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

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  3. 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)

    Reply

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