# Postulates of quantum mechanics: momentum

References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Sections 4.1 – 4.2; Exercises 4.2.2 – 4.2.3.

One of the postulates of quantum mechanics is that the momentum operator ${P}$ in position space is given by

$\displaystyle \left\langle x\left|P\right|x^{\prime}\right\rangle =-i\hbar\delta^{\prime}\left(x-x^{\prime}\right) \ \ \ \ \ (1)$

By using the properties of the derivative of the delta function, we can find the eigenfunctions of ${P}$. We have

 $\displaystyle \left\langle x\left|P\right|\psi\right\rangle$ $\displaystyle =$ $\displaystyle \int\left\langle x\left|P\right|x^{\prime}\right\rangle \left\langle x^{\prime}\left|\psi\right.\right\rangle dx^{\prime}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -i\hbar\int\delta^{\prime}\left(x-x^{\prime}\right)\left\langle x^{\prime}\left|\psi\right.\right\rangle dx^{\prime}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -i\hbar\frac{d}{dx}\left\langle x\left|\psi\right.\right\rangle \ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -i\hbar\frac{d\psi\left(x\right)}{dx} \ \ \ \ \ (5)$

The eigenvector of ${P}$ is ${\left|p\right\rangle }$ and has the property that

$\displaystyle P\left|p\right\rangle =p\left|p\right\rangle \ \ \ \ \ (6)$

If we project this onto position space and use 5 we get

 $\displaystyle \left\langle x\left|P\right|\psi\right\rangle$ $\displaystyle =$ $\displaystyle p\left\langle x\left|p\right.\right\rangle \ \ \ \ \ (7)$ $\displaystyle -i\hbar\frac{d\psi_{p}\left(x\right)}{dx}$ $\displaystyle =$ $\displaystyle p\psi_{p}\left(x\right) \ \ \ \ \ (8)$

where

$\displaystyle \psi_{p}\left(x\right)\equiv\left\langle x\left|p\right.\right\rangle \ \ \ \ \ (9)$

Solving this differential equation and normalizing so that ${\left\langle p^{\prime}\left|p\right.\right\rangle =\delta\left(p-p^{\prime}\right)}$ we get

$\displaystyle \psi_{p}\left(x\right)=\frac{1}{\sqrt{2\pi\hbar}}e^{ipx/\hbar} \ \ \ \ \ (10)$

For an arbitrary wave function ${\left|\psi\right\rangle }$, if we know its position-space form, we can find its momentum-space version as follows:

 $\displaystyle \left\langle p\left|\psi\right.\right\rangle$ $\displaystyle =$ $\displaystyle \int\left\langle p\left|x\right.\right\rangle \left\langle x\left|\psi\right.\right\rangle dx\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int\psi_{p}^*\left(x\right)\left\langle x\left|\psi\right.\right\rangle dx\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2\pi\hbar}}\int e^{-ipx/\hbar}\psi\left(x\right)dx \ \ \ \ \ (13)$

This has an interesting consequence if the position-space function ${\psi\left(x\right)}$ is real. The probability density for finding a particle in a state with momentum ${p}$ is ${\left|\left\langle p\left|\psi\right.\right\rangle \right|^{2}}$, which we can write as

 $\displaystyle \left|\left\langle p\left|\psi\right.\right\rangle \right|^{2}$ $\displaystyle =$ $\displaystyle \left\langle p\left|\psi\right.\right\rangle ^*\left\langle p\left|\psi\right.\right\rangle \ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2\pi\hbar}\int\int e^{ip\left(x-x^{\prime}\right)/\hbar}\psi\left(x\right)\psi\left(x^{\prime}\right)dx\;dx^{\prime}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2\pi\hbar}\int\int e^{-ip\left(x^{\prime}-x\right)/\hbar}\psi\left(x\right)\psi\left(x^{\prime}\right)dx\;dx^{\prime}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2\pi\hbar}\int\int e^{-ip\left(x-x^{\prime}\right)/\hbar}\psi\left(x^{\prime}\right)\psi\left(x\right)dx\;dx^{\prime}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left|\left\langle -p\left|\psi\right.\right\rangle \right|^{2} \ \ \ \ \ (18)$

In the fourth line, since ${x}$ and ${x^{\prime}}$ are dummy integration variables, both of which are integrated over the same range, we can simply swap them without changing anything. Note that the derivation relies on ${\psi\left(x\right)}$ being real, since if it were complex we would have

 $\displaystyle \left|\left\langle p\left|\psi\right.\right\rangle \right|^{2}$ $\displaystyle =$ $\displaystyle \left\langle p\left|\psi\right.\right\rangle ^*\left\langle p\left|\psi\right.\right\rangle \ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2\pi\hbar}\int\int e^{ip\left(x-x^{\prime}\right)/\hbar}\psi\left(x\right)\psi^*\left(x^{\prime}\right)dx\;dx^{\prime}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2\pi\hbar}\int\int e^{-ip\left(x^{\prime}-x\right)/\hbar}\psi\left(x\right)\psi^*\left(x^{\prime}\right)dx\;dx^{\prime}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2\pi\hbar}\int\int e^{-ip\left(x-x^{\prime}\right)/\hbar}\psi\left(x^{\prime}\right)\psi^*\left(x\right)dx\;dx^{\prime}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle \ne$ $\displaystyle \left|\left\langle -p\left|\psi\right.\right\rangle \right|^{2} \ \ \ \ \ (23)$

since

$\displaystyle \left|\left\langle -p\left|\psi\right.\right\rangle \right|^{2}=\frac{1}{2\pi\hbar}\int\int e^{-ip\left(x-x^{\prime}\right)/\hbar}\psi\left(x\right)\psi^*\left(x^{\prime}\right)dx\;dx^{\prime} \ \ \ \ \ (24)$

That is, for ${\left|\left\langle -p\left|\psi\right.\right\rangle \right|^{2}}$ the position ${x^{\prime}}$ that is the argument of the ${\psi^*\left(x^{\prime}\right)}$ factor appears as the positive term ${ipx^{\prime}}$ in the exponential, but in 22 the argument of the complex conjugate wave function is ${x}$, which appears as the negative term ${-ipx}$ in the exponential.

