I found this problem, and so far I am stumped. I was wondering if anyone wanted to solve it with me, or help me calculate eigenvectors, or just give insight on my questions.

Consider a system of two spin-1/2 particles interacting through the Hamitonian $$H = A(S_x^2 − S_y^2) + BS_z^2,$$ where $A$ and $B$ are constants and $S_x$, $S_y$ and $S_z$ are the three components of the total spin of the system. Find the energy spectrum and the corresponding eigenvectors.

It's been awhile since I did QM, and I know you can't have simultaneous eigenvectors of more than one measurement of spin along an axis, but how does that relate to this problem where there Sx^2 and Sy^2 are in the hamiltonian? I've been trying to work out 4x4 matrices, but I can't find good eigenvectors, but even then I am confused how applying a pauli matrix (say for x or y) transforms the eigenvector to another eigenvector, which I think is another example of not being able to define simultaneous eigenvectors etc. How would one start this problem? I will break it down to S^2 and Sz terms, but there will be remaining terms, will they just not contribute to the energy spectrum? If someone much more knowledgable than me could solve this problem easily and post it, that would clarify so much to me. For now, I'll try to read Feynman's lecture on the subject.


Here I basically do what joshphysics has already mentioned, just in a little more detail, and in a bit more intuitive basis (which makes effectively no difference). So definitely not a slick way.

I use the basis $|m_1\rangle\otimes|m_2\rangle$, where $m_i=\pm1/2$. Keeping only the sign we denote the basis as $\{|++\rangle,|+-\rangle,|-+\rangle,|--\rangle\}$.

Now for $S_z=S_{1z}+S_{2z}$ you can check that $$ S_z|++\rangle=\left(\frac{\hbar}{2}+\frac{\hbar}{2}\right)|++\rangle $$ and similarly $$ S_z|--\rangle=\left(-\frac{\hbar}{2}-\frac{\hbar}{2}\right)|--\rangle $$ so that you and up with the matrix $$ S_z\rightarrow\hbar\left[\begin{array}{cccc}1&0&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&-1\end{array}\right] $$


$$ S_z^2\rightarrow\hbar^2\left[\begin{array}{cccc}1&0&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&1\end{array}\right] $$

For $S_x$ and $S_y$ you use raising and lowering operators as josh suggested and obtain the matrix elements by acting on each state in the basis and seeing what has a non-zero dot product with the result, e.g.

$$ \begin{align} S_y|+-\rangle=\frac{1}{2i}\left(S_+-S_-\right)|+-\rangle&=\\\frac{1}{2i}\left(S_{1+}+S_{2+}-S_{1-}-S_{2-}\right)|+-\rangle&=\frac{1}{2i}\left(\hbar|++\rangle-\hbar|--\rangle\right) \end{align} $$

thus $\langle++|S_y|+-\rangle=\frac{\hbar}{2i}$ and $\langle--|S_y|+-\rangle=-\frac{\hbar}{2i}$. Once you do this you can get

$$ S_x\rightarrow\hbar\left[\begin{array}{cccc}0&1&1&0\\1&0&0&1\\1&0&0&1\\0&1&1&0\end{array}\right] $$


$$ S_y\rightarrow\hbar\left[\begin{array}{cccc}0&1&1&0\\-1&0&0&1\\-1&0&0&1\\0&-1&-1&0\end{array}\right] $$

Finally the Hamiltonian looks like $$ H=\left[\begin{array}{cccc}\hbar^2B&0&0&\hbar^2A\\0&0&0&0\\0&0&0&0\\\hbar^2A&0&0&\hbar^2B\end{array}\right] $$

Therefore $|+-\rangle$ and $|-+\rangle$ are eigenvectors with eigenvalue $0$, and you only need diagonalize $2\times2$ matrix to find that the eigenvalue $\hbar^2\left(B+A\right)$ corresponds to the eigenvector $\left(|++\rangle+|--\rangle\right)/\sqrt{2}$, and $\hbar^2\left(B-A\right)$ corresponds to the eigenvector $\left(|++\rangle-|--\rangle\right)/\sqrt{2}$.

