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?