# Concerning the momentum operator

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

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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.

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 $$U$$

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.

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).