right\rangle }$ is a product of ${N}$ single-particle states in a certain order, where the order of the state in the product determines its spatial coordinate. For example, if ${N=3}$ then one permutation is ${P=3,1,2}$ so for that permutation

$\displaystyle \prod_{i=1}^{3}\left|\psi_{P\left(i\right)}\right\rangle =\left|\psi_{3}\left(\mathbf{x}_{1}\right)\right\rangle \left|\psi_{1}\left(\mathbf{x}_{2}\right)\right\rangle \left|\psi_{2}\left(\mathbf{x}_{3}\right)\right\rangle \ \ \ \ \ (5)$

The factor ${\xi}$ is ${+1}$ for bosons and ${-1}$ for fermions, and the sum over ${P}$ sums over all ${N!}$ possible permutations of the integers ${1,\dots,N}$, so the state ${\left|\psi_{1},\dots,\psi_{N}\right\rangle }$ consists of ${N!}$ terms, each of which contains a product of ${N}$ different single particle states. [The fact that the states are all different isn’t mentioned in L&B’s book, but it seems to me that this is a necessary condition.]

For the purposes of using ${P}$ as an exponent in ${\xi^{P}}$, ${P}$ can be regarded as the number of swaps of integers in the original sequence ${1,\dots,N}$ are required to get the permutation ${P}$. Thus, to get 3,1,2 from 1,2,3 we swap 1 with 2, then 2 with 3, so there are 2 swaps. Permutations requiring an even (odd) number of swaps are called even (odd) permutations.

Now suppose we have a different ${N}$-particle state given by

$\displaystyle \left|\chi_{1},\dots,\chi_{N}\right\rangle =\frac{1}{\sqrt{N!}}\sum_{Q}\xi^{Q}\prod_{j=1}^{N}\left|\chi_{Q\left(j\right)}\right\rangle \ \ \ \ \ (6)$

The single-particle states ${\left|\chi_{i}\right\rangle }$ also form a complete orthonormal set, but they could be a different such set from the ${\left|\psi_{i}\right\rangle }$. If we take the inner product of these two ${N}$-particle states we get the rather horrible expression

$\displaystyle \left\langle \chi_{1},\dots,\chi_{N}\left|\psi_{1},\dots,\psi_{N}\right.\right\rangle =\frac{1}{N!}\sum_{P,Q}\xi^{P+Q}\prod_{i=1}^{N}\left\langle \chi_{Q\left(i\right)}\left|\psi_{P\left(i\right)}\right.\right\rangle \ \ \ \ \ (7)$

We need only one product since we are summing over both permutations ${P}$ and ${Q}$, so we get all possible inner products between terms from 4 and 6.

Example 1 For example, if ${N=2}$, then for fermions

 $\displaystyle \left|\psi_{1}\psi_{2}\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(\left|\psi_{1}\left(\mathbf{x}_{1}\right)\right\rangle \left|\psi_{2}\left(\mathbf{x}_{2}\right)\right\rangle -\left|\psi_{1}\left(\mathbf{x}_{2}\right)\right\rangle \left|\psi_{2}\left(\mathbf{x}_{1}\right)\right\rangle \right)\ \ \ \ \ (8)$ $\displaystyle \left|\chi_{1}\chi_{2}\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(\left|\chi_{1}\left(\mathbf{x}_{1}\right)\right\rangle \left|\chi_{2}\left(\mathbf{x}_{2}\right)\right\rangle -\left|\chi_{1}\left(\mathbf{x}_{2}\right)\right\rangle \left|\chi_{2}\left(\mathbf{x}_{1}\right)\right\rangle \right) \ \ \ \ \ (9)$

On the RHS, we can form inner products only between single-particle states that use the same spatial coordinate and, since ${\left\langle \chi_{i}\left(\mathbf{x}_{1}\right)\left|\psi_{j}\left(\mathbf{x}_{1}\right)\right.\right\rangle =\left\langle \chi_{i}\left(\mathbf{x}_{2}\right)\left|\psi_{j}\left(\mathbf{x}_{2}\right)\right.\right\rangle }$ (since we’re integrating over all space on both sides, the integration coordinate doesn’t matter) we get

 $\displaystyle \left\langle \chi_{1}\chi_{2}\left|\psi_{1}\psi_{2}\right.\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[2\left\langle \chi_{1}\left|\psi_{1}\right.\right\rangle \left\langle \chi_{2}\left|\psi_{2}\right.\right\rangle -2\left\langle \chi_{1}\left|\psi_{2}\right.\right\rangle \left\langle \chi_{2}\left|\psi_{1}\right.\right\rangle \right]\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle \chi_{1}\left|\psi_{1}\right.\right\rangle \left\langle \chi_{2}\left|\psi_{2}\right.\right\rangle -\left\langle \chi_{1}\left|\psi_{2}\right.\right\rangle \left\langle \chi_{2}\left|\psi_{1}\right.\right\rangle \ \ \ \ \ (11)$

