# Showing there are no eigenvectors of the annhilation operator

1. Nov 25, 2013

### Hakkinen

1. The problem statement, all variables and given/known data
Show there are no eigenvectors of $a^{\dagger}$ assuming the ground state |0> is the lowest energy state of the system.

2. Relevant equations
Coherent states of the SHO satisfy:
a|z> = z|z>

3. The attempt at a solution
Based on the hint that was given (assume there is such an eigenvector like the coherent state above and expand the state in the basis |n>) I tried this, but it seems too simple.

$a^{\dagger}|0> = k|0>$ (is this expansion in the |n> basis? or |0> basis?)

then conjugate transpose both sides

$a<0|=\bar{k}<0|$

the lhs vanishes as a is acting on the ground state ket

Is this correct? Any help is greatly appreciated

2. Nov 26, 2013

### Oxvillian

Hello Hakkinen!

You are not taking the conjugate transpose correctly - the left hand side should read $\langle 0 \lvert a$. That is, it should look like a row vector multiplying a matrix. The operator $a$ "acts to the left" as a creation operator, not as an annihilation operator.

3. Nov 26, 2013

### Hakkinen

Thanks for the reply! sorry for the simple mistakes

So is this not the correct approach to show this? If it is, should I start with $a|0> = k|0>$ and then conjugate, leaving the annihilation operator acting on the ground state bra?

so $0 = <0|\bar{k}$

but this doesn't necessarily show there are no eigenvectors of the ann. operator does it?

4. Nov 26, 2013

### Oxvillian

Well $a\lvert 0 \rangle = 0$, since $\lvert 0 \rangle$ is the ground state. We need a new name for the hypothetical eigenstate of $a^{\dagger}$, let's just call it $\lvert \psi \rangle$. Then

$a^{\dagger}\lvert \psi \rangle = \lambda \lvert \psi \rangle,$

where $\lambda$ is some number (the eigenvalue). Try expanding $\lvert \psi \rangle$ in terms of the $\lvert n \rangle$ and see what you get.

5. Nov 26, 2013

### Hypersphere

Note that people usually call $a$ the annihilation operator, and $a^\dagger$ the creation operator. The reason it is worth to avoid confusion here, is that the annihilation operator $a$ (which satisfies $a|0\rangle=0$, where $|0$ is the ground state) actually has an eigenstate, while the creation operator does not.