# Operator dependent functions in QM

1. Oct 13, 2009

### Void123

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

Show that $$\widehat{A}\Psi(x) = \Psi(x + b)$$, where $$b$$ is a constant.

2. Relevant equations

Given: $$\widehat{A} = exp(b[d/dx])$$

3. The attempt at a solution

I know I'm supposed to write out the function as a power series expansion, though I'm not sure what I am exactly to do after this. What is (d/dx) of? What information do the above expressions convey?

2. Oct 13, 2009

### Dick

Write out a taylor expansion of psi(x+b) around x. Now compare that with the power series expansion of A applied to psi(x). The derivative operators will operate on psi(x). Don't you see a similarity in the two sides?

3. Oct 13, 2009

### Void123

Pardon my ignorance, but what would I be expanding since I am not explicitly given a function for $$\Psi(x)$$?

4. Oct 13, 2009

### Dick

Write them out without assuming you know psi(x). The first derivative of psi(x) is psi'(x), the second is psi''(x) etc etc.

5. Oct 14, 2009

### Void123

Just to make sure my work is correct, would I get {[(b^n)/n!]*(d/dx)^n}*$$\Psi(x)$$?

6. Oct 14, 2009

### Dick

Well, yeah. Isn't that the Taylor series of psi(x+b) expanded around x?

7. Oct 14, 2009

### Void123

Okay, so I did it right. Though, I am curious about $$\Psi(x + b)$$. If $$\Psi(x)$$ is an eigenstate of $$\widehat{A}$$, then the former is equal to some factor times $$\Psi(x)$$ right?

8. Oct 14, 2009

### Dick

That's the definition of an eigenstate alright, A(psi(x))=k*psi(x)=psi(x+b).