For any given permutation of the ${\psi_{i}}$ or ${\chi_{i}}$, the position coordinates can be distributed among the ${N}$ single-particle states in ${N!}$ ways. If we choose a permutation ${Q}$ for ${\left|\chi_{1},\dots,\chi_{N}\right\rangle }$ and ${P}$ for ${\left|\psi_{1},\dots,\psi_{N}\right\rangle }$, then the product ${\prod_{i=1}^{N}\left\langle \chi_{Q\left(i\right)}\left|\psi_{P\left(i\right)}\right.\right\rangle }$ occurs ${N!}$ times because of the ${N!}$ ways of assigning positions. For example, for ${N=2}$, we can choose ${Q=1,2}$ and ${P=1,2}$ so that

$\displaystyle \prod_{i=1}^{2}\left\langle \chi_{Q\left(i\right)}\left|\psi_{P\left(i\right)}\right.\right\rangle =\left\langle \chi_{1}\left|\psi_{1}\right.\right\rangle \left\langle \chi_{2}\left|\psi_{2}\right.\right\rangle \ \ \ \ \ (12)$

This combination can occur with ${\chi_{1}}$ and ${\psi_{1}}$ functions of ${\mathbf{x}_{1}}$ and ${\chi_{2}}$ and ${\psi_{2}}$ functions of ${\mathbf{x}_{2}}$ or ${\chi_{1}}$ and ${\psi_{1}}$ functions of ${\mathbf{x}_{2}}$ and ${\chi_{2}}$ and ${\psi_{2}}$ functions of ${\mathbf{x}_{1}}$. In general, for any pairing of ${\chi_{i}}$ states with ${\psi_{i}}$ states, there are ${N!}$ ways of distributing the position coordinates, each of which gives the same product of single-particle inner products. We can rewrite this by always ordering the ${\chi_{i}}$ states in their original order ${1,\dots,N}$ and pairing this ordering with each permutation ${P}$ of ${\psi_{i}}$ states. That is

 $\displaystyle \left\langle \chi_{1},\dots,\chi_{N}\left|\psi_{1},\dots,\psi_{N}\right.\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{N!}\sum_{P}\xi^{P}N!\prod_{i=1}^{N}\left\langle \chi_{i}\left|\psi_{P\left(i\right)}\right.\right\rangle \ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{P}\xi^{P}\prod_{i=1}^{N}\left\langle \chi_{i}\left|\psi_{P\left(i\right)}\right.\right\rangle \ \ \ \ \ (14)$

For fermions (with ${\xi=-1}$) this is actually the definition of the determinant of a matrix (we’ll accept this mathematical result…a proof would take too long):

$\displaystyle \left\langle \chi_{1},\dots,\chi_{N}\left|\psi_{1},\dots,\psi_{N}\right.\right\rangle _{\mbox{fermions}}=\left|\begin{array}{cccc} \left\langle \chi_{1}\left|\psi_{1}\right.\right\rangle & \left\langle \chi_{1}\left|\psi_{2}\right.\right\rangle & \ldots & \left\langle \chi_{1}\left|\psi_{N}\right.\right\rangle \\ \left\langle \chi_{2}\left|\psi_{1}\right.\right\rangle & \left\langle \chi_{2}\left|\psi_{2}\right.\right\rangle & \ldots & \left\langle \chi_{2}\left|\psi_{N}\right.\right\rangle \\ \vdots & \vdots & \ddots & \vdots\\ \left\langle \chi_{N}\left|\psi_{1}\right.\right\rangle & \left\langle \chi_{N}\left|\psi_{2}\right.\right\rangle & \ldots & \left\langle \chi_{N}\left|\psi_{N}\right.\right\rangle \end{array}\right| \ \ \ \ \ (15)$

For bosons, the equivalent structure is called the permanent of a matrix. A permanent is the same as a determinant except all the minus signs are replaced by plus signs. It doesn’t seem to have its own notation so we’ll just write it as ‘perm’.

$\displaystyle \left\langle \chi_{1},\dots,\chi_{N}\left|\psi_{1},\dots,\psi_{N}\right.\right\rangle _{\mbox{bosons}}=\mbox{perm}\left[\begin{array}{cccc} \left\langle \chi_{1}\left|\psi_{1}\right.\right\rangle & \left\langle \chi_{1}\left|\psi_{2}\right.\right\rangle & \ldots & \left\langle \chi_{1}\left|\psi_{N}\right.\right\rangle \\ \left\langle \chi_{2}\left|\psi_{1}\right.\right\rangle & \left\langle \chi_{2}\left|\psi_{2}\right.\right\rangle & \ldots & \left\langle \chi_{2}\left|\psi_{N}\right.\right\rangle \\ \vdots & \vdots & \ddots & \vdots\\ \left\langle \chi_{N}\left|\psi_{1}\right.\right\rangle & \left\langle \chi_{N}\left|\psi_{2}\right.\right\rangle & \ldots & \left\langle \chi_{N}\left|\psi_{N}\right.\right\rangle \end{array}\right] \ \ \ \ \ (16)$

Now we can apply a creation operator ${a_{\phi}^{\dagger}}$ to the state ${\left|\psi_{1},\dots,\psi_{N}\right\rangle }$ (as far as I can tell, the state ${\phi}$ can be any state, including one that is a linear combination of the ${\psi_{i}}$). This gives

