# Fermions and bosons in the infinite square well

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 10, Exercise 10.3.4.

Suppose we have two identical particles in an infinite square well. The energy levels in a well of width ${L}$ are

$\displaystyle E=\frac{\left(\pi n\hbar\right)^{2}}{2mL^{2}} \ \ \ \ \ (1)$

where ${n=1,2,3,\ldots}$ The corresponding wave functions are given by

$\displaystyle \psi_{n}\left(x\right)=\sqrt{\frac{2}{L}}\sin\frac{n\pi x}{L} \ \ \ \ \ (2)$

If the total energy of the two particles is ${\pi^{2}\hbar^{2}/mL^{2}}$, the only possible configuration is for both particles to be in the ground state ${n=1}$. This means the particles must be bosons, so the state vector is

$\displaystyle \left|x_{1},x_{2}\right\rangle =\frac{2}{L}\sin\frac{\pi x_{1}}{L}\sin\frac{\pi x_{2}}{L} \ \ \ \ \ (3)$

If the total energy is ${5\pi^{2}\hbar^{2}/2mL^{2}}$, then one particle is in the state ${n=1}$ and the other is in ${n=2}$. Since the states are different, the particles can be either bosons or fermions. For bosons, the state vector is

 $\displaystyle \left|x_{1},x_{2}\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left[\frac{2}{L}\sin\frac{\pi x_{1}}{L}\sin\frac{2\pi x_{2}}{L}+\frac{2}{L}\sin\frac{2\pi x_{1}}{L}\sin\frac{\pi x_{2}}{L}\right]\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\sqrt{2}}{L}\left[\sin\frac{\pi x_{1}}{L}\sin\frac{2\pi x_{2}}{L}+\sin\frac{2\pi x_{1}}{L}\sin\frac{\pi x_{2}}{L}\right] \ \ \ \ \ (5)$

For fermions, the state must be antisymmetric, so we have

$\displaystyle \left|x_{1},x_{2}\right\rangle =\frac{\sqrt{2}}{L}\left[\sin\frac{\pi x_{1}}{L}\sin\frac{2\pi x_{2}}{L}-\sin\frac{2\pi x_{1}}{L}\sin\frac{\pi x_{2}}{L}\right] \ \ \ \ \ (6)$

# Invariance of symmetric and antisymmetric states; exchange operators

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 10, Exercise 10.3.5.

In a system with two particles, the state in the ${X}$ basis is given by ${\left|x_{1},x_{2}\right\rangle }$ where ${x_{i}}$ is the position of particle ${i}$. We can define the exchange operator ${P_{12}}$ as an operator that swaps the two particles, so that

$\displaystyle P_{12}\left|x_{1},x_{2}\right\rangle =\left|x_{2},x_{1}\right\rangle \ \ \ \ \ (1)$

To find the eigenvalues and eigenvectors of ${P_{12}}$ we have

$\displaystyle P_{12}\left|\psi\left(x_{1},x_{2}\right)\right\rangle =\alpha\left|\psi\left(x_{1},x_{2}\right)\right\rangle =\psi\left(x_{2},x_{1}\right) \ \ \ \ \ (2)$

where ${\alpha}$ is the eigenvalue and ${\left|\psi\left(x_{1},x_{2}\right)\right\rangle }$ is the eigenvector. Using the same argument as before, we can write

 $\displaystyle \left|\psi\left(x_{1},x_{2}\right)\right\rangle$ $\displaystyle =$ $\displaystyle \beta\left|x_{1},x_{2}\right\rangle +\gamma\left|x_{2},x_{1}\right\rangle \ \ \ \ \ (3)$ $\displaystyle \left|\psi\left(x_{2},x_{1}\right)\right\rangle$ $\displaystyle =$ $\displaystyle \beta\left|x_{2},x_{1}\right\rangle +\gamma\left|x_{1},x_{2}\right\rangle \ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \alpha\left[\beta\left|x_{1},x_{2}\right\rangle +\gamma\left|x_{2},x_{1}\right\rangle \right] \ \ \ \ \ (5)$

Equating coefficients in the first and third lines, we arrive at

$\displaystyle \alpha=\pm1 \ \ \ \ \ (6)$

which gives the same symmetric and antisymmetric eigenfunctions that we had before:

 $\displaystyle \psi_{S}\left(x_{1},x_{2}\right)$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(\left|x_{1},x_{2}\right\rangle +\left|x_{2},x_{1}\right\rangle \right)\ \ \ \ \ (7)$ $\displaystyle \psi_{A}\left(x_{1},x_{2}\right)$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(\left|x_{1},x_{2}\right\rangle -\left|x_{2},x_{1}\right\rangle \right) \ \ \ \ \ (8)$

We can derive a couple of other properties of the exchange operator by noting that if it is applied twice in succession, we get the original state back, so that

 $\displaystyle P_{12}^{2}$ $\displaystyle =$ $\displaystyle I\ \ \ \ \ (9)$ $\displaystyle P_{12}$ $\displaystyle =$ $\displaystyle P_{12}^{-1} \ \ \ \ \ (10)$

Thus the operator is its own inverse.

Consider also the two states ${\left|x_{1}^{\prime},x_{2}^{\prime}\right\rangle }$ and ${\left|x_{1},x_{2}\right\rangle }$. Then

 $\displaystyle \left\langle x_{1}^{\prime},x_{2}^{\prime}\left|P_{12}^{\dagger}P_{12}\right|x_{1},x_{2}\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle P_{12}x_{1}^{\prime},x_{2}^{\prime}\left|P_{12}x_{1},x_{2}\right.\right\rangle \ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle x_{2}^{\prime},x_{1}^{\prime}\left|x_{2},x_{1}\right.\right\rangle \ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\left\langle x_{2}^{\prime}\right|\otimes\left\langle x_{1}^{\prime}\right|\right)\left(\left|x_{2}\right\rangle \otimes\left|x_{1}\right\rangle \right)\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \delta\left(x_{2}^{\prime}-x_{2}\right)\delta\left(x_{1}^{\prime}-x_{1}\right) \ \ \ \ \ (14)$

However, the last line is just equal to the inner product of the original states, that is

$\displaystyle \left\langle x_{1}^{\prime},x_{2}^{\prime}\left|x_{1},x_{2}\right.\right\rangle =\delta\left(x_{2}-x_{2}^{\prime}\right)\delta\left(x_{1}-x_{1}^{\prime}\right)=\delta\left(x_{2}^{\prime}-x_{2}\right)\delta\left(x_{1}^{\prime}-x_{1}\right) \ \ \ \ \ (15)$

This means that

 $\displaystyle P_{12}^{\dagger}P_{12}$ $\displaystyle =$ $\displaystyle I\ \ \ \ \ (16)$ $\displaystyle P_{12}^{\dagger}$ $\displaystyle =$ $\displaystyle P_{12}^{-1}=P_{12} \ \ \ \ \ (17)$

Thus ${P_{12}}$ is both Hermitian and unitary.

