# 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

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

## 4 thoughts on “Postulates of quantum mechanics: momentum”

1. Arpon Paul

I think, in equation (12) & (13), there is a mistake.
Here, psi_p (x) =
So, in equation (12), it should be psi_p * (x), and in equation (13), it should be exp(-ipx/h)