show the functions are eigenfunctions of the hamiltonian

rmjmu507

Given the hamiltonian in this form: H=[itex]\hbar[/itex][itex]\omega[/itex]([itex]b^{+}[/itex]b+.5)

b[itex]\Psi_{n}[/itex]=[itex]\sqrt{n}[/itex][itex]\Psi_{n-1}[/itex]
[itex]b^{+}[/itex][itex]\Psi_{n}[/itex]=[itex]\sqrt{n+1}[/itex][itex]\Psi_{n+1}[/itex]

Attempt:

H[itex]\Psi_{n}[/itex]=[itex]\hbar[/itex][itex]\omega[/itex]([itex]b^{+}[/itex]b+.5)[itex]\Psi_{n}[/itex]

I get to

H[itex]\Psi_{n}[/itex]=[itex]\hbar[/itex][itex]\omega[/itex][itex]\sqrt{n}[/itex]([itex]b^{+}[/itex][itex]\Psi_{n-1}[/itex]+.5[itex]\Psi_{n-1}[/itex])


But now I'm stuck. Where can I go from here?

dextercioby

It's not correct, you have to split the Hamiltonian as it should be split:

[tex] H = \hbar\omega b^{\dagger}b + \frac{1}{2}\hbar\omega \hat{1} [/tex]

then act on an arbitrary vector.

rmjmu507

I still end up with a similar problem though...I will have the raising operator acting on [itex]\Psi_{n-1}[/itex]


H= ℏω[itex]\sqrt{n}[/itex]([itex]b^{+}[/itex][itex]\Psi_{n-1}[/itex])+[itex]\frac{1}{2}[/itex]ℏω[itex]\Psi_{n}[/itex]

dextercioby

Excelent. You need to do a trick on the relation given, namely realize that the <n> can be replaced by other values. Which substitution is useful ?

P.S. Always post your HW questions here, in this forum.

rmjmu507

Can I say that if [itex]b^{+}[/itex][itex]\Psi_{n}[/itex]=[itex]\sqrt{n+1}[/itex][itex]\Psi_{n+1}[/itex] then [itex]b^{+}[/itex][itex]\Psi_{n-1}[/itex]=[itex]\sqrt{n}[/itex][itex]\Psi_{n}[/itex]

which would allow me to write the Hamiltonian as

H=ℏω(n+[itex]\frac{1}{2}[/itex])[itex]\Psi_{n}[/itex]

and because ℏω(n+[itex]\frac{1}{2}[/itex]) is just a number, then

b[itex]\Psi_{n}[/itex]=[itex]\sqrt{n}[/itex][itex]\Psi_{n-1}[/itex] and
[itex]b^{+}[/itex][itex]\Psi_{n}[/itex]=[itex]\sqrt{n+1}[/itex][itex]\Psi_{n+1}[/itex]

are eigenfunctions of the Hamiltonian?