5
$\begingroup$

Wikipedia indicates that the following relation is "easily shown": $[x_i, F(\vec p)] = i \hbar \frac {\partial F(\vec p)}{\partial p_i}$, however I'm having some trouble showing it. I think I'm just messing up the multivariable Taylor expansion (of $F(\vec p)$). Can one of you walk me through it or link me to site that will? Thanks.


Edit: Here's what I get (without using $x=i\hbar \frac \partial {\partial p}$ which I haven't proven yet):

$$F(\vec p) = F(\vec 0) + \sum^3_{j=1} \frac {\partial F(\vec 0)}{\partial p_j}p_j + \frac 12 \sum^3_{k=1} \sum^3_{j=1} \frac {\partial^2 F(\vec 0)}{\partial p_k \partial p_j} p_j p_k + \dots$$ so $$(x_iF(\vec p) - F(\vec p)x_i)\psi$$ $$= x_i[F(\vec 0)\psi -i\hbar \sum^3_{j=1} \frac {\partial F(\vec 0)}{\partial p_j}(\nabla \psi)_j + \hbar^2\frac 12 \sum^3_{k=1} \sum^3_{j=1} \frac {\partial^2 F(\vec 0)}{\partial p_k \partial p_j} (\nabla \psi)_j (\nabla \psi)_k + \dots]$$ $$- [F(\vec 0)x_i\psi -i\hbar \sum^3_{j=1} \frac {\partial F(\vec 0)}{\partial p_j}(\nabla x_i\psi)_j + \hbar^2\frac 12 \sum^3_{k=1} \sum^3_{j=1} \frac {\partial^2 F(\vec 0)}{\partial p_k \partial p_j} (\nabla x_i\psi)_j (\nabla x_i\psi)_k + \dots]$$

where $$(\nabla x_i\psi)_j=\frac {\partial x_i}{\partial x_j}\psi + x_i\frac {\partial \psi}{\partial x_j} = \delta_{ij}\psi + x_i\frac {\partial \psi}{\partial x_j}$$ and $$(\nabla x_i\psi)_j(\nabla x_i\psi)_k=(\delta_{ij}\psi + x_i\frac {\partial \psi}{\partial x_j})(\delta_{ik}\psi + x_i\frac {\partial \psi}{\partial x_k}) = \delta_{jk}\psi^2 + x_j \psi \frac {\partial \psi}{\partial x_k} + x_k \frac {\partial \psi}{\partial x_j} \psi + x_i^2 \frac {\partial^2 \psi}{\partial x_j \partial x_k}$$

Thus: $$(x_iF(\vec p) - F(\vec p)x_i)\psi$$ $$= x_i[F(\vec 0)\psi -i\hbar \sum^3_{j=1} \frac {\partial F(\vec 0)}{\partial p_j}\frac {\partial \psi}{\partial x_j} + \hbar^2\frac 12 \sum^3_{k=1} \sum^3_{j=1} \frac {\partial^2 F(\vec 0)}{\partial p_k \partial p_j} \frac {\partial \psi}{\partial x_j} \frac {\partial \psi}{\partial x_k} + \dots]$$ $$- [F(\vec 0)x_i\psi -i\hbar \sum^3_{j=1} \frac {\partial F(\vec 0)}{\partial p_j}(\delta_{ij}\psi + x_i\frac {\partial \psi}{\partial x_j}) + \hbar^2\frac 12 \sum^3_{k=1} \sum^3_{j=1} \frac {\partial^2 F(\vec 0)}{\partial p_k \partial p_j} (\delta_{jk}\psi^2 + x_j \psi \frac {\partial \psi}{\partial x_k} + x_k \frac {\partial \psi}{\partial x_j} \psi + x_i^2 \frac {\partial^2 \psi}{\partial x_j \partial x_k}) + \dots]$$

From here it doesn't look like those higher order terms are all going to cancel out.

$\endgroup$
8
  • 2
    $\begingroup$ It's basically analogous to this physics.stackexchange.com/q/87038 $\endgroup$
    – Bubble
    Oct 7, 2014 at 11:49
  • $\begingroup$ @Bubble That one is much easier (for me) because $F(x)$ is just a function whereas $F(p)$ is an operator and thus has to be Taylor expanded (I think?). $\endgroup$
    – Bob Dylan
    Oct 7, 2014 at 11:52
  • $\begingroup$ Actually in this case, I'm not even sure that $p_i$ makes sense. Because $p$ isn't actually a vector is it? It's $-i\hbar \nabla$. $\endgroup$
    – Bob Dylan
    Oct 7, 2014 at 11:54
  • $\begingroup$ $\nabla f$ for a function $f$ is a vector. $\nabla$ itself isn't a vector, but you can get away with pretending that it is. When people write $p_i = -i\hbar \nabla_i$ they really mean $p_i f = -i\hbar(\nabla f)_i$, which makes perfect sense. As long as you remember that $\nabla$ eventually acts on something, and that produces a vector, you can get away with pretending that $\nabla$ is a vector. $\endgroup$ Oct 7, 2014 at 12:01
  • $\begingroup$ Hi @Bob Dylan. Comments: 1. Note that the position operator $\hat{x}^i$ and momentum operator $\hat{p}_j$ (up to a minus sign) enter on equal footing in the CCR, cf. above comment by Bubble. 2. Also note that the sought-for formula is independent of operator representation. E.g. one may work in the Fourier transformed momentum representation, or one may work in a formalism manifestly independent of representation. $\endgroup$
    – Qmechanic
    Oct 7, 2014 at 12:28

