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Just wondering if there is a faster or better way of doing this question.

I have 3 matrices: $A = {1\over2}\hbar\left( \begin{smallmatrix} 0&-i\\ i&0 \end{smallmatrix} \right), B = {1\over2}\hbar\left( \begin{smallmatrix} 0&1\\ 1&0 \end{smallmatrix} \right), C = {1\over2}\hbar\left( \begin{smallmatrix} 1&0\\ 0&-1 \end{smallmatrix} \right), $

Let $D = A^2+B^2+C^2$ and that $v$ is an eigenvector of $C$, (with eigenvalue $-{1\over2}\hbar$).

Question: Show that $v$ is an eigenvector of $D$ with eigenvalue ${1\over2}({1\over2}+1)\hbar^2$.

My take: $D={3\over4}\hbar^2\left( \begin{smallmatrix} 1&0\\ 0&1 \end{smallmatrix} \right)$. $Cv=-{1\over2}\hbar v \implies C^2v={1\over4}\hbar^2 v$.

Since $C^2={1\over4}\hbar^2\left( \begin{smallmatrix} 1&0\\ 0&1 \end{smallmatrix} \right)=B^2=A^2$, Therefore $Dv = 3\times {1\over4}\hbar^2v$. So we are done.

Though I do get the numerical answer, it is not immediately in the form asked for. Could anyone suggest another way of doing this? Thank you.

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Are you aware that these matrices describe spin $1/2$, are proportional to the Pauli matrices and form a basis of the Lie algebra $\mathfrak{su}(2)$ of $SU(2)$? If you can use the machinery associated with that, the result is immediate because an eigenvector of $C$ must have eigenvalue $s(s+1)\hbar^2$ with $s=1/2$ under the squared total spin operator $D$. –  joriki Sep 4 '11 at 7:37
    
@joriki: That sounds very interesting! :) Unfortunately I have not met spin before... Would you mind enlightening me on that? I am acquainted with the fundamental Schrodinger equation, but not with Pauli matrices or Lie algebra... –  ans Sep 4 '11 at 7:48
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I'd be keen to know how you bumped into this problem without having met spin before :-). Of course it makes sense as a purely mathematical problem, but then one wouldn't usually include a factor $\hbar$ in the matrices -- that seems to indicate that the problem is from a physics context, in which this would be part of a treatment of a spin $1/2$ particle. I'll try to explain some of this in an answer. –  joriki Sep 4 '11 at 7:52
    
@joriki: I think my teacher likes to set pure questions that he can later on point out, as a side note -- "now this is the ... equation/relation/whatnot." :-) And thanks sooo much for elaborating! –  ans Sep 4 '11 at 8:08
    
@ans: Just out of curiosity, in view of your comment, is this a probiem out of the homework?? If so, I think you ought to tag it. Just a suggestion. –  awllower Sep 4 '11 at 9:40

1 Answer 1

up vote 3 down vote accepted

The matrices $A$, $B$, $C$ are multiples of the Pauli matrices. Their commutation relations are

$$[B,A]=\mathrm i\hbar C$$

and cyclic permutations thereof. These are the commutation relations of angular momentum operators, or, mathematically speaking, of generators of $SU(2)$, which form a basis of $\mathfrak{su}(2)$, the Lie algebra of $SU(2)$. They provide an irreducible representation of $\mathfrak{su}(2)$ (the defining representation). The Casimir invariant given by the sum of the squares of the generators acts as a multiple of the identity within an irreducible representation, the factor being $j(j+1)$, where $j$ labels the irreducible representations and may be either integer or half-integer. The defining representation has $j=\frac12$, so $j(j+1)=\frac12(\frac12+1)=\frac34$. So not only eigenvectors of $v$ are eigenvectors of $D$; all vectors are eigenvectors of $D$ with the same eigenvalue. (This is only true within an irreducible representation; a reducible representation will generally contain eigenvectors of the squared angular momentum operator corresponding to different values of $j$.)

There's obviously a lot more to be said about all this; I hope I've given you some pointers what to look at if you're interested in the background of this question.

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Thanks a lot! A quick question: What are (ir)reducible representations? I have looked them up on Wiki, but it is a bit too technical. Is there a simpler/more intuitive definition? –  ans Sep 4 '11 at 8:47
    
@ans: A representation of a group $G$ assigns to every $g\in G$ an automorphism $\rho(g)$ of a vector space $V$ (or a square matrix if you prefer) such that $\rho(g_1)\rho(g_2)=\rho(g_1g_2)$ for all $g_1,g_2\in G$. A representation is reducible if $V$ can be written as the sum of two subspaces $V_1$ and $V_2$ such that $\rho(g)(V_i)=V_i$ for all $g\in G$. Basically that means that the group elements are represented by square matrices that preserve the multiplication law of $G$, and for a reducible representation there's a similarity transform that makes the matrices block-diagonal. –  joriki Sep 4 '11 at 8:59
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@ans: Every representation on a vector space over the complex numbers can be decomposed into irreducible representations, so it's enough to understand those to understand all representations. This is an area that may sound rather abstract when you first encounter it, but when you get the hang of it it can be very simple and intuitive, and it's a very powerful instrument both in math and in physics. For instance, in quantum chemistry, some results can be derived purely from symmetry using representation theory, without actually evaluating any integrals or the like. –  joriki Sep 4 '11 at 9:03
    
Thanks, joriki :-) –  ans Sep 4 '11 at 9:17

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