# How to derive $[x_i, F(\vec p)] = i \hbar \frac {\partial F(\vec p)}{\partial p_i}$

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.

• It's basically analogous to this physics.stackexchange.com/q/87038 Oct 7, 2014 at 11:49
• @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?). Oct 7, 2014 at 11:52
• 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 $-i\hbar \nabla$. Oct 7, 2014 at 11:54
• $\nabla f$ 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. Oct 7, 2014 at 12:01
• 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. Oct 7, 2014 at 12:28

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 $$\begin{eqnarray} \left[\hat{x}_i,F\left(\hat{\bf p}\right)\right] &=& \left[i \hbar \frac{\partial}{\partial p_i},F\left({\bf p}\right)\right] \\ &=& \frac{i \hbar}{f}\left[ \frac{\partial}{\partial p_i},F\left({\bf p}\right)\right] f \\ &=& \frac{i \hbar}{f}\left( \frac{\partial}{\partial p_i}\left[F\left({\bf p}\right)f\right] - F\left({\bf p}\right)\frac{\partial f}{\partial p_i}\right) \\ &=& i \hbar \frac{\partial F\left({\bf p}\right)}{\partial p_i} \\ \end{eqnarray}$$

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

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

• 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. Oct 7, 2014 at 13:36