Does the Creation Operator Have Eigenvalues?

[carllacan]

Homework Statement

Prove that the creation operator a_+ has no eigenvalues, for instance in the \vert n \rangle.

Homework Equations

Action of a_+ in a harmonic oscillator eigenket \vert n \rangle:
a_+\vert n \rangle =\vert n +1\rangle

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.

 

[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 $|\Psi>=|n>+|n-1>+...+|n-m>$?

 

[carllacan]

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

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.

 

[bloby]

(and it is, right?)

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

$a_+|\Psi\rangle=|n+1>+|n\rangle+...+|n-m+1\rangle$

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

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?

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.

 

[carllacan]

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

What do you mean with the bounds?

 

[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?

 

[vela]

Try writing out the first few terms of each summation.

 

[carllacan]

Sorry for the delay.

Wow, why do you apologyze? You're helping me! I really appreciate it, no matter how long does it take.

Your result $a c_n=c_{n-1}$ cannot hold for all n, do you see why?

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.

 

[bloby]

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

 

[carllacan]

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

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

 

[bloby]

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