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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!

Qmechanic
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Tomas Libutti
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2 Answers2

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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.

Sofia
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  • 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 01 '15 at 21:55
  • @TomasLibutti I am writing the answer to what you ask. – Sofia Feb 01 '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 01 '15 at 22:03
  • But the first way isnt just an evaluation of the diagonal matrix elements? – Tomas Libutti Feb 01 '15 at 22:04
  • @TomasLibutti : no, I have to introduce a modification, look at it. – Sofia Feb 01 '15 at 22:11
  • Im looking but I still don't see where I'm wrong or if everything is fine then how the equality between both ways came up. Thanks for your time Sofia! – Tomas Libutti Feb 01 '15 at 22:14
  • @TomasLibutti : look now! – Sofia Feb 01 '15 at 22:18
  • @TomasLibutti : the result of the 1st integral is zero in we guarantee that $|- \infty| = |+ \infty|$ i.e. that the two limits are equal in absolute value, although they are infinite. But, can we guarantee this? – Sofia Feb 01 '15 at 22:20
  • Indeed a tricky question Sofia. I think that the "correct" way to do it is the second one. Seems more legit dont you think? – Tomas Libutti Feb 01 '15 at 22:23
  • @TomasLibutti , no, it is even more unsure. Anyway, what you have in the 2nd case for $p' = p_0$ is the derivative of a constant (for $p' = p_0$ the delta is a constant, but equal to infinite). Don't use this thing, don't deal with the derivative of an infinite constant. – Sofia Feb 01 '15 at 22:28
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wich I think is nonsense

Formally, shouldn't the 2nd equation be

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

Alfred Centauri
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