Expectation value in the ground state of simple harmonic oscillator

Question

Motivated by a problem in chapter 2 of Sakurai's book Modern Quantum Mechanics, I'm interested in confirming something about the simple harmonic oscillator in quantum mechanics, I have found that the quantity $\langle m| \hat{p} \hat{x} |n \rangle$, where $|n \rangle$ is the n-th energy eigenstate of the Hamiltonian $H = \frac{P^2}{2m} + \frac{m \omega^2 x^2}{2}$, is given by $$\langle m| \hat{p}\hat{x}|n \rangle = \bigg[\frac{-i\hbar}{2}\bigg(\sqrt{m+1}\langle m+1| - \sqrt{m}\langle m-1|)(\sqrt{n}|n-1 \rangle + \sqrt{n+1}|n+1 \rangle\bigg)\bigg] \\= -\frac{i \hbar}{2}\bigg[\sqrt{(m+1)n} \langle m+1|n-1 \rangle + \sqrt{(m+1)(n+1)}\langle m+1| n+1 \rangle - \sqrt{mn} \langle m-1| n-1 \rangle - \sqrt{m(n+1)}\langle m-1|n+1 \rangle\bigg]$$

Hence I get $\langle 0| \hat{p}\hat{x}|0 \rangle = \frac{-i \hbar}{2}$, I believe the correct result is $\langle 0| \hat{p}\hat{x}|0 \rangle = 0$, does anyone know where I went wrong?

Answer 1

The result should actually be nonzero, because $x \sim a + a^\dagger$, $p \sim a - a^\dagger$ so that $$\langle 0 | px |0 \rangle \sim \langle 0 | (a - a^\dagger) (a + a^\dagger) |0 \rangle = \langle 0 | a a^\dagger |0 \rangle = 1.$$ Now, to get the exact value, note that $[x, p] = i \hbar$, so $$\langle 0 | [x, p] | 0 \rangle = \langle 0 | i \hbar | 0 \rangle = i \hbar.$$ But the left-hand side is also equal to $(\langle 0 | px | 0 \rangle)^* - \langle 0 | px | 0 \rangle$, so therefore $$\langle 0 | px | 0 \rangle = - \frac{i\hbar}{2}$$ which is exactly as you calculated.