# Expectation value with plane waves [duplicate]

This question already has an answer here:

Hey guys Im a little confused with the concept of plane waves and how to perform an expectation value. Let me show you by an example. Suppose you have a wave function of the form

$\psi_{\boldsymbol{p}_{0}}(x)=f(p_{0})e^{\frac{i}{\hbar}\boldsymbol{p}_{0}\cdot\boldsymbol{x}}$

where $\boldsymbol{p}_{0}=(0,0,p_{0})$ and suppose you want to perform an expectation value of the position of the particle, that is

$<x>=f^{2}(p_{0})\int\,d^{3}\boldsymbol{x}\,x\,e^{\frac{i}{\hbar}\boldsymbol{p}_{0}\cdot\boldsymbol{x}}e^{-\frac{i}{\hbar}\boldsymbol{p}_{0}\cdot\boldsymbol{x}}=f^{2}(p_{0})\int\,d^{3}\boldsymbol{x}\,x$

wich I think is nonsense. But if you define an arbitrary momentum vector $\boldsymbol{p}'=(p_{1}',p_{2}',p_{3}')$, and perform the transition probability

\begin{align} \left<\psi_{\boldsymbol{p}'}(\boldsymbol{x})|\,x\,|\psi_{\boldsymbol{p}_{0}}(\boldsymbol{x})\right>&=f(p')f(p_{0})\int\,d^{3}\boldsymbol{x}\,xe^{-\frac{i}{\hbar}(\boldsymbol{p}'-\boldsymbol{p}_{0})\cdot\boldsymbol{x}}=f(p')f(p_{0})\int\,d^{3}\boldsymbol{x}\left( i\hbar\frac{\partial}{\partial p_{x}'}\right)e^{-\frac{i}{\hbar}(\boldsymbol{p}'-\boldsymbol{p}_{0})\cdot\boldsymbol{x}}= \\ &=i\hbar f(p')f(p_{0})\frac{\partial}{\partial p_{x}'}\delta^{3}(\boldsymbol{p}'-\boldsymbol{p}_{0})=-i\hbar\delta^{3}(\boldsymbol{p}'-\boldsymbol{p}_{0})\frac{\partial}{\partial p_{x}'}\left( f(p')\right)f(p_{0}) \end{align}

where I made use of the property $f(x)\delta'(x)=-f'(x)\delta(x)$. So now with my new expresion I have a meaningful result and I can evaluate for $\boldsymbol{p}'=\boldsymbol{p}_{0}$ and get a result that I wasnt able to get with the first method. What I'm doing wrong or the the second way is the correct way to do it? Thanks!

## marked as duplicate by ACuriousMind♦, Brandon Enright, Kyle Kanos, Pranav Hosangadi, JamalSFeb 2 '15 at 7:44

Nothing will help, a plane wave occupies all the space, and the mean value of the position doesn't make much sense, but in your case is zero, because the integrand in the 1st calculus is anti-symmetrical. But this, on one condition, namely if we write the integral as

$lim _{a \to \infty} f^2(p_0) \int _a^a x d^3 x$

Otherwise it's hard to say what is the value of your integral.

The second integration is something else than the 1st one, it evaluates the matrix elements of the quantity $x$ in the base of the Fourier functions. The matrix element that is equivalent with the mean value of $x$, is the diagonal one for $p' = p_0$.

I am not sure whether you can interchange between the derivative and the integral. Up to the point where you introduced the derivative (and including that step), the integral is zero for $p' = p_0$, in the condition that we guarantee the equality in absolute value of the limits. But even if the interchange were allowed, in your before-last expression, $\delta (p' - p_0)$ for $p' = p_0$ is infinite, and you try to calculate the derivative of an infinite constant. It's not advisable to use this way.

• That was my first answer when this problem came up to me. But you can tell the same about the second result? Both have to be the same when p'=p0 – Tomas Libutti Feb 1 '15 at 21:55
• @TomasLibutti I am writing the answer to what you ask. – Sofia Feb 1 '15 at 22:00
• @TomasLibutti : no, I don't think that the answer is as you said, because your second way is not the same as the first. In your second calculus you evaluate matrix elements. – Sofia Feb 1 '15 at 22:03
• But the first way isnt just an evaluation of the diagonal matrix elements? – Tomas Libutti Feb 1 '15 at 22:04
• @TomasLibutti : no, I have to introduce a modification, look at it. – Sofia Feb 1 '15 at 22:11

wich I think is nonsense

Formally, shouldn't the 2nd equation be

$$\langle x \rangle = \frac{\langle\psi_{\mathbf p_0}|\hat x|\psi_{\mathbf p_0}\rangle}{\langle\psi_{\mathbf p_0}|\psi_{\mathbf p_0}\rangle} = \frac{f^2(p_0)\int d\mathbf x^3 x}{f^2(p_0)\int d\mathbf x^3} = \frac{\int d\mathbf x^3 x}{\int d\mathbf x^3} = \lim_{\tau \rightarrow \infty}\frac{\int_{-\tau}^{\tau} d\mathbf x^3 x}{\int_{-\tau}^{\tau} d\mathbf x^3} = \lim_{\tau \rightarrow \infty}\frac{0}{2\tau} = 0$$