4
$\begingroup$

I have been trying for hours and cannot figure it out. I am not asking anyone to do it for me, but to understand how to proceed.

We have the relations $$[L_i,p_j] ~=~ i\hbar\; \epsilon_{ijk}p_k,$$ $$[L_i,r_j] ~=~ i\hbar\; \epsilon_{ijk}r_k,$$ $$[L_i,L_j] ~=~ i\hbar\; \epsilon_{ijk}L_k.$$

Now I am trying to calculate

$$[L_i,(p\times L)_j].$$

I do not know how to reduce it. When I try I am left with too many dummy indices that I don't know what to do. For example, using

$$[AB,C] = A[B,C]+[A,C]B,$$

and expanding the cross product in terms of Levi-Civita symbols

$$[L_i,\epsilon_{jmn}p_m L_n],$$

but I don't know how to proceed correctly from here. For example, I tried,

$$ =~ \epsilon_{jmn}(\;p_m[L_i,L_n]+[L_i,p_m]L_n ).$$

Is this correct? If so, in the next step I used the known commutation relations

$$ =~i\hbar\; \epsilon_{jmn}(\; \epsilon_{ink}p_mL_k+\epsilon_{imk}p_kL_n).$$

Once again, I am stuck and do not know how to evaluate this further. Could someone tell me what I am doing wrong, or if not, how to proceed?

$\endgroup$
  • 1
    $\begingroup$ I've deleted a variety of comments on answers to this question. Please keep it civil. $\endgroup$ – dmckee Dec 19 '11 at 23:02
6
$\begingroup$

You should use the Levi-Civita reduction formula

$$\epsilon_{ijk}\epsilon_{ilm}=\delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl}$$

and using the fact that $(a\times b)_i=\epsilon_{ijk}a_jb_k$ you should be done.

$\endgroup$
3
$\begingroup$

It is possible to continue OP's calculation as follows

$$ =~i\hbar\; \epsilon_{jmn}(\epsilon_{ink}p_m L_k+\epsilon_{imk}p_k L_n) ~=~-i\hbar( \epsilon_{jkm} \epsilon_{min}+\epsilon_{kim} \epsilon_{mjn})p_k L_n $$ $$ \stackrel{\rm Jac.Id.}{=}~i\hbar\; \epsilon_{ijm} \epsilon_{mkn}p_k L_n ~=~i\hbar\; \epsilon_{ijm}(p\times L)_m ,$$

where one uses the Jacobi identity

$$\sum_{{\rm cycl.}~i,j,k} \epsilon_{ijm} \epsilon_{mkn}~=~0 $$

for the Levi-Civita symbols.

$\endgroup$
0
$\begingroup$

Let's see:

$\textbf{L}=\textbf{r}\times \textbf{p}$

$\textbf{p}\times\textbf{L}=\textbf{p}\times (\textbf{r}\times \textbf{p})=\textbf{p}^2 \textbf{r}-(\textbf{p}\cdot \textbf{r})\textbf{p}$

The second term is not zero ($\textbf{p}\cdot \textbf{r}\neq 0$) as in classical mechanics because of $[x_j,p_j]=i\hbar$. Therefore

$[L_i,(\textbf{p}\times\textbf{L})_j]=[L_i,\textbf{p}^2r_j]-[L_i,(\textbf{p}\cdot \textbf{r}) p_j]$

However

$[L_i,\textbf{p}^2r_j]=\textbf{p}^2[L_i,r_j]=i\hbar \textbf{p}^2\epsilon_{ijk}r_k$

because $[L_i,\textbf{p}^2]=[L_i,p_m p_m]=[L_i,p_m]p_m+p_m[L_i,p_m]=i\hbar\epsilon_{ijk}(p_k p_m+p_m p_k)=0$

and

$[L_i,(\textbf{p}\cdot \textbf{r}) p_j]=[L_i,p_m r_m p_j]=[L_i,p_m]r_m p_j+p_m[L_i,r_m]p_j+p_m r_m[L_i,p_j]$

but $[L_i,p_m]r_m p_j+p_m[L_i,r_m]p_j=i\hbar\epsilon_{iml}p_l r_m p_j+i\hbar\epsilon_{iml} p_m r_l p_j=0$

and $p_m r_m[L_i,p_j]=i\hbar(\textbf{p}\cdot \textbf{r})\epsilon_{ijk}p_k$

Therefore

$[L_i,(\textbf{p}\times\textbf{L})_j]=i\hbar \textbf{p}^2\epsilon_{ijk}r_k-i\hbar(\textbf{p}\cdot \textbf{r})\epsilon_{ijk}p_k=i\hbar\epsilon_{ijk}[\textbf{p}^2r_k-(\textbf{p}\cdot \textbf{r})p_k]=i\hbar\epsilon_{ijk}(p\times L)_k$

This is true for any $A_j$: $[L_i,A_j]=i\hbar\epsilon_{ijk}A_k$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.