Concerning the momentum operator

[naggy]

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)

I sense Fourier analysis

 

[xepma]

It's matter of Taylor expanding the right hand side. By definition this is just a sum of Psi and its derivatives. You can pull the Psi out of this Taylor series, and what's left is an operator (which is a sum of derivatives) acting on Psi(x). You'll recognise this operator as the series expansion of the exponent U.

 

[naggy]

It's matter of Taylor expanding the right hand side. By definition this is just a sum of Psi and its derivatives. You can pull the Psi out of this Taylor series, and what's left is an operator (which is a sum of derivatives) acting on Psi(x). You'll recognise this operator as the series expansion of the exponent U.

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

 

[pellman]

naggy, there might be some connection with your last post.

But what xepma is pointing out is that

U=exp(-iaP/h)=exp[-a(d/dx)]=\sum{\frac{(-a)^n}{n!}\frac{d^n}{dx^n}}.

Applying U to \Psi(x) is identical to the Taylor expansion of \Psi(x-a) around x.

That's all there is to it.

 

[Count Iblis]

A problem with this proof is that the operator identity is valid on L^2, and thus remains valid when applied to non-analytic psi(x) for which the Taylor expansion around x does not converge to psi(x-a). So, you have to fix this (which is not difficult).