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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?

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I've deleted a variety of comments on answers to this question. Please keep it civil. –  dmckee Dec 19 '11 at 23:02
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3 Answers

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.

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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 $A_j$: $[L_i,A_j]=i\hbar\epsilon_{ijk}A_k$

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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.

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