A question on eigenstates and operators

[vertices]

Why can we say that:

<x'|e^{i\hat{x}}|x>=e^{ix'}\delta(x'-x)

where where \hat{x} is an operator?

I mean if

\hat{x}|x>=x|x>

we may write <x'|\hat{x}|x>=x<x'|x>=x\delta(x'-x)

but in the expression at the top, we have an exponential operator (something I've never come across before) - is |x> an eigenstate of this operator (it seems to be), and why is the eigenvalue of the operator exactly the same form as the operator?

Thanks.

 

[meopemuk]

You can prove it by using power series expansion

e^{i\hat{x}} = 1 + i\hat{x} - \hat{x}^2/2! + \ldots

Eugene.

 

[vertices]

You can prove it by using power series expansion

e^{i\hat{x}} = 1 + i\hat{x} - \hat{x}^2/2! + \ldots

Eugene.

Aaahh thanks Eugene. So in general if {\hat{Z}} is any operator and |Z> the eigenfunctions of this operator, we can also say that |Z> are also the eigenfunctions of e^{i\hat{Z}}.

 

[meopemuk]

Aaahh thanks Eugene. So in general if {\hat{Z}} is any operator and |Z> the eigenfunctions of this operator, we can also say that |Z> are also the eigenfunctions of e^{i\hat{Z}}.

That's right. |Z> is an eigenfunction of any function f(\hat{Z}) of the operator \hat{Z}. The corresponding eigenvalue is f(Z).

Eugene.