Solving for $\hat{A}\psi(x)=\psi(x+b)$

[CalcYouLater]

Homework Statement

Given operator $$\hat{A}=exp({b}\frac{d}{dx})$$ where b is a constant, show that:

$$\hat{A}\psi(x)=\psi(x+b)$$

Homework Equations

The Attempt at a Solution

I have no idea with this one. If I understand correctly the operator, A, says take the derivative of (something) with respect to x, multiply the result by b, and then exponentiate it. A pointer would be much appreciated here.

 

[vela]

It's usually a good idea to consider an infinitesimal change first and then deduce the result for a finite change.

 

[diazona]

You haven't quite gotten the meaning of the exponential notation correct, which is why you're confused A function of an operator is defined by its series expansion,
$$f(\hat{D}) = f_0 + f_1 \hat{D} + f_2 \hat{D}^2 + \cdots$$
Typically the function f is one that can be expanded in a Maclaurin series (Taylor series around 0), so in that case the coefficients $f_n$ are just the same coefficients that appear in the Maclaurin series:
$$f_n = \frac{1}{n!}f^{(n)}(0)$$
For the exponential function, the series is
$$e^x = 1 + x + \frac{x^2}{2} + \cdots$$
so the operator $$e^{\hat{D}}$$ is defined as
$$e^{\hat{D}} = 1 + \hat{D} + \frac{\hat{D}^2}{2} + \cdots$$
In your case, $\hat{D}$ is multiplication by a constant composed with the derivative operator $\mathrm{d}/\mathrm{d}x$, so the notation $\hat{D}^n$ is effectively telling you to take the n'th derivative and then multiply by $b^n$. This would of course be applied to whatever the operator is acting on - for example, in
$$e^{b\mathrm{d}/\mathrm{d}x}\psi(x)$$
each term would be produced by differentiating ψ(x) some number of times and then multiplying by b that many times.

 

[CalcYouLater]

Oh I see! Thank you both for responding. Interesting that it yields the series expansion for psi of (x+b). Learning operator basics and notation is turning out harder than I expected.