How to Show \widehat{A}\Psi(x) = \Psi(x + b)?

[Void123]

Homework Statement

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

Homework Equations

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

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?

 

[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?

 

[Void123]

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

 

[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.

 

[Void123]

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

 

[Dick]

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

 

[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?

 

[Dick]

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?

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