Thus for any real wave function, the probability of the particle having momentum ${+p}$ is equal to the probability of it having ${-p}$, so for such wave functions, the mean momentum is always ${\left\langle P\right\rangle =0}$.

As another example, suppose we have a wave function ${\psi\left(x\right)}$ with a mean momentum ${\bar{p}}$, so that

$\displaystyle \left\langle \psi\left|P\right|\psi\right\rangle =\bar{p} \ \ \ \ \ (25)$

If we now multiply ${\psi}$ by ${e^{ip_{0}x/\hbar}}$ where ${p_{0}}$ is a constant momentum, we can calculate the new mean momentum using 5:

 $\displaystyle \left\langle P\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle e^{ip_{0}x/\hbar}\psi\left|P\right|e^{ip_{0}x/\hbar}\psi\right\rangle \ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -i\hbar\int e^{-ip_{0}x/\hbar}\psi^*\left(x\right)\frac{d}{dx}\left(e^{ip_{0}x/\hbar}\psi\left(x\right)\right)dx\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -i\hbar\int e^{-ip_{0}x/\hbar}\psi^*\left[\frac{ip_{0}}{\hbar}e^{ip_{0}x/\hbar}\psi\left(x\right)+e^{ip_{0}x/\hbar}\frac{d}{dx}\psi\left(x\right)\right]dx\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int p_{0}\psi^*\psi dx-i\hbar\int\psi^*\left(x\right)\frac{d}{dx}\psi\left(x\right)dx\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle p_{0}+\bar{p} \ \ \ \ \ (30)$

The first integral in the fourth line uses the fact that ${p_{0}}$ is constant and ${\psi}$ is normalized so that

$\displaystyle \int\psi^*\psi dx=1 \ \ \ \ \ (31)$

# Momentum of particles in a Dirac field

Reference: References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 4, Problems 4.26 – 4.27.

To work out the total 3-momentum of the particles (and antiparticles) in a Dirac field, we can start with the 3-momentum density

$\displaystyle p^{i}=-\pi_{r}\frac{\partial\phi^{r}}{\partial x^{i}} \ \ \ \ \ (1)$

where the ${r}$ index is summed, and for the Dirac field, ${\phi^{1}=\psi}$ and ${\phi^{2}=\bar{\psi}}$. The conjugate momenta for the Dirac field are

 $\displaystyle \pi_{1}$ $\displaystyle =$ $\displaystyle \pi^{1/2}=i\psi^{\dagger}\ \ \ \ \ (2)$ $\displaystyle \pi_{2}$ $\displaystyle =$ $\displaystyle \bar{\pi}^{1/2}=0 \ \ \ \ \ (3)$

The two field solutions are

 $\displaystyle \psi$ $\displaystyle =$ $\displaystyle \sum_{r=1}^{2}\sum_{\mathbf{p}}\sqrt{\frac{m}{VE_{\mathbf{p}}}}\left[c_{r}\left(\mathbf{p}\right)u_{r}\left(\mathbf{p}\right)e^{-ipx}+d_{r}^{\dagger}\left(\mathbf{p}\right)v_{r}\left(\mathbf{p}\right)e^{ipx}\right]\ \ \ \ \ (4)$ $\displaystyle \bar{\psi}$ $\displaystyle =$ $\displaystyle \sum_{r=1}^{2}\sum_{\mathbf{p}}\sqrt{\frac{m}{VE_{\mathbf{p}}}}\left[d_{r}\left(\mathbf{p}\right)\bar{v}_{r}\left(\mathbf{p}\right)e^{-ipx}+c_{r}^{\dagger}\left(\mathbf{p}\right)\bar{u}_{r}\left(\mathbf{p}\right)e^{ipx}\right] \ \ \ \ \ (5)$

The latter can be converted into the Hermitian conjugate by post-multiplying the equation by ${\gamma^{0}}$, which gives

$\displaystyle \psi^{\dagger}=\sum_{r=1}^{2}\sum_{\mathbf{p}}\sqrt{\frac{m}{VE_{\mathbf{p}}}}\left[d_{r}\left(\mathbf{p}\right)v_{r}^{\dagger}\left(\mathbf{p}\right)e^{-ipx}+c_{r}^{\dagger}\left(\mathbf{p}\right)u_{r}^{\dagger}\left(\mathbf{p}\right)e^{ipx}\right] \ \ \ \ \ (6)$

