Harmonic oscillator eigenvector/eigenvalue spectrum

[jtaa]

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

 

[George Jones]

What is the commutator of H and a^\dagger?

 

[jtaa]

N=a†a

[N,a†]=a†
[N,a]=-a

 

[George Jones]

What about H?

 

[jtaa]

[H,a†] = \hbar\omegaa†

[H,a] = -\hbar\omegaa

 

[George Jones]

\phi_{n+1} = \left( n + 1 \right)^{-\frac{1}{2}} a^\dagger \phi_n gives H \phi_{n+1} = \left( n + 1 \right)^{-\frac{1}{2}} H a^\dagger \phi_n. Now use the commutator to reorder H a^\dagger.

 

[jtaa]

re-order how?

by using: Ha†=\hbar\omegaa† + a†H
?

 

[George Jones]

Yes.

 

[jtaa]

i then get:

=(n+1)^-1/2(a†\hbar\omega+a†H)\phi_n
=\hbar\omega\phi_n+1 + (n+1)^-1/2a†H\phi_n

not sure what to do from here..

 

[George Jones]

The inductive step assumes what about H\phi_n?

 

[jtaa]

it assumes that Hϕ_n = nϕ_n

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


=> = (\hbar\omega+n)ϕ_n+1

?

 

[George Jones]

it assumes that Hϕ_n = nϕ_n

i.e. that ϕ_n is an eigenvector of H.

The inductive step assumes that \phi_n is an eigenvector of H, but it doesn't assume that the associated eigenvalue is n. Energies are eigenvalues of the Hamiltonian, so call the the eigenvalue E_n. Maybe E_n = n, but maybe it doesn't. Let's find out!

i can prove it for the basis step n=0.

What is the eigenvalue of H associated with the eigenvector \phi_0?

 

[jtaa]

Hϕ₀=\frac{1}{2}ℏωϕ₀

So, E₀=\frac{1}{2}ℏω

=> Hϕ_n+1 = ℏωϕ_n+1+E_nϕ_n+1
?

 

[George Jones]

Right.

 

[jtaa]

can i simply say after that:
the energies are:

E_n=hw(n+\frac{1}{2})

?