Shankar asks us to show that, for a general basis vector ${\left|\omega_{1},\omega_{2}\right\rangle }$, ${P_{12}\left|\omega_{1},\omega_{2}\right\rangle =\left|\omega_{2},\omega_{1}\right\rangle }$. One argument could be that, since the ${X}$ basis spans the space, we can express any other vector such as ${\left|\omega_{1},\omega_{2}\right\rangle }$ as a linear combination of the ${\left|x_{1},x_{2}\right\rangle }$ vectors, so that applying ${P_{12}}$ to ${\left|\omega_{1},\omega_{2}\right\rangle }$ means applying it to a sum of ${\left|x_{1},x_{2}\right\rangle }$ vectors, which swaps the two particles in every term. I’m not sure if this is a rigorous result. In any case, if we accept this result it shows that if we start in a state that is totally symmetric (that is, a boson state), this state is an eigenvector of ${P_{12}}$ with eigenvalue ${+1}$. Similarly, if we start in an antisymmetric (fermion) state, this state is an eigenvector of ${P_{12}}$ with eigenvalue ${-1}$.

Now we can look at some other properties of ${P_{12}}$. Consider

 $\displaystyle P_{12}X_{1}P_{12}\left|x_{1},x_{2}\right\rangle$ $\displaystyle =$ $\displaystyle P_{12}X_{1}\left|x_{2},x_{1}\right\rangle \ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle x_{2}P_{12}\left|x_{2},x_{1}\right\rangle \ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle x_{2}\left|x_{1},x_{2}\right\rangle \ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle X_{2}\left|x_{1},x_{2}\right\rangle \ \ \ \ \ (21)$

This follows because the operator ${X_{1}}$ operates on the first particle in the state ${\left|x_{2},x_{1}\right\rangle }$ which on the RHS of the first line is at position ${x_{2}}$. Thus ${X_{1}\left|x_{2},x_{1}\right\rangle =x_{2}\left|x_{2},x_{1}\right\rangle }$, that is, ${X_{1}}$ returns the numerical value of the position of the first particle, which is ${x_{2}}$. This means that in terms of the operators alone

 $\displaystyle P_{12}X_{1}P_{12}$ $\displaystyle =$ $\displaystyle X_{2}\ \ \ \ \ (22)$ $\displaystyle P_{12}X_{2}P_{12}$ $\displaystyle =$ $\displaystyle X_{1}\ \ \ \ \ (23)$ $\displaystyle P_{12}P_{1}P_{12}$ $\displaystyle =$ $\displaystyle P_{2}\ \ \ \ \ (24)$ $\displaystyle P_{12}P_{2}P_{12}$ $\displaystyle =$ $\displaystyle P_{1} \ \ \ \ \ (25)$

In the last two lines, the operator ${P_{i}}$ is the momentum of particle ${i}$, and the result follows by applying the operators to the momentum basis state ${\left|p_{1},p_{2}\right\rangle }$.

For some general operator which can be expanded in a power series of terms containing powers of ${X_{i}}$ and/or ${P_{i}}$, we can use 10 to insert ${P_{12}P_{12}}$ between every factor of ${X_{i}}$ or ${P_{i}}$. For example

 $\displaystyle P_{12}P_{1}X_{2}^{2}X_{1}P_{12}$ $\displaystyle =$ $\displaystyle P_{12}P_{1}P_{12}P_{12}X_{2}P_{12}P_{12}X_{2}P_{12}P_{12}X_{1}P_{12}\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle P_{2}X_{1}^{2}X_{2} \ \ \ \ \ (27)$

That is, for any operator ${\Omega\left(X_{1},P_{1};X_{2},P_{2}\right)}$ we have

$\displaystyle P_{12}\Omega\left(X_{1},P_{1};X_{2},P_{2}\right)P_{12}=\Omega\left(X_{2},P_{2};X_{1},P_{1}\right) \ \ \ \ \ (28)$

The Hamiltonian for a system of two identical particles must be symmetric under exchange of the particles, since it represents an observable (the energy), and this observable must remain unchanged if we swap the particles. (In the case of two fermions, the wave function is antisymmetric, but the wave function itself is not an observable. The wave function gets multiplied by ${-1}$ if we swap the particles, but the square modulus of the wave function, which contains the physics, remains the same.) Thus we have

$\displaystyle P_{12}H\left(X_{1},P_{1};X_{2},P_{2}\right)P_{12}=H\left(X_{2},P_{2};X_{1},P_{1}\right)=H\left(X_{1},P_{1};X_{2},P_{2}\right) \ \ \ \ \ (29)$

[Note that this condition doesn’t necessarily follow if the two particles are not identical, since exchanging them in this case leads to an observably different system. For example, exchanging the proton and electron in a hydrogen atom leads to a different system.]

The propagator is defined as

$\displaystyle U\left(t\right)=e^{-iHt/\hbar} \ \ \ \ \ (30)$

and the propagator dictates how a state evolves according to

$\displaystyle \left|\psi\left(t\right)\right\rangle =U\left(t\right)\left|\psi\left(0\right)\right\rangle \ \ \ \ \ (31)$

Since the only operator on which ${U}$ depends is ${H}$, then ${U}$ is also invariant, so that

$\displaystyle P_{12}U\left(X_{1},P_{1};X_{2},P_{2}\right)P_{12}=U\left(X_{2},P_{2};X_{1},P_{1}\right)=U\left(X_{1},P_{1};X_{2},P_{2}\right) \ \ \ \ \ (32)$

Multiplying from the left by ${P_{12}}$ and subtracting, we get the commutator

$\displaystyle \left[U,P_{12}\right]=0 \ \ \ \ \ (33)$

For a symmetric state ${\left|\psi_{S}\right\rangle }$ or antisymmetric state ${\left|\psi_{A}\right\rangle }$, we have

 $\displaystyle UP_{12}\left|\psi_{S}\left(0\right)\right\rangle$ $\displaystyle =$ $\displaystyle U\left|\psi_{S}\left(0\right)\right\rangle =\left|\psi_{S}\left(t\right)\right\rangle =P_{12}U\left|\psi_{S}\left(0\right)\right\rangle \ \ \ \ \ (34)$ $\displaystyle UP_{12}\left|\psi_{A}\left(0\right)\right\rangle$ $\displaystyle =$ $\displaystyle -U\left|\psi_{A}\left(0\right)\right\rangle =-\left|\psi_{A}\left(t\right)\right\rangle =P_{12}U\left|\psi_{A}\left(0\right)\right\rangle \ \ \ \ \ (35)$

This means that states that begin as symmetric or antisymmetric remain symmetric or antisymmetric for all time. In other words, a system that starts in an eigenstate of ${P_{12}}$ remains in the same eigenstate as time passes.

# Compound systems of fermions and bosons

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 10, Exercise 10.3.6.

In a system of identical particles, we’ve seen that if the particles are bosons, the state vector is symmetric with respect to the exchange of any two particles (that is, ${\psi\left(a,b\right)=\psi\left(b,a\right)}$ where ${a}$ and ${b}$ are any two of the particles in the system), while for fermions, the state vector is antisymmetric, meaning that ${\psi\left(a,b\right)=-\psi\left(a,b\right)}$. What happens if we have a compound object such as a hydrogen atom that is composed of a collection of fermions and/or bosons?

