Given the one dimensional harmonic oscillator $$\Psi(x) = \frac{1}{\sqrt{14}}[\psi_{0}(x) + 2i \psi_{1}(x) + 3 \psi_{2}(x)]$$

I'm told to determine the expectation value for the momentum. I use the ladder operator $$p = i \sqrt{\frac{\hbar \omega m}{2}}(a^{+}-a)$$

Which gives me $$\langle p \rangle =\langle \Psi(x)|i \sqrt{\frac{\hbar \omega m}{2}}(a^{+}-a)| \Psi(x) \rangle = i \sqrt{\frac{\hbar \omega m}{2}}[\langle \Psi(x)| a^{+}|\Psi(x)\rangle - \langle \Psi(x)|a|\Psi(x)\rangle] = i \sqrt{\frac{\hbar \omega m}{2}}[-2i + 6\sqrt{2}i -2i + 6\sqrt{2}i]\frac{1}{14} = - \frac{4}{7}\sqrt{\frac{\hbar \omega m}{2}}$$

Is it possible for the expectation value of $p$ to be negative or is there a mistake in my calculation?

  • 5
    $\begingroup$ Why would you think it would not be possible? Is a negative value for (directional!) momentum unusual? $\endgroup$ – ACuriousMind Apr 15 '18 at 13:36
  • 1
    $\begingroup$ Yes, negative values are even valid in classical mechanics. $\endgroup$ – Mauricio Apr 15 '18 at 13:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.