The 3-momentum density is therefore

 $\displaystyle p^{i}$ $\displaystyle =$ $\displaystyle -i\sum_{r=1}^{2}\sum_{\mathbf{p}}\sqrt{\frac{m}{VE_{\mathbf{p}}}}\left[d_{r}\left(\mathbf{p}\right)v_{r}^{\dagger}\left(\mathbf{p}\right)e^{-ipx}+c_{r}^{\dagger}\left(\mathbf{p}\right)u_{r}^{\dagger}\left(\mathbf{p}\right)e^{ipx}\right]\times\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle$ $\displaystyle i\sum_{s=1}^{2}\sum_{\mathbf{q}}\sqrt{\frac{m}{VE_{\mathbf{q}}}}\left[\left(-q_{i}\right)c_{s}\left(\mathbf{q}\right)u_{s}\left(\mathbf{q}\right)e^{-iqx}+q_{i}d_{s}^{\dagger}\left(\mathbf{q}\right)v_{s}\left(\mathbf{q}\right)e^{iqx}\right]\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{r=1}^{2}\sum_{\mathbf{p}}\sqrt{\frac{m}{VE_{\mathbf{p}}}}\left[d_{r}\left(\mathbf{p}\right)v_{r}^{\dagger}\left(\mathbf{p}\right)e^{-ipx}+c_{r}^{\dagger}\left(\mathbf{p}\right)u_{r}^{\dagger}\left(\mathbf{p}\right)e^{ipx}\right]\times\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle$ $\displaystyle \sum_{s=1}^{2}\sum_{\mathbf{q}}\sqrt{\frac{m}{VE_{\mathbf{q}}}}\left[q^{i}c_{s}\left(\mathbf{q}\right)u_{s}\left(\mathbf{q}\right)e^{-iqx}-q^{i}d_{s}^{\dagger}\left(\mathbf{q}\right)v_{s}\left(\mathbf{q}\right)e^{iqx}\right] \ \ \ \ \ (10)$

The total momentum is the integral of ${p^{i}}$ over the volume ${V}$:

$\displaystyle P^{i}=\int p^{i}d^{3}x \ \ \ \ \ (11)$

and, as usual, any terms with an exponential term will integrate to zero due to the requirement that an integral number of wavelengths must fit into the volume. This means that in the product of the first term of 9 with the first term of 10, and in the product of the last term of 9 with the last term of 10, we must have ${\mathbf{q}=-\mathbf{p}}$. However, due to the orthogonality of the spinors we have

$\displaystyle v_{r}^{\dagger}\left(\mathbf{p}\right)u_{s}\left(-\mathbf{p}\right)=u_{r}^{\dagger}\left(\mathbf{p}\right)v_{s}\left(-\mathbf{p}\right)=0 \ \ \ \ \ (12)$

so these terms all equal zero. The remaining terms result from the product of the first term of 9 with the last term of 10, and the product of the last term of 9 with the first term of 10, where in both cases we must have ${\mathbf{p}=\mathbf{q}}$. Further, because

$\displaystyle u_{r}^{\dagger}\left(\mathbf{p}\right)u_{s}\left(\mathbf{p}\right)=v_{r}^{\dagger}\left(\mathbf{p}\right)v_{s}\left(\mathbf{p}\right)=\frac{E_{\mathbf{p}}}{m}\delta_{rs} \ \ \ \ \ (13)$

the double sum over spins ${r}$ and ${s}$ collapses to a single sum over ${r}$. Because ${\mathbf{p}=\mathbf{q}}$, the exponential terms cancel out and using ${\int d^{3}x=V}$ we get

$\displaystyle P^{i}=\sum_{r,\mathbf{p}}p^{i}\left[c_{r}^{\dagger}\left(\mathbf{p}\right)c_{r}\left(\mathbf{p}\right)-d_{r}\left(\mathbf{p}\right)d_{r}^{\dagger}\left(\mathbf{p}\right)\right] \ \ \ \ \ (14)$

Using the anticommutators

$\displaystyle \left[d_{r}\left(\mathbf{p}\right),d_{s}^{\dagger}\left(\mathbf{p}^{\prime}\right)\right]_{+}=\delta_{rs}\delta_{\mathbf{pp}^{\prime}} \ \ \ \ \ (15)$

we get

$\displaystyle P^{i}=\sum_{r,\mathbf{p}}p^{i}\left[c_{r}^{\dagger}\left(\mathbf{p}\right)c_{r}\left(\mathbf{p}\right)+d_{r}^{\dagger}\left(\mathbf{p}\right)d_{r}\left(\mathbf{p}\right)-1\right] \ \ \ \ \ (16)$

Because the sum over ${\mathbf{p}}$ includes both negative and positive momenta, the ${-1}$ in the sum cancels out and we’re left with

 $\displaystyle P^{i}$ $\displaystyle =$ $\displaystyle \sum_{r,\mathbf{p}}p^{i}\left[c_{r}^{\dagger}\left(\mathbf{p}\right)c_{r}\left(\mathbf{p}\right)+d_{r}^{\dagger}\left(\mathbf{p}\right)d_{r}\left(\mathbf{p}\right)\right]\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{r,\mathbf{p}}p^{i}\left(N_{r}\left(\mathbf{p}\right)+\bar{N}_{r}\left(\mathbf{p}\right)\right) \ \ \ \ \ (18)$

For the full vector 3-momentum, we thus have

$\displaystyle \mathbf{P}=\sum_{r,\mathbf{p}}\mathbf{p}\left(N_{r}\left(\mathbf{p}\right)+\bar{N}_{r}\left(\mathbf{p}\right)\right) \ \ \ \ \ (19)$

That is, the total momentum is just the sum of the momenta over all the particles. For a state ${\left|\psi_{r_{1}\mathbf{p}_{1}}\psi_{r_{2}\mathbf{p}_{2}}\psi_{r_{1}\mathbf{p}_{3}}\bar{\psi}_{r_{1}\mathbf{p}_{1}}\right\rangle }$ the momentum is

$\displaystyle \mathbf{P}\left|\psi_{r_{1}\mathbf{p}_{1}}\psi_{r_{2}\mathbf{p}_{2}}\psi_{r_{1}\mathbf{p}_{3}}\bar{\psi}_{r_{1}\mathbf{p}_{1}}\right\rangle =2\mathbf{p}_{1}+\mathbf{p}_{2}+\mathbf{p}_{3} \ \ \ \ \ (20)$

Using a similar derivation, we can find the total charge in a state to be (see Klauber’s section 4.7.2)

$\displaystyle Q=-e\sum_{r,\mathbf{p}}\left(N_{r}\left(\mathbf{p}\right)-\bar{N}_{r}\left(\mathbf{p}\right)\right) \ \ \ \ \ (21)$

where ${-e}$ is the electron charge. Note that the antiparticle number operator ${\bar{N}_{r}\left(\mathbf{p}\right)}$ contributes a charge of ${+e}$ for each antiparticle. Klauber’s derivation uses the conserved probability current and integrates it using a method similar to the above.

