# Creation operator and eigenvalues

1. Jun 28, 2014

### carllacan

1. The problem statement, all variables and given/known data
Prove that the creation operator $a_+$ has no eigenvalues, for instance in the $\vert n \rangle$.

2. Relevant equations
Action of $a_+$ in a harmonic oscillator eigenket $\vert n \rangle$:
$a_+\vert n \rangle =\vert n +1\rangle$

3. The attempt at a solution
Calling a the eigenvalues of $a_+$
$a_+ \vert \Psi \rangle = a \vert \Psi \rangle = a \sum c_n \vert n \rangle = \sum a c_n \vert n \rangle$
$a_+ \vert \Psi \rangle = a_+ \sum c_n \vert n \rangle = \sum c_n a_+ \vert n \rangle = \sum c_n\vert n+1\rangle = \sum c_{n-1}\vert n\rangle$

Equating both
$a_+ \vert \Psi \rangle = \sum a c_n \vert n \rangle= \sum c_{n-1}\vert n\rangle$

We have
$a c_n = c_{n-1}$.

I think I can take the a factor out and then claim that eigenkets have to be linearly dependent, so their coefficients cannot be proportional to each other.

However, I am not sure that this does prove that the creation operatro has no eigenvalues.

Last edited: Jun 28, 2014
2. Jun 28, 2014

### bloby

Hi,
$a=1=c_n=c_{n-1}=...$ is a solution of the equations for the coefficients. What happens when $a_+$ is applied on $|\Psi>=|n>+|n-1>+...+|n-m>$?

3. Jun 28, 2014

### carllacan

We get $a_+|\Psi\rangle=|n+1>+|n\rangle+...+|n-m+1\rangle$. If the set of eigenvalues for the oscilator Hamiltonian is infinite (and it is, right?) the sum remains the same except for $\vert 0 \rangle$, which becomes $\vert 1 \rangle$. Since this is the new ground state we have transformed the oscillator eigenkets into eigenkets for an oscillator with $\hbar \omega$ more energy. Is that so?

What now? I don't know what do you mean.

4. Jun 28, 2014

### bloby

Yes but you are more interested in eigenvector to express $|\Psi>$.

Yes and what is $a|\Psi>$ with a=1 in the basis $\{|i>\}$?

No, it's just some algebra to show that there is no vector $|\Psi>$ such that $a_+|\Psi>=a|\Psi>$. The example with a=1 is just a simple example to see what happens.

If you add the bounds to the sums in your attempt at a solution you will see that there is a problem.

5. Jun 28, 2014

### carllacan

What do you mean with the bounds?

6. Jun 28, 2014

### bloby

Sorry for the delay. I will try another way: you expended $|\Psi>$ on the basis $\{|n>:n=0,1,...\}$. On one hand you multiplied it by a and you obtained $a|\Psi>$ on that basis. On the other hand you applied $a_+$ on it and you obtained $a_+|\Psi>$ on the same basis $\{|n>:n=0,1,...\}$. For the two vectors obtained to be equal the coefficients must be equal for all basis vectors. Your result $a c_n=c_{n-1}$ cannot hold for all n, do you see why?

7. Jun 28, 2014

### vela

Staff Emeritus
Try writing out the first few terms of each summation.

8. Jun 28, 2014

### carllacan

Wow, why do you apologyze? You're helping me! I really appreciate it, no matter how long does it take.
I've just realized that $\hat{a}^{\dagger}\vert n \rangle = \sqrt{n} \vert n \rangle$, so my recursion formula should've been $c_n a = c_{n-1} \sqrt{n}$.

Anyway, I don't see what's the deal with it. Is it with the ground state? It is 0 in one expansion and $c_0 a$ in the other one.

9. Jun 28, 2014

### bloby

Exactly. And they have to be the same, so...?

10. Jun 28, 2014

### carllacan

So $c_0$ is zero and so are all others. Thank you!

11. Jun 28, 2014

### bloby

Some clean up is needed(what if a=0? what if $c_0$,...,$c_i$=0?) but you get it.