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Suppose $|k\rangle =\exp(k \hat O^\dagger)|0\rangle$ where $c_k\in \mathbb C$ and $|0\rangle$ is normalized.

I wish to evaluate $\langle a|b\rangle$.

Here is what I think, but the result is not right, could someone please tell me what is wrong with it?

It is not hard to show that $\hat O|k\rangle=k|k\rangle$

So $\langle a|b\rangle= \langle 0|(1+a^*\hat O+{1\over 2}(a^*)^2(\hat O)^2+...)|b\rangle$

$=\exp(a^*b)\langle 0|b\rangle$

$=\exp(a^*b)\langle 0| \exp(b \hat O^\dagger)|0\rangle$

$=\exp(a^*b)\langle \exp(b^* \hat O)(0)|0\rangle$

because I think $\langle \exp(b^* \hat O)(0)|=\exp(b \hat O^\dagger)|0\rangle^\dagger$ ---$(*)$

$=\exp(a^*b)\langle 0|0\rangle$


This cannot be right. Since for one thing, it is strange that $\langle 0|b\rangle=\langle 0|0\rangle$.

Also, I expected that the "vectors" should be orthonormal!

Thank you!

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I believe everything is fine: as $O$ is not selfadjoint, you cannot expect the eigenvectors to be orthogonal. –  Fabian Oct 1 '12 at 11:28
@Fabian: Thanks, so it's true that $\langle 0|b\rangle=\langle 0|0\rangle$? –  George Oct 1 '12 at 11:37
I guess so. I did not check the calculation in detail. Maybe this wiki page helps. As far as I understand, you do not have normalized states so the normalization factor $\exp[-(|a|^2+|b|^2)/2]$ present on the wiki-page is absent in your case. –  Fabian Oct 1 '12 at 11:49
I think you may be using some notation particular to some linear algebra stuff in physics. It may be not many here understand what's that (I, for one, don't, and I all my friends in university were physicists, at least at graduate school). Try to ask in a physics forum or "translate" your symbols (I think there are some inner product things and stuff...) –  DonAntonio Oct 1 '12 at 14:51

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