Commutation relations for angular momentum operator

[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?

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

 

[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

$=-\epsilon_{ikl}\epsilon_{jmn}[x_{k}x_{m}\nabla_{l}\nabla_{n}+x_{k}\nabla_{n}\nabla_{l}x_{m}-x_{m}x_{k}\nabla_{n}\nabla_{l}-x_{m}\nabla_{l}\nabla_{n}x_{k}]$

to

$=-\epsilon_{ikl}\epsilon_{jmn}[x_{k}x_{m}\nabla_{l}\nabla_{n}+x_{k}\nabla_{n}\delta_{lm}-x_{m}x_{k}\nabla_{n}\nabla_{l}-x_{m}\nabla_{l}\delta_{nk}]$

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.

 

[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?

 

[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!

 

[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.

 

[PeroK]

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.

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.

 

[blue_leaf77]

$=-(\delta_{in}\delta_{kj}-\delta_{ij}\delta_{kn})x_{k}(\nabla_{n}\psi)+(\delta_{lj}\delta_{im}-\delta_{lm}\delta_{ij})x_{m}(\nabla_{l}\psi)$
$=-x_{i}(\nabla_{j}\psi)+x_{k}(\nabla_{k}\psi)+x_{i}(\nabla_{j}\psi)-x_{k}(\nabla_{k}\psi)$

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.

 

[samalkhaiat]

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.

$$[J_{i}, J_{m}] = \epsilon_{ijk} \ \epsilon_{mnr}[x_{j} \ p_{k}, x_{n} \ p_{r}] .$$
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
$$\begin{align*}<br /> [J_{i},J_{m}] &= i\epsilon_{ijk} \ \epsilon_{mrk} \ x_{j} p_{r} - i \epsilon_{ikj} \ \epsilon_{mnj} \ x_{n} p_{k} \\<br /> &= ix_{j} p_{r} \left( \delta_{im} \delta_{jr} - \delta_{ir} \delta_{jm} \right) - ix_{n} p_{k} \left( \delta_{im} \delta_{kn} - \delta_{in} \delta_{km} \right) \\<br /> &= i\left( x_{r} \ p_{r} \ \delta_{im} -x_{m} \ p_{i} \right) - i \left( x_{k} \ p_{k} \ \delta_{im} - x_{i} \ p_{m} \right) \\<br /> &= i \left( x_{i} \ p_{m}-x_{m} \ p_{i} \right) \\<br /> &=i \ \epsilon_{imn} \ J_{n} .<br /> \end{align*}$$
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
$$\begin{align*}<br /> \big\{ J_{i},J_{m} \big\} &= \left( \epsilon_{mnl} \epsilon_{ikl} - \epsilon_{inl} \epsilon_{mkl} \right) x_{k} \ p_{n} \\<br /> &= x_{i} \ p_{m} -x_{m} \ p_{i} \\<br /> &= \epsilon_{imn} \ J_{n} .<br /> \end{align*}$$