Suppose we look at the hydrogen atom in particular. It is composed of a proton and an electron, both of which are fermions. The proton and electron are not, of course, identical particles, but now suppose we have two hydrogen atoms. The two protons are identical fermions, just as are the two electrons. However, when analyzing a system of two hydrogen atoms, the relevant question is what happens to the state vector if we exchange the two atoms. In doing so, we exchange both the two protons and the two electrons. Each exchange multiplies the state vector by ${-1}$, so the net effect of exchanging both protons and both electrons is to multiply the state vector by ${\left(-1\right)^{2}=1}$. In other words, a hydrogen atom acts as a boson, even though it is composed of two fermions.

In general, if we have a compound object containing ${n}$ fermions, then the state vector for a system of two such objects is multiplied by ${\left(-1\right)^{n}}$ when these two objects are exchanged. That is, a compound object containing an even number of fermions behaves as a boson, while if it contains an odd number of fermions, it behaves as a fermion.

A compound object consisting entirely of bosons will always behave as a boson, no matter how many such bosonic particles it contains, since interchanging all ${n}$ bosons just multiplies the state vector by ${\left(+1\right)^{n}=1}$.

# Identical particles – bosons and fermions revisited

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 10, Exercises 10.3.1 – 10.3.3.

Although we’ve looked at the quantum treatment of identical particles as done by Griffiths, it’s worth summarizing Shankar’s treatment of the topic as it provides a few more insights.

In classical physics, suppose we have two identical particles, where ‘identical’ here means that all their physical properties such as mass, size, shape, charge and so on are the same. Suppose we do an experiment in which these two particles collide and rebound in some way. Can we tell which particle ends up in which location? We’re not allowed to label the particles by writing on them, for example, since then they would no longer be identical. In classical physics, we can determine which particle is which by tracing their histories. For example, if we start with particle 1 at position ${\mathbf{r}_{1}}$ and particle 2 at position ${\mathbf{r}_{2}}$, then let them collide, and finally measure their locations at some time after the collision, we might find that one particle ends up at position ${\mathbf{r}_{3}}$ and the other at position ${\mathbf{r}_{4}}$. If we videoed the collision event, we would see the two particles follow well-defined paths before and after the collision, so by observing which particle followed the path that leads from ${\mathbf{r}_{1}}$ to the collision and then out again, we can tell whether it ends up at ${\mathbf{r}_{3}}$ or ${\mathbf{r}_{4}}$. That is, the identification of a particle depends on our ability to watch it as it travels through space.

In quantum mechanics, because of the uncertainty principle, a particle does not have a well-defined trajectory, since in order to define such a trajectory, we would need to specify its position and momentum precisely at each instant of time as it travels. In terms of our collision experiment, if we measured one particle to be at starting position ${\mathbf{r}_{1}}$ at time ${t=0}$ then we know nothing about its momentum, because we specified the position exactly. Thus we can’t tell what trajectory this particle will follow. If we measure the two particles at positions ${\mathbf{r}_{1}}$ and ${\mathbf{r}_{2}}$ at ${t=0}$, and then at ${\mathbf{r}_{3}}$ and ${\mathbf{r}_{4}}$ at some later time, we have no way of knowing which particle ends up at ${\mathbf{r}_{3}}$ and which at ${\mathbf{r}_{4}}$. In terms of the state vector, this means that the physics in the state vector must be the same if we exchange the two particles within the wave function. Since multiplying a state vector ${\psi}$ by some complex constant ${\alpha}$ leaves the physics unchanged, this means that we require

$\displaystyle \psi\left(a,b\right)=\alpha\psi\left(b,a\right) \ \ \ \ \ (1)$

where ${a}$ and ${b}$ represent the two particles.

For a two-particle system, the vector space is spanned by a direct product of the two one-particle vector spaces. Thus the two basis vectors in this vector space that can describe the two particle ${a}$ and ${b}$ are ${\left|ab\right\rangle }$ and ${\left|ba\right\rangle }$. If these two particles are identical, then ${\psi}$ must be some linear combination of these two vectors that satisfies 1. That is

 $\displaystyle \psi\left(b,a\right)$ $\displaystyle =$ $\displaystyle \beta\left|ab\right\rangle +\gamma\left|ba\right\rangle \ \ \ \ \ (2)$ $\displaystyle \psi\left(a,b\right)$ $\displaystyle =$ $\displaystyle \alpha\psi\left(b,a\right)\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \alpha\left(\beta\left|ab\right\rangle +\gamma\left|ba\right\rangle \right) \ \ \ \ \ (4)$

However, ${\psi\left(a,b\right)}$ is also just ${\psi\left(b,a\right)}$ with ${a}$ swapped with ${b}$, that is

$\displaystyle \psi\left(a,b\right)=\beta\left|ba\right\rangle +\gamma\left|ab\right\rangle \ \ \ \ \ (5)$

Since ${\left|ab\right\rangle }$ and ${\left|ba\right\rangle }$ are independent, we can equate their coefficients in the last two equations to get

 $\displaystyle \alpha\beta$ $\displaystyle =$ $\displaystyle \gamma\ \ \ \ \ (6)$ $\displaystyle \alpha\gamma$ $\displaystyle =$ $\displaystyle \beta \ \ \ \ \ (7)$

Inserting the second equation into the first, we get

 $\displaystyle \alpha^{2}\gamma$ $\displaystyle =$ $\displaystyle \gamma\ \ \ \ \ (8)$ $\displaystyle \alpha^{2}$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (9)$ $\displaystyle \alpha$ $\displaystyle =$ $\displaystyle \pm1 \ \ \ \ \ (10)$

Thus the two possible state functions 1 are combinations of ${\left|ab\right\rangle }$ and ${\left|ba\right\rangle }$ such that

$\displaystyle \psi\left(a,b\right)=\pm\psi\left(b,a\right) \ \ \ \ \ (11)$

The plus sign gives the symmetric state, which can be written as

$\displaystyle \psi\left(ab,S\right)=\frac{1}{\sqrt{2}}\left(\left|ab\right\rangle +\left|ba\right\rangle \right) \ \ \ \ \ (12)$

and the minus sign gives the antisymmetric state

$\displaystyle \psi\left(ab,A\right)=\frac{1}{\sqrt{2}}\left(\left|ab\right\rangle -\left|ba\right\rangle \right) \ \ \ \ \ (13)$

The ${\frac{1}{\sqrt{2}}}$ factor normalizes the states so that

 $\displaystyle \left\langle \psi\left(ab,S\right)\left|\psi\left(ab,S\right)\right.\right\rangle$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (14)$ $\displaystyle \left\langle \psi\left(ab,A\right)\left|\psi\left(ab,A\right)\right.\right\rangle$ $\displaystyle =$ $\displaystyle 1 \ \ \ \ \ (15)$

This follows because the basis vectors ${\left|ab\right\rangle }$ and ${\left|ba\right\rangle }$ are orthonormal vectors.

