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The book I am reading takes the unjustified step $$e^{-\frac{i}{\hbar}\vec{p}\cdot\vec{r}}f(\vec{p}) = f(\frac{\hbar}{i}\vec{\nabla})e^{-\frac{i}{\hbar}\vec{p}\cdot\vec{r}}$$

and similarly, he uses elsewhere $$e^{-\frac{i}{\hbar}\vec{p}\cdot\vec{r}}g(\vec{r}) = g(i\hbar\vec{\nabla}_p)e^{-\frac{i}{\hbar}\vec{p}\cdot\vec{r}}$$

I have a gut feeling this follows somehow from the commutation relation $[r^i,p_j] = i\hbar\delta^i_j$ but I cannot justify this.

Anyhow, the book has not introduced the standard commutation relations, and is working only with the fact that $\phi(\vec{p},t)$ and $\psi(\vec{r},t)$ are Fourier transforms of each other, and $f(\vec{p})$ and $g(\vec{r})$ are analytic.

Could someone give me a hint?

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1 Answer 1

up vote 5 down vote accepted

It follows from the rule

$$\frac{\partial}{\partial x^i} e^{a_j x^j}= a_i e^{a_j x^j},$$

which by repeated use leads to

$$f(\frac{\partial}{\partial x^i}) e^{a_j x^j}= f(a_i)e^{a_j x^j},$$

where $f$ is a polynomial. More general classes of functions $f$ can be reached by taking appropriate limits.

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So am I to take that if a function of an operator is analytic, it can be expanded in a power series of that operator, i.e $f(\nabla)$ can be expressed as a series in $O(\nabla^n)$ –  Approximist Apr 17 '11 at 20:18
    
@Approximist: yep, exactly. This is the method by which a function of an operator is defined. –  David Z Apr 17 '11 at 21:29
    
@David @Qmechanic thanks. –  Approximist Apr 17 '11 at 22:01
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@Approximist: you may also want to consult this wikipedia page on functional calculus and links therein. –  Willie Wong Apr 17 '11 at 22:37
    
@Willie Wong Thanks. The name of the field where all this stuff happens is something what I've been looking for. –  Approximist Apr 17 '11 at 23:10
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