# Commutation relations for angular momentum operator

1. Dec 6, 2015

### spaghetti3451

I would like to prove that the angular momentum operators $\vec{J} = \vec{x} \times \vec{p} = \vec{x} \times (-i\vec{\nabla})$ can be used to obtain the commutation relations $[J_{i},J_{j}]=i\epsilon_{ijk}J_{k}$.

Something's gone wrong with my proof below. Can you point out the mistake?

$[J_{i},J_{j}]$
$=J_{i}J_{j}-J_{j}J_{i}$
$=(-i\epsilon_{ikl}x_{k}\nabla_{l})(-i\epsilon_{jmn}x_{m}\nabla_{n})-(-i\epsilon_{jmn}x_{m}\nabla_{n})(-i\epsilon_{ikl}x_{k}\nabla_{l})$
$=-\epsilon_{ikl}\epsilon_{jmn}[x_{k}\nabla_{l}(x_{m}\nabla_{n})-x_{m}\nabla_{n}(x_{k}\nabla_{l})$
$=-\epsilon_{ikl}\epsilon_{jmn}[x_{k}x_{m}\nabla_{l}\nabla_{n}+x_{k}\nabla_{n}\nabla_{l}x_{m}-x_{m}x_{k}\nabla_{n}\nabla_{l}-x_{m}\nabla_{l}\nabla_{n}x_{k}]$
$=-\epsilon_{ikl}\epsilon_{jmn}[x_{k}x_{m}\nabla_{l}\nabla_{n}+x_{k}\nabla_{n}\delta_{lm}-x_{m}x_{k}\nabla_{n}\nabla_{l}-x_{m}\nabla_{l}\delta_{nk}]$
$=-\epsilon_{ikl}\epsilon_{jmn}[x_{k}x_{m}\nabla_{l}\nabla_{n}+0-x_{m}x_{k}\nabla_{n}\nabla_{l}-0]$
$=0$

2. Dec 6, 2015

### blue_leaf77

I have not reviewed my old lecture on the use of the Levi-Civita symbol so I can't tell, at least for now, whether or not you made a mistake in that symbol related operation, but I can tell that in the transition from
to
You forgot that this operator will act on an arbitrary wavefunction. Therefore you have to pretend that the derivatives will also act on the product between $x_l$ and this dummy wavefunction.

3. Dec 6, 2015

### spaghetti3451

Let me post my calculation with the wavefunction plugged in explicitly.

$[J_{i},J_{j}] \psi$
$=J_{i}J_{j}\psi-J_{j}J_{i}\psi$
$=(-i\epsilon_{ikl}x_{k}\nabla_{l})(-i\epsilon_{jmn}x_{m}\nabla_{n})\psi-(-i\epsilon_{jmn}x_{m}\nabla_{n})(-i\epsilon_{ikl}x_{k}\nabla_{l})\psi$
$=-\epsilon_{ikl}\epsilon_{jmn}[x_{k}\nabla_{l}(x_{m}\nabla_{n}\psi)-x_{m}\nabla_{n}(x_{k}\nabla_{l}\psi)$
$=-\epsilon_{ikl}\epsilon_{jmn}[x_{k}x_{m}\nabla_{l}(\nabla_{n}\psi)+x_{k}(\nabla_{n}\psi)\nabla_{l}x_{m}-x_{m}x_{k}\nabla_{n}(\nabla_{l}\psi)-x_{m}(\nabla_{l}\psi)\nabla_{n}x_{k}]$
$=-\epsilon_{ikl}\epsilon_{jmn}[x_{k}x_{m}\nabla_{l}(\nabla_{n}\psi)+x_{k}(\nabla_{n}\psi)\delta_{lm}-x_{m}x_{k}\nabla_{n}(\nabla_{l}\psi)-x_{m}(\nabla_{l}\psi)\delta_{nk}]$
$=-\epsilon_{ikm}\epsilon_{jmn}x_{k}(\nabla_{n}\psi)+\epsilon_{inl}\epsilon_{jmn}x_{m}(\nabla_{l}\psi)$
$=-\epsilon_{mik}\epsilon_{mnj}x_{k}(\nabla_{n}\psi)+\epsilon_{nli}\epsilon_{njm}x_{m}(\nabla_{l}\psi)$
$=-(\delta_{in}\delta_{kj}-\delta_{ij}\delta_{kn})x_{k}(\nabla_{n}\psi)+(\delta_{lj}\delta_{im}-\delta_{lm}\delta_{ij})x_{m}(\nabla_{l}\psi)$
$=-x_{i}(\nabla_{j}\psi)+x_{k}(\nabla_{k}\psi)+x_{i}(\nabla_{j}\psi)-x_{k}(\nabla_{k}\psi)$
$=0$