# Momentum of a free scalar Klein-Gordon field

Reference: References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 3, Problems 3.13 – 3.14.

We can work out the 3-momentum of a free scalar Klein-Gordon field in much the same way as we worked out the total Hamiltonian, by integrating the 3-momentum density over the total volume.

The 3-momentum density is given by

$\displaystyle p^{i}=\pi_{r}\frac{\partial\phi^{r}}{\partial x_{i}}=-\pi_{r}\frac{\partial\phi^{r}}{\partial x^{i}} \ \ \ \ \ (1)$

where the conjugate momentum is

$\displaystyle \pi_{r}=\frac{\partial\mathcal{L}}{\partial\dot{\phi}^{r}} \ \ \ \ \ (2)$

For the Klein-Gordon free field, the conjugate momentum is

 $\displaystyle \pi$ $\displaystyle =$ $\displaystyle \dot{\phi}^{\dagger}\ \ \ \ \ (3)$ $\displaystyle \pi^{\dagger}$ $\displaystyle =$ $\displaystyle \dot{\phi} \ \ \ \ \ (4)$

Thus the total momentum is given by (recall that there is a sum over ${r}$, and that the field and its conjugate are considered as independent fields):

 $\displaystyle P^{i}$ $\displaystyle =$ $\displaystyle \int p^{i}d^{3}x\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\int\pi_{r}\frac{\partial\phi^{r}}{\partial x^{i}}d^{3}x\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\int\left[\pi\frac{\partial\phi}{\partial x^{i}}+\pi^{\dagger}\frac{\partial\phi^{\dagger}}{\partial x^{i}}\right]d^{3}x\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\int\left[\dot{\phi}^{\dagger}\frac{\partial\phi}{\partial x^{i}}+\dot{\phi}\frac{\partial\phi^{\dagger}}{\partial x^{i}}\right]d^{3}x \ \ \ \ \ (8)$

The discrete fields are

 $\displaystyle \phi\left(x\right)$ $\displaystyle =$ $\displaystyle \sum_{\mathbf{k}}\frac{1}{\sqrt{2V\omega_{\mathbf{k}}}}a\left(\mathbf{k}\right)e^{-ikx}+\sum_{\mathbf{k}}\frac{1}{\sqrt{2V\omega_{\mathbf{k}}}}b^{\dagger}\left(\mathbf{k}\right)e^{ikx}\ \ \ \ \ (9)$ $\displaystyle \phi^{\dagger}\left(x\right)$ $\displaystyle =$ $\displaystyle \sum_{\mathbf{k}^{\prime}}\frac{1}{\sqrt{2V\omega_{\mathbf{k}^{\prime}}}}a^{\dagger}\left(\mathbf{k}^{\prime}\right)e^{ik^{\prime}x}+\sum_{\mathbf{k}^{\prime}}\frac{1}{\sqrt{2V\omega_{\mathbf{k}^{\prime}}}}b\left(\mathbf{k}^{\prime}\right)e^{-ik^{\prime}x} \ \ \ \ \ (10)$

The integrals over ${x}$ are all of the form ${\int e^{ix\left(k\pm k^{\prime}\right)}d^{3}x}$ which, because of the condition that an integral number of wavelengths must fit into the volume, works out to zero unless the exponent of the integrand is zero, in which case the integral just gives ${V}$. In multiplying out the terms in 8, we get four sums. Consider the term containing products of ${a}$ and ${b}$ operators. In this case, since the exponential terms in 9 and 10 are ${e^{-ikx}}$ and ${e^{-ik^{\prime}x}}$, the only non-zero terms in the integral are when ${k^{\prime}=-k}$, so we get (recall ${\omega_{\mathbf{k}}=\omega_{-\mathbf{k}}}$ and ${\partial e^{ikx}/\partial x^{i}=ik_{i}e^{ikx}=-ik^{i}e^{ikx}}$):

$\displaystyle P_{ab}^{i}=\frac{1}{2V}\sum_{\mathbf{k}}\int\left[\frac{-i\omega_{\mathbf{k}}\left(-ik^{i}\right)}{\omega_{\mathbf{k}}}b\left(-\mathbf{k}\right)a\left(\mathbf{k}\right)+\frac{-i\omega_{\mathbf{k}}\left(ik^{i}\right)}{\omega_{\mathbf{k}}}a\left(\mathbf{k}\right)b\left(-\mathbf{k}\right)\right]d^{3}x \ \ \ \ \ (11)$

Since ${\left[a,b\right]=0}$, the two terms in the integrand cancel and ${P_{ab}^{i}=0}$. The same is true of the term containing products of ${a^{\dagger}}$ and ${b^{\dagger}}$, so ${P_{a^{\dagger}b^{\dagger}}^{i}=0}$ and we’re left with the terms containing products of ${a^{\dagger}}$and ${a}$, and products of ${b^{\dagger}}$ and ${b}$. In both these cases, since the exponential terms in 9 and 10 are ${e^{\pm ikx}}$ and ${e^{\mp ik^{\prime}x}}$, the only non-zero terms in the integral are when ${k^{\prime}=k}$. We get, using ${\left[a,a^{\dagger}\right]=\left[b,b^{\dagger}\right]=1}$:

