# Finding eigenstate for the annhilation operator

1. Oct 5, 2015

### Sandrasa

1. The problem statement, all variables and given/known data
Find the eigenvector of the annhilation operator a.

2. Relevant equations
$a|n\rangle = \sqrt{n}|{n-1}\rangle$

3. The attempt at a solution
Try to show this for an arbitrary wavefunction:
$|V\rangle = \sum_{n=1}^\infty c_{n}|n\rangle$

$a|V\rangle = a\sum_{n=1}^\infty c_{n}|n\rangle = \sum_{n=1}^\infty \sqrt{n} c_{n}|n-1\rangle$

Define a new constant, k = n - 1, and put this in:

- > $\sum_{k=0}^\infty \sqrt{k + 1} c_{k + 1}|k\rangle = \sum_{n=0}^\infty \sqrt{n + 1} c_{n + 1}|n\rangle$

This was a trick I found in my notes from a lecture, but I do not quite understand how you can say that k = n in the last part. Is it just because k and n are an arbitrary notation for the basis set? Or is it due to the fact that since the sum goes to infity, the difference of 1 does not matter (since I really defined k to be n - 1)?

When I continue with this, I get:

$a( c_{1}|1\rangle + c_{2}|2\rangle + c_{3}|3\rangle + ...) = c( \sqrt{1}c_{1}|0\rangle\ + \sqrt{2}c_{2}|1\rangle + \sqrt{3}c_{3}|2\rangle + \sqrt{4}c_{4}|3\rangle)$

I am not really sure where to go next. I tried setting the coeffecient before each basisvector equal, like this:

$c\sqrt{1}c_{1} = 0$
$c_{1} = \sqrt{2}c_{2}c$
$c_{2} = \sqrt{3}c_{3}c$
$c_{3} = \sqrt{4}c_{4}c$

The problem her is that since the first part does not contain the vector $0|\rangle$, either c or $c_{1}$ has to be zero in the first equation, and then everything becomes zero, but I know that the annihilation operator has eigenstates.

2. Oct 5, 2015

### strangerep

Is there some reason you didn't start this sum at n=0 ?

3. Oct 6, 2015

### Sandrasa

Yes, because when the annihilation operator works on the state zero, we will get zero by definition according to my book so I thought that there was no point in taking it. But I can see that that would get me on the right track again, but I don't know how I can do it?

But by taking the sum from n = 0, I end up with $c_n = c_{n-1}\frac{c'}{\sqrt{n}}$, where c' = $\frac{1}{c}$. And now I'm not really sure what to do next. Can I just put this in my expression for $|V\rangle$ and call that an eigenstate with c' as the constant? It doesn't look like the expression I found for the eigenstate, which was $|V\rangle$ = $c_{0}$ $\sum\limits_{n=0}^\infty$ $\frac{c^n}{\sqrt{n!}}$$|n\rangle$ in my notation. I tried writting the expression I found out, but then I got something like:
$c_0 + \frac{c'c_0}{\sqrt{1}} + \frac{c_0c'^2}{\sqrt{2}} + \frac{c_0c'^3}{\sqrt{6}}$. It sort of looks like the equations I got in the start, but I'm stuck in trying to get to one place from the other. Any hints?

And thanks for the help!

4. Oct 6, 2015

### strangerep

In your "relevant equations", you didn't write out the action of the creation operator $a^*$. E.g.,
$$a^*|0\rangle ~=~ ~?~$$$$(a^*)^2 |0\rangle ~=~ ~?~$$etc. Try to express each term in your sum for $|V\rangle$ in terms of $a^*$ acting on the vacuum.