$\displaystyle a_{\phi}^{\dagger}\left|\psi_{1},\dots,\psi_{N}\right\rangle =\left|\phi,\psi_{1},\dots,\psi_{N}\right\rangle \ \ \ \ \ (17)$

If we want to discover the action of an annihilation operator, things are a bit more complicated, since we can choose to annihilate a linear combination of the basis states rather than just a single basis state. Again, as far as I can tell, this annihilation operation works only on the original ${N}$-particle state ${\left|\psi_{1},\dots,\psi_{N}\right\rangle }$. We want to find ${a_{\phi}\left|\psi_{1},\dots,\psi_{N}\right\rangle }$ so we take the inner product with some other state ${\left|\chi_{1},\dots,\chi_{N-1}\right\rangle }$ (we use an ${N-1}$ particle state so that the number of particles match up on both sides):

 $\displaystyle \left\langle \chi_{1},\dots,\chi_{N-1}\left|a_{\phi}\right|\psi_{1},\dots,\psi_{N}\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle \psi_{1},\dots,\psi_{N-1}\left|a_{\phi}^{\dagger}\right|\chi_{1},\dots,\chi_{N}\right\rangle ^*\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle \psi_{1},\dots,\psi_{N}\left|\phi,\chi_{1},\dots,\chi_{N-1}\right.\right\rangle ^*\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left|\begin{array}{cccc} \left\langle \psi_{1}\left|\phi\right.\right\rangle & \left\langle \psi_{1}\left|\chi_{1}\right.\right\rangle & \ldots & \left\langle \psi_{1}\left|\chi_{N-1}\right.\right\rangle \\ \left\langle \psi_{2}\left|\phi\right.\right\rangle & \left\langle \psi_{2}\left|\chi_{1}\right.\right\rangle & \ldots & \left\langle \psi_{2}\left|\chi_{N-1}\right.\right\rangle \\ \vdots & \vdots & \ddots & \vdots\\ \left\langle \psi_{N}\left|\phi\right.\right\rangle & \left\langle \psi_{N}\left|\chi_{1}\right.\right\rangle & \ldots & \left\langle \psi_{N}\left|\chi_{N-1}\right.\right\rangle \end{array}\right|_{\xi}^* \ \ \ \ \ (20)$

where the subscript ${\xi}$ on the determinant means to use the permanent if we’re talking about bosons so that ${\xi=1}$. From here on, I mean ‘determinant or permanent’ whenever I say ‘determinant’. We can expand the determinant about the first column to get

 $\displaystyle \left\langle \chi_{1},\dots,\chi_{N-1}\left|a_{\phi}\right|\psi_{1},\dots,\psi_{N}\right\rangle$ $\displaystyle =$ $\displaystyle \sum_{k=1}^{N}\xi^{k-1}\left\langle \psi_{k}\left|\phi\right.\right\rangle ^*\left\langle \psi_{1},\dots\left(\mbox{no }\psi_{k}\right),\ldots,\psi_{N}\left|\chi_{1},\dots,\chi_{N-1}\right.\right\rangle ^*\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{k=1}^{N}\xi^{k-1}\left\langle \phi\left|\psi_{k}\right.\right\rangle \left\langle \chi_{1},\dots,\chi_{N-1}\left|\psi_{1},\dots\left(\mbox{no }\psi_{k}\right),\ldots,\psi_{N}\right.\right\rangle \ \ \ \ \ (22)$

where the state ${\left|\psi_{1},\dots\left(\mbox{no }\psi_{k}\right),\ldots,\psi_{N}\right\rangle }$ is the state ${\left|\psi_{1},\dots,\psi_{N-1}\right\rangle }$ without ${\psi_{k}}$. The clever thing about this form is that we made no assumptions about the state ${\left|\chi_{1},\dots,\chi_{N-1}\right\rangle }$ so we can remove it from both sides to get

$\displaystyle a_{\phi}\left|\psi_{1},\dots,\psi_{N}\right\rangle =\sum_{k=1}^{N}\xi^{k-1}\left\langle \phi\left|\psi_{k}\right.\right\rangle \left|\psi_{1},\dots\left(\mbox{no }\psi_{k}\right),\ldots,\psi_{N}\right\rangle \ \ \ \ \ (23)$

Example 2 Suppose we have a 3-boson system and

$\displaystyle \left|\phi\right\rangle =\frac{\sqrt{2}}{3}\left|\psi_{1}\right\rangle +\frac{\sqrt{6}}{3}\left|\psi_{2}\right\rangle +\frac{1}{3}\left|\psi_{3}\right\rangle \ \ \ \ \ (24)$