 $\displaystyle P_{a^{\dagger}a}^{i}$ $\displaystyle =$ $\displaystyle \frac{1}{2V}\sum_{\mathbf{k}}\int\left[\frac{i\omega_{\mathbf{k}}\left(-ik^{i}\right)}{\omega_{\mathbf{k}}}a^{\dagger}\left(\mathbf{k}\right)a\left(\mathbf{k}\right)+\frac{-i\omega_{\mathbf{k}}\left(ik^{i}\right)}{\omega_{\mathbf{k}}}a\left(\mathbf{k}\right)a^{\dagger}\left(\mathbf{k}\right)\right]d^{3}x\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{\mathbf{k}}k^{i}\left[a^{\dagger}\left(\mathbf{k}\right)a\left(\mathbf{k}\right)+\frac{1}{2}\right]\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{\mathbf{k}}k^{i}\left[N_{a}\left(\mathbf{k}\right)+\frac{1}{2}\right]\ \ \ \ \ (14)$ $\displaystyle P_{b^{\dagger}b}^{i}$ $\displaystyle =$ $\displaystyle \frac{1}{2V}\sum_{\mathbf{k}}\int\left[\frac{-i\omega_{\mathbf{k}}\left(ik^{i}\right)}{\omega_{\mathbf{k}}}b\left(\mathbf{k}\right)b^{\dagger}\left(\mathbf{k}\right)+\frac{i\omega_{\mathbf{k}}\left(-ik^{i}\right)}{\omega_{\mathbf{k}}}b^{\dagger}\left(\mathbf{k}\right)b\left(\mathbf{k}\right)\right]d^{3}x\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{\mathbf{k}}k^{i}\left[b^{\dagger}\left(\mathbf{k}\right)b\left(\mathbf{k}\right)+\frac{1}{2}\right]\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{\mathbf{k}}k^{i}\left[N_{b}\left(\mathbf{k}\right)+\frac{1}{2}\right] \ \ \ \ \ (17)$

At first glance, because of the ${\frac{1}{2}}$ terms in each sum, it looks like the momentum is going to suffer from the same infinity that the total energy does. However, in this case, the sum is over ${k^{i}}$ which takes on both positive and negative values, so fortunately here the sums cancel out to zero and we’re left with

 $\displaystyle P^{i}$ $\displaystyle =$ $\displaystyle P_{a^{\dagger}a}^{i}+P_{b^{\dagger}b}^{i}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{\mathbf{k}}k^{i}\left[N_{a}\left(\mathbf{k}\right)+N_{b}\left(\mathbf{k}\right)\right] \ \ \ \ \ (19)$

Thus the total momentum is just the sum of the momenta of the individual particles (assuming each particle is in an eigenstate ${\left|\phi_{\mathbf{k}}\right\rangle }$). For example, in the state ${\left|2\phi_{\mathbf{k}_{1}},3\bar{\phi}_{\mathbf{k}_{1}},\bar{\phi}_{\mathbf{k}_{2}}\right\rangle }$, the expectation value is

$\displaystyle \left\langle 2\phi_{\mathbf{k}_{1}},3\bar{\phi}_{\mathbf{k}_{1}},\bar{\phi}_{\mathbf{k}_{2}}\left|P^{i}\right|2\phi_{\mathbf{k}_{1}},3\bar{\phi}_{\mathbf{k}_{1}},\bar{\phi}_{\mathbf{k}_{2}}\right\rangle =\left(2+3\right)k_{1}^{i}+k_{2}^{i}=5k_{1}^{i}+k_{2}^{i} \ \ \ \ \ (20)$

# Conservation of momentum

References: Anthony Zee, Einstein Gravity in a Nutshell, (Princeton University Press, 2013) – Chapter I.2, problem 1.

The nature of the dependence of a force or potential on the underlying position coordinates can determine certain conservation laws. In his chapter I.2, Zee shows that a central force (a force that is always directed towards its source, such as the Earth’s gravity or a point charge’s electrostatic field) conserves angular momentum. Actually his derivation is a generalization to any number ${D\ge2}$ dimensions of the more familiar proof in 3-d, which goes like this:

The angular momentum of a mass ${m}$ is defined as

$\displaystyle \mathbf{L}=\mathbf{r}\times\mathbf{p} \ \ \ \ \ (1)$

where ${\mathbf{p}=m\mathbf{v}=m\dot{\mathbf{r}}}$ is the linear momentum. Taking the time derivative we get

 $\displaystyle \dot{\mathbf{L}}$ $\displaystyle =$ $\displaystyle \dot{\mathbf{r}}\times\mathbf{p}+\mathbf{r}\times\dot{\mathbf{p}}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle m\dot{\mathbf{r}}\times\dot{\mathbf{r}}+\mathbf{r}\times\mathbf{F}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0+0\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (5)$

where the third line uses the fact that ${\mathbf{r}\parallel\mathbf{F}}$ for a central force, so their cross product is zero. Thus ${\mathbf{L}}$ doesn’t change with time.