3 Answers 3

5
$\begingroup$

As @Qmechanic pointed out in a comment, we are free to use any operator representation. In momentum space, $\hat{\bf x} = + i \hbar \ \partial/\partial {\bf p} $ and $\hat{\bf p} = {\bf p}$, so $$ \begin{eqnarray} \left[\hat{x}_i,F\left(\hat{\bf p}\right)\right] &=& \left[i \hbar \frac{\partial}{\partial p_i},F\left({\bf p}\right)\right] \\ &=& \frac{i \hbar}{f}\left[ \frac{\partial}{\partial p_i},F\left({\bf p}\right)\right] f \\ &=& \frac{i \hbar}{f}\left( \frac{\partial}{\partial p_i}\left[F\left({\bf p}\right)f\right] - F\left({\bf p}\right)\frac{\partial f}{\partial p_i}\right) \\ &=& i \hbar \frac{\partial F\left({\bf p}\right)}{\partial p_i} \\ \end{eqnarray} $$

$\endgroup$
4
$\begingroup$

Choose the momentum representation,

$$x_i = i \hbar \frac{\partial}{\partial p_i}$$

distribute $i \hbar$ and act the commutator on vector $\psi$,

$$[x_i, F(\mathbf p)] \psi = i \hbar \left(\frac{\partial}{\partial p_i}(F(\mathbf{p}) \space \psi) -F(\mathbf p) \frac{\partial }{\partial p_i} \psi \right)$$

and apply the product rule:

$$= i \hbar \left(\frac{\partial F(\mathbf p )}{\partial p_i} \psi + F(\mathbf p) \frac{\partial \psi}{\partial p_i} - F(\mathbf p) \frac{\partial \psi}{\partial p_i} \right)$$

$$= i \hbar\frac{\partial F(\mathbf p )}{\partial p_i} \psi.$$

We left $\psi$ unspecified, so: $$ [ x_i, F(\mathbf p ) ] = i \hbar \frac{\partial F( \mathbf p )}{\partial p_i}$$

$\endgroup$
2
$\begingroup$

The commutation of two variables, in some cases, can be related to Poisson Bracket via $$ \left[\hat A,\,\hat B\right]=i\hbar\left\{\hat A,\,\hat B\right\} $$ Thus, $$ \left[\hat A,\,\hat B\right]=i\hbar\sum_i\left(\frac{\partial A}{\partial q_i}\frac{\partial B}{\partial p_i}-\frac{\partial A}{\partial p_i}\frac{\partial B}{\partial q_i}\right)\tag{1} $$ Formally $\hat A=A(\hat q,\,\hat p)$ and $\hat B= B(\hat q,\,\hat p)$. You should be able to use (1) to solve your problem. Note, though, that this is not a solution that works in all cases (cf., this and this that Qmechanic pointed out) as it is an approximation that only holds under certain cases.

A general solution involves the Moyal bracket, $$ \left[\hat A,\,\hat B\right]\sim\{\{\hat A\,,\hat B\}\}\sim \hat A\star \hat B-\hat B\star\hat A $$ where $\star$ denotes the Moyal star-product (see answers on either this post or this post for more on the Moyal product). The above then can be written as $$ \{\{\hat A,\,\hat B\}\}=\{\hat A,\,\hat B\}+\mathcal{O}(\hbar^2) $$ where $\{\cdot,\,\cdot\}$ here is the above Poisson bracket and $\mathcal{O}(\hbar^2)$ are corrections (referred to as deformations if the Poisson bracket).

Thus, Equation (1) becomes $$ \left[\hat A,\,\hat B\right]=i\hbar\sum_i\left(\frac{\partial A}{\partial q_i}\frac{\partial B}{\partial p_i}-\frac{\partial A}{\partial p_i}\frac{\partial B}{\partial q_i}\right)+\mathcal{O}(\hbar^2)\tag{2} $$

$\endgroup$
1
  • 2
    $\begingroup$ Comment to the answer (v1): The method of replacing quantum commutators $[\hat{f},\hat{g}]$ with classical Poisson brackets $\{f,g\}_{PB}$ may miss higher quantum corrections, cf. e.g. this, this and this Phys.SE posts. For OP's question the Poisson bracket happens to give the precise answer. $\endgroup$
    – Qmechanic
    Oct 7, 2014 at 13:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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