  • $\begingroup$ Thank you very very much for your detailed walkthrough. I am going to work through this later tonight and post if I have any questions. $\endgroup$ – walczyk Jun 22 '13 at 16:03
  • $\begingroup$ My only question (I think) is how do you explicitly decide what column vectors represent each state? I have read before about two basis eigenvectors will have eigenvalues of zero, but then is that a justification that linear combinations of the other two states will be states of the system? Also, (h-bar taken to be unity), what is the reasoning why there is no eigenvalue (A-B)? Thank you again, I'm still working it out. $\endgroup$ – walczyk Jun 22 '13 at 16:14
  • $\begingroup$ You're welcome. In principle your only constraint is to choose a ortho-normal basis; the ordering of the vectors is your free choice (I hope I understood your question correctly). $\endgroup$ – mgphys Jun 22 '13 at 18:57
  • $\begingroup$ The fact that you have zero eigenvalues doesn't have much to do with the fact that you need to diagonalize the $\{|++\rangle,|--\rangle\}$ subspace. For example if you would calculate $S^2$ you would find no zero eigenvalues, but would again need to diagonalize a subspace, this time $\{|+-\rangle,|-+\rangle\}$. My reasoning for the lack of the eigenvalue $A-B$ is the lack of the symmetry of the Hamiltonian and the appropriate eigenvector. Anyhow I recommend you take a look at Shankar's Principle of Quantum Mechanics, Section 15 - Addition of Angular Momenta it might be helpful. $\endgroup$ – mgphys Jun 22 '13 at 19:00
  • $\begingroup$ I will find a digital copy of that book and read that section, when you say diagonalize the subspace does that mean generate a basis that spans the subspace? I am unfamiliar with the concept of diagonalizing anything more than a matrix! Also, what is the physical meaning of $\langle--|S_y|+-\rangle=-\frac{\hbar}{2i}$ ? I thought expectation values need to have the same bra and ket eigenvector, and if they are different eigenvectors that represents a transition probability? I definitely have a lot of reading to do tonight! Thank you again. $\endgroup$ – walczyk Jun 22 '13 at 19:17

It's not clear to me at this point that there is a tricky way to do this problem, but here is a systematic way that would work for any hamiltonian written in terms of the components of the total spin operator.

As you have indicated, the Hamiltonian acts on a $4$-dimensional Hilbert space $\mathcal H$, so it can be written as a $4$-by-$4$ matrix in a given basis. The eigenvalues of the matrix don't depend on the basis you choose, so provided you can determine the matrix representation of $H$ in some basis, you can just find the eigenvalues of this matrix, and you'll be done.

Is there a convenient basis to choose? Well, I'd say that the basis $|s, m\rangle$ consisting of simultaneous eigenvectors of $\mathbf S^2$ and $S_z$, the total spin squared and total $z$-component of spin operators, is best because we'll easily be able to compute the matrix elements of the Hamiltonian in this basis as you'll see in a moment; $$ \mathbf S^2|s,m\rangle = \hbar^2s(s+1)|s,m\rangle, \qquad S_z |s,m\rangle = \hbar m|s,m\rangle $$ Since the system consists of two spin-$1/2$ particles, this basis contains three spin-$1$ states and one spin-$0$ state; $\{|1,1\rangle,|1,0\rangle, |1,-1\rangle, |0,0\rangle\}$ . These are the so-called triplet and singlet states that arise from appropriately "adding" the two spins quantum mechanically.

Now, of course the $S_z^2$ term is particularly easy to deal with in this basis, in fact, the matrix representation of $S_z^2$ in this basis is diagonal. The trick is to therefore compute the matrix representation of $S_x^2 - S_y^2$ in this basis. This can be done systematically using raising and lowering operators. Recall that $$ S_+ =S_x + iS_y, \qquad S_- = S_x - iS_y $$ so that $$ S_x = \frac{1}{2}(S_+ + S_-), \qquad S_y = \frac{1}{2i}(S_+ - S_-) $$ and therefore for example $$ S_x^2 = \frac{1}{4}(S_+^2 + 2S_+S_- + S_-^2) $$ and similarly for $S_y^2$. Now you just need to use these expressions to determine the matrix representation of the Hamiltonian in this basis and find its eigenvalues.

I strongly suspect that there is a slicker way of doing this problem, but it's just not coming to me at the moment.

  • $\begingroup$ Thanks for your solution, I think what was holding me back from explicitly writing 4x4 matrices is choosing an eigenvector basis that makes sense. For isntance how do you decide what column matrix |+-> should be represented by? $\endgroup$ – walczyk Jun 22 '13 at 16:08
  • $\begingroup$ @walczyk If you are only looking for the spectrum of this Hamiltonian, then you don't need the column matrix representation of any vector to do this since you only need the matrix representation of the Hamiltonian in the given basis whose entries can be computed (for the basis I refer to) as $\langle m,s|H|m',s'\rangle$. As a secondary issue, however, if you are given any vector $|\psi\rangle$ in the Hilbert space, then its column vector representation in the basis I gave would just have as entries the four numbers $\langle m,s|\psi\rangle$. $\endgroup$ – joshphysics Jun 22 '13 at 17:59
  • $\begingroup$ Sorry the last symbol in that response should have been written as $\langle s, m|\psi\rangle$. $\endgroup$ – joshphysics Jun 22 '13 at 18:14
  • $\begingroup$ you said $\langle s,m | \psi \rangle$ is represented by four numbers? s would be 1 in this case, and m would be -1, 0, or 1 no? And then the other two numbers would basically be a spinor ? Sorry, I know that an explicit column vector is unncessary, but I guess I like the explicit nature of matrix algebra, so I am having trouble giving up explicit eigenvectors in the derivation. $\endgroup$ – walczyk Jun 22 '13 at 19:07
  • $\begingroup$ Given two spin-$1/2$ particles, the possible combinations of $(s, m)$, are $(1,1),(1,0),(1,-1),(0,0)$; there are four of them. This gives four possible values for the inner product $\langle s,m|\psi\rangle$, and these four numbers are the entries of $|\psi\rangle$ in the $|s,m\rangle$ basis. $\endgroup$ – joshphysics Jun 22 '13 at 22:43

If one is able to have an explicit 4 * 4 matrix for the hamiltonian, the work to find eigenvalues and eigenvectors is a basic math problem. So I will suppose you know that.

