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).