2
$\begingroup$

Is it possible to derive: \begin{equation} \hat{L}=\hat{r}\times \hat{p} \end{equation} from the angular momentum algebra: \begin{equation} [\hat{L}_i,\hat{L}_j]=i\ \hbar\ \epsilon_{ijk}\hat{L}_k\ ? \end{equation} I have read somewhere that commutation relations of the form \begin{equation} [a_i,b_j]=\epsilon_{ijk} c_k \end{equation} admit a "natural rewriting in terms of cross products", but there weren't any details about this statement.

$\endgroup$
6
$\begingroup$

It's not possible to derive the orbital angular momentum $L = r \times p$ from the $\mathfrak{so}(3)$ commutation relations alone, since the spin operator $S$ also fulfills the same commutation relations, but certainly is different from $r \times p$.

$\endgroup$
  • $\begingroup$ That's exactly what I would have written, but I don't know if more needs to be said to explain why or how we know that spin is different. $\endgroup$ – Timaeus May 2 '15 at 23:41
6
$\begingroup$

I have read somewhere that commutation relations of the form \begin{equation} [a_i,b_j]=\epsilon_{ijk} c_k \end{equation} admit a "natural rewriting in terms of cross products", but there weren't any details about this statement.

This "natural rewriting" of the canonical commutation relations for angular momenta in term of cross products is: $$ \vec J \times \vec J = i\vec J $$

I.e., the "natural rewriting" refers to rewriting the commutation relations themselves in terms of a cross product, not rewriting a single angular momentum in terms of cross products of other quantities.

Explicitly, for angular momentum: $$ \vec J \times \vec J = i\vec J $$ Thus: $$ \epsilon_{ijk}J_jJ_k=iJ_i $$ Thus: $$ \epsilon_{ilm}\epsilon_{ijk}J_jJ_k=i\epsilon_{ilm}J_i $$ Thus: $$ J_lJ_m-J_mJ_l=i\epsilon_{ilm}J_i $$ I.e.: $$ [J_l,J_m]=i\epsilon_{lmi}J_i $$

In terms of your "a", "b", and "c", the analogous "natural rewriting" is: $$ \vec a\times \vec b + \vec b \times \vec a=2\vec c $$

You can see this by considering: $$ (\vec a \times \vec b)_i =\epsilon_{ijk}a_jb_k=\epsilon_{ijk}(b_ka_j+[a_j,b_k]) $$ $$ =-\epsilon_{ikj}b_ka_j+\epsilon_{ijk}[a_j,b_k] $$ $$ =-(\vec b \times \vec a)_i+\epsilon_{ijk}\epsilon_{jkl}c_l $$ $$ =-(\vec b\times\vec a)_i+2\delta_{il}c_l $$ $$ =(-\vec b\times \vec a +2c)_i $$

$\endgroup$
  • $\begingroup$ How is $[a_l,b_m]=a_lb_m-a_mb_l$? $\endgroup$ – Nephente May 2 '15 at 17:48
  • $\begingroup$ whoops, good question... screwed that up somehow... Let me fix it. $\endgroup$ – hft May 2 '15 at 17:49
0
$\begingroup$

$$\begin{align} \left[\hat{A}_{i}, \hat{B}_{j} \right] & = \epsilon_{ijk}\hat{C}_{k}, \\[3mm] \hat{A}_{i}\hat{B}_{j} - \hat{B}_{j}\hat{A}_{i} & = \epsilon_{ijk}\hat{C}_{k}, \\[3mm] \epsilon_{ijn}\left( \hat{A}_{i}\hat{B}_{j} - \hat{B}_{j}\hat{A}_{i}\right) & = \epsilon_{ijn}\epsilon_{ijk}\hat{C}_{k}, \\[3mm] \epsilon_{ijn}\hat{A}_{i}\hat{B}_{j} - \epsilon_{ijn}\hat{B}_{j}\hat{A}_{i} & = 2\hat{C}_{n}, \\[3mm] \left(\hat{\vec{A}} \times \hat{\vec{B}}\right)_{n} + \left(\hat{\vec{B}} \times \hat{\vec{A}}\right)_{n} & = 2\hat{C}_{n}, \\[3mm] \frac{1}{2}\left(\hat{\vec{A}} \times \hat{\vec{B}} + \hat{\vec{B}} \times \hat{\vec{A}} \right) & = \hat{\vec{C}}. \end{align}$$ due to the fact that $\epsilon_{ijn}\epsilon_{ijk} = 2\delta_{nk}$.

$\endgroup$
  • $\begingroup$ What is wrong with this derivation? $\endgroup$ – WoofDoggy May 3 '15 at 8:15
  • $\begingroup$ Besides the fact that sometime you explicitly write the summations and sometimes they are implicit, there's nothing wrong with the derivation. Probably people are down-voting because there is no explanatory text to go along with the equations... or maybe the explicit/implicit thing is confusing... $\endgroup$ – hft May 3 '15 at 20:15

protected by Qmechanic May 3 '15 at 0:18

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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