Particles with symmetric states are called bosons and particles with antisymmetric states are called fermions. The Pauli exclusion principle for fermions follows directly from 13, since if we set the state variables of the two particles to be the same, that is, ${a=b}$, then

$\displaystyle \psi\left(aa,A\right)=\frac{1}{\sqrt{2}}\left(\left|aa\right\rangle -\left|aa\right\rangle \right)=0 \ \ \ \ \ (16)$

The symmetry or antisymmetry rules apply to all the properties of the particle taken as an aggregate. That is, the labels ${a}$ and ${b}$ can refer to the particle’s location plus its other quantum numbers such as spin, charge, and so on. In order for two fermions to be excluded, the states of the two fermions must be identical in all their quantum numbers, so that two fermions with the same orbital location (as two electrons in the same orbital within an atom, for example) are allowed if their spins are different.

Example 1 Suppose we have 2 identical bosons that are measured to be in states ${\left|\phi\right\rangle }$ and ${\left|\chi\right\rangle }$ where ${\left\langle \phi\left|\chi\right.\right\rangle \ne0}$. What is their combined state vector? Since they are bosons, their state vector must be symmetric, so we must have

 $\displaystyle \psi\left(\phi,\chi\right)$ $\displaystyle =$ $\displaystyle A\left|\phi\chi\right\rangle +B\left|\chi\phi\right\rangle \ \ \ \ \ (17)$

Because ${\psi}$ must be symmetric, we must have ${A=B}$, so that ${\psi\left(\phi,\chi\right)=\psi\left(\chi,\phi\right)}$. The 2-particle states can be written as direct products, so we have

$\displaystyle \psi\left(\phi,\chi\right)=A\left(\left|\phi\right\rangle \otimes\left|\chi\right\rangle +\left|\chi\right\rangle \otimes\left|\phi\right\rangle \right) \ \ \ \ \ (18)$

To normalize, we have, assuming that ${\left|\phi\right\rangle }$ and ${\left|\chi\right\rangle }$ are normalized:

 $\displaystyle \left|\psi\right|^{2}$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left|A\right|^{2}\left(\left\langle \phi\right|\otimes\left\langle \chi\right|+\left\langle \chi\right|\otimes\left\langle \phi\right|\right)\left(\left|\phi\right\rangle \otimes\left|\chi\right\rangle +\left|\chi\right\rangle \otimes\left|\phi\right\rangle \right)\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left|A\right|^{2}\left(1+1+\left|\left\langle \phi\left|\chi\right.\right\rangle \right|^{2}+\left|\left\langle \chi\left|\phi\right.\right\rangle \right|^{2}\right)\ \ \ \ \ (21)$ $\displaystyle A$ $\displaystyle =$ $\displaystyle \frac{\pm1}{\sqrt{2\left(1+\left|\left\langle \phi\left|\chi\right.\right\rangle \right|^{2}\right)}} \ \ \ \ \ (22)$

Thus the normalized state vector is (choosing the + sign):

 $\displaystyle \psi\left(\phi,\chi\right)$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2\left(1+\left|\left\langle \phi\left|\chi\right.\right\rangle \right|^{2}\right)}}\left(\left|\phi\chi\right\rangle +\left|\chi\phi\right\rangle \right) \ \ \ \ \ (23)$

Notice that this reduces to 12 if ${\left\langle \phi\left|\chi\right.\right\rangle =0}$.

For more than 2 particles, we need to form state vectors that are either totally symmetric or totally antisymmetric.

Example 2 Suppose we have 3 identical bosons, and they are measured to be in states 3, 3 and 4. Since two of them are in the same state, there are 3 possible combinations, which we can write as ${\left|334\right\rangle ,}$ ${\left|343\right\rangle }$ and ${\left|433\right\rangle }$. Assuming these states are orthonormal, the full normalized state vector is

$\displaystyle \psi\left(3,3,4\right)=\frac{1}{\sqrt{3}}\left(\left|334\right\rangle +\left|343\right\rangle +\left|433\right\rangle \right) \ \ \ \ \ (24)$

The ${\frac{1}{\sqrt{3}}}$ ensures that ${\left|\psi\left(3,3,4\right)\right|^{2}=1}$.

Incidentally, for ${N\ge3}$ particles, it turns out to be impossible to construct a linear combination of the basis states such that the overall state vector is symmetric with respect to the interchange of some pairs of particles and antisymmetric with respect to the interchange of other pairs. A general proof for all ${N}$ requires group theory, but for ${N=3}$ we can show this by brute force. There are ${3!=6}$ basis vectors

$\displaystyle \left|123\right\rangle ,\left|231\right\rangle ,\left|312\right\rangle ,\left|132\right\rangle ,\left|321\right\rangle ,\left|213\right\rangle \ \ \ \ \ (25)$

Suppose we require the compound state vector to be symmetric with respect to exchanging 1 and 2. We then must have

$\displaystyle \psi=A\left(\left|123\right\rangle +\left|213\right\rangle \right)+B\left(\left|231\right\rangle +\left|132\right\rangle \right)+C\left(\left|312\right\rangle +\left|321\right\rangle \right) \ \ \ \ \ (26)$

If we now try to make ${\psi}$ antisymmetric with respect to exchanging 2 and 3, we must have

$\displaystyle \psi=D\left(\left|123\right\rangle -\left|132\right\rangle \right)+E\left(\left|231\right\rangle -\left|321\right\rangle \right)+F\left(\left|312\right\rangle -\left|213\right\rangle \right) \ \ \ \ \ (27)$

Comparing the two, we see that

 $\displaystyle A$ $\displaystyle =$ $\displaystyle D=-F\ \ \ \ \ (28)$ $\displaystyle B$ $\displaystyle =$ $\displaystyle E=-D\ \ \ \ \ (29)$ $\displaystyle C$ $\displaystyle =$ $\displaystyle F=-E \ \ \ \ \ (30)$

Eliminating ${A,B}$, and ${C}$ we have, combining the 3 equations:

$\displaystyle D=-E=F \ \ \ \ \ (31)$

But from the first equation, we have ${D=-F}$, so ${F=-F=0}$. From the other equations, this implies that ${D=-F=0}$ and ${E=-F=0}$, and thus that ${A=B=C=0}$. So there is no non-trivial solution that allows both a symmetric and antisymmetric particle exchange within the same state vector.

Example 3 Suppose we have 3 particles and only 3 distinct states that each particle can have. If the particles are distinguishable (not identical) the total number of states is found by considering the possibilities. If all 3 particles are in different states, then there are ${3!=6}$ possible overall states. If two particles are in one state and one particle in another, there are ${\binom{3}{2}=3}$ ways of choosing the two states, for each of which there are 2 ways of partitioning these two states (that is, which state has 2 particles and which has the other one), and for each of those there are 3 possible configurations, so there are ${3\times2\times3=18}$ possible configurations. Finally, if all 3 particles are in the same state, there are 3 possibilities. Thus the total for distinguishable particles is ${6+18+3=27}$.

