# Concerning the momentum operator

1. Sep 22, 2009

### naggy

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

If $$U$$ is an operator so $$U\Psi(x)$$ = $$\Psi(x-a)$$.

How can I show that $$exp(-iaP/h) = U$$

where P is the momentum operator $$P = -ih(d/dx)$$

2. Relevant equations

Not sure

3. The attempt at a solution

What I do know that if I have a function F of an operator then

$$F(P)\psi$$ = $$\sum_{i} c_iF(\lambda_i)\psi_i$$

where $$\lambda_i$$ are the eigenvalues of $$P$$

and $$c_i = <\psi_i,\psi>$$

can I somehow relate all of this to the operator $$U$$

2. Sep 22, 2009

### gabbagabbahey

I think what you'll want to do is combine the fact that (by the definition of the exponential of an operator)

$$e^{-iaP/\hbar}=\sum_{n=0}^{\infty}\frac{(-1)^na^nP^n}{\hbar^n n!}$$

with your expression for $P$ in the x-basis to calculate $e^{-iaP/\hbar}\Psi(x)$....your result should remind you of a Taylor series!

3. Sep 22, 2009

### naggy

Ok so:

$$e^{-iaP/\hbar}= e^{-aD_x} = 1 - aD_x + \frac{a^2D_x^2}{2!} - \frac{a^3D_x^3}{3!} + ...$$

and applying this to $$\psi(x)$$ I can´t really say that I recognise the series, but I suppose it is the series for shifted $$\psi(x)$$ about a