Why do I get negative expectationvalues when I use ladder operators? [closed]

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.

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Martin, Kyle Kanos, Qmechanic
If this question can be reworded to fit the rules in the help center, please edit the question.

• 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 $a_{-}a_{+}\psi_0 = \psi_0$. – QuantumMechanic Feb 25 '16 at 19:59
• Third line is wrong. – DanielSank Feb 25 '16 at 20:26
• Why? I don't see it. @DanielSank – QuantumMechanic Feb 25 '16 at 20:34
• Ooops, I was wrong. Forget it. – DanielSank Feb 25 '16 at 20:45
• By the way, the easy way to do this is use $[a_-, a_+] = 1$. Then $\langle 0 | a_- a_+ | 0 \rangle = \langle | a_+ a_- | 0 \rangle + \langle 0 | 0 \rangle = 1$. – DanielSank Feb 25 '16 at 20:48

Your mistake is that $$\langle 0|\hat{a}^{\dagger} \neq \langle 1 |$$ In fact, $$\langle 0 |\hat{a}^{\dagger} = (\hat{a}|0\rangle)^{\dagger} = 0$$