I will answer to how get this hamiltonian matrix.

1) a basis, for the 2-spin state is $|0>\otimes|0>, ~|0>\otimes|1>, ~|1>\otimes|0>, ~|1>\otimes|1>$

2) The meaning of $S_i$ is in fact $S_i = ((s_1)_i \otimes \mathbb{Id_2} + \mathbb{Id_1} \otimes (s_2)_i)$,

where $(s_1)_i$ is a $2 * 2$ matrix (spin operator) acting on the first particle, and $(s_2)_i$ is a $2 * 2$ matrix (spin operator) acting on the second particle, and where $\mathbb{Id_1}$ and $\mathbb{Id_2}$ are $2 * 2$ matrix acting respectively on particules #1 and #2.

Note, then you have $(s_1)_i = (s_2)_i = \frac{1}{2} \sigma_i$, where $\sigma_i$ is the Pauli matrice.

Note, then, that you have $((s_1)_i)^2 = ((s_2)_i)^2 = \frac{1}{4} \mathbb{Id}$

3) Now, you can simplify the hamiltonian : $H = A(S_x^2 − S_y^2) + BS_z^2,$

For instance $S_x^2 = ((s_1)_x \otimes \mathbb{Id_2} + \mathbb{Id_1} \otimes (s_2)_x)^2 = \frac{1}{2} \mathbb{Id_1} \otimes \mathbb{Id_2} + \frac{1}{2}\sigma_x \otimes \sigma_x$.

So, you will have :

$H =\frac{B}{2} \mathbb{Id} + \frac{1}{2} (A(\sigma_x \otimes \sigma_x - \sigma_y \otimes \sigma_y) + B \sigma_z \otimes \sigma_z)$

Here $\mathbb{Id} = \mathbb{Id_1} \otimes \mathbb{Id_2}$ is the $4*4$ identity matrix.

4) To get the hamiltonian matrix, now just apply H to the basic vectors, for instance :

$$H ~(|0>\otimes|0>) = \frac{B}{2} |0>\otimes|0> + \frac{1}{2}(A(|1>\otimes|1> - i^2 |1>\otimes|1>) + B|0>\otimes|0>)$$

That is : $$H ~(|0>\otimes|0>) = B|0>\otimes|0> + A |1>\otimes|1>$$

By doing the same thing with the other basis vectors, you will get the full $4*4$ hamiltonian matrix.

Note: if you understand the notation $|0>\otimes|0>$, you can use the notation $|0>|0>$ which is simpler.

  • $\begingroup$ Hi! Thanks for your answer, I need to use tensor products more frequently to get a better understanding of it. I will work through this tonight. $\endgroup$ – walczyk Jun 22 '13 at 16:02
  • $\begingroup$ There is no difficulty with the tensorial product.The basic idea, when you have a operator $A \otimes B$ acting on a state $|a> \otimes |b>$, the result is :$A|a> \otimes B|b>$. And you may consider the basis $|0>\otimes|0>, ~|0>\otimes|1>, ~|1>\otimes|0>, ~|1>\otimes|1>$ like any standard orthonormal basis. $\endgroup$ – Trimok Jun 22 '13 at 18:05
  • $\begingroup$ If you make an operator product, it gives simply $(A \otimes B) * (C \otimes D) = (AC \otimes BD) $ $\endgroup$ – Trimok Jun 22 '13 at 18:12
  • $\begingroup$ Interesting. The issue I have with representing tensor products with matrices is how elements that refer to particle #1 are distinguished from the elements that refer to particle #2. Is it implicit in the design of the matrix? That the 4x4 identity matrix, while it looks like a standard identity matrix, actually represents two 2x2 identity matrices. I think I need to practice going from tensor products to explicit matrix representations and seeing how the elements stay separated. The tensor product in this case would be a kronecker product (direct product) right? $\endgroup$ – walczyk Jun 22 '13 at 19:22
  • $\begingroup$ The formalism and rules I gave you, is precisely the way to not consider in detail how the tensorial matrix is built, and so, it is simpler. However, If you want to manipulate matrix directly, let me give you an example : For instance suppose a $2*2$ matrix $A = \left[\begin{array}{cccc}a_{11} &a_{12}\\\\a_{21}&a_{22}&\end{array}\right]$, and a $2*2$ matrix $B$. What is $A \otimes B$? It is simply the $4*4$ matrix: $\left[\begin{array}{cccc}a_{11} B&a_{12}B\\\\a_{21}B&a_{22}B&\end{array}\right]$ $\endgroup$ – Trimok Jun 22 '13 at 19:34

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