Then

 $\displaystyle \left\langle \phi\left|\psi_{1}\right.\right\rangle$ $\displaystyle =$ $\displaystyle \frac{\sqrt{2}}{3}\ \ \ \ \ (25)$ $\displaystyle \left\langle \phi\left|\psi_{2}\right.\right\rangle$ $\displaystyle =$ $\displaystyle \frac{\sqrt{6}}{3}\ \ \ \ \ (26)$ $\displaystyle \left\langle \phi\left|\psi_{3}\right.\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{3} \ \ \ \ \ (27)$

so

$\displaystyle a_{\phi}\left|\psi_{1}\psi_{2}\psi_{3}\right\rangle =\frac{\sqrt{2}}{3}\left|\psi_{2}\psi_{3}\right\rangle +\frac{\sqrt{6}}{3}\left|\psi_{1}\psi_{3}\right\rangle +\frac{1}{3}\left|\psi_{1}\psi_{2}\right\rangle \ \ \ \ \ (28)$

Now suppose we apply a creation operator:

 $\displaystyle a_{\alpha}^{\dagger}a_{\phi}\left|\psi_{1},\dots,\psi_{N}\right\rangle$ $\displaystyle =$ $\displaystyle \sum_{k=1}^{N}\xi^{k-1}\left\langle \phi\left|\psi_{k}\right.\right\rangle a_{\alpha}^{\dagger}\left|\psi_{1},\dots\left(\mbox{no }\psi_{k}\right),\ldots,\psi_{N}\right\rangle \ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{k=1}^{N}\xi^{k-1}\left\langle \phi\left|\psi_{k}\right.\right\rangle \left|\alpha,\psi_{1},\dots\left(\mbox{no }\psi_{k}\right),\ldots,\psi_{N}\right\rangle \ \ \ \ \ (30)$

For the fermion case, we can swap ${\alpha}$ with all the states ${\psi_{1},\dots,\psi_{k-1}}$ and since swapping rows in a determinant changes the sign, this results in a factor of ${\xi^{k-1}}$ which eliminates the other ${\xi^{k-1}}$ (since ${\xi^{2\left(k-1\right)}=1}$ always), so we get the final form

$\displaystyle a_{\alpha}^{\dagger}a_{\phi}\left|\psi_{1},\dots,\psi_{N}\right\rangle =\sum_{k=1}^{N}\left\langle \phi\left|\psi_{k}\right.\right\rangle \left|\psi_{1},\dots,\psi_{k-1},\alpha,\psi_{k+1},\ldots,\psi_{N}\right\rangle \ \ \ \ \ (31)$

For bosons, swapping columns in a permanent makes no difference to the result and since ${\xi=1}$ in this case we can just ignore the ${\xi^{k-1}}$ factor.

After all this, we can get back to our original operator ${\hat{\mathcal{A}}}$. Since it’s a single-particle operator, we can assume that its multi-particle equivalent’s effect on a multi-particle state is the sum of the single-particle operator’s effects on each individual particle within the state. That is, from 2 the inner product of the state ${\left\langle \beta\right|}$ is taken with each ${\left|\psi_{k}\right\rangle }$ in turn and the result summed over ${k}$. Calling the multi-particle operator ${\hat{A}}$ we have

 $\displaystyle \hat{A}\left|\psi_{1},\dots,\psi_{N}\right\rangle$ $\displaystyle =$ $\displaystyle \sum_{k}\sum_{\alpha,\beta}\left|\alpha\right\rangle \mathcal{A}_{\alpha\beta}\left\langle \beta\left|\psi_{k}\right.\right\rangle \left|\psi_{1},\dots\left(\mbox{no }\psi_{k}\right),\ldots,\psi_{N}\right\rangle \ \ \ \ \ (32)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{\alpha,\beta}\mathcal{A}_{\alpha\beta}\left\langle \beta\left|\psi_{k}\right.\right\rangle \left|\psi_{1},\dots,\psi_{k-1},\alpha,\psi_{k+1},\ldots,\psi_{N}\right\rangle \ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{\alpha,\beta}\mathcal{A}_{\alpha\beta}a_{\alpha}^{\dagger}a_{\beta}\left|\psi_{1},\dots,\psi_{N}\right\rangle \ \ \ \ \ (34)$

In the second line we inserted the state ${\left|\alpha\right\rangle }$ into ${\left|\psi_{1},\dots\left(\mbox{no }\psi_{k}\right),\ldots,\psi_{N}\right\rangle }$ at the position occupied by ${\psi_{k}}$ since ${\hat{\mathcal{A}}}$ is a single-particle operator so it operates on the same coordinate throughout (that is ${\left\langle \beta\right|}$ operates on the same coordinate as ${\left|\alpha\right\rangle }$). Therefore the multi-particle operator is

$\displaystyle \hat{A}=\sum_{\alpha,\beta}\mathcal{A}_{\alpha\beta}a_{\alpha}^{\dagger}a_{\beta} \ \ \ \ \ (35)$

We can think of this as the operator looking for a particle (or component of a particle) in each state ${\beta}$, removing that particle and operating on it with the single-particle operator ${\hat{\mathcal{A}}}$ and then reinserting the particle in state ${\left|\alpha\right\rangle }$.

Field operators for the infinite square well

Reference: Tom Lancaster and Stephen J. Blundell, Quantum Field Theory for the Gifted Amateur, (Oxford University Press, 2014), Section 4.1.

