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I would like to describe the following situation:

We have two spin systems: Spin 1 ($S_1$) and Spin 1/2 ($S_2$).

Now imagine you somehow change their interaction so that you can fine-tune the coupling $J$ between them in the form:

$$H = \mathbf{S_1} \cdot \mathbf{J_{12}} \cdot \mathbf{S_2}$$

where $\mathbf{J}$ is a matrix describing this interaction.

Now my question is how do I write this in matrix form in order to calculate the different eigenstates of this coupled system for different coupling strenghts $J$?

Should I assume a spin 3/2 system (4x4 Matrix) or an entangled Hilbert space with spin 1/2 and spin 1 (6x6 Matrix)?

Also, what if I still want to include effects on the spin 1 system such as Zeeman splitting in a magnetic field $B_z$, how could I include this?


Update

So let's make the situation a bit more simple, just a magnetic field $B_z$ acting on the spin-1 and only an isotropic ferromagnetic coupling between the spin-1 and the spin-1/2:

$$H = g\mu_B * B_z * S_z + J * \mathbf{S_{Spin1}} \cdot \mathbf{S_{Spin1/2}}$$

So I know my spin matrices for the spin-1/2 (Pauli matrices) and for the spin-1. My approach now would be to take the tensorproduct of these operators to create the new operators for the above Hamiltonian, i.e.:

$$S_x^{both} = S_x^{spin1} \otimes S_x^{spin1/2}$$ as well as for $y$ and $z$.

With these I construct the new Hamiltonian, I think these operators are correct for the spin coupling term, for the magnetic field B_z that should only act on the spin-1 I need to project it on the subspace of the spin-1 system I think?

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  • $\begingroup$ Dear Matthias. I plan to come back to my answer to tidy it a bit and make it more general some time. I have made some notes at the end as to how you might tackle the problem: I suspect the best way is to use Schur's lemma in some way. You could also find the subspace which is common to all three of the nullspaces of the three $36\times 36$ matrices $1_{36\times 36}\otimes \Sigma_j - \Sigma_j^T\otimes 1_{36\times 36}$ in Mathematica or Matlab, but I suspect there is a much eleganter method. $\endgroup$ – WetSavannaAnimal Apr 16 '15 at 1:03
  • $\begingroup$ Also, please add your own answer if you work it out: I'm actually quite interested in this myself now. My answer could probably reformulate the question so that it could be asked on Maths SE. $\endgroup$ – WetSavannaAnimal Apr 16 '15 at 1:05
  • $\begingroup$ Dear Rod, thanks for the detailed answer but I'm afraid that this is a bit too complicated for me. I thought it must be easier by making a few simplifications, e.g. we only care about the isotropic part of the coupling and assume ferromagnetic coupling. So let's say in this case I want to apply an external magnetic field B_z to the spin-1 system and the two spins are connected by $H = J * \vec{S_1} \cdot \vec{S_2}$. So our total Hamiltonian would be: $H = g\mu B_z*S_z + J * \vec{S_1} \cdot \vec{S_2}$. Can't I just take the tensorproducts of the individual operators and that's it? $\endgroup$ – Mike May 16 '15 at 17:05
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I haven't thought about this one before, so here is an approach that will work if you work hard enough at it.

Before I begin banging on, point number 1:

Should I assume a spin 3/2 system (4x4 Matrix) or an entangled Hilbert space with spin 1/2 and spin 1 (6x6 Matrix)?

Unquestionably the latter. It is a bipartite system and its state space is the tensor product of the two particle spaces. It simply cannot be anything else.

The basic principle here is conservation of angular momentum, so your basic procedure to solve your problem is:

  1. Work out the matrices for the observables for the three nett angular momentum components (the three nett angular momentum operators);

  2. Find the most general Hamiltonian which commutes for all of these three as commutation with the Hamiltonian is equivalent to invariance with time of all the moments of probability distributions of the measurements.

Part 1: The Three Angular Momentum Operators

The $x$-AM component observable for the spin half particle,

$$\sigma_x=\left(\begin{array}{cc}0&1\\1&0\end{array}\right)$$

has AM eigenvectors:

$$\psi_+=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\1\end{array}\right);\quad\psi_-=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\-1\end{array}\right)$$

and AM eigenvalues $\lambda_+=+\frac{1}{2}$ and $\lambda_-=-\frac{1}{2}$, respectively.