Now suppose that a force ${F}$ is the negative gradient of a potential function ${V}$ so that by Newton’s law:

$\displaystyle m_{a}\frac{d^{2}x_{a}^{i}}{dt^{2}}=-\frac{\partial V\left(x\right)}{\partial x_{a}^{i}} \ \ \ \ \ (6)$

where the index ${a}$ refers to particle ${a}$ in a collection of ${N}$ interacting particles, and ${i}$ is the component of the coordinate ${x}$. Note that the ${x}$ in ${V\left(x\right)}$ represents all ${D}$ components of ${x}$ (if we’re doing the calculation in ${D}$-dimensional space) and not just the magnitude of the distance.

Now suppose that ${V}$ is a function only of the coordinate differences ${x_{a}^{i}-x_{b}^{i}}$ between particles ${a}$ and ${b}$, where ${a,b=1,\ldots,N}$ with ${a\ne b}$. In this case, the total linear momentum is given by

$\displaystyle p^{i}=\sum_{a}m_{a}\frac{dx_{a}^{i}}{dt} \ \ \ \ \ (7)$

The time derivative is

 $\displaystyle \dot{p}^{i}$ $\displaystyle =$ $\displaystyle \sum_{a}m_{a}\frac{d^{2}x_{a}^{i}}{dt^{2}}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\sum_{a}\frac{\partial V\left(x\right)}{\partial x_{a}^{i}} \ \ \ \ \ (9)$

Since ${V}$ is a function of the set of all possible differences ${x_{a}^{i}-x_{b}^{i}}$, the terms in the sum 9 cancel in pairs. For example, for 3 particles if ${V=f\left(\left(x_{a}^{i}-x_{b}^{i}\right),\left(x_{a}^{i}-x_{c}^{i}\right),\left(x_{b}^{i}-x_{c}^{i}\right)\right)}$, then ${\frac{\partial V}{\partial x_{a}^{i}}}$ will contain a term equivalent to ${\frac{\partial V}{\partial\left(x_{a}^{i}-x_{b}^{i}\right)}}$ (plus another term resulting from ${\frac{\partial V}{\partial\left(x_{a}^{i}-x_{c}^{i}\right)}}$). However, ${\frac{\partial V}{\partial x_{b}^{i}}}$ will contain a term equivalent to ${-\frac{\partial V}{\partial\left(x_{a}^{i}-x_{b}^{i}\right)}}$ which cancels the first term. That is

 $\displaystyle \frac{\partial V}{\partial x_{a}^{i}}$ $\displaystyle =$ $\displaystyle \frac{\partial f}{\partial\left(x_{a}^{i}-x_{b}^{i}\right)}+\frac{\partial f}{\partial\left(x_{a}^{i}-x_{c}^{i}\right)}\ \ \ \ \ (10)$ $\displaystyle \frac{\partial V}{\partial x_{b}^{i}}$ $\displaystyle =$ $\displaystyle -\frac{\partial f}{\partial\left(x_{a}^{i}-x_{b}^{i}\right)}+\frac{\partial f}{\partial\left(x_{b}^{i}-x_{c}^{i}\right)}\ \ \ \ \ (11)$ $\displaystyle \frac{\partial V}{\partial x_{c}^{i}}$ $\displaystyle =$ $\displaystyle -\frac{\partial f}{\partial\left(x_{a}^{i}-x_{c}^{i}\right)}-\frac{\partial f}{\partial\left(x_{b}^{i}-x_{c}^{i}\right)} \ \ \ \ \ (12)$

so adding up all the derivatives causes the terms to cancel in pairs. The argument is fairly easily extended to ${N}$ particles.

Thus ${\dot{p}^{i}=0}$ and linear momentum is conserved. [I realize this isn’t a very elegant or mathematical way of proving it; there is probably a better way of writing it down, but hopefully you get the idea.]

# Uncertainty principle in three dimensions

Required math: calculus

Required physics: Schrödinger equation

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 4.1.

In three dimensions, the position and momentum operators are generalizations of their one-dimensional form:

 $\displaystyle r_{x}$ $\displaystyle =$ $\displaystyle x\ \ \ \ \ (1)$ $\displaystyle r_{y}$ $\displaystyle =$ $\displaystyle y\ \ \ \ \ (2)$ $\displaystyle r_{z}$ $\displaystyle =$ $\displaystyle z\ \ \ \ \ (3)$ $\displaystyle p_{x}$ $\displaystyle =$ $\displaystyle -i\hbar\frac{\partial}{\partial x}\ \ \ \ \ (4)$ $\displaystyle p_{y}$ $\displaystyle =$ $\displaystyle -i\hbar\frac{\partial}{\partial y}\ \ \ \ \ (5)$ $\displaystyle p_{z}$ $\displaystyle =$ $\displaystyle -i\hbar\frac{\partial}{\partial z} \ \ \ \ \ (6)$

All the position operators commute with each other since they simply multipliers. Thus

$\displaystyle \left[r_{i},r_{j}\right]=0 \ \ \ \ \ (7)$

Similarly, all the components of momentum commute with each other, since they are derivatives, and there are no occurrences of the variables on which they operate

$\displaystyle \left[p_{i},p_{j}\right]=0 \ \ \ \ \ (8)$

Mixtures of momentum and position will interact the same way as in one dimension if the two components are along the same axis. Mixtures of momentum and position that lie along different axes will commute, since the derivative in the momentum is with respect to a component not found in the position. Thus

$\displaystyle \left[r_{i},p_{j}\right]=i\hbar\delta_{ij} \ \ \ \ \ (9)$

Earlier, we derived an equation for the rate of change of an observable ${Q}$:

$\displaystyle \frac{d}{dt}\left\langle Q\right\rangle =\frac{i}{\hbar}\left\langle \left[H,Q\right]\right\rangle +\left\langle \frac{\partial Q}{\partial t}\right\rangle \ \ \ \ \ (10)$

where ${H}$ is the hamiltonian.