If the particles are bosons, then if all 3 are in different states, there is only 1 symmetric combination of the 6 basis states. If two particles are in one state and one particle in another, there are ${3\times2=6}$ ways of partitioning the states, each of which contributes only one symmetric overall state. Finally, if all 3 particles are in the same state, there are 3 possibilities. Thus the total for bosons is ${1+6+3=10}$.

For fermions, all three particles must be in different states, so there is only 1 possibility.

# 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.

# Occupation number representation; delta function as a series

References: Tom Lancaster and Stephen J. Blundell, Quantum Field Theory for the Gifted Amateur, (Oxford University Press, 2014) – Problem 3.1.

We can write the hamiltonian for the harmonic oscillator in terms of the creation and annihilation operators as

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

 $\displaystyle a\left|n\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{n}\left|n-1\right\rangle \ \ \ \ \ (2)$ $\displaystyle a^{\dagger}\left|n\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{n+1}\left|n+1\right\rangle \ \ \ \ \ (3)$

so the combined operator ${a^{\dagger}a}$ acts as a number operator, giving the number of quanta in a state:

 $\displaystyle a^{\dagger}a\left|n\right\rangle$ $\displaystyle =$ $\displaystyle a^{\dagger}\sqrt{n}\left|n-1\right\rangle \ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{n}a^{\dagger}\left|n-1\right\rangle \ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle n\left|n\right\rangle \ \ \ \ \ (6)$

We can generalize this to a collection of independent oscillators where oscillator ${k}$ has frequency ${\omega_{k}}$. In that case

$\displaystyle \hat{H}=\hbar\sum_{k}\omega_{k}\left(a_{k}^{\dagger}a_{k}+\frac{1}{2}\right) \ \ \ \ \ (7)$

where ${a_{k}^{\dagger}}$ and ${a_{k}}$ are the creation and annihilation operators for one quantum in oscillator ${k}$. For the harmonic oscillator, the energy levels are all equally spaced, with a spacing of ${\hbar\omega_{k}}$ so if we redefine the zero point of energy to be ${\frac{1}{2}\hbar\omega_{k}}$ for oscillator ${k}$, then the hamiltonian above can be rewritten as

$\displaystyle \hat{H}=\sum_{k}n_{k}\hbar\omega_{k} \ \ \ \ \ (8)$

where ${n_{k}}$ is the number of quanta in oscillator ${k}$. An eigenstate of this hamiltonian is a state containing ${N}$ oscillators with oscillator ${k}$ containing ${n_{k}}$ quanta, which we can write as ${\left|n_{1}n_{2}\ldots n_{N}\right\rangle }$. This is called the occupation number representation since rather than writing out a complex wave function describing all ${N}$ oscillators, we just list the number of quanta contained within each oscillator.

The application of this to quantum field theory is that we can interpret each quantum in oscillator ${k}$ as a particle with a momentum ${p_{k}}$. We’re not saying that a particle is an oscillator; rather we’re noting that we can use the same notation to refer to both particles and oscillators. So if we have a number of momentum states ${p_{k}}$ available in our system, then we can define creation and annihilation operators ${a_{p_{k}}^{\dagger}}$ and ${a_{p_{k}}}$ for that momentum state and write the hamiltonian as

$\displaystyle \hat{H}=\sum_{k}E_{p_{k}}a_{p_{k}}^{\dagger}a_{p_{k}} \ \ \ \ \ (9)$

In order for creation operators to work properly when creating elementary particles, we need to recall that there are two fundamental types of particles: fermions and bosons. The wave function for two bosons is, in position space:

$\displaystyle \psi\left(\mathbf{r}_{a},\mathbf{r}_{b}\right)=A\left[\psi_{1}\left(\mathbf{r}_{a}\right)\psi_{2}\left(\mathbf{r}_{b}\right)+\psi_{2}\left(\mathbf{r}_{a}\right)\psi_{1}\left(\mathbf{r}_{b}\right)\right] \ \ \ \ \ (10)$

If we interchange the two particles by swapping ${\mathbf{r}_{a}}$ and ${\mathbf{r}_{b}}$, the compound wave function ${\psi\left(\mathbf{r}_{a},\mathbf{r}_{b}\right)}$ doesn’t change, so that ${\psi\left(\mathbf{r}_{a},\mathbf{r}_{b}\right)=\psi\left(\mathbf{r}_{b},\mathbf{r}_{a}\right)}$

If we have two fermions, on the other hand, the wave function is

$\displaystyle \psi\left(\mathbf{r}_{a},\mathbf{r}_{b}\right)=A\left[\psi_{1}\left(\mathbf{r}_{a}\right)\psi_{2}\left(\mathbf{r}_{b}\right)-\psi_{2}\left(\mathbf{r}_{a}\right)\psi_{1}\left(\mathbf{r}_{b}\right)\right] \ \ \ \ \ (11)$

and now if we swap the particles we get ${\psi\left(\mathbf{r}_{a},\mathbf{r}_{b}\right)=-\psi\left(\mathbf{r}_{b},\mathbf{r}_{a}\right)}$.

If we use two creation operators operating on the vacuum state ${\left|0\right\rangle }$ to create a state containing two particles, the resulting state must behave properly under the exchange of the two particles. Another way of putting this is that if we swap the order in which the particles are created we must get exactly the same state if the particles are bosons, but the negative of the original state if the particles are fermions. That is, for bosons

$\displaystyle a_{p_{1}}^{\dagger}a_{p_{2}}^{\dagger}=a_{p_{2}}^{\dagger}a_{p_{1}}^{\dagger} \ \ \ \ \ (12)$

or in terms of commutators

$\displaystyle \left[a_{p_{1}}^{\dagger},a_{p_{2}}^{\dagger}\right]=0 \ \ \ \ \ (13)$

For fermions, we’ll use the symbols ${c^{\dagger}}$ and ${c}$ for creation and annihilation operators, and in this case we must have

$\displaystyle c_{p_{1}}^{\dagger}c_{p_{2}}^{\dagger}=-c_{p_{2}}^{\dagger}c_{p_{1}}^{\dagger} \ \ \ \ \ (14)$

For fermions we define an anticommutator as

$\displaystyle \left\{ c_{p_{1}}^{\dagger},c_{p_{2}}^{\dagger}\right\} \equiv c_{p_{1}}^{\dagger}c_{p_{2}}^{\dagger}+c_{p_{2}}^{\dagger}c_{p_{1}}^{\dagger} \ \ \ \ \ (15)$

so we have

$\displaystyle \left\{ c_{p_{1}}^{\dagger},c_{p_{2}}^{\dagger}\right\} =0 \ \ \ \ \ (16)$

For the harmonic oscillator, the creation and annihilation operators satisfied the commutation relation

$\displaystyle \left[a_{p_{1}},a_{p_{2}}^{\dagger}\right]=\delta_{p_{1}p_{2}} \ \ \ \ \ (17)$