The creation and annihilation operators ${\hat{a}_{\mathbf{p}}^{\dagger}}$ and ${\hat{a}_{\mathbf{p}}}$ create and annihilate a particle in a specific momentum state so, because of the uncertainty principle, such states are completely unlocalized in position. We can construct analogous operators that create and annihilate a particle at a specific position. Such operators are called field operators. As you would expect, a field operator, operating at a precise position, must be completely unlocalized in momentum.

The creation and annihilation field operators for a particle in a 3-d infinite square well are defined as

 $\displaystyle \hat{\psi}^{\dagger}\left(\mathbf{x}\right)$ $\displaystyle \equiv$ $\displaystyle \frac{1}{\sqrt{\mathcal{V}}}\sum_{\mathbf{p}}\hat{a}_{\mathbf{p}}^{\dagger}e^{-i\mathbf{p}\cdot\mathbf{x}}\ \ \ \ \ (1)$ $\displaystyle \hat{\psi}\left(\mathbf{x}\right)$ $\displaystyle \equiv$ $\displaystyle \frac{1}{\sqrt{\mathcal{V}}}\sum_{\mathbf{p}}\hat{a}_{\mathbf{p}}e^{i\mathbf{p}\cdot\mathbf{x}} \ \ \ \ \ (2)$

where ${\mathcal{V}}$ is the volume of the square well.

To see that they actually do create and annihilate a particle at position ${\mathbf{x}}$ we’ll have a look at their effect when operating on particular states.

First, we’ll apply ${\hat{\psi}^{\dagger}\left(x\right)}$ to the vacuum state.

 $\displaystyle \hat{\psi}^{\dagger}\left(\mathbf{x}\right)\left|0\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{\mathcal{V}}}\sum_{\mathbf{p}}e^{-i\mathbf{p}\cdot\mathbf{x}}\hat{a}_{\mathbf{p}}^{\dagger}\left|0\right\rangle \ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{\mathcal{V}}}\sum_{\mathbf{p}}e^{-i\mathbf{p}\cdot\mathbf{x}}\left|\mathbf{p}\right\rangle \ \ \ \ \ (4)$

We can now insert the unit operator in the form

$\displaystyle 1=\int d^{3}\mathbf{y}\left|\mathbf{y}\right\rangle \left\langle \mathbf{y}\right| \ \ \ \ \ (5)$

and we get

style=”text-align:center”> $\displaystyle =$
 $\displaystyle \hat{\psi}^{\dagger}\left(\mathbf{x}\right)\left|0\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{\mathcal{V}}}\int d^{3}\mathbf{y}\sum_{\mathbf{p}}e^{-i\mathbf{p}\cdot\mathbf{x}}\left|\mathbf{y}\right\rangle \left\langle \mathbf{y}\left|\mathbf{p}\right.\right\rangle \ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{\mathcal{V}}}\int d^{3}\mathbf{y}\sum_{\mathbf{p}}e^{-i\mathbf{p}\cdot\mathbf{x}}\left|\mathbf{y}\right\rangle \left[\frac{1}{\sqrt{\mathcal{V}}}e^{i\mathbf{p}\cdot\mathbf{y}}\right]\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int d^{3}\mathbf{y}\left[\frac{1}{\mathcal{V}}\sum_{\mathbf{p}}e^{i\mathbf{p}\cdot\left(\mathbf{y}-\mathbf{x}\right)}\right]\left|\mathbf{y}\right\rangle \ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int d^{3}\mathbf{y}\delta^{\left(3\right)}\left(\mathbf{y}-\mathbf{x}\right)\left|\mathbf{y}\right\rangle \ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle \left|\mathbf{x}\right\rangle \ \ \ \ \ (10)$

So the creation operator ${\hat{\psi}^{\dagger}\left(\mathbf{x}\right)}$ operating on the vacuum state creates a single particle at position ${\mathbf{x}}$.

We can now try the annihilation operator ${\hat{\psi}\left(\mathbf{y}\right)}$ operating on a one-particle state ${\left|\mathbf{x}\right\rangle }$. We get

 $\displaystyle \hat{\psi}\left(\mathbf{y}\right)\left|\mathbf{x}\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{\mathcal{V}}}\sum_{\mathbf{p}}\hat{a}_{\mathbf{p}}e^{i\mathbf{p}\cdot\mathbf{y}}\left|\mathbf{x}\right\rangle \ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{\mathcal{V}}}\sum_{\mathbf{p}.\mathbf{q}}\hat{a}_{\mathbf{p}}e^{i\mathbf{p}\cdot\mathbf{y}}\left|\mathbf{q}\right\rangle \left\langle \mathbf{q}\left|\mathbf{x}\right.\right\rangle \ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{\mathcal{V}}}\sum_{\mathbf{p}.\mathbf{q}}\hat{a}_{\mathbf{p}}e^{i\mathbf{p}\cdot\mathbf{y}}\left|\mathbf{q}\right\rangle \left[\frac{1}{\sqrt{\mathcal{V}}}e^{-i\mathbf{q}\cdot\mathbf{x}}\right]\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\mathcal{V}}\sum_{\mathbf{p},\mathbf{q}}e^{i\left(\mathbf{p}\cdot\mathbf{y}-\mathbf{q}\cdot\mathbf{x}\right)}\hat{a}_{\mathbf{p}}\left|\mathbf{q}\right\rangle \ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\mathcal{V}}\sum_{\mathbf{p},\mathbf{q}}e^{i\left(\mathbf{p}\cdot\mathbf{y}-\mathbf{q}\cdot\mathbf{x}\right)}\delta_{\mathbf{pq}}\left|0\right\rangle \ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\mathcal{V}}\sum_{\mathbf{p}}e^{i\mathbf{p}\cdot\left(\mathbf{y}-\mathbf{x}\right)}\left|0\right\rangle \ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \delta^{\left(3\right)}\left(\mathbf{y}-\mathbf{x}\right)\left|0\right\rangle \ \ \ \ \ (17)$

