Creation operator and eigenvalues

[carllacan]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

Calling a the eigenvalues of [itex]a_+ [/itex]
[itex]a_+ \vert \Psi \rangle = a \vert \Psi \rangle = a \sum c_n \vert n \rangle = \sum a c_n \vert n \rangle[/itex]
[itex]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[/itex]

Equating both
[itex]a_+ \vert \Psi \rangle = \sum a c_n \vert n \rangle= \sum c_{n-1}\vert n\rangle[/itex]

We have
[itex]a c_n = c_{n-1}[/itex].

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.