# 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.

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#### 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.

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