where in the fourth line the momentum annihilation operator ${\hat{a}_{\mathbf{p}}}$ operating on ${\left|\mathbf{q}\right\rangle }$ produces the vacuum state only if ${\mathbf{p}=\mathbf{q}}$; otherwise it produces zero.

Thus ${\hat{\psi}\left(\mathbf{y}\right)\left|\mathbf{x}\right\rangle }$ produces the vacuum state if ${\mathbf{y}=\mathbf{x}}$ and zero otherwise.

Creation and annihilation operators for the 3-d harmonic oscillator

Reference: Tom Lancaster and Stephen J. Blundell, Quantum Field Theory for the Gifted Amateur, (Oxford University Press, 2014), Problem 3.3.

For the 3-d harmonic oscillator, we can write the hamiltonian as

$\displaystyle \hat{H}=\frac{1}{2m}\left(\hat{p}_{1}^{2}+\hat{p}_{2}^{2}+\hat{p}_{3}^{2}\right)+\frac{m\omega^{2}}{2}\left(\hat{x}_{1}^{2}+\hat{x}_{2}^{2}+\hat{x}_{3}^{2}\right) \ \ \ \ \ (1)$

This is effectively three independent oscillators, one in each of the three coordinate directions, so using the 1-d form of the hamiltonian in terms of creation and annihilation operators, we have

$\displaystyle \hat{H}=\hbar\omega\sum_{i=1}^{3}\left(\frac{1}{2}+\hat{a}_{i}^{\dagger}\hat{a}_{i}\right) \ \ \ \ \ (2)$

Using the position and momentum operators expressed as

 $\displaystyle \hat{x}_{i}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{\hbar}{2m\omega}}\left(\hat{a}_{i}^{\dagger}+\hat{a}_{i}\right)\ \ \ \ \ (3)$ $\displaystyle \hat{p}_{i}$ $\displaystyle =$ $\displaystyle i\sqrt{\frac{\hbar m\omega}{2}}\left(\hat{a}_{i}^{\dagger}-\hat{a}_{i}\right) \ \ \ \ \ (4)$

we can write the components of angular momentum in terms of creation and annihilation operators. For example, the ${z}$ component can be written as

 $\displaystyle L^{3}$ $\displaystyle =$ $\displaystyle \hat{x}_{1}\hat{p}_{2}-\hat{x}_{2}\hat{p}_{1}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{i\hbar}{2}\left[\left(\hat{a}_{1}^{\dagger}+\hat{a}_{1}\right)\left(\hat{a}_{2}^{\dagger}-\hat{a}_{2}\right)-\left(\hat{a}_{2}^{\dagger}+\hat{a}_{2}\right)\left(\hat{a}_{1}^{\dagger}-\hat{a}_{1}\right)\right]\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{i\hbar}{2}\left[-\hat{a}_{1}^{\dagger}\hat{a}_{2}+\hat{a}_{1}\hat{a}_{2}^{\dagger}-\hat{a}_{2}\hat{a}_{1}^{\dagger}+\hat{a}_{2}^{\dagger}\hat{a}_{1}\right]\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -i\hbar\left(\hat{a}_{1}^{\dagger}\hat{a}_{2}-\hat{a}_{2}^{\dagger}\hat{a}_{1}\right) \ \ \ \ \ (8)$

where we’ve used the fact that all operators of one coordinate commute with all operators of the other two coordinates. If we work out the other two coordinates we get the general formula

$\displaystyle L^{i}=-i\hbar\epsilon^{ijk}\hat{a}_{j}^{\dagger}\hat{a}_{k} \ \ \ \ \ (9)$

where ${\epsilon^{ijk}}$ is the Levi-Civita symbol, which is +1 if ${ijk}$ is a even permutation of 1,2,3, ${-1}$ if ${ijk}$ is an odd permutation of 1,2,3 and 0 if any two of ${ijk}$ are equal, and repeated indices are summed.