Examining the derivation of this equation shows that nothing depends on the calculation being done in one, two or three dimensions (the wave function and hamiltonian in the derivation could be in any number of dimensions), so it can be applied to each component of ${\mathbf{r}}$ and ${\mathbf{p}}$ separately. Because of this we can use the results we worked out earlier for the rates of change of position and momentum in one dimension, applied to each of the three axes, and get the corresponding result (Ehrenfest’s theorem) in three dimensions:

 $\displaystyle \frac{d\langle\mathbf{r}\rangle}{dt}$ $\displaystyle =$ $\displaystyle \frac{\langle\mathbf{p}\rangle}{m}\ \ \ \ \ (11)$ $\displaystyle \frac{d\langle\mathbf{p}\rangle}{dt}$ $\displaystyle =$ $\displaystyle \left\langle -\nabla V\right\rangle \ \ \ \ \ (12)$

To work out the uncertainty principle in three dimensions, we can use the relation derived earlier:

$\displaystyle \sigma_{A}^{2}\sigma_{B}^{2}\ge\left(\frac{1}{2i}\left\langle [\hat{A},\hat{B}]\right\rangle \right)^{2} \ \ \ \ \ (13)$

Again, the derivation of this equation does not depend on the number of dimensions. We can therefore use it to calculate the uncertainty principle for the three components of position and momentum. From the commutators above, we have

$\displaystyle \sigma_{r_{i}}\sigma_{p_{j}}\ge\frac{\hbar}{2}\delta_{ij} \ \ \ \ \ (14)$

The ${\delta_{ij}}$ indicates that a position component and a perpendicular momentum component can both be measured precisely at the same time, since these components commute, as we saw above. For position and momentum components along the same direction, the uncertainty principle is the same as it is in one dimension.

# The uncertainty principle

Required math: calculus, complex numbers

Required physics: basics of quantum mechanics

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Sec 3.5.

The uncertainty principle is probably the most famous of the predictions of quantum mechanics, and it is usually known in the specific case of position and momentum: it is impossible to measure both position and momentum exactly at the same time.

In fact, this is just one example of a wider uncertainty principle, which can be derived algebraically. The principle relies on the generalized statistical interpretation of the wave function, which is that, for an operator ${\hat{A}}$, the expectation value (that is, the average value over a large (essentially infinite) number of measurements) is given by the integral

$\displaystyle \left\langle A\right\rangle =\int\Psi^*\hat{A}\Psi dx \ \ \ \ \ (1)$

where the integral extends over all space (or at least over all space accessible to the system). This integral is the one-dimensional case, but the three-dimensional case is easily written down by integrating over all three spatial coordinates.

This integral is usually written in the bra-ket notation to save space:

$\displaystyle \int\Psi^*\hat{A}\Psi dx\equiv\left\langle \Psi|\hat{A}\Psi\right\rangle \ \ \ \ \ (2)$

Any function placed in the ‘bra’ side (the left part of the bra-ket) is the complex conjugate of what’s written there, while the function in the ‘ket’ side is the unmodified function.

For an observable, the operator ${\hat{A}}$ is Hermitian, which means that the integral above is equivalent to

 $\displaystyle \int\Psi^*\hat{A}\Psi dx$ $\displaystyle =$ $\displaystyle \int(\hat{A}\Psi)^*\Psi dx\ \ \ \ \ (3)$ $\displaystyle \left\langle \Psi|\hat{A}\Psi\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle \hat{A}\Psi|\Psi\right\rangle \ \ \ \ \ (4)$

Using this principle, we can write down an expression for the variance of an observable. In statistics, the variance is defined as the average of the square of the difference from the mean. That is

$\displaystyle \sigma_{A}^{2}\equiv\left\langle (\hat{A}-\left\langle A\right\rangle )^{2}\right\rangle \ \ \ \ \ (5)$

Note that here the angle brackets denote the average, as opposed to the bra-ket notation where the angle brackets denote an integral. Averages can be distinguished from bra-kets since in the latter there is always a vertical bar in the middle to separate the bra from the ket.

By assumption, this can be calculated as

 $\displaystyle \sigma_{A}^{2}$ $\displaystyle =$ $\displaystyle \left\langle \Psi|(\hat{A}-\left\langle A\right\rangle )^{2}\Psi\right\rangle \ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle (\hat{A}-\left\langle A\right\rangle )\Psi|(\hat{A}-\left\langle A\right\rangle )\Psi\right\rangle \ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle \equiv$ $\displaystyle \left\langle f|f\right\rangle \ \ \ \ \ (8)$

where the function ${f}$ is defined by this equation.

Similarly, we can define the variance for another observable ${\hat{B}}$:

 $\displaystyle \sigma_{B}^{2}$ $\displaystyle =$ $\displaystyle \left\langle \Psi|(\hat{B}-\left\langle B\right\rangle )^{2}\Psi\right\rangle \ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle (\hat{B}-\left\langle B\right\rangle )\Psi|(\hat{B}-\left\langle B\right\rangle )\Psi\right\rangle \ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle \equiv$ $\displaystyle \left\langle g|g\right\rangle \ \ \ \ \ (11)$

Using the integral form of the Schwarz inequality (the proof of which would take us too far afield here), we can write

 $\displaystyle \sigma_{A}^{2}\sigma_{B}^{2}$ $\displaystyle =$ $\displaystyle \left\langle f|f\right\rangle \left\langle g|g\right\rangle \ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle \ge$ $\displaystyle |\left\langle f|g\right\rangle |^{2} \ \ \ \ \ (13)$

For any complex number ${z=x+iy}$, we have

 $\displaystyle \left|z\right|^{2}$ $\displaystyle =$ $\displaystyle x^{2}+y^{2}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle \ge$ $\displaystyle y^{2}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\frac{1}{2i}(z-z^*)\right]^{2} \ \ \ \ \ (16)$

Letting ${z=\left\langle f|g\right\rangle }$, we can combine this with the Schwarz inequality and get

$\displaystyle \sigma_{A}^{2}\sigma_{B}^{2}\ge\left[\frac{1}{2i}(\left\langle f|g\right\rangle -\left\langle g|f\right\rangle )\right]^{2} \ \ \ \ \ (17)$

Now we need to work out the two bra-ket terms in terms of the original operators. Remember that the mean value ${\left\langle A\right\rangle }$ of an operator is just a number, so it can be taken outside the bra-ket.