The $x$-AM component observable for the spin 1 particle,

$$S_x = \left(\begin{array}{ccc}0&0&0\\0&0&i\\0&-i&0\end{array}\right)$$

has AM eigenvectors:

$$\Psi_+=\frac{1}{\sqrt{2}}\left(\begin{array}{c}0\\i\\1\end{array}\right);\quad\Psi_-=\frac{1}{\sqrt{2}}\left(\begin{array}{c}0\\1\\i\end{array}\right);\quad\Psi_0=\left(\begin{array}{c}1\\0\\0\end{array}\right)$$

and AM eigenvalues $\Lambda_+=+1$, $\Lambda_-=-1$ and $\Lambda_0=0$, respectively. So now, for the two particle system, the six $x$-AM eigenstates are:

  1. $\psi_+\otimes\Psi_+$ with AM eigenvalue $\frac{1}{2}+1=\frac{3}{2}$
  2. $\psi_+\otimes\Psi_0$ with AM eigenvalue $\frac{1}{2}+0=\frac{1}{2}$
  3. $\psi_+\otimes\Psi_-1$ with AM eigenvalue $\frac{1}{2}-1=-\frac{1}{2}$
  4. $\psi_-\otimes\Psi_+$ with AM eigenvalue $-\frac{1}{2}+1=\frac{1}{2}$
  5. $\psi_-\otimes\Psi_0$ with AM eigenvalue $-\frac{1}{2}+0=-\frac{1}{2}$
  6. $\psi_-\otimes\Psi_-1$ with AM eigenvalue $-\frac{1}{2}-1=-\frac{3}{2}$

and so, if we order the eigenstates as above, the eigenvectors as columns are $\mathrm{vec}(\psi_+\otimes\Psi_+),\,\mathrm{vec}(\psi_+\otimes\Psi_0)\cdots$ (see the Wikipedia Vectorization Page) and so at last we get as the total $x$-AM component observable $\Sigma_X = P_X \Lambda_X P_X^\dagger$ where

$$P_X=\left( \begin{array}{cccccc} 0 & \frac{1}{\sqrt{2}} & 0 & 0 & \frac{1}{\sqrt{2}} & 0 \\ 0 & \frac{1}{\sqrt{2}} & 0 & 0 & -\frac{1}{\sqrt{2}} & 0 \\ \frac{i}{2} & 0 & \frac{1}{2} & \frac{i}{2} & 0 & \frac{1}{2} \\ \frac{i}{2} & 0 & \frac{1}{2} & -\frac{i}{2} & 0 & -\frac{1}{2} \\ \frac{1}{2} & 0 & \frac{i}{2} & \frac{1}{2} & 0 & \frac{i}{2} \\ \frac{1}{2} & 0 & \frac{i}{2} & -\frac{1}{2} & 0 & -\frac{i}{2} \\ \end{array} \right)$$

and $\Lambda_X =\mathrm{diag}\left(\frac{3}{2},\,\frac{1}{2},\,\frac{-1}{2},\,\frac{1}{2},\,\frac{-1}{2},\,\frac{3}{2}\right)$. The result is:

$$\Sigma_X=\left( \begin{array}{cccccc} 0 & \frac{1}{2} & 0 & 0 & 0 & 0 \\ \frac{1}{2} & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & \frac{3}{4} & -\frac{1}{4} & \frac{i}{4} & \frac{3 i}{4} \\ 0 & 0 & -\frac{1}{4} & \frac{3}{4} & \frac{3 i}{4} & \frac{i}{4} \\ 0 & 0 & -\frac{i}{4} & -\frac{3 i}{4} & \frac{3}{4} & -\frac{1}{4} \\ 0 & 0 & -\frac{3 i}{4} & -\frac{i}{4} & -\frac{1}{4} & \frac{3}{4} \\ \end{array} \right)$$

From here on it should be conceptually clear how to go, although tedious. You do the same for the $y$-AM observables:

$$\sigma_y=\left(\begin{array}{cc}0&-i\\i&0\end{array}\right)$$ $$S_y = \left(\begin{array}{ccc}0&0&-i\\0&0&0\\i&0&0\end{array}\right)$$

to find the total system $y$-AM observable $\Sigma_Y$ and for the $z$-AM observables:

$$\sigma_z=\left(\begin{array}{cc}i&0\\0&-i\end{array}\right)$$ $$S_z = \left(\begin{array}{ccc}0&i&0\\-i&0&0\\0&0&0\end{array}\right)$$

to get the total system $z$-AM observable $\Sigma_Z$.