We can define some new creation and annihilation operators as follows:

 $\displaystyle \hat{b}_{1}^{\dagger}$ $\displaystyle \equiv$ $\displaystyle -\frac{1}{\sqrt{2}}\left(\hat{a}_{1}^{\dagger}+i\hat{a}_{2}^{\dagger}\right)\ \ \ \ \ (10)$ $\displaystyle \hat{b}_{0}^{\dagger}$ $\displaystyle \equiv$ $\displaystyle \hat{a}_{3}^{\dagger}\ \ \ \ \ (11)$ $\displaystyle \hat{b}_{-1}^{\dagger}$ $\displaystyle \equiv$ $\displaystyle \frac{1}{\sqrt{2}}\left(\hat{a}_{1}^{\dagger}-i\hat{a}_{2}^{\dagger}\right) \ \ \ \ \ (12)$
 $\displaystyle \hat{b}_{1}$ $\displaystyle \equiv$ $\displaystyle -\frac{1}{\sqrt{2}}\left(\hat{a}_{1}-i\hat{a}_{2}\right)\ \ \ \ \ (13)$ $\displaystyle \hat{b}_{0}$ $\displaystyle \equiv$ $\displaystyle \hat{a}_{3}\ \ \ \ \ (14)$ $\displaystyle \hat{b}_{-1}$ $\displaystyle \equiv$ $\displaystyle \frac{1}{\sqrt{2}}\left(\hat{a}_{1}+i\hat{a}_{2}\right) \ \ \ \ \ (15)$

From the commutation relations for ${\hat{a}_{i}^{\dagger}}$ and ${\hat{a}_{i}}$:

$\displaystyle \left[\hat{a}_{i},\hat{a}_{j}^{\dagger}\right]=\delta_{ij} \ \ \ \ \ (16)$

we get

 $\displaystyle \left[\hat{b}_{0},\hat{b}_{j}^{\dagger}\right]$ $\displaystyle =$ $\displaystyle \delta_{0,j}\ \ \ \ \ (17)$ $\displaystyle \left[\hat{b}_{1},\hat{b}_{1}^{\dagger}\right]$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(\left[\hat{a}_{1},\hat{a}_{1}^{\dagger}\right]+\left(-i\right)i\left[\hat{a}_{2},\hat{a}_{2}^{\dagger}\right]\right)\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(1+1\right)\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (20)$ $\displaystyle \left[\hat{b}_{1},\hat{b}_{-1}^{\dagger}\right]$ $\displaystyle =$ $\displaystyle -\frac{1}{2}\left(\left[\hat{a}_{1},\hat{a}_{1}^{\dagger}\right]+i^{2}\left[\hat{a}_{2},\hat{a}_{2}^{\dagger}\right]\right)\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (22)$ $\displaystyle \left[\hat{b}_{-1},\hat{b}_{-1}^{\dagger}\right]$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(\left[\hat{a}_{1},\hat{a}_{1}^{\dagger}\right]+\left(-i\right)i\left[\hat{a}_{2},\hat{a}_{2}^{\dagger}\right]\right)\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (24)$ $\displaystyle \left[\hat{b}_{-1},\hat{b}_{1}^{\dagger}\right]$ $\displaystyle =$ $\displaystyle -\frac{1}{2}\left(\left[\hat{a}_{1},\hat{a}_{1}^{\dagger}\right]+i^{2}\left[\hat{a}_{2},\hat{a}_{2}^{\dagger}\right]\right)\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (26)$

so in general

$\displaystyle \left[\hat{b}_{i},\hat{b}_{j}^{\dagger}\right]=\delta_{ij} \ \ \ \ \ (27)$

To express the hamiltonian 2 in terms of the new operators, we need

 $\displaystyle \hat{b}_{1}^{\dagger}\hat{b}_{1}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\hat{a}_{1}^{\dagger}\hat{a}_{1}+\hat{a}_{2}^{\dagger}\hat{a}_{2}+i\left(\hat{a}_{2}^{\dagger}\hat{a}_{1}-\hat{a}_{1}^{\dagger}\hat{a}_{2}\right)\right]\ \ \ \ \ (28)$ $\displaystyle \hat{b}_{-1}^{\dagger}\hat{b}_{-1}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\hat{a}_{1}^{\dagger}\hat{a}_{1}+\hat{a}_{2}^{\dagger}\hat{a}_{2}-i\left(\hat{a}_{2}^{\dagger}\hat{a}_{1}-\hat{a}_{1}^{\dagger}\hat{a}_{2}\right)\right]\ \ \ \ \ (29)$ $\displaystyle \hat{b}_{1}^{\dagger}\hat{b}_{1}+\hat{b}_{-1}^{\dagger}\hat{b}_{-1}$ $\displaystyle =$ $\displaystyle \hat{a}_{1}^{\dagger}\hat{a}_{1}+\hat{a}_{2}^{\dagger}\hat{a}_{2}\ \ \ \ \ (30)$ $\displaystyle \hat{H}$ $\displaystyle =$ $\displaystyle \hbar\omega\sum_{i=-1}^{+1}\left(\frac{1}{2}+\hat{b}_{i}^{\dagger}\hat{b}_{i}\right) \ \ \ \ \ (31)$

Finally, we can write ${\hat{L}^{3}}$ in terms of the new operators.

