Consider the $l$ component of vector position $\vec{r}$, $r_l$, and the $i$ component of angular momentum $\vec{L}$, $L_i$.

We have that

$$L_i=[r\times p]_{i}=\varepsilon_{ijk}r_jp_k$$

$\varepsilon_{ijk}$ is the Levi-Civita symbol, that has the following properties:

  • $\varepsilon_{ijk}$=1 for $i=1,j=2,k=3$ and for the cyclic permutation of this indeces.
  • $\varepsilon_{ijk}$=0 for repeated indeces.
  • $\varepsilon_{ijk}$=-1 for any non cyclic permutation of $i=1,j=2,k=3$

So, for the component $x\equiv 1$ of $L$ we have $L_1=\varepsilon_{123}r_2p_3+ \varepsilon_{132}r_3p_2=r_2p_3-r_3p_2$

Analysing the commutation between $r_l$ and $L_i$:

$$[r_l, L_i]=\varepsilon_{ijk}\left(r_j[r_l,p_k]+[r_l,r_j]p_k \right)=i\hbar \varepsilon_{ijk}r_j\delta_{lk}$$

If $\varepsilon_{ijk}\delta_{lk}=\varepsilon_{ilk}$, then:

$$[r_l, L_i]=i\hbar \varepsilon_{ilk}r_j$$

Now, consider:


The problem is the following:

$$[r_l^2,L_i]=2i\hbar\varepsilon_{ijl}r_jr_l=[r\times r]_{i}=0$$

In other hand, if, for example, $l=1$ and $i=2$:

$$[r_1^2,L_2]=2i\hbar\varepsilon_{132}r_3r_1=-2i\hbar r_3r_1$$

So, I know that I misconcepted this indices notation. What I need to correct here? Is $\varepsilon_{ijk} \delta_{jl}=\varepsilon_{ilk}$ true?


1 Answer 1


The issue here is whether or not you have a sum over $l$. If you want $r_l^2$ to mean any of $r_1^2, r_2^2, r_3^2$, then when you write $r_l^2 = r_l r_l$ you should not be summing over $l$. So $[r_l^2,L_i] = 2i\hbar \epsilon_{ijl}r_jr_l$ is correct as long as you sum over $j$ but not over $l$.

On the other hand, if in $r_lr_l$ you sum over $l$ you get the $r^2=r_1^2+r_2^2+r_3^2$ operator. This operator is a scalar and so it commutes with $L_i$, as you are correctly calculating.


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