Matrix elements of the operator $\hat{x} \hat{p}$ in position and momentum basis - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-07-20T05:15:03Z https://physics.stackexchange.com/feeds/question/121645 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://physics.stackexchange.com/q/121645 2 Matrix elements of the operator $\hat{x} \hat{p}$ in position and momentum basis Spine Feast https://physics.stackexchange.com/users/23937 2014-06-24T16:05:58Z 2014-06-26T13:42:15Z <p>I want to calculate the matrix elements of the operator $\hat{x} \hat{p}$ in momentum and position basis, that is the two quantities ($p,q$ - momenta, $x,y$ - positions):</p> <p>$$\langle p|\hat{x} \hat{p}|q\rangle$$ $$\langle x|\hat{x} \hat{p}|y\rangle$$</p> <p>I don't know how to do this. I write $\hat{p}|q\rangle = q | q \rangle$. And $\hat{x} |q \rangle = -i\hbar \frac{d}{dp} | q\rangle$, so </p> <p>$$\langle p|\hat{x} \hat{p}|q\rangle = -i\hbar q \frac{d}{dp} \delta(p-q)$$</p> <p>This is nonsensical.</p> <p>How do I proceed?</p> https://physics.stackexchange.com/questions/121645/-/121674#121674 3 Answer by jwimberley for Matrix elements of the operator $\hat{x} \hat{p}$ in position and momentum basis jwimberley https://physics.stackexchange.com/users/22148 2014-06-24T19:48:27Z 2014-06-26T13:42:15Z <p>It's not non-sensical at all, except that there shouldn't be a minus sign (as mentioned in the comments) and that you took an operator outside of an expectation value, which I think worked out OK in this case but in general should be avoided. More conservatively,</p> <p>$$\hat x = i \hbar \frac{d}{d \hat p}$$</p> <p>it follows that</p> <p>$$\langle q \mid \hat x \hat p \mid q' \rangle = q' \langle q \mid \hat x \mid q' \rangle = i \hbar q' \langle q \mid \frac{d}{d \hat p} \mid q' \rangle = i \hbar q' \langle q \mid \frac{d}{d q'} \mid q' \rangle$$</p> <p>Remember, an actual physical state is never an idealized momentum (or position) eigenket. A slightly more realistic description of a physical state is a wave packet with a narrow width around some momentum:</p> <p>$$\mid p, \sigma \rangle = \int \frac{dq}{2 \pi \hbar} \, \left( \frac{2 \pi \hbar^2}{\sigma^2}\right)^{1/4} e^{- \frac{1}{4} \left( \frac{q-p}{\sigma} \right)^2} \mid q \rangle$$</p> <p>You can check that this is normalized to unity, because</p> <p>$$\langle p, \sigma \mid p, \sigma \rangle = 1$$</p> <p>Remember that $\langle p \mid q \rangle = 2 \pi \hbar \delta(p-q)$. Now, the expectation value of $\hat x \hat p$ for this physical state is</p> <p>$$\langle p, \sigma \mid \hat x \hat p \mid p, \sigma \rangle = \frac{1}{\sqrt{2 \pi} \, \sigma } \int dq \, dq' \, e^{- \frac{1}{4} \left( \frac{q-p}{\sigma} \right)^2} e^{- \frac{1}{4} \left( \frac{q'-p}{\sigma} \right)^2} \frac{\langle q \mid \hat x \hat p \mid q' \rangle}{2 \pi \hbar}$$</p> <p>Using the above result, we can integrate the $q'$ integral by parts, yielding</p> <p>$$\frac{-i\hbar}{\sqrt{2 \pi} \, \sigma } \int dq \, dq' \, e^{- \frac{1}{4} \left( \frac{q-p}{\sigma} \right)^2} \frac{d}{dq'} \left( q' e^{- \frac{1}{4} \left( \frac{q'-p}{\sigma} \right)^2} \right) \frac{\langle q \mid q' \rangle}{2 \pi \hbar}$$</p> <p>Since $\langle q \mid q' \rangle = 2 \pi \hbar \delta(q-q')$, we can integrate out q (and then relabel q' back to q), making this a single integral</p> <p>$$\frac{-i\hbar}{\sqrt{2 \pi} \, \sigma} \int dq \, e^{- \frac{1}{4} \left( \frac{q-p}{\sigma} \right)^2} \frac{d}{dq} \left( q e^{- \frac{1}{4} \left( \frac{q-p}{\sigma} \right)^2} \right)$$</p> <p>We can integrate by parts again, ending up with</p> <p>$$\frac{i\hbar}{\sqrt{2 \pi} \, \sigma} \int dq \, q e^{- \frac{1}{4} \left( \frac{q-p}{\sigma} \right)^2} \frac{d}{dq} \left( e^{- \frac{1}{4} \left( \frac{q-p}{\sigma} \right)^2} \right)$$</p> <p>which is a bit easier to evaluate since we don't have to use the product rule. The derivative brings down a factor $-(q-p)/2\sigma^2$, and the two exponentials combine, leaving us with</p> <p>$$\frac{-i\hbar}{ 2\sqrt{2 \pi} \, \sigma^3} \int dq \, q(q-p) e^{- \frac{1}{2} \left( \frac{q-p}{\sigma} \right)^2}$$</p> <p>The first $q$ in the integrand can be rewritten as $(q-p) + p$. Then there are two terms, the latter of which has an odd integrand (of the form $(q-p) \exp(\alpha (q-p)^2)$ and vanishes. So, changing the integrand to $x = (q-p)/\sigma$, we are left with</p> <p>$$\frac{-i\hbar}{ 2\sqrt{2 \pi}} \int dx \, x^2 e^{- \frac{1}{2} x^2}$$</p> <p>The remaining integral is a textbook Gaussian integral, equal to </p> <p>$$\int dx \, x^2 e^{-\frac{1}{2} x^2} = \sqrt{2\pi}$$</p> <p>So, the expectation value of the operator $\hat x \hat p$ for the wave packet is simply</p> <p>$$\frac{-i\hbar}{2}$$</p>