4
$\begingroup$

The canonical commutation relation $$[x,p] = i\hbar$$ can be generalized to $$[p_i,F(\vec{x})] = -i\hbar\frac{\partial F(\vec{x})}{\partial x_i}, \ [x_i, F(\vec{p})] = i\hbar\frac{\partial F(\vec{p})}{\partial p_i}$$ according to Sakurai's Modern Quantum Mechanics.

The book requires readers to prove it in one problem, probably using power series method, since Sakurai claims it can be proved by repeatedly using $[A,BC]=[A,B]C+B[A,C]$.

I know the proof would be straightforward if $F$ is an entire function, which can be expressed as a power series. But the situation seems to be not very clear if $F$ isn't an entire function?

$\endgroup$
  • 3
    $\begingroup$ How would you even define a function $F$ of an operator $x$ if not by a power series? $\endgroup$ – ACuriousMind Oct 28 '14 at 14:03
  • 4
    $\begingroup$ ACuriousmind: by diagonalizing the operator and applying the function to each eigenvalue, right? It can easily be non-smooth. $\endgroup$ – Luboš Motl Oct 28 '14 at 14:05
  • 1
    $\begingroup$ Note also that Sakurai requires you to prove the two statements in Problem 29 of Chapter 1. $\endgroup$ – Kyle Kanos Oct 28 '14 at 14:05
  • $\begingroup$ @KyleKanos I did that problem using power series. $\endgroup$ – Qianyi Guo Oct 28 '14 at 14:08
  • 1
    $\begingroup$ Related: physics.stackexchange.com/q/87038/2451 and links therein. $\endgroup$ – Qmechanic Oct 28 '14 at 14:17
9
$\begingroup$

Generally speaking functions of operators are defined as spectral functions, making use of the spectral theorem machinery.

The approaches based of power series usually have not a rigorous basis: Even identities like $$e^A = \sum_n \frac{1}{n!}A^n$$ are generally false if $A$ is an unbounded operator.

Nevertheless in most relevant physical cases some of the results obtained by formal (non-rigorous) manipulations can be obtained following alternative rigorous ways.

However the case you are considering is very easy. Suppose that $F$ is a smooth, generally non-analytic, complex-valued function bounded by some polynomial of $\vec{x}$. In this case both $F(\vec{x})$ and $p_i$ admit the space of Schwartz functions ${\cal S}(\mathbb R^3)$ as invariant space, so that $[p_i, F(\vec{x})]\psi$ is well defined for $\psi \in {\cal S}(\mathbb R^3)$. On that space $p_i$ coincides to $-i \hbar\frac{\partial}{\partial x_i}$. By direct computation: $$[p_i, F(\vec{x})]\psi = -i \hbar\frac{\partial}{\partial x_i} F(\vec{x}) \psi(x) + i \hbar F(\vec{x})\frac{\partial \psi}{\partial x_i}= -i\hbar \frac{\partial F}{\partial x_i} \psi\:.$$ We have obtained that $$\left([p_i, F(\vec{x})] + i\hbar \frac{\partial F}{\partial x_i} \right) \psi =0 \quad \forall \psi \in {\cal S}(\mathbb R^3)\:.$$ In other words, at least on the domain ${\cal S}(\mathbb R^3)$:
$$[p_i, F(\vec{x})] = -i\hbar \frac{\partial F}{\partial x_i} $$ In general, this identity holds on larger domains and, in fact, it can be extended exploiting some further know property of $F$ and some other properties like the fact that $\cal S(\mathbb R^3)$ is a core for $p_i$...

One could assume weaker hypotheses. The self-adjointness domain of $p_j$ is so made. $$D(p_j) := \{\psi \in L^2(\mathbb R^3)\:|\: \exists\: w\mbox{-}\partial_{x_j}\psi \in L^2(\mathbb R^3)\}$$ where $w\mbox{-}\partial_{x_j}\psi$ denotes the weak partial $j$-derivative of $\psi$. On that domain $p_j = w\mbox{-}\partial_{x_j}$ as expected.

If $F$ is, for instance, just $C^1$ and compactly supported, all above reasoning can be re-implemented with $\psi \in D(p_j)$, obtaining exactly the same result, using the fact that the weak derivative of a compactly supported $C^1$ function $F$ coincides with the standard one.

$\endgroup$
  • $\begingroup$ So what about the $[x_i,F(\vec{p})]$ relation? It is almost the same in power series method, but seems to be not very straightforward for direct computation. $\endgroup$ – Qianyi Guo Oct 30 '14 at 14:15
  • 1
    $\begingroup$ Its treatment is identical, since there is a unitary map $U$ with $Ux_iU^\dagger=p$ and thus $U^\dagger F(\vec{p})U = F(\vec{x})$ and one reduces to the already discussed case. $\endgroup$ – Valter Moretti Oct 30 '14 at 15:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.