That is, the annihilation operator commutes with the creation operator if they refer to different oscillators; otherwise the commutator is 1. To complete the analogy between particles and oscillators, we just define the commutation relations between creation and annihilation operators for particles as

 $\displaystyle \left[a_{p_{1}},a_{p_{2}}^{\dagger}\right]$ $\displaystyle =$ $\displaystyle \delta_{p_{1}p_{2}}\ \ \ \ \ (18)$ $\displaystyle \left\{ c_{p_{1}},c_{p_{2}}^{\dagger}\right\}$ $\displaystyle =$ $\displaystyle \delta_{p_{1}p_{2}} \ \ \ \ \ (19)$

Example The commutation relations can be inserted into a formula which gives a new form of the Dirac delta function. For two different momentum states ${\mathbf{p}}$ and ${\mathbf{q}}$ we have, for a pair of bosons

$\displaystyle \left[a_{p},a_{q}^{\dagger}\right]=\delta_{pq} \ \ \ \ \ (20)$

Suppose that the system is enclosed in a cube of side length ${L}$. Then we can construct the sum

$\displaystyle \frac{1}{\mathcal{V}}\sum_{p,q}e^{i\left(\mathbf{p}\cdot\mathbf{x}-\mathbf{q}\cdot\mathbf{y}\right)}\left[a_{p},a_{q}^{\dagger}\right]=\frac{1}{\mathcal{V}}\sum_{p}e^{i\mathbf{p}\cdot\left(\mathbf{x}-\mathbf{y}\right)} \ \ \ \ \ (21)$

What can we make of the sum on the RHS? To see what it is, suppose we have some function ${f\left(x\right)}$ defined for ${-\pi\le x\le\pi}$. We can expand it in a Fourier series as
follows:

$\displaystyle f\left(x\right)=\sum_{n=-\infty}^{\infty}c_{n}e^{inx} \ \ \ \ \ (22)$

where the coefficients are

$\displaystyle c_{n}=\frac{1}{2\pi}\int_{-\pi}^{\pi}f\left(x\right)e^{-inx}dx \ \ \ \ \ (23)$

We can write the Fourier series for the function at a particular point ${x=a}$ as

 $\displaystyle f\left(a\right)$ $\displaystyle =$ $\displaystyle \frac{1}{2\pi}\sum_{n}e^{ina}\int_{-\pi}^{\pi}f\left(x\right)e^{-inx}dx\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{-\pi}^{\pi}f\left(x\right)\left[\frac{1}{2\pi}\sum_{n}e^{in\left(a-x\right)}\right]dx \ \ \ \ \ (25)$

The term in brackets in the last line behaves exactly like ${\delta\left(x-a\right)}$ so we can take it as another definition of the Dirac delta function

$\displaystyle \delta\left(x-a\right)=\frac{1}{2\pi}\sum_{n}e^{in\left(a-x\right)}=\frac{1}{2\pi}\sum_{n}e^{in\left(x-a\right)} \ \ \ \ \ (26)$

where we can change the exponent in the last term because the sum over ${n}$ extends from ${-\infty}$ to ${\infty}$ so we can replace ${n}$ by ${-n}$ and get the same sum.

Now if the function ${f\left(x\right)}$ extends from 0 to ${L}$ instead of from ${-\pi}$ to ${\pi}$ we can replace ${x}$ by ${\xi\equiv Lx/2\pi}$ (and ${a}$ by ${\xi_{a}\equiv La/2\pi}$) to get

 $\displaystyle f\left(\xi_{a}\right)$ $\displaystyle =$ $\displaystyle \int_{0}^{L}f\left(\xi\right)\left[\frac{1}{2\pi}\frac{2\pi}{L}\sum_{n}e^{i2\pi n\left(\xi_{a}-\xi\right)/L}\right]d\xi\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{0}^{L}f\left(\xi\right)\left[\frac{1}{L}\sum_{p}e^{ip\left(\xi-\xi_{a}\right)}\right]d\xi \ \ \ \ \ (28)$

where

$\displaystyle p\equiv\frac{2\pi n}{L} \ \ \ \ \ (29)$

Obviously, the same argument works for the ${y}$ and ${z}$ directions, so in 3-d

 $\displaystyle f\left(\mathbf{a}\right)$ $\displaystyle =$ $\displaystyle \int_{\mathcal{V}}f\left(\mathbf{r}\right)\left[\frac{1}{L^{3}}\sum_{p}e^{i\mathbf{p}\cdot\left(\mathbf{r}-\mathbf{a}\right)}\right]d^{3}\mathbf{r}\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{\mathcal{V}}f\left(\mathbf{r}\right)\left[\frac{1}{\mathcal{V}}\sum_{p}e^{i\mathbf{p}\cdot\left(\mathbf{r}-\mathbf{a}\right)}\right]d^{3}\mathbf{r} \ \ \ \ \ (31)$

so the 3-d delta function is

$\displaystyle \delta^{\left(3\right)}\left(\mathbf{x}-\mathbf{y}\right)=\frac{1}{\mathcal{V}}\sum_{p}e^{i\mathbf{p}\cdot\left(\mathbf{x}-\mathbf{y}\right)} \ \ \ \ \ (32)$

From 21 we get

$\displaystyle \frac{1}{\mathcal{V}}\sum_{p,q}e^{i\left(\mathbf{p}\cdot\mathbf{x}-\mathbf{q}\cdot\mathbf{y}\right)}\left[a_{p},a_{q}^{\dagger}\right]=\delta^{\left(3\right)}\left(\mathbf{x}-\mathbf{y}\right) \ \ \ \ \ (33)$

# Bosons in the infinite square well

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 6.3.

An example of perturbation theory applied to the interaction between two particles. We place two bosons in an infinite square well, where the single particle wave functions are given by

$\displaystyle \psi_{n}=\sqrt{\frac{2}{a}}\sin\frac{n\pi x}{a} \ \ \ \ \ (1)$

and energies by

$\displaystyle E_{n}=\frac{\left(n\pi\hbar\right)^{2}}{2ma^{2}} \ \ \ \ \ (2)$

Since there is no exclusion principle for bosons, both particles can be in the ground state, with wave function

$\displaystyle \psi_{11}=\frac{2}{a}\sin\frac{\pi x_{1}}{a}\sin\frac{\pi x_{2}}{a} \ \ \ \ \ (3)$

The energy of this state is

$\displaystyle E_{11,0}=2E_{1,0}=\frac{\pi^{2}\hbar^{2}}{ma^{2}} \ \ \ \ \ (4)$

If one boson is in the ground state and the other in the first excited state, the overall wave function must be symmetric with respect to interchange of the particles, so we have

$\displaystyle \psi_{12}=\frac{\sqrt{2}}{a}\left[\sin\frac{2\pi x_{1}}{a}\sin\frac{\pi x_{2}}{a}+\sin\frac{\pi x_{1}}{a}\sin\frac{2\pi x_{2}}{a}\right] \ \ \ \ \ (5)$

Applying the two-particle hamiltonian to this wave function yields the total energy as the sum of the two individual energies:

$\displaystyle E_{12,0}=E_{1,0}+E_{2,0}=\frac{5}{2}\frac{\pi^{2}\hbar^{2}}{ma^{2}} \ \ \ \ \ (6)$

Now suppose we introduce an interaction between the particles of form

$\displaystyle V=-aV_{0}\delta\left(x_{1}-x_{2}\right) \ \ \ \ \ (7)$

That is, there is a delta function well when the particles occupy the same location. Using first order perturbation theory, the effect on the energy ${E_{11,0}}$ is

 $\displaystyle E_{11,1}$ $\displaystyle =$ $\displaystyle \left\langle 11,0\right|V\left|11,0\right\rangle \ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -aV_{0}\frac{4}{a^{2}}\int_{0}^{a}\int_{0}^{a}\sin^{2}\frac{\pi x_{1}}{a}\sin^{2}\frac{\pi x_{2}}{a}\delta\left(x_{1}-x_{2}\right)dx_{1}dx_{2}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{4V_{0}}{a}\int_{0}^{a}\sin^{4}\frac{\pi x_{1}}{a}dx_{1}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{3}{2}V_{0} \ \ \ \ \ (11)$

For ${E_{12,0}}$ we have

 $\displaystyle E_{12,1}$ $\displaystyle =$ $\displaystyle \left\langle 12,0\right|V\left|12,0\right\rangle \ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -aV_{0}\frac{2}{a^{2}}\int_{0}^{a}4\sin^{2}\frac{\pi x_{1}}{a}\sin^{2}\frac{2\pi x_{1}}{a}dx_{1}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -2V_{0} \ \ \ \ \ (14)$

We used Maple for the integral.

# Fermions and bosons: counting states

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Chapter 5, Post 33.

A simple example of counting available states. We have three available single-particle states, and three particles to fit into these states.

For distinguishable particles, each particle can be in any of the three states, so there are a total of ${3^{3}=27}$ possible combinations.

For identical bosons, the total state must be symmetric. We can have all three particles in the same state (3 ways, one for each state), or all three in different states (1 way, since the combination must be symmetric), or two in one state and one in another (3 choices for the first state and 2 for the other state, so a total of 6 possible combinations). The total is thus ${3+1+6=10}$ possibilities.

For identical fermions, there is only one possible state, that being the totally antisymmetric combination of the 3 states.

# Statistical mechanics in quantum theory: Bose condensation

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 5.29.

We’ve seen the distribution of particles for the most probable state in the cases of distinguishable particles and fermions so to complete the set we need to look at bosons. Following the same technique as in the other two cases, we start with the total number of states for bosons:

$\displaystyle S_{b}\left(\left\{ n_{j}\right\} \right)=\prod_{j=1}^{m}\binom{n_{j}+d_{j}-1}{n_{j}} \ \ \ \ \ (1)$

Taking the log of this and using Lagrange multipliers to add in the constraints, we get the function

 $\displaystyle G$ $\displaystyle =$ $\displaystyle \ln S_{b}+\alpha\left(N-\sum_{j=1}^{\infty}n_{j}\right)+\beta\left(E-\sum_{j=1}^{\infty}n_{j}E_{j}\right)\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{j=1}^{\infty}\left[\ln\left(n_{j}+d_{j}-1\right)!-\ln n_{j}!-\ln\left(d_{j}-1\right)!-\alpha n_{j}-\beta n_{j}E_{j}\right]+\alpha N+\beta E \ \ \ \ \ (3)$

Using Stirling’s approximation again we get

$\displaystyle G\approx\sum_{j=1}^{\infty}\left[\left(n_{j}+d_{j}-1\right)\ln\left(n_{j}+d_{j}-1\right)-\left(n_{j}+d_{j}-1\right)-n_{j}\ln n_{j}+n_{j}-\ln\left(d_{j}-1\right)!-\alpha n_{j}-\beta n_{j}E_{j}\right]+\alpha N+\beta E \ \ \ \ \ (4)$

We can now take the derivative to get ${n_{j}}$:

 $\displaystyle \frac{\partial G}{\partial n_{j}}$ $\displaystyle =$ $\displaystyle \ln\left(n_{j}+d_{j}-1\right)-\ln n_{j}-\alpha-\beta E_{j}=0\ \ \ \ \ (5)$ $\displaystyle n_{j}$ $\displaystyle =$ $\displaystyle \left(n_{j}+d_{j}-1\right)e^{-\alpha-\beta E_{j}}\ \ \ \ \ (6)$ $\displaystyle n_{j}$ $\displaystyle =$ $\displaystyle \frac{\left(d_{j}-1\right)e^{-\alpha-\beta E_{j}}}{1-e^{-\alpha-\beta E_{j}}}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{d_{j}-1}{e^{\alpha+\beta E_{j}}-1} \ \ \ \ \ (8)$

Since ${d_{j}\gg1}$ (usually), we can safely drop the ${-1}$ in the numerator to get the number distribution for bosons:

$\displaystyle n_{j}=\frac{d_{j}}{e^{\alpha+\beta E_{j}}-1} \ \ \ \ \ (9)$

At this stage, we can try to find ${\alpha}$ and ${\beta}$ by evaluating the total number of particles and the total energy for a particular potential, such as the infinite square well. Using the same technique as before, we get

 $\displaystyle N$ $\displaystyle =$ $\displaystyle \int_{0}^{\infty}\frac{d\left(k\right)}{e^{\alpha+\beta E_{j}}-1}dk\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{V}{2\pi^{2}}\int_{0}^{\infty}\frac{k^{2}}{e^{\alpha+\hbar^{2}k^{2}\beta/2m}-1}dk \ \ \ \ \ (11)$

This integral does not appear to have a closed form, even if we try to find some special functions. If we use the same definitions for ${\alpha}$ and ${\beta}$ as before, we get

$\displaystyle N=\frac{V}{2\pi^{2}}\int_{0}^{\infty}\frac{k^{2}}{e^{\left(\hbar^{2}k^{2}/2m-\mu\right)/k_{B}T}-1}dk \ \ \ \ \ (12)$

We can use this formula to derive a few conclusions. First, since the integrand is the product of two quantities (number at energy level ${k}$ and degeneracy), the integrand must always be non-negative. This means that the exponent in the denominator must be non-negative, so

$\displaystyle \mu<\frac{\hbar^{2}k^{2}}{2m} \ \ \ \ \ (13)$

for all possible values of ${k}$. Since ${\mu}$ is a constant for given values of ${N}$, ${V}$ and ${T}$, it must be less than the minimum energy in the system.

For the ideal gas, the minimum energy is zero, so ${\mu<0}$ for all values of ${N}$, ${V}$ and ${T}$. If ${N}$ and ${V}$ are fixed, then from 12, the integral must remain constant as ${T}$ is varied. This means that, for a given value of ${k}$, ${\left(\hbar^{2}k^{2}/2m-\mu\right)/k_{B}T}$ must remain constant as ${T}$ varies. In particular, as ${T}$ decreases, ${\hbar^{2}k^{2}/2m-\mu}$ must also decrease and since ${\mu<0}$, this means that ${\mu}$ must increase towards zero.

