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How can I find the eigenstates of creation and annihilation operator in QM?

My attempt:

Such eigenstate will obey: $$ a^{\dagger} |\psi \rangle = \alpha |\psi \rangle. $$

We can expand $|\psi \rangle$ in terms of the quantum SHM eigenstates: $|\psi \rangle = \sum_{n=0}^{\infty} c_n |n\rangle $.

Knowing the action of the creation operation of quantum SHM eigenmodes ($a^{\dagger}|n\rangle = \sqrt{n+1}|n+1\rangle)$:

$$ a^{\dagger} |\psi \rangle = a^{\dagger} \sum_{n=0}^{\infty} c_n |n\rangle = \sum_{n=0}^{\infty} c_n \sqrt{n+1}|n+1\rangle$$

from which the state $|0\rangle$ is now missing, so it will never be equal to the RHS of the first expression, $\alpha |\psi\rangle = \sum_{n=0}^{\infty} \alpha c_n |n\rangle $.

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    $\begingroup$ The creation operator cannot possibly have eigenstates. The eigenstates of the annihilation operator are called "coherent states". Google it :) $\endgroup$
    – DanielSank
    Dec 30 '14 at 23:52
  • $\begingroup$ This is no surprise, as $a$ and $a^\dagger$ are not self-adjoint $\endgroup$
    – Phoenix87
    Dec 30 '14 at 23:57
  • $\begingroup$ …but see this $\endgroup$ Sep 1 at 17:40
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Write an arbitrary state as

$$|\Psi\rangle = \sum_{n=0}^{\infty} c_n |n\rangle \,.$$

Now apply the raising operator

$$ \begin{align} a^\dagger |\Psi\rangle &= a^\dagger \sum_{n=0}^{\infty} c_n |n\rangle \\ &= \sum_{n=0}^{\infty} c_n \sqrt{n+1} |n+1\rangle \\ &= \sum_{n=1}^{\infty} c_{n-1} \sqrt{n} |n\rangle \end{align} $$

If $|\Psi\rangle$ is an eigenstate of $a^\dagger$ with eigenvalue $\alpha$ then we have

$$\sum_{n=0}^{\infty}c_n|n\rangle = \sum_{n=1}^\infty c_{n-1} \sqrt{n}|n \rangle \, .$$

You already got this far. Indeed, the only solution to this equation is $c_n=0$ for all $n$. Therefore, there is no eigenstate of $a^\dagger$.

The eigenstates of $a$, which are called "coherent states" are given by

$$ |\alpha \rangle = e^{-|\alpha|^2/2}\sum_{n=0}^\infty \frac{\alpha^n}{\sqrt{n!}}|n\rangle \, . $$ You can check easily by applying $a$ to $|\alpha \rangle$ that $|\alpha \rangle$ is an eigenstate of $a$.

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    $\begingroup$ Can you please explain why $c_{n}$ =0 for all n, is the only solution for creation operator case $\endgroup$
    – Draco_1125
    Jan 23 '18 at 18:25
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    $\begingroup$ @SSP_user5275 The right hand side of the equation has no $|0\rangle$ term, so the left hand side must also have no $|0\rangle$ term, and so $c_0=0$. Now consider the $|1\rangle$ term and we get $c_1 = c_0 \sqrt{1} = 0$. Repeat this argument for all terms. $\endgroup$
    – DanielSank
    Jan 23 '18 at 21:39
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Another way to obtain the coherent state $|\alpha\rangle$ is by acting the displacement operator

$$\hat{D}(\alpha) = \exp(\alpha \hat{a}^\dagger - \alpha^* \hat{a})$$

on the vacuum state $|0\rangle$.

$$|\alpha\rangle = \hat{D}(\alpha) |0\rangle$$

You can verify that

$$\hat{a} |\alpha\rangle = \alpha |\alpha\rangle$$

This looks deceptively simple, and you do not need to know the general expression for all $|n\rangle$ to use this method. But the difficult part is evaluating the exponential of an operator. The exponential is to be interpreted in terms of the taylor series

$$e^X = I + X + \frac{1}{2}X^2 + \frac{1}{6}X^3 + \ldots$$

and be careful that

$$e^{A-B} \neq e^A e^{-B}$$

in general. One intermediate useful result is

$$\hat{D}^\dagger(\alpha) \hat{a} \hat{D}(\alpha) = \hat{a} + \alpha I$$

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