Harmonic oscillator eigenvector/eigenvalue spectrum

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jtaa
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the problem is attached as an image.

im having troubles with the question. I'm assuming this is an induction question?
i can prove it for the basis step n=0.

but I am having trouble as to what i have to do for n+1 (inductive step).

any help or hints would be great!thanks
 

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What is the commutator of [itex]H[/itex] and [itex]a^\dagger[/itex]?
 
N=a†a

[N,a†]=a†
[N,a]=-a
 
What about [itex]H[/itex]?
 
[H,a†] = [itex]\hbar[/itex][itex]\omega[/itex]a†

[H,a] = -[itex]\hbar[/itex][itex]\omega[/itex]a
 
[itex]\phi_{n+1} = \left( n + 1 \right)^{-\frac{1}{2}} a^\dagger \phi_n[/itex] gives [itex]H \phi_{n+1} = \left( n + 1 \right)^{-\frac{1}{2}} H a^\dagger \phi_n[/itex]. Now use the commutator to reorder [itex]H a^\dagger[/itex].
 
re-order how?

by using: Ha†=[itex]\hbar\omega[/itex]a† + a†H
?
 
i then get:

=(n+1)-1/2(a†[itex]\hbar\omega[/itex]+a†H)[itex]\phi[/itex]n
=[itex]\hbar\omega[/itex][itex]\phi[/itex]n+1 + (n+1)-1/2a†H[itex]\phi[/itex]n

not sure what to do from here..
 
The inductive step assumes what about [itex]H\phi_n[/itex]?
 
it assumes that Hϕn = nϕn

i.e. that ϕn is an eigenvector of H. n being the eigenvalue


=> = ([itex]\hbar\omega[/itex]+n)ϕn+1

?
 
jtaa said:
it assumes that Hϕn = nϕn

i.e. that ϕn is an eigenvector of H.

The inductive step assumes that [itex]\phi_n[/itex] is an eigenvector of [itex]H[/itex], but it doesn't assume that the associated eigenvalue is [itex]n[/itex]. Energies are eigenvalues of the Hamiltonian, so call the the eigenvalue [itex]E_n[/itex]. Maybe [itex]E_n = n[/itex], but maybe it doesn't. Let's find out!
jtaa said:
i can prove it for the basis step n=0.

What is the eigenvalue of [itex]H[/itex] associated with the eigenvector [itex]\phi_0[/itex]?
 
0=[itex]\frac{1}{2}[/itex]ℏωϕ0

So, E0=[itex]\frac{1}{2}[/itex]ℏω

=> Hϕn+1 = ℏωϕn+1+Enϕn+1
?
 
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can i simply say after that:
the energies are:

En=hw(n+[itex]\frac{1}{2}[/itex])

?