The condition is, for a function ${A\left(k\right)}$:

$\displaystyle \frac{\hbar^{2}k^{2}}{2m}-A\left(k\right)k_{B}T=\mu\left(T\right) \ \ \ \ \ (14)$

That is, ${A}$ depends only on ${k}$ and ${\mu}$ depends only on ${T}$. Since the LHS must always be less than zero, there is a critical temperature ${T_{c}}$ where the LHS becomes zero, which occurs at

$\displaystyle T_{c}=\frac{\hbar^{2}k^{2}}{2mA\left(k\right)k_{B}} \ \ \ \ \ (15)$

Of course, we can’t use this formula to determine ${T_{c}}$ because we don’t know ${A\left(k\right)}$. However, we can try to do the integral above in the special case where ${\mu=0}$. That is, we have

$\displaystyle \frac{2\pi^{2}N}{V}=\int_{0}^{\infty}\frac{k^{2}}{e^{\hbar^{2}k^{2}/2mk_{B}T_{c}}-1}dk \ \ \ \ \ (16)$

Using Maple, the integral comes out to

$\displaystyle \int_{0}^{\infty}\frac{k^{2}}{e^{\hbar^{2}k^{2}/2mk_{B}T_{c}}-1}dk=\left(mk_{B}T_{c}\right)^{3/2}\frac{\sqrt{\pi}}{\sqrt{2}\hbar^{3}}\zeta\left(\frac{3}{2}\right) \ \ \ \ \ (17)$

where ${\zeta}$ is the Riemann zeta function. Solving for ${T_{c}}$ we get

$\displaystyle T_{c}=\frac{2\pi\hbar^{2}}{mk_{B}}\left(\frac{N}{V\zeta\left(\frac{3}{2}\right)}\right)^{2/3} \ \ \ \ \ (18)$

For liquid ${^{4}\mbox{He}}$ the mass density is ${150\mbox{ kg m}^{-3}}$ which gives it a number density of

$\displaystyle \frac{N}{V}=\frac{150}{6.645\times10^{-27}}=2.2573\times10^{28}\mbox{ m}^{-3} \ \ \ \ \ (19)$

Plugging in the numbers then gives

$\displaystyle T_{c}=3.21\mbox{ K} \ \ \ \ \ (20)$

The experimental value is ${T_{c}=2.17\mbox{ K}}$ and at this point, helium becomes a superfluid. The phenomenon that occurs at the critical temperature is known as Bose condensation. The ideal gas model cannot, of course, explain superfluidity, but it’s interesting that we can predict the existence of a critical temperature even in this simple model.

# Statistical mechanics in quantum theory: counting boson states

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 5.25.

When working out the number of possible configurations in which ${N}$ identical bosons can be placed into a set of energy states, each of which can be degenerate, we came to the intermediate result of the number of ways of distributing ${n}$ bosons among the ${d}$ degenerate levels in a single energy state:

$\displaystyle S_{b}(n)=\binom{n+d-1}{n} \ \ \ \ \ (1)$

Although we proved this by a simple combinatorial argument, it is also possible to prove this formula by a slightly more involved route. We start with the lowest values of ${n}$ and attempt to deduce a pattern.

With ${n=1}$, there are ${d}$ ways of assigning them in ${d}$ bins, so

$\displaystyle S_{b}\left(1\right)=d=\binom{d}{1} \ \ \ \ \ (2)$

With ${n=2}$, we could put both bosons in the same bin (${d=\binom{d}{1}}$ ways) or one in each of two bins ${\binom{d}{2}}$ ways. Thus the total is

$\displaystyle S_{b}\left(2\right)=\binom{d}{1}+\binom{d}{2} \ \ \ \ \ (3)$

With ${n=3}$, we could put all 3 in the same bin (${d=\binom{d}{1}}$ ways), or 2 in one bin and 1 in another (${2\binom{d}{2}}$ ways, since if we choose 2 bins, swapping the bin that contains 2 with the bin that contains 1 is a different configuration), or all 3 in separate bins (${\binom{d}{3}}$ ways). The total is then

$\displaystyle S_{b}\left(3\right)=\binom{d}{1}+2\binom{d}{2}+\binom{d}{3} \ \ \ \ \ (4)$

For ${n=4}$, we can have all 4 in the same bin (${\binom{d}{1}}$ ways), 3 in one bin and 1 in another (${2\binom{d}{2}}$ ways), 2 in one bin and 2 in another (${\binom{d}{2}}$ ways), 2 in one bin and the other 2 in separate bins (${3\binom{d}{3}}$ ways) or all 4 in separate bins (${\binom{d}{4}}$ ways) making a total of

$\displaystyle S_{b}\left(4\right)=\binom{d}{1}+3\binom{d}{2}+3\binom{d}{3}+\binom{d}{4} \ \ \ \ \ (5)$

The coefficients of each term on the RHS appear to be the binomial coefficients ${\binom{n-1}{j-1}}$ for ${j=1,\ldots,n}$. That is, the general form appears to be

$\displaystyle S_{b}\left(n\right)=\sum_{j=1}^{n}\binom{d}{j}\binom{n-1}{j-1} \ \ \ \ \ (6)$

We can verify this by the following argument. First we need to work out the number of ways of dividing ${n}$ bosons into ${j}$ groups. We can do this by arranging the bosons in a line and placing a partition between each pair of particles, so there are ${n-1}$ partitions. To divide the bosons into ${j}$ groups, we select ${j-1}$ of these partitions and ignore the rest. The number of ways we can select the partitions is then ${\binom{n-1}{j-1}}$. Once we have selected the groups of bosons, we need to distribute them among the ${d}$ degenerate states, and this can be done in ${\binom{d}{j}}$ ways. Combining these two gives the result above.

It might seem that something is a bit wonky with this argument. The formula for the number of ways of partitioning ${n}$ bosons assumes, for example, that if we had 3 bosons, putting the partition between bosons 1 and 2 is different from putting it between 2 and 3, whereas both partitions divide the bosons into a group of 2 and a group of 1. Since bosons are identical particles, these two groupings should be the same, so it looks like we’re overcounting.

The point is that the other factor of ${\binom{d}{j}}$ undercounts, since it assumes that the order in which the members of the partition are assigned to degenerate states doesn’t matter, whereas because all the degenerate states are different, the order does matter. Thus retaining the order in the original partitioning compensates for neglecting it when assigning groups to degenerate states, and the total result is in fact correct.

By using the symmetry of the binomial coefficient:

$\displaystyle \binom{n-1}{j-1}=\binom{n-1}{n-j} \ \ \ \ \ (7)$

we can use Vandermonde’s identity to say that

 $\displaystyle S_{b}\left(n\right)$ $\displaystyle =$ $\displaystyle \sum_{j=1}^{n}\binom{d}{j}\binom{n-1}{n-j}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \binom{d+n-1}{n} \ \ \ \ \ (9)$

As far as I know, there is no induction proof of Vandermonde’s identity, but for a couple of other proofs, see the link above.