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Does anybody knows how to show that the position expectation value $$\langle x\rangle = \int_{-\infty}^{\infty} x|\psi(x)|^2dx$$

can be expressed in terms of momentum?

$$\langle x\rangle = i\hbar\int_{-\infty}^{\infty} \tilde{\psi}^*(p)\frac{\partial \tilde{\psi}(p)}{\partial p}dp$$

My attempts:

\begin{eqnarray} \langle x \rangle &=&\int \langle \psi |\hat{x}| \psi \rangle \int dp |p\rangle \langle p | \\ &=& \int \langle \psi |\hat{p}|p\rangle\langle p |\psi \rangle dp \\ &=& \int \psi^*(p) x \psi(p) dp \end{eqnarray}

I am stuck at this part ... any help will be appreciated.

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  • $\begingroup$ x^just means x hat $\endgroup$ – john Apr 13 '18 at 5:14
  • $\begingroup$ Also I am not sure how to get rid of x to get the desired expression in this case $\endgroup$ – john Apr 13 '18 at 5:15
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You just need to heed the definitions of your book, $\langle p|\psi\rangle=\tilde{\psi}(p)$, $\langle \psi|x\rangle= \psi^* (x)$, $\langle x|p\rangle=e^{ixp/\hbar}/\sqrt{2\pi\hbar}$, insert two complete states and integrate by parts, $$\langle x\rangle = \langle \psi |\hat{x}| \psi \rangle =\int dx dp ~~ \langle \psi |\hat{x}|x\rangle \langle x |p\rangle \langle p |\psi \rangle \\= \int dx dp ~ \psi^* (x) ~x~ \frac{e^{ixp/\hbar}}{\sqrt{2\pi\hbar} } \tilde{\psi}(p) \\= \int dx dp ~ \psi^* (x)\frac{\hbar}{i} \frac{\partial_p e^{ixp/\hbar}}{\sqrt{2\pi\hbar} } \tilde{\psi}(p) \\= i\hbar \int dp ~\left (dx~ \psi^* (x) \frac{ e^{ixp/\hbar}}{\sqrt{2\pi\hbar} }\right )~ \partial_p\tilde{\psi}(p) \\ =i\hbar\int dp~ \tilde{\psi}^*(p)\frac{\partial \tilde{\psi}(p)}{\partial p} ~.$$

So, as often, a variable is gotten rid of in favor of its conjugate through the standard Fourier trick.

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  • $\begingroup$ Dear Dr. Zachos, thanks for the beautiful answer. It has helped me to compute the expectation value of a non-local potential in momentum space. Thanks again! $\endgroup$ – rainman Jan 27 at 4:48
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This looks a little bit like homework which is generally not answered here. I'll give you a few hints, so that you can do it yourself.

You are trying to get the expectation value of the position operator. In bra-ket notation: $$ \begin{equation} \langle \hat x\rangle = \langle \psi|\hat x | \psi\rangle \end{equation} $$ Now be careful what you do. Your attempts were not fully correct but you are right - you'll need to insert a "1" in the form of $|p\rangle \langle p|$: $$ \begin{equation} \langle \hat x\rangle = \langle \psi|\hat x | \psi\rangle \\ = \langle \psi|p \rangle \langle p |\hat x | \bar{p}\rangle \langle \bar{p}|\psi\rangle \end{equation} $$

You know that $\langle \bar{p}|\psi\rangle = \psi(\bar{p})$ and $\langle \psi|p \rangle = \psi^* (p)$ so in order to solve your question, you'll need to evaluate the expression $\langle p |\hat x | \bar{p}\rangle$. Can you take it from here?

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Bracket algebra alone will never give you the result. The answer is written in a particular basis, so the explicit form of the operator and functions involved in that basis is needed. How is this expectation value written in the position basis? You probably know that you can change basis by performing a Fourier transform. How can you write the result in the form of a scalar product involving an operator? Now be very careful to distinguish between operators and scalars.

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protected by Qmechanic Apr 13 '18 at 19:46

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