# A problem with Ladder operators

1. Oct 29, 2009

### <---

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

The problem is to show that,

$$\hat{a_{+}}|\alpha>=A_{\alpha}|\alpha+1>$$

using

$$\hat{a_{+}}\hat{a_{-}}|\alpha>=\alpha|\alpha>$$

It's not hard to manipulate $$\hat{a_{+}}\hat{a_{-}}|\alpha>=\alpha|\alpha>$$ into the form,

$$\hat{a_{+}}\hat{a_{-}}[{\hat{a_{+}}|\alpha>}]=(1+\alpha)[\hat{a_{+}}|\alpha>]$$

But I am unable to make the connection from this to,

$$\hat{a_{+}}|\alpha>=A_{\alpha}|\alpha+1>$$

I know it's just using the properties of the eigenfunctions/values of a Hermatian operator at this point, but I seem to be missing exactly what that is.

What am I missing?

2. Oct 29, 2009

### gabbagabbahey

Well, this is an eigenvalue equation for the operator $\hat{a}_+\hat{a}_-$, with eigenvalue $\alpha+1$ and eigenstate $\hat{a}_+|\alpha\rangle$...but compare this to your original eigenvalue equation for this operator....surely if $\alpha+1$ is the eigenvalue, the eigenstate must be $|\alpha+1\rangle$ (or at least a scalar multiple of it).....doesn't that tell you everything you need to know about $\hat{a}_+|\alpha\rangle$?

3. Oct 29, 2009

### jdwood983

From what you've done, using both ladder operators on the ket, what does that say about $N=\hat{a}_+\hat{a}_-$ and $\hat{a}_\pm$??

4. Oct 29, 2009

### <---

Thanks very much for the replys.

gabba, That did occur to me, but I wasn't willing to make the concession that,

$$\hat{a_{+}}\hat{a_{-}}|\alpha>=\alpha|\alpha>$$

Was a general property and not $$\alpha$$ specific.Is this a property of NORMALIZED eigenvectors(for which I should have specified |alpha> is defined as)? If so I suppose that would explain the $$A_{\alpha}$$ as a normalization constant.

jd, I know $$\hat{a_{+}}\hat{a_{-}}$$ is Hermatian although neither are individually..... I'm not sure if that's what you mean.

5. Oct 29, 2009

### jdwood983

I was trying to guide you with less words than what gabba said: If

$$N\hat{a}_\pm|n\rangle=(n\pm1)\hat{a}_\pm|n\rangle$$

and $N|n\rangle=n|n\rangle$, then $\hat{a}_\pm|n\rangle$ are multiplicative eigenstates of $|n\pm1\rangle$.

6. Oct 29, 2009

### <---

Thank you that helps, I'll have to stare at that for awhile to let it sink in.