Part 2: Find most general Hamiltonian

Your most general Hamiltonian will be defined by the three commutator relationships expressing conservation of AM:

$$[\hat{H},\,\Sigma_j]=0;\;j=X,\,Y,\,Z$$

You'll need to work out the invariant spaces of the three $\Sigma$s to do this. You'll get a linear space of possible $\hat{H}$s: in the two coupled spin half particles case there is essentially only one possible Hamiltonian that falls out of this approach and that is one proportional to $\sigma_x\otimes\sigma_x+\sigma_y\otimes\sigma_y+\sigma_z\otimes\sigma_z$ (plus a term proportional to the $4\times4$ identity matrix expressing the shift in the ground state energy) but this six dimensional case things will be a bit more complicated. Now as I said, I've never done this before, so I daresay there is a more systematic and less cumbersome way to work this all out. But any method is going to rest on the first principles expressed above.

Magnetic Field

What are the terms for the influence of the magnetic field. Well that's an easy one: in the ordering we have studied above, the uncoupled Hamiltonian will be:

$$\hat{H} = \gamma_{\frac{1}{2}}\left(\sigma_x\,B_x + \sigma_y\,B_y+ \sigma_z\,B_z\right)\otimes 1_{3\times3} + \gamma_1\,I_{2\times2}\otimes\left(S_x\,B_x + S_y\,B_y+ S_z\,B_z\right)$$

where $\gamma_{\frac{1}{2}}$ and $\gamma_1$ are the respective gyromagnetic ratios.


Notes on completing the method. You can also represent a bipartite state $\Phi=\psi\otimes\Psi$ as the literal $2\times 3$ matrix that is the outer product $\Phi=\psi Psi^T$ of the $2\times 1$ and $3\times 1$ column vectors. Then the operator on the first space act on the left and the operators on the second act on the right. So our $x$-component observable would be the linear, homogeneous transformation:

$$\Phi\mapsto \sigma_x\,\Phi\,S_x^T$$

and the vectorization operator (See Vectorization Wiki Page), which reorders our states into a $6\times 1$ column vectors as in my answer, writes this as

$$\mathrm{vec}(\Phi) \mapsto S_x\otimes\sigma_x\,\mathrm{vec}(\Phi)$$

Using the standard formula $\mathrm{vec}(A\,B\,C) = C^T\otimes A \,\mathrm{vec}(B)$. By dint of the formula $(A\otimes B)\, (C\otimes D) = (A\,C)\otimes(B\,D)$, and using the fact that inverse, complex conjugate, Hermitian conjugate and transpose operations distribute over the Kronecker produt, we can diagonalize $S_x\otimes\sigma_x$ inside the Kronecker product and find that the coupled system’s eigenstates are $\Pi_x\otimes \pi_x$, where $P_X,\,p_x$ are the matrices of eigenvectors of the individual multiplicands written as columns. So this will let you calculate $\Sigma_j,\,j=X,\,Y,\,Z$ systematically and fast.

Now to find the most general Hamiltonian, you need to find the invariant space of the group of matrices generated by the three matrices $\exp(i\,\Sigma_j)$ and find the irreducible representation of it: equivalently the smallest vector subspace of $\mathbb{C}^6$ left invariant by the group: by Schur’s lemma, any matrix commuting with all three must be proportional to the identity operator when restricted to this subspace. The scaling factor is possibly nought – i.e. the operator could possibly be the zero endomorphism. This completely characterizes the most general Hamiltionian: it can be any operator which is proportional to the identity when restricted to this irreducible subspace.

You could also find the subspace which is common to all three of the nullspaces of the three $36\times 36$ matrices $1_{36\times 36}\otimes \Sigma_j - \Sigma_j^T\otimes 1_{36\times 36}$ in Mathematica or Matlab, but I suspect there is a much eleganter method grounded on Schur's lemma!

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