How Can I Show That \( \exp(-iaP/\hbar) = U \)?

[naggy]

Homework Statement

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)

Homework Equations

Not sure

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

 

[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!

 

[naggy]

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!

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