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Intro (you may skip this if you're an expert, I'm including this for completeness):

Say I have two bases for two systems,

The first is a spin-1/2 system $|+\rangle = \left(\begin{array}{c} 1\\0 \end{array}\right),|-\rangle=\left(\begin{array}{c} 0\\1 \end{array}\right)$

The second is a spin-1 system, with $|1_+\rangle=\left(\begin{array}{c} 1\\0\\0 \end{array}\right),|1_0\rangle=\left(\begin{array}{c} 0\\1\\0 \end{array}\right),|1_-\rangle=\left(\begin{array}{c} 0\\0\\1 \end{array}\right)$

Now for the first system, I can use the Pauli matrix

$$\hat{S_z}=\frac{1}{2}\hbar\hat{\sigma}_z = \left( \begin{array}{cc} -\frac{\hbar }{2} & 0 \\ 0 & \frac{\hbar }{2} \\ \end{array} \right)$$

in order to get the projection of my state on the z-axis. Likewise, I could use the projection matrix $$\hat{J}_z=\hbar\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \\ \end{array} \right)$$

To project the other state on the z-axis. Those operators will act on my basis in the following way:

$$\hat{S}_z|+\rangle=\frac{\hbar}{2}|+\rangle$$ $$\hat{J}_z|1_+\rangle=\hbar|1_+\rangle$$

Problem: (here comes the question)

So far everything is good! Now the problem comes when I introduce a space for the composite system, so I'm getting the basis

$$|S_z\rangle\otimes |J_z\rangle\rightarrow\left\{ |+,1_+\rangle,|+,1_0\rangle,|+,1_-\rangle,|-,1_+\rangle,|-,1_0\rangle,|-,1_-\rangle\right\}$$

Now the question is: how do I use the matrix formalism to have such operations just like I had them before in the single systems:

$$S_z|+,1_+\rangle=\frac{\hbar}{2}|+,1_+\rangle$$ $$S_z|+,1_-\rangle=\frac{\hbar}{2}|+,1_-\rangle$$ $$J_z|+,1_0\rangle=0|+,1_0\rangle$$ $$J_z|+,1_-\rangle=-\hbar|+,1_-\rangle$$

In other words, how do I write the state-kets and the operators in the composite system in matrix formalism (just like I showed in the beginning) to give results compatible with what I would expect in the examples?

Is this wrong in some way?

Every time I try to do this with Kronecker Product (like $\hat{S}_z \otimes \hat{J}_z$) I arrive at a mess, and I get terms proportional to $\hbar^2$, and I don't get the eigen-values I expect, and I'm not sure what I'm doing wrong. Could you please show me how to do this?

Thank you.

The Quantum Physicist
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    What operation are you trying to do with $\hat{S}z \otimes \hat{J}_z$? You could, potentially, do the simple projection you're talking about with $\hat{S}_z \otimes \mathbf{1}{3x3}$ and $\mathbf{1}_{2x2} \otimes \hat{J}_z$, but in general you're going to need to deal with adding the angular momenta of the states. – Flavin Nov 05 '13 at 17:05
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    A tensorial product of 2 systems creates a new system with 2 sub-systems which are independent. The total action is the sum of the actions of the sub-systems. A consequence of this is that quantities like ordinary momenta, and (total, orbital, spin) angular momenta, are then additive properties (because momenta are linear relatively to actions).

    Any additive property $P$ could then be written $P = P_1 \otimes \mathbb{Id_2} + \mathbb{Id_1} \otimes P_2$, and not $P = P_1 \otimes P_2$

     – Trimok Nov 05 '13 at 18:05
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    Read this. It's $V_1$ and $V_2$ spaces are your spin 1/2 and spin 1 systems. Pay special attention to the part where it says "the total angular momentum operators are defined by ..." If you have any questions after reading this article might want to edit your question to ask for clarification. – Brian Moths Nov 05 '13 at 18:53
  • Also @Trimok has his finger on why you're getting your "nonlinearity" ($\hbar^2$). Read his answer and my answer to a problem very like yours (not putting integer and half integer spin states together, though, but the principles are the right same). – Selene Routley Nov 06 '13 at 01:20

1 Answers1

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Just a remark on the introduction. $\hat{S}_z$ does not "project" states on the $z$ direction in spin space. Indeed $\hat{S}_z ^2 = \hbar/2 \;\text{Id}_{2\times 2}\neq \hat{S}_z$, while a projector $P$ verifies $P^2=P$. Actually, $\hat{S}_z$ rotates states around the $z$ axis in spin space ($e^{i\theta \hat{S}_z}$ rotates states by an angle $\theta$ around the $z$ axis).

Back to the main problem, I have two equivalent answers

  • The matrix elements of $\hat{S}_z\otimes \hat{J}_z$ are given by \begin{equation} \langle \pm , 1_{+,0,-}|\hat{S}_z\otimes \hat{J}_z |\pm ', 1'_{+,0,-}\rangle =\langle \pm| \hat{S}_z|\pm '\rangle \langle 1_{+,0,-}|\hat{J}_z |1'_{+,0,-}\rangle \end{equation} where $\pm$, and $\pm'$ refers to the eigenstates of $S_z$, while $1_{+,0,-}$ and $1'_{+,0,-}$ are the eigenstates of $J_z$. Hence you can compute all the matrix elements separately. For instance, in your basis the top left element of the matrix is \begin{equation} \langle + , 1_+|\hat{S}_z\otimes \hat{J}_z |+, 1_+\rangle =\langle +| \hat{S}_z|+ \rangle \langle 1_+|\hat{J}_z |1_+\rangle = \frac{\hbar}{2}\times \hbar = \frac{1}{2}\hbar^2 \end{equation} and the other elements have to be computed in the same way (most of them are zero).

  • Here you deal with direct product operators, since $\hat{S}_\alpha$ does not act on $\hat{J}_\beta$ eigenstates (whatever $\alpha$ and $\beta$) and vice versa. Thus, you get block diagonal matrices like \begin{equation} \hat{S}_z \otimes \hat{J}_z = \begin{pmatrix} \frac{\hbar}{2}J_z & 0_{3\times 3} \\ 0_{3\times 3} & -\frac{\hbar}{2}J_z \end{pmatrix}\quad . \end{equation} where $0_{3\times 3}$ is the 3 by 3 matrix with all elements set to zero and $J_z$ is the 3 by 3 matrix given by \begin{equation} J_z = \hbar \begin{pmatrix} 1 & 0 & 0 \\ 0 &0 & 0\\ 0 & 0 & -1 \end{pmatrix} \end{equation} in your basis.

The other matrices your are interested in, like $S_x\otimes J_y$ (as far as I understood) can be similarly obtained by one of those two ways.

Hope this helps!

Wajsbrot
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