# Commutation relations of angular momentum with position, momentum.

1. Dec 5, 2011

### B L

1. The problem statement, all variables and given/known data
Using the position space representation, prove that:
$\left[L_i, x_j\right] = i\hbar\epsilon_{ijk}x_k$.
Similarly for $\left[L_i, p_j\right]$.

2. Relevant equations
Presumably, $L_i = \epsilon_{ijk}x_jp_k$.
$\left[x_i, p_j\right] = i\hbar\delta_{ij}$.

3. The attempt at a solution
$\left[L_i, x_j\right] = \epsilon_{ijk}\left[x_jp_k, x_j\right] = \epsilon_{ijk}\left(x_jp_kx_j - x_jx_jp_k\right) = \epsilon_{ijk}x_j\left(p_kx_j - x_jp_k\right) = \epsilon_{ijk}x_j\left[p_k, x_j\right] = -i\hbar\epsilon_{ijk}x_j\delta_{jk}$
which is where I become confused - it seems to me that the right hand side is always zero (if the Kronecker delta is nonzero, the Levi-Civita symbol is zero, and vice-versa).

Any help is much appreciated.

2. Dec 5, 2011

### dextercioby

You should respect the laws of tensor algebra. Don't use the same index 3 times in a tensor product.

So $[L_i,x_j]=\epsilon_{ikl}[x_k p_l, x_j] = ...$

3. Dec 5, 2011

### dextercioby

I can't edit my post ??? Ok...

It's not true that epsilon times (in a tensor way) delta =0. This happens iff there's a double contracted tensor product among them.

4. Dec 5, 2011

### B L

That's the thing: when I asked my TA, he said that we are not to assume the summation convention. That's the source of my confusion - I have no idea what this question even means without the summation convention.

(I'm literally the only person in the course who knows what a tensor is).

5. Dec 5, 2011

### dextercioby

You need the summation convention and know a little about cartesian vectors/tensors, if you're approaching the identity/equality from the all-components-at-the-same-time perspective