Are Eigenkets for the Creation Operator Valid?

[Swatch]

Homework Statement

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

Homework Equations

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

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

 

[olgranpappy]

Homework Statement

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

Homework Equations

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

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

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

 

[Swatch]

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

 

[olgranpappy]

By "vacuum" I just mean the state |0\rangle that satisfies
<br /> a|0\rangle=0<br />

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

 

[olgranpappy]

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

 

[Swatch]

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

 

[olgranpappy]

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

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

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

 

[Swatch]

I believe the overlap of |\lambda&gt; with the one particle state to be zero

 

[borgwal]

I believe the overlap of |\lambda&gt; with the one particle state to be zero

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

 

[Swatch]

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

 

[olgranpappy]

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

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&gt; must be zero--i.e., there are no eigenstate of a^\dagger with non-zero eigenvalues.

 

[Swatch]

Thank you all for your help