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If $\langle\hat x\rangle_{\Psi} = 0$ can we conclude by the Ehrenfest theorem that $\langle\hat p\rangle_{\Psi}=0$ ?

I've seen that in this video but I'd rather think that an expectation value is like a function (it's derivative can be non-zero when the function itself is zero). What do you think?

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  • $\begingroup$ Have you understood Schroedinger's wavepacket ? $\endgroup$ May 24 at 14:13
  • $\begingroup$ I have studied the harmonic oscillator but not Schrödinger wavepackets, is it compulsory for my question? $\endgroup$
    – niobium
    May 24 at 14:44
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    $\begingroup$ It is, absolutely, if you were interested in the general answer... $\endgroup$ May 24 at 14:52

1 Answer 1

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Consider a harmonic oscillator with mass $m$ and frequency $\omega$, and the state : $$|\psi\rangle = \frac 1{\sqrt{2}}\left(|0\rangle + i|1\rangle\right)$$ In this state we have : \begin{align} \langle\hat x\rangle_\psi &= \sqrt{\frac{\hbar}{2m\omega}}\langle\psi|a+a^\dagger|\psi\rangle\\ &= \frac 12 \sqrt{\frac{\hbar}{2m\omega}} (\langle 0 |-i\langle1|)(a+a^\dagger)(|0\rangle + i|1\rangle) \\ &= \frac 12 \sqrt{\frac{\hbar}{2m\omega}} (\langle 0 |-i\langle1|)(i|0\rangle + |1\rangle + i\sqrt 2|2\rangle)\\ &=\frac 12 \sqrt{\frac{\hbar}{2m\omega}}(i-i)\\ &= 0 \end{align}

and : $$|\psi\rangle = \frac 1{\sqrt{2}}\left(|0\rangle + i|1\rangle\right)$$ In this state we have : \begin{align} \langle\hat p\rangle_\psi &=-i \sqrt{\frac{\hbar m\omega}{2}}\langle\psi|a-a^\dagger|\psi\rangle\\ &= \frac 1{2i} \sqrt{\frac{\hbar m\omega}{2}}(\langle 0 |-i\langle1|)(a-a^\dagger)(|0\rangle + i|1\rangle) \\ &= \frac 1{2i} \sqrt{\frac{\hbar m\omega}{2}}(\langle 0 |-i\langle1|)(i|0\rangle - |1\rangle -i\sqrt 2|2\rangle)\\ &=\frac 1{2i}\sqrt{\frac{\hbar m\omega}{2}}(i+i)\\ &\neq 0 \end{align}

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