 $\displaystyle \left\langle f|g\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle (\hat{A}-\left\langle A\right\rangle )\Psi|(\hat{B}-\left\langle B\right\rangle )\Psi\right\rangle \ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle \Psi|(\hat{A}-\left\langle A\right\rangle )(\hat{B}-\left\langle B\right\rangle )\Psi\right\rangle \ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle \Psi|(\hat{A}\hat{B}-\hat{A}\left\langle B\right\rangle -\left\langle A\right\rangle \hat{B}+\left\langle A\right\rangle \left\langle B\right\rangle )\Psi\right\rangle \ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle \hat{A}\hat{B}\right\rangle -\left\langle A\right\rangle \left\langle B\right\rangle -\left\langle A\right\rangle \left\langle B\right\rangle +\left\langle A\right\rangle \left\langle B\right\rangle \ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle \hat{A}\hat{B}\right\rangle -\left\langle A\right\rangle \left\langle B\right\rangle \ \ \ \ \ (22)$

By the same argument, we can work out ${\left\langle g|f\right\rangle }$, or we can obtain it merely by switching ${A}$ and ${B}$ in the result above.

$\displaystyle \left\langle g|f\right\rangle =\left\langle \hat{B}\hat{A}\right\rangle -\left\langle A\right\rangle \left\langle B\right\rangle \ \ \ \ \ (23)$

Note that ${\left\langle \hat{B}\hat{A}\right\rangle }$ is not necessarily the same as ${\left\langle \hat{A}\hat{B}\right\rangle }$ since in general the two operators do not commute. In fact, this is nub of the argument, since plugging these results back into the Schwarz inequality we get

 $\displaystyle \sigma_{A}^{2}\sigma_{B}^{2}$ $\displaystyle \ge$ $\displaystyle \left[\frac{1}{2i}\left(\left\langle f|g\right\rangle -\left\langle g|f\right\rangle \right)\right]^{2}\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\frac{1}{2i}\left(\left\langle \hat{A}\hat{B}\right\rangle -\left\langle \hat{B}\hat{A}\right\rangle \right)\right]^{2} \ \ \ \ \ (25)$

In terms of the commutator of the two operators:

$\displaystyle \left[\hat{A},\hat{B}\right]\equiv\hat{A}\hat{B}-\hat{B}\hat{A} \ \ \ \ \ (26)$

we have the generalized uncertainty principle:

$\displaystyle \sigma_{A}^{2}\sigma_{B}^{2}\ge\left(\frac{1}{2i}\left\langle [\hat{A},\hat{B}]\right\rangle \right)^{2} \ \ \ \ \ (27)$

Note that since the mean value of a commutator is the difference between a quantity ${\left\langle f|g\right\rangle }$ and its complex conjugate ${\left\langle g|f\right\rangle }$, it is always a pure imaginary number with zero real part, so the quantity ${\frac{1}{2i}\left\langle [\hat{A},\hat{B}]\right\rangle }$ is always real, and its square is therefore always non-negative (it could be zero if the two operators commute). Thus this inequality says that for any two observable operators that do not commute, there will always be a lower limit on the accuracy with which they can be measured simultaneously.

As an example, we can work out the most fundamental commutator of all: that of position and momentum: ${x}$ and ${p}$. Since ${p}$ is a differential operator, we need a dummy function for it to operate on.

 $\displaystyle \left[x,p\right]f$ $\displaystyle =$ $\displaystyle x\frac{\hbar}{i}\frac{df}{dx}-\frac{\hbar}{i}\frac{d(xf)}{dx}\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{i}\left(x\frac{df}{dx}-x\frac{df}{dx}-f\right)\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar f \ \ \ \ \ (30)$

Thus the commutator on its own is

$\displaystyle \left[x,p\right]=i\hbar \ \ \ \ \ (31)$

Plugging this into the uncertainty principle, we get the well-known result

 $\displaystyle \sigma_{x}^{2}\sigma_{p}^{2}$ $\displaystyle \ge$ $\displaystyle \left(\frac{1}{2i}\left\langle [x,p]\right\rangle \right)^{2}\ \ \ \ \ (32)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar^{2}}{4} \ \ \ \ \ (33)$

or in terms of the standard deviation (the square root of the variance):

$\displaystyle \sigma_{x}\sigma_{p}\ge\frac{\hbar}{2} \ \ \ \ \ (34)$

Thus Planck’s constant (divided by ${4\pi}$) servers as a lower bound on the accuracy with which position and momentum can be measured at the same time.

Similar relations exist for any pair of observables that don’t commute, and in any attempt to conceptualize a result in quantum mechanics, it must be remembered that any such pair cannot be visualized simultaneously. In the case of angular momentum, for example, any pair of its components does not commute, which means that the three components of angular momentum cannot be visualized, meaning that there is no such thing as a strict angular momentum vector.