# Find the eigenvalues of the Hamiltonian - Harmonic Oscillator

1. Nov 2, 2013

### Jalo

1. The problem statement, all variables and given/known data

Find the eigenvalues of the following Hamiltonian.

$Ĥ = ħwâ^{†}â + \alpha(â + â^{†}) , \alpha \in |R$

2. Relevant equations

$â|\phi_{n}>=\sqrt{n}|\phi_{n-1}>$
$â^{†}|\phi_{n}>=\sqrt{n+1}|\phi_{n+1}>$

3. The attempt at a solution

By applying the Hamiltonian to a random state n I get:

$Ĥ |\phi_{n}> = E_{n}|\phi_{n}>$
$Ĥ |\phi_{n}>= ħwâ^{†}â|\phi_{n}> + \alpha(â|\phi_{n}> + â^{†}|\phi_{n}>)$
$Ĥ |\phi_{n}>= ħw\sqrt{n}\sqrt{n}|\phi_{n}> + \alpha(\sqrt{n}|\phi_{n-1}> + \sqrt{n+1}|\phi_{n+1}> )$
$E_{n} |\phi_{n}> = ħwn + \alpha(\sqrt{n}|\phi_{n-1}> + \sqrt{n+1}|\phi_{n+1}>)$

This is where my problem arrives. I don't know how to prove that

$\alpha(\sqrt{n}|\phi_{n-1}> + \sqrt{n+1}|\phi_{n+1}>) = 0$

Any help would be highly appreciated!
Thanks.

2. Nov 2, 2013

### fzero

That linear combination doesn't vanish and the states $|n\rangle$ are not eigenstates of that Hamiltonian. The eigenstates will be infinite linear combinations of the $| n \rangle$. However, constructing these eigenstates is certainly not the easiest way to compute the eigenvalues of this operator. I would suggest defining a new operator $b = a + c$, where $c$ is a number to be determined by requiring that $\hat{H} = \hbar \omega b^\dagger b + C$, where $C$ is another constant. Using the commutation relations for $b,b^\dagger$, you should be able to compute the eigenvalues in the same way as for the regular harmonic oscillator.

3. Nov 3, 2013

### Jalo

And how can I find the operator b?

4. Nov 3, 2013

### fzero

You solve the equation

$$\hbar \omega (a+c)^\dagger (a+c) + C = \hbar \omega a^\dagger a + \alpha (a + a^\dagger)$$

for $c$ and $C$. This is a linear equation, since the $a^\dagger a$ terms cancel.

5. Nov 6, 2013

### Jalo

I just realized I forgot to thank you! Accept my apologies.

Daniel