# Eigenkets of the creation operator

1. Oct 28, 2008

### Swatch

1. The problem statement, all variables and given/known data

The problem is to find the eigenkets for the creation operator ,$$a^{\dagger}$$ if they exist

2. Relevant equations

$$a^{\dagger}|\Psi>=\lambda|\Psi>$$
$$a^{\dagger}=\frac{1}{\sqrt{2*\hbar*m*\omega}}*(-\hbar*\frac{d}{dx}+m*\omega*x)$$

3. The attempt at a solution
I use the expression for the creation operator above and set up the differential equation.
The boundary conditions for the wavefunction should be that it is zero when x goes to +- infinity.
When I solve it I get this result.
$$0=\lambda*x\right|_{-\infty}^{+\infty} - \frac{1}{2}\sqrt{\frac{m*\omega}{2*\hbar}}*x^{2} \right|_{-\infty}^{+\infty}$$

I know the creation operator is not hermitian and it is not an observable but could this answer be correct?
Could someone please comment on this

2. Oct 28, 2008

### olgranpappy

If eigenkets of $a^\dagger$ exist. What is their overlap with the vacuum?

3. Oct 28, 2008

### Swatch

What is their overlap with vaccum?
I don't understand!!

4. Oct 28, 2008

### olgranpappy

By "vacuum" I just mean the state $|0\rangle$ that satisfies
$$a|0\rangle=0$$

By "overlap with the vacuum" I mean, what is
$$\langle 0 |\lambda\rangle$$
equal to?

5. Oct 28, 2008

### olgranpappy

...where $|\lambda\rangle$ is the eigenket of $a^\dagger$ with eigenvalue $\lambda$

6. Oct 29, 2008

### Swatch

The overlap with the vacuum is zero
but how does this help me?

7. Oct 29, 2008

### olgranpappy

So, the overlap with the vacuum is zero. Good.

Now, what is the overlap with the "one-particle" state
$$|1\rangle = a^\dagger |0\rangle\;$$
?

8. Oct 29, 2008

### Swatch

I believe the overlap of $$|\lambda>$$ with the one particle state to be zero

9. Oct 29, 2008

### borgwal

So, given that, what's the overlap of $$|\lambda>$$ with the 2-particle state |2>? I hope you realize what the nth question is going to be.

10. Oct 29, 2008

### Swatch

I think I understand this.
since the overlap of $$\lambda$$ and the vaccum is zero and also the overlap with the one, two and so one particle functions is zero, $$\lambda$$ must be zero.
Am I right?

11. Oct 29, 2008

### olgranpappy

The notation you are using above confuses the state with it's eigenvalue.

What we have just seen is that if the *eigenvalue* $\lambda$ is non-zero then the *state* $|\lambda>$ must be zero--i.e., there are no eigenstate of $a^\dagger$ with non-zero eigenvalues.

12. Oct 30, 2008

### Swatch

Thank you all for your help