 $\displaystyle -\hat{b}_{-1}^{\dagger}\hat{b}_{-1}+0\times\hat{b}_{0}^{\dagger}\hat{b}_{0}+\hat{b}_{1}^{\dagger}\hat{b}_{1}$ $\displaystyle =$ $\displaystyle i\left(\hat{a}_{2}^{\dagger}\hat{a}_{1}-\hat{a}_{1}^{\dagger}\hat{a}_{2}\right)\ \ \ \ \ (32)$ $\displaystyle \hat{L}^{3}$ $\displaystyle =$ $\displaystyle \hbar\sum_{m=-1}^{+1}m\hat{b}_{m}^{\dagger}\hat{b}_{m} \ \ \ \ \ (33)$

Creation and annihilation operators in the harmonic oscillator: a few theorems

Reference: Tom Lancaster and Stephen J. Blundell, Quantum Field Theory for the Gifted Amateur, (Oxford University Press, 2014), Problem 3.2.

For the harmonic oscillator, we’ve seen that the effects of the creation and annihilation (raising and lowering) operators are

 $\displaystyle \hat{a}^{\dagger}\left|n\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{n+1}\left|n+1\right\rangle \ \ \ \ \ (1)$ $\displaystyle \hat{a}\left|n\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{n}\left|n-1\right\rangle \ \ \ \ \ (2)$

Therefore

 $\displaystyle \left\langle m\left|\hat{a}^{\dagger}\right|n\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{n+1}\left\langle m\left|n+1\right.\right\rangle \ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{n+1}\delta_{m,n+1}\ \ \ \ \ (4)$ $\displaystyle \left\langle m\left|\hat{a}\right|n\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{n}\left\langle m\left|n-1\right.\right\rangle \ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{n}\delta_{m,n-1} \ \ \ \ \ (6)$

From the commutation relation

$\displaystyle \left[\hat{a},\hat{a}^{\dagger}\right]=1 \ \ \ \ \ (7)$

we’d like to prove that

$\displaystyle \left[\hat{a},\left(\hat{a}^{\dagger}\right)^{n}\right]=n\left(\hat{a}^{\dagger}\right)^{n-1} \ \ \ \ \ (8)$

We can prove this using mathematical induction. The formula is true for ${n=1}$, so we assume it’s true for ${n-1}$ and then show it’s true for ${n}$.

We get

 $\displaystyle \left[\hat{a},\left(\hat{a}^{\dagger}\right)^{n}\right]$ $\displaystyle =$ $\displaystyle \hat{a}\left(\hat{a}^{\dagger}\right)^{n}-\left(\hat{a}^{\dagger}\right)^{n}\hat{a}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(1+\hat{a}^{\dagger}\hat{a}\right)\left(\hat{a}^{\dagger}\right)^{n-1}-\left(\hat{a}^{\dagger}\right)^{n}\hat{a}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\hat{a}^{\dagger}\right)^{n-1}+\hat{a}^{\dagger}\left[\hat{a},\left(\hat{a}^{\dagger}\right)^{n-1}\right] \ \ \ \ \ (11)$

We can now use our assumption that ${\left[\hat{a},\left(\hat{a}^{\dagger}\right)^{n-1}\right]=\left(n-1\right)\left(\hat{a}^{\dagger}\right)^{n-2}}$ and we get

 $\displaystyle \left[\hat{a},\left(\hat{a}^{\dagger}\right)^{n}\right]$ $\displaystyle =$ $\displaystyle \left(\hat{a}^{\dagger}\right)^{n-1}+\hat{a}^{\dagger}\left(n-1\right)\left(\hat{a}^{\dagger}\right)^{n-2}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle n\left(\hat{a}^{\dagger}\right)^{n-1} \ \ \ \ \ (13)$

QED.

We can also work out

 $\displaystyle \left\langle 0\left|\hat{a}^{m}\left(\hat{a}^{\dagger}\right)^{n}\right|0\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\left(1!\right)}\left\langle 0\left|\hat{a}^{m}\left(\hat{a}^{\dagger}\right)^{n-1}\right|1\right\rangle \ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\left(2!\right)}\left\langle 0\left|\hat{a}^{m}\left(\hat{a}^{\dagger}\right)^{n-2}\right|2\right\rangle \ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \ldots\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{n!}\left\langle 0\left|\hat{a}^{m}\right|n\right\rangle \ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle n\sqrt{\left(n-1\right)!}\left\langle 0\left|\hat{a}^{m-1}\right|n-1\right\rangle \ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \ldots \ \ \ \ \ (19)$

If ${m=n}$, the sequence of annihilation operators will eventually reduce the right-hand state to ${\left|0\right\rangle }$ with a factor of ${n!}$ out front. If ${m then ${\hat{a}^{m}}$ will reduce the right-hand state to ${\left|n-m\right\rangle }$ and ${\left\langle 0\left|n-m\right.\right\rangle =\delta_{nm}=0}$. If ${m>n}$, then after applying ${\hat{a}^{n}}$ we’ll still have ${\hat{a}^{m-n}}$ left over so we’ll have an annihilation operator acting on ${\left|0\right\rangle }$ which gives zero. Therefore

$\displaystyle \left\langle 0\left|\hat{a}^{m}\left(\hat{a}^{\dagger}\right)^{n}\right|0\right\rangle =n!\delta_{nm} \ \ \ \ \ (20)$