Expectation value of an operator

Question

Suppose we have: $$ \hat{Q}|\psi_1\rangle=q_1|\psi_1\rangle \\ \hat{Q}|\psi_2\rangle=q_2|\psi_2\rangle $$ with $q_1 \neq q_2$. Then consider the state: $$ |\Psi\rangle=\frac{1}{\sqrt{2}}(|\psi_1\rangle-i|\psi_2\rangle) $$ I want to calculate the uncertainty in $\hat{Q}$. Then first we must compute $\langle\hat{Q^2}\rangle$ and $\langle\hat{Q}\rangle^2$. I started with $\langle\hat{Q^2}\rangle$. First, $$ \hat{Q}|\Psi\rangle=\frac{1}{\sqrt{2}}(q_1|\psi_1\rangle-iq_2|\psi_2\rangle) $$ Then: $$ \langle\Psi|\hat{Q}=\frac{1}{\sqrt{2}}(q_1^{*}\langle\psi_1|+iq_2^{*}\langle\psi_2|) $$ It follows that: $$ \langle\Psi|\hat{Q^2}|\Psi\rangle = (\langle\Psi|\hat{Q})(\hat{Q}|\Psi\rangle) =\frac{1}{2}(q_1^{*}\langle \psi_1|+iq_2^{*}\langle \psi_2|)(q_1|\psi_1\rangle -iq_2|\psi_2\rangle ) $$ Using the fact that: $$ \langle A|bB+cC\rangle = b\langle A|B\rangle + c\langle A|C\rangle \text{ and}\\ \langle bB+cC|A\rangle = b^{*}\langle B|A\rangle + c^{*}\langle C|A\rangle $$ I found: $$ \langle\Psi|\hat{Q^2}|\Psi\rangle = \frac{1}{2}(q_1^2-q_2 q_2^{*}) $$ Note that $q_1^2 \neq |q_1|^2$. On the other hand, if I act on the ket vector twice to compute $\hat{Q^2}|\Psi\rangle$ I find: $$ \hat{Q^2}|\Psi\rangle = \frac{1}{\sqrt{2}}(q_1^2|\psi_1\rangle -iq_2^2|\psi_2\rangle) $$ Then using this expression to compute $\langle\Psi|\hat{Q^2}|\Psi\rangle$ yields: $$ \langle\Psi|\hat{Q^2}|\Psi\rangle = \frac{1}{2}(q_1^2 - q_2^2) $$ Now using the representation in the basis: $$ \hat{Q}|\Psi\rangle = \frac{1}{\sqrt{2}} \left[\begin{matrix}q_1 \\ -iq_2\end{matrix}\right] \\ \langle\Psi|\hat{Q} = \frac{1}{\sqrt{2}}[q_1^{*} \text{ }iq_2^{*}] $$ Which gives: $$ \langle\Psi|\hat{Q^2}|\Psi\rangle = q_1^{*}q_1 + q_2^{*}q_2 $$ Why am I getting different results?

Answer 1

Assuming $|\psi_1\rangle$ and $|\psi_2\rangle$ are orthonormal, your equation after you write "I found" is incorrect, it should be \begin{align} (\langle \Psi|\hat Q)(\hat Q|\Psi\rangle) &= \frac{1}{2}\Big(q_1^*q_1\langle\psi_1|\psi_1\rangle - i q_1^*q_2 \langle\psi_1|\psi_2\rangle +iq_2^*q_1\langle\psi_2|\psi_1\rangle -i^2q_2^*q_2\langle\psi_2|\psi_2\rangle\Big) \\ &= \frac{1}{2}(|q_1|^2+|q_2|^2) \end{align}