Schrödinger: Coherent states A coherent state is called
$\Psi_{{\alpha}} \left( x,t=0 \right)$
and is defined by:
$a_{{{\it \_}}}\Psi_{{\alpha}} \left( x \right) =\alpha\,\Psi_{{\alpha}} \left( x \right)  $
where $a_{{{\it \_}}}$ represent the lowering operator. 
I must show that
$\Psi_{{\alpha}} \left( x,t=0 \right) ={{\rm e}^{-1/2\, \left(  \left| \alpha \right|
 \right) ^{2}}}\sum _{n=0}^{\infty }{\frac {{\alpha}^{n}\psi_{{n}}
 \left( x \right) }{\sqrt {n!}}}$
is a coherent state. 
How do I go about doing this? I understand that the above right-hand side is a normalization constant multiplied with a general solution to the time-independent Schrödinger equation, but I don't know how to show that it is a coherent state. I believe I am to use the eigenvalue, however this is as far as I am able to come at this hour. I would appreciate any help and hints.
 A: As you said, a coherent state is defined by the equation
$$a \Psi_{\alpha} = \alpha \Psi_{\alpha} \, .$$
Therefore, to check whether a particular expression is a coherent state, just act $a$ on it and see if you get the same thing back multiplied by a complex number.
Let's try it
$$
\begin{align}
a \Psi_{\alpha} &=
a \left[ e^{-|\alpha|^2/2} \sum_{n=0}^\infty \frac{\alpha^n \psi_n}{\sqrt{n!}} \right] \\
&= e^{-|\alpha|^2/2} \sum_{n=0}^\infty \frac{\alpha^n a \psi_n}{\sqrt{n!}} \\
&= e^{-|\alpha|^2/2} \sum_{n=1}^\infty \frac{\alpha^n \sqrt{n}\psi_{n-1}}{\sqrt{n!}} \\
\text{factor out one power of }\alpha \qquad
&= \alpha e^{-|\alpha|^2/2} \sum_{n=1}^\infty \frac{\alpha^{n-1} \psi_{n-1}}{\sqrt{(n-1)!}} \\
\text{let }m=n-1\qquad
&= \alpha \left[ e^{-|\alpha|^2/2} \sum_{m=0}^\infty \frac{\alpha^m \psi_m}{\sqrt{m!}} \right] \, .
\end{align}
$$
If you look carefully at what you've got there, you'll see the resolution of your question.
A: Assume $\Psi_\alpha = \sum_n c_n \psi_n$. This is always true because $\{\psi_n\}_{n=0,1,2,\ldots}$ is a Hilbertian basis.
Then uses the definition of $\Psi_\alpha$ and the one of $a$, assuming that $a$ and the symbol of sum can be swapped:
$$\sum_n \alpha  c_n \psi_n  = \alpha \Psi_\alpha = a \Psi_\alpha =  \sum_n c_n a\psi_n = \sum_n c_n \sqrt{n}\psi_{n-1}\:.$$
Comparing the first and the last sum,
$$\sum_{n=0}^{+\infty} \alpha c_n  \psi_n = \sum_{m=0}^{+\infty} c_{m+1} \sqrt{m+1}\psi_{m}$$
using the uniqueness of the coefficients of the decomposition of a vector along a Hilbert basis, you see that
$$\alpha c_n = c_{n+1}\sqrt{n+1}\:.$$
Assuming $c_0=\alpha$ (and this value can be fixed arbitrarily as it can be fixed up to a phase by the final normalization requirement, if $\alpha \neq 0$ otherwise we have the trivial vector) you have 
$$c_0=\alpha\:, \quad c_2 = \frac{\alpha^2}{\sqrt{2}}\:, \quad c_3 = \frac{\alpha^3}{\sqrt{3!}}\:, \ldots$$
In general
$$c_n = \frac{\alpha^n}{\sqrt{n!}}\:.$$
We have the candidate
