Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

On page 203 3rd edition of Schiff we are given the angular momentum matrices ${J}$ for $j=1$.

I am curious as to how these relate to orbital angular momentum for $j = 1$. If we take the corresponding 3x3 matrices for spin given on page 198. Lets just use

$$S_x = i \hbar \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \\ \end{pmatrix}\,\,\,\mathrm{and}\,\,\,S_y = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \\ \end{pmatrix}$$

With

$$J_x = \frac{\hbar}{\sqrt{2}} \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \\ \end{pmatrix}\,\,\,\mathrm{and}\,\,\,J_y = \frac{\hbar}{\sqrt{2}}\begin{pmatrix} 0 & -i & 1 \\ i & 0 & -i \\ 0 & i & 0 \\ \end{pmatrix}$$

Using $\vec{J} = \vec{L} + \vec{S}$ we can solve for $L_x$ and $L_y$ by subtracting $J$ from $S$.

$$L_x = J_x-S_x = \begin{pmatrix} 0 & \frac{\hbar}{\sqrt{2}} & 0 \\ \frac{\hbar}{\sqrt{2}} & 0 & i\hbar + \frac{\hbar}{\sqrt{2}} \\ 0 & -i\hbar + \frac{\hbar}{\sqrt{2}} & 0 \\ \end{pmatrix} $$ $$ L_y = J_y-S_y = \begin{pmatrix} 0 & -\frac{i\hbar}{\sqrt{2}} & -i \hbar \\ \frac{i \hbar}{\sqrt{2}} & 0 & -\frac{i\hbar}{2} \\ i \hbar & \frac{i \hbar}{\sqrt{2}} & 0 \\ \end{pmatrix}$$

$$L_z = J_z - S_z = \begin{pmatrix} \hbar & i \hbar& 0\\ -i \hbar & 0 & 0 \\ 0 & 0 & -\hbar \\ \end{pmatrix}$$ My question is that when I do this, they don't obey the standard commutation relations $[L_i,L_j] = i \hbar \epsilon_{ijk} L_k$. More curious is the 2x2 case for $j=1/2$ on page 203 again, you get null orbital angular momentum matrices. What went wrong here and what am I missing?

share|improve this question
    
Hi John M. If you haven't already done so, please take a minute to read the definition of when to use the homework tag, and the Phys.SE policy for homework-like problems. –  Qmechanic Oct 25 '13 at 0:19
add comment

2 Answers 2

up vote 3 down vote accepted

Your mistake is taking $\vec{J}=\vec{L}+\vec{S}$.

A more complete statement is $\vec{J} = \vec{L}\otimes\mathbb{1}_S + \mathbb{1}_L\otimes\vec{S}$. The operators $\vec{L}$ and $\vec{S}$ act on independent components of the Hilbert space of the particle. The complete Hilbert space is given as the direct product of space and spin components: $\mathcal{H}\simeq\mathcal{H}_L\otimes\mathcal{H}_S$.

In short, the computation that you describe has no meaning.

share|improve this answer
add comment

See the discussion at the top of page 205. The two sets of matrices are "essentially the same", related "by a unitary transformation that merely has the effect of regrouping the components of the vector wave function".

In group-speak, the two sets of matrices are different representations of the same Lie algebra, with the same dimension (3), but taken with respect to different bases.

Btw, does your copy of Schiff have an unfortunate type-setting error on page 199, where 4 lines of text ("of the infinitesimal ... the identity element") are repeated (once above "Commutation Relations for the Generators", and again below)? I'd love to know the correct wording...

share|improve this answer
    
Yes I also have the typo. –  John M Oct 29 '13 at 19:35
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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