Creation operator and eigenvalues

samedi 28 juin 2014

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

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



2. Relevant equations

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

[itex] a_+\vert n \rangle =(E_n + \hbar \omega\vert n +1\rangle [/itex]



3. 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 a c_n(E_n + \hbar \omega)\vert n+1\rangle = \sum a c_{n-1}(E_{n-1} + \hbar \omega)\vert n\rangle[/itex]



Equating both

[itex]a_+ \vert \Psi \rangle = \sum a c_n \vert n \rangle= \sum a c_{n-1}(E_{n-1} + \hbar \omega)\vert n\rangle[/itex]



We have

[itex]a c_n = a c_{n-1}(E_{n-1} + \hbar \omega)[/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.





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