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For a coherent state $$ |\alpha\rangle=e^{-\frac{|\alpha|^{2}}{2}}\sum_{n}\frac{\alpha^{n}}{\sqrt{n!}}|n\rangle $$ I can't solve the eigenvalue problem for $\hat{a}^{\dagger}|\alpha\rangle$ where $\hat{a}^{\dagger}$ is the creation operator. I can only get this far

$$ \begin{align} \hat{a}^{\dagger}|\alpha\rangle&=e^{-\frac{|\alpha|^{2}}{2}}\sum_{n}\frac{\alpha^{n}}{\sqrt{n!}}\hat{a}^{\dagger}|n\rangle\\ &=e^{-\frac{|\alpha|^{2}}{2}}\sum_{n}\frac{\alpha^{n}}{\sqrt{n!}}\sqrt{n+1}|n+1\rangle \end{align} $$

Ultimately, I want to calculate $\langle \alpha |a\hat{a}^{\dagger}|\alpha\rangle$, but I don't know $\hat{a}^{\dagger}|\alpha\rangle$.

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closed as off-topic by DanielSank, user10851, Danu, ACuriousMind, Martin Sep 10 '15 at 12:06

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  • $\begingroup$ You know the result that a creation operator acts on an eigenstate $|n>$, then, just sum over all the results. $\endgroup$ – qfzklm Sep 9 '15 at 3:51
  • $\begingroup$ I did. But then, I stuck from there afterward. $\endgroup$ – TBBT Sep 9 '15 at 3:55
  • $\begingroup$ @qfzklm does the result simplify much? e.g. is the resulting sum a well-known one? $\endgroup$ – innisfree Sep 9 '15 at 4:02
  • $\begingroup$ @TBBT what result do you expect to find? $|\alpha\rangle$ isn't an eigenfunction of the creation operator - you won't find something $\propto |\alpha\rangle$. Do you want to know whether the sum simplifies? $\endgroup$ – innisfree Sep 9 '15 at 4:11
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    $\begingroup$ @TBBT try using commutation relations to get an $a |\alpha \rangle$? $\endgroup$ – zeldredge Sep 9 '15 at 4:38
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A coherent state is, amongst other interesting things, an eigenstate of the annihilation operator. It is not an eigenstate of the creation operator; hence, I'm not sure this "eigenvalue problem" makes much sense.

This is easy to realize. You can quickly see that $\langle0|a^\dagger|\alpha\rangle=0$, whereas $\langle0|\alpha\rangle\neq0$.

If you really want to find $\langle \alpha | a a^\dagger|\alpha\rangle $ in e.g. $$ \langle x^2 \rangle \propto \langle \alpha | (a+a^\dagger)(a+a^\dagger)|\alpha\rangle $$ you can commute the operators $a$ and $a^\dagger$ with the rule $[a,a^\dagger] = 1$, such that $$ \langle \alpha | a a^\dagger|\alpha\rangle = \langle \alpha | a^\dagger a|\alpha\rangle + 1 = 1 + |\alpha|^2 $$ You can also verify this the long way around by acting the the operators.

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  • $\begingroup$ You are correct. However, I can't still calculate, $\langle\alpha|aa^{\dagger}|\alpha\rangle$. I have to solve something in my homework that look like this,$\langle\alpha|aa|\alpha\rangle+\langle\alpha|a^{\dagger}a^{\dagger}|\alpha\rangle+\langle\alpha|aa^{\dagger}|\alpha\rangle+\langle\alpha|a^{\dagger}a|\alpha\rangle$ $\endgroup$ – TBBT Sep 9 '15 at 4:30
  • $\begingroup$ I can solve three out of four terms above, but not the $\langle\alpha|aa^{\dagger}|\alpha\rangle$ $\endgroup$ – TBBT Sep 9 '15 at 4:35
  • $\begingroup$ I think that's quite straight-forward - just commute the operators. See my edited answer. $\endgroup$ – innisfree Sep 9 '15 at 4:43
  • $\begingroup$ Wow, can't believe I couldn't see that. Thank you so much! $\endgroup$ – TBBT Sep 9 '15 at 6:12
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To add to Innisfree's correct answer, I'd like to emphasize something that the OP does not seem to know and that is that the creation operator has no eigenvectors (nor, therefore, eigenvalues). It is easy to see this: write a general state as a row vector $(\psi_0,\,\psi_1,\,\cdots)$ of superposition weights for the number states $|0\rangle,\,|1\rangle,\,\cdots$ and in this notation, our eigenvalue equation (in $\lambda$) for $a^\dagger$ is:

$$a^\dagger (\psi_0,\,\psi_1,\,\cdots) =(0,\,\psi_0,\,\sqrt{2} \psi_1,\,\sqrt{3} \psi_2,\,\cdots) = \lambda (\psi_0,\,\psi_1,\,\cdots)$$

whence we get $\lambda \,\psi_n=\sqrt{n}\psi_{n-1}$ and $\lambda\,\psi_0=0$. If $\lambda = 0$ it follows straight away that $\psi_n=0\,\forall\,n\in\mathbb{N}$. If $\lambda\neq 0$, then $\psi_0=0$, whence (by induction through $\psi_n = \sqrt{n}\psi_{n-1}/\lambda$) $\psi_n=0\,\forall\,n\in\mathbb{N}$. There is therefore no normalizable superposition of number states that is an eigenvector for $a^\dagger$. It's therefore not surprising that the OP was having difficulty!

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    $\begingroup$ This is also in my answer ;) it is the same as the comment "You can quickly see..." $\endgroup$ – innisfree Sep 9 '15 at 5:27
  • $\begingroup$ @innisfree Ah, sorry, I missed that. I just wanted to emphasize a bit more to the OP (because I can recall at one stage assuming that the creation operator would have an eigenvector). Doubtless he/she will remember the nonexistence of creation operator eigenvectors after today. $\endgroup$ – WetSavannaAnimal Sep 9 '15 at 5:50
  • $\begingroup$ No worries, it is helpful that you stressed it a bit more than I did. $\endgroup$ – innisfree Sep 9 '15 at 7:23
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    $\begingroup$ "you can quickly see" is good, but sometimes it's helpful to have things spelled out fully. I found this answer useful. $\endgroup$ – Dom Sep 9 '15 at 10:16
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Using the definition of the creation operator, $a^\dagger = c(m\omega \hat x - i\hat p)$ where $c$ is a constant, and $\hat p = -i\hbar\partial_x$, you can write the eigenvalue problem in the position representation as $$(m\omega x - \hbar\partial_x)\psi = \alpha\psi.$$ You can solve this differential equation to find $$\psi = C\exp(m\omega x^2/\hbar - \alpha x/\hbar)$$ which is clearly not normalizable. Hence the creation operator has no normalizable eigenstates.

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