Where's the mistake now?

4. Dec 6, 2015

### PeroK

I would simply do it explicitly for $J_1$ and $J_2$ and use symmetry of the cross product. Yes, you should be able to crank it out using the L-C and delta symbols, but somehow it always seems to go wrong!

5. Dec 6, 2015

### spaghetti3451

I know I can do it in your way, but I am trying to practice my Levi-Civita and tensor manipulations, hence I would like to do it in the hard way.

6. Dec 6, 2015

### PeroK

Expand it out fully (for $J_1$ and $J_2$, say) and check each line of the symbolic solution to each line of the specific solution to find where you've gone wrong.

I would get rid of the $-i$ as well. You can always put that back in at the end.

7. Dec 6, 2015

### blue_leaf77

In the upper line $\delta_{ij} = 0$ if $i \neq j$, and also check again the first term in the second line, you made a mistake there.

8. Dec 6, 2015

### samalkhaiat

$$[J_{i}, J_{m}] = \epsilon_{ijk} \ \epsilon_{mnr}[x_{j} \ p_{k}, x_{n} \ p_{r}] .$$
To avoid confusion, always use the identity
$$[AB,C] = A[B,C] + [A,C]B$$
Applying this to the bracket on the RHS together with the fundamental commutation relations,
$$[x_{i},x_{j}] = [p_{i},p_{j}] = 0, \ \ \ [x_{i},p_{j}] = i \delta_{ij} ,$$
you get
$$[x_{j} \ p_{k},x_{n} \ p_{r}] = -i \delta_{nk} \ x_{j} \ p_{r} + i \delta_{jr} \ x_{n} \ p_{k} .$$
Thus
\begin{align*} [J_{i},J_{m}] &= i\epsilon_{ijk} \ \epsilon_{mrk} \ x_{j} p_{r} - i \epsilon_{ikj} \ \epsilon_{mnj} \ x_{n} p_{k} \\ &= ix_{j} p_{r} \left( \delta_{im} \delta_{jr} - \delta_{ir} \delta_{jm} \right) - ix_{n} p_{k} \left( \delta_{im} \delta_{kn} - \delta_{in} \delta_{km} \right) \\ &= i\left( x_{r} \ p_{r} \ \delta_{im} -x_{m} \ p_{i} \right) - i \left( x_{k} \ p_{k} \ \delta_{im} - x_{i} \ p_{m} \right) \\ &= i \left( x_{i} \ p_{m}-x_{m} \ p_{i} \right) \\ &=i \ \epsilon_{imn} \ J_{n} . \end{align*}
You can also use the Poisson Bracket
$$\big\{ J_{i},J_{m} \big\} = \frac{\partial J_{i}}{\partial x_{l}} \frac{\partial J_{m}}{\partial p_{l}} - \frac{\partial J_{m}}{\partial x_{l}}\frac{\partial J_{i}}{\partial p_{l}} .$$
Using
$$\frac{\partial J_{i}}{\partial x_{l}} = \epsilon_{ilk}p_{k} , \ \ \frac{\partial J_{i}}{\partial p_{l}} = \epsilon_{ijl}x_{j} ,$$
you get
\begin{align*} \big\{ J_{i},J_{m} \big\} &= \left( \epsilon_{mnl} \epsilon_{ikl} - \epsilon_{inl} \epsilon_{mkl} \right) x_{k} \ p_{n} \\ &= x_{i} \ p_{m} -x_{m} \ p_{i} \\ &= \epsilon_{imn} \ J_{n} . \end{align*}