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I'm trying to find the expectationvalue for $p^2$ where $p = i\sqrt{\frac{hmw}{2}}(a_{+} - a_{-})$ and i end up with the following result \begin{align*} \langle \psi_0|p^2|\psi_0\rangle &= -\frac{\hbar mw}{2}\langle\psi_0|(a_{+} - a_{-})^2|\psi_0\rangle\\ &= -\frac{\hbar mw}{2}\langle\psi_0|a_{+}^2 - a_{+}a_{-} - a_{-}a_{+} + (a_{-})^2|\psi_0\rangle\\ \Rightarrow \langle\psi_0|p^2|\psi_0\rangle &= -\frac{\hbar mw}{2}(\langle\psi_0|a_{+}^2|\psi_0\rangle -\langle\psi_0|a_{-}a_{+}|\psi_0\rangle)\\ &= -\frac{\hbar mw}{2}(\langle a_{+}\psi_0|a_{+}\psi_0\rangle - \langle a_{-}\psi_0|a_{+}\psi_0\rangle)\\ &= -\frac{\hbar mw}{2}(\langle\psi_1|\psi_1\rangle - \langle0|0\rangle)\\ &= -\frac{\hbar m w}{2} \end{align*} Where i've used the fact that $a_{+}a_{-}\psi_0 = 0$ and $a_{-}a_{+}\psi_0 = \psi_0$. I can see that the minus sign appears because to the imaginary number, but i must be missing something because the result is not supposed to be negative.

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closed as off-topic by Martin, Kyle Kanos, Qmechanic Feb 26 '16 at 13:01

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  • $\begingroup$ I see now that i made another mistake when it comes to $\langle \psi_0|a_{-}a_{+}|\psi_0\rangle$ which $\neq 0$. It's quite an odd mistake because i stated that i used the fact that $a_{-}a_{+}\psi_0 = \psi_0$. $\endgroup$ – QuantumMechanic Feb 25 '16 at 19:59
  • $\begingroup$ Third line is wrong. $\endgroup$ – DanielSank Feb 25 '16 at 20:26
  • $\begingroup$ Why? I don't see it. @DanielSank $\endgroup$ – QuantumMechanic Feb 25 '16 at 20:34
  • $\begingroup$ Ooops, I was wrong. Forget it. $\endgroup$ – DanielSank Feb 25 '16 at 20:45
  • $\begingroup$ By the way, the easy way to do this is use $[a_-, a_+] = 1$. Then $\langle 0 | a_- a_+ | 0 \rangle = \langle | a_+ a_- | 0 \rangle + \langle 0 | 0 \rangle = 1$. $\endgroup$ – DanielSank Feb 25 '16 at 20:48
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Your mistake is that $$ \langle 0|\hat{a}^{\dagger} \neq \langle 1 | $$ In fact, $$ \langle 0 |\hat{a}^{\dagger} = (\hat{a}|0\rangle)^{\dagger} = 0 $$

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  • $\begingroup$ Thanks! I also made a quite embarrassing mistake in the other term. $\endgroup$ – QuantumMechanic Feb 25 '16 at 19:54

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