## show the functions are eigenfunctions of the hamiltonian

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?
 Blog Entries: 9 Recognitions: Homework Help Science Advisor 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.
 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}$

Blog Entries: 9
Recognitions:
Homework Help
 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?