Show the functions are eigenfunctions of the hamiltonian

[rmjmu507]

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

b\Psi_{n}=\sqrt{n}\Psi_{n-1}
b^{+}\Psi_{n}=\sqrt{n+1}\Psi_{n+1}

Attempt:

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

I get to

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


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:

H = \hbar\omega b^{\dagger}b + \frac{1}{2}\hbar\omega \hat{1}

then act on an arbitrary vector.

 

[rmjmu507]

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


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

 

[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 b^{+}\Psi_{n}=\sqrt{n+1}\Psi_{n+1} then b^{+}\Psi_{n-1}=\sqrt{n}\Psi_{n}

which would allow me to write the Hamiltonian as

H=ℏω(n+\frac{1}{2})\Psi_{n}

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

b\Psi_{n}=\sqrt{n}\Psi_{n-1} and
b^{+}\Psi_{n}=\sqrt{n+1}\Psi_{n+1}

are eigenfunctions of the Hamiltonian?