# Operator dependent functions in QM

## 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?

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Dick
Homework Helper
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?

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

Dick
Homework Helper
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.

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

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

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?