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Given a stationary 1-D wave function $\psi(x)$, how is the derivative in the momentum operator interpreted?

$$ \int_{-\infty}^\infty \psi^*(x) \hat{p} \psi(x) dx = \int_{-\infty}^\infty \psi^*(x) (-i\hbar\nabla) \psi(x) dx $$

Should the integral be interpreted as

$$-i\hbar\int_{-\infty}^\infty \psi^*(x) \psi'(x) dx$$

where $\psi'(x)$ is the derivative with respect to $x$?

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Do you have any other interpretation in mind? –  Qmechanic Mar 30 '12 at 13:25
@Qmechanic: No. I'm just not used to the notation, so I wanted to verify that I'm interpreting it correctly. –  Per Mar 30 '12 at 13:29
The answer is yes. –  Qmechanic Mar 30 '12 at 13:29
When I see this kind of "implied multiplication" I usually assume that it's commutative, which doesn't seem to be the case in much of QM. –  Per Mar 30 '12 at 13:30
Never assume commutativity in QM unless you're sure of it. –  P O'Conbhui Mar 31 '12 at 0:56

1 Answer 1

up vote 2 down vote accepted

Yes. In mathematics, the symbol $f'$ means $$ f'(x) = \frac{{\rm d}f}{{\rm d}x} = \lim_{\epsilon\to 0} \frac{f(x+\epsilon)-f(x)}{\epsilon} $$ This notation using ${\rm d}$ was introduced by Leibniz; the notation with the prime was introduced by Lagrange.

You also ask whether there is some problem with noncommutativity. There is absolutely nothing noncommutative in the integrals you write down. They're integrals with ordinary functions and their derivatives. The only objects that don't commute in quantum mechanics are observables i.e. operators such as $\hat x$ and $\hat p$.

However, the derivative $\psi'$ isn't really an operator. More precisely, it is proportional to the operator $\hat p$ acting on the wave function $\psi$ but $\psi$ isn't an operator so you can't really move it to the left side from the derivative.

If we have things like $V'(x)$ in quantum mechanics, the derivative of the potential energy, then the potential energy $V(x)$ may be interpreted as an operator. Then $V'(x)$ may be rewritten as $$ V'(x) = [\frac{\rm d}{{\rm d}x},V(x)] = \frac{i}{\hbar} [\hat p,\hat V(x)] $$ However, when we choose the explicit integrals over $x$ etc., they always mean the operations with ordinary commuting functions as they do in the calculus. One only has noncommuting objects when we write the actions of the operators more abstractly, in a way that doesn't depend on the representation of the wave function.

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A good answer but the phrase «there is absolutely nothing noncommutative in the integrals you write down» is not completely correct and could be a little misleading, since it contradicts the trenchant comment by @QMechanic . Look at the integrand in the right hand side of the first equality, basically $\psi ^*D \psi$. Now the Laplacian does not commute with functions...this isnt the same as $D \psi^*\psi$, and note, it isn't associative either. I know that to contradict this is not what you meant, but it's what you said, so some editing is in order. (Editing and posting aren't commutative.. –  joseph f. johnson Feb 14 '13 at 21:48

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