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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?

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  • $\begingroup$ Notice that $\left \langle a \psi + b \phi \right| \neq a\left\langle\psi\right| + b\left\langle\phi\right|$, since $\left\langle a\psi\right| = \left| a \psi \right\rangle^\dagger = (a\left|\psi\right\rangle)^\dagger = a^*\left\langle\psi\right| $. $\endgroup$
    – George G
    Jun 5, 2014 at 18:17
  • $\begingroup$ Is $\hat{Q}$ Hermitian? $\endgroup$
    – user26143
    Jun 5, 2014 at 20:25

1 Answer 1

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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}

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  • $\begingroup$ Does this follow from: $\frac{1}{2}(\langle q_1^{*}\psi_1+i q_2^{*}\psi_2|q_1\psi_1-iq_2\psi_2\rangle)$? $\endgroup$
    – Thiago
    Jun 5, 2014 at 18:22
  • $\begingroup$ @Ali Yes; just use the properties of the inner product that you wrote down, and you'll get precisely that. The cross terms cancel due to orthogonality of $|\psi_1\rangle$ and $|\psi_2\rangle$. $\endgroup$ Jun 5, 2014 at 18:34
  • $\begingroup$ Did you assume that $\langle A+B|C+D\rangle=\langle A|B \rangle+\langle A|D \rangle +\langle B |C \rangle+\langle B|D \rangle$? Because I can get to what you posted if I use this. But I can't see why this is true. $\endgroup$
    – Thiago
    Jun 5, 2014 at 18:36
  • $\begingroup$ @Ali Yes (but the first term on the right should be $\langle A|C\rangle$). You can prove that by applying the inner product properties you wrote down twice, once for the first "slot" and once for the second "slot." $\endgroup$ Jun 5, 2014 at 18:38
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    $\begingroup$ @Ali It seems your confusion stems from mixing legitimate bra-ket notation with inner product notation used more commonly in mathematics. In legit bra-ket notation, $|\psi\rangle$ is a vector, the $\psi$ inside is just a label, and $\langle\psi|$ is a corresponding dual vector. In mathy notation, $\psi$ is a vector, the symbol $|\psi\rangle$ has no independent meaning, and $\langle \psi|\phi\rangle$ is the inner product of two vectors $\psi$ and $\phi$. I'd recommending reading up on this stuff; this might help: physics.stackexchange.com/a/57793/19976 $\endgroup$ Jun 5, 2014 at 19:04

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