$$\Psi_\alpha = c \sum_{n=0}^{+\infty} \frac{\alpha^n}{\sqrt{n!}} \psi_n\:.$$
By direct inspection, since the elements $\psi_n$ are orthonormal,
$$||\Psi_\alpha||^2 = |c|^2 \sum_{n=0}^{+\infty} \frac{(|\alpha|^2)^n}{n!}= |c|^2 e^{|\alpha|^2}\:.$$
The simplest choice for obtaining $||\Psi_\alpha||=1$ is $c = e^{-\frac{|\alpha|^2}{2}}$. Hence
$$\Psi_\alpha = e^{-\frac{|\alpha|^2}{2}} \sum_{n=0}^{+\infty} \frac{\alpha^n}{\sqrt{n!}} \psi_n\:.$$
Finally, on the mathematical side, one should check if
$$a\left(  \sum_{n=0}^{+\infty} \frac{\alpha^n}{\sqrt{n!}} \psi_n\right)=
\sum_{n=0}^{+\infty} \frac{\alpha^n}{\sqrt{n!}} a\psi_n\tag{1}$$
This fact is not obvious because $a$ is not continuous. However it is a closed (actually closable) operator. It implies that (1) holds if both $$ \sum_{n=0}^{N} \frac{\alpha^n}{\sqrt{n!}} \psi_n$$
and
 $$\sum_{n=0}^{N} \frac{\alpha^n}{\sqrt{n!}} a\psi_n = \sum_{n=1}^{N} \frac{\alpha^n}{\sqrt{(n-1)!}} \psi_{n-1}$$
converge in the Hilbert space to some vectors, as $N \to +\infty$. This is true because, for every $N=0,1,\ldots$
$$  \sum_{n=0}^{N} \left|\frac{\alpha^n}{\sqrt{n!}}\right|^2 =\sum_{n=0}^{N} \frac{|\alpha|^{2n}}{n!} \leq e^{|\alpha|^2} +\infty$$
and
$$  \sum_{n=1}^{N} \left|\frac{\alpha^n}{\sqrt{(n-1)!}}\right|^2 =\sum_{n=0}^{N-1} |\alpha|^2\frac{|\alpha|^{2n}}{n!} \leq |\alpha|^2e^{|\alpha|^2}<+\infty\:.$$
A: You can find the answer in Wikipedia http://en.wikipedia.org/wiki/Coherent_states.
It's written there that the coherent state is generated by the displacement operator $D(\alpha)$
$$|\alpha\rangle = D(\alpha)|0\rangle = \exp(\alpha a^\dagger - \alpha^*a)|0\rangle \, .$$
Now everything is simple. Expand $\exp(\alpha a^† - \alpha^*a)$ as you expand every exponential.
$$
\begin{align}
\exp(\alpha a^† - \alpha^*a) &=
1 + \alpha a^† - \alpha^* a + (1/2) (\alpha a^† - \alpha^*a)(\alpha a^† - \alpha^* a) \\
&+ (1/3!)(\alpha a^† - \alpha^* a)(\alpha a^† - \alpha^* a)(\alpha a^† - \alpha^* a) + \ldots
\end{align}
$$
and apply this operator to $|0\rangle$ and find the general term in the series.
Let me help you with a few first terms:
$$
\begin{align}
D(a)|0\rangle &= |0\rangle + \alpha|1\rangle + (\alpha/2) (\alpha a^† - \alpha^*a)|1\rangle \\
&+ (\alpha/3!)(\alpha a^† - \alpha^* a)(\alpha a^† - \alpha^*a)|1\rangle + \ldots\\
&= |0\rangle + \alpha|1\rangle + (\alpha^2 \sqrt{2}|2\rangle - \alpha α^*|0\rangle)/2 \\
&+ (\alpha/3!)(\alpha a^† - a \alpha^* a)(\alpha^2 \sqrt{2}|2\rangle - \alpha \alpha^*|0\rangle) + \ldots
\end{align}
$$
Use whenever needed the expressions of the raising and lowering operators on Fock states $|n\rangle$, i.e. $a|n\rangle = \sqrt{n}|n-1\rangle$, and $a^†|n\rangle = \sqrt{n+1}|n+1\rangle$, and also the fact that $a|0\rangle = 0$.
After obtaining the series, and, most important, the general term, you can normalize the series by using the Poisson distribution norm.
Try your hand!
Good luck !
