I read a textbook today on quantum mechanics regarding the Pauli spin matrices for two particles, it gives the Hamiltonian as $$ H = \alpha[\sigma_z^1 + \sigma_z^2] + \gamma\vec{\sigma}^1\cdot\vec{\sigma}^2 $$ where $\vec{\sigma}^1$ and $\vec{\sigma}^2$ are the Pauli spin matrices for two particles separately. I think $\sigma_z$ is the z component, I found that $$ \sigma_z = \left( \begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right) $$ which is 2x2 matrix. I am wondering if the $\sigma_z$ is the same for particle 1 and 2? if so,

$$ \sigma_z^1 + \sigma_z^2 = 2\left( \begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right) $$ Is that right? The most confusing part is $\vec{\sigma}^1\cdot\vec{\sigma}^2$, there are two matrices involved, so how does the dot product work? I am trying solve for the eigenvalues of H, it looks like to me that each $\sigma_z^1$ and $\sigma_z^2$ is 2x2, so there are two eigenvalues, is that correct?

up vote 16 down vote accepted

The expression you wrote for $\sigma_z^1 + \sigma_z^2$ is not quite right, but it's not surprising that you're unsure of how to proceed because the notation is somewhat obscuring the real math behind all of this. What's actually going on here is manipulations with tensor products of Hilbert spaces.

The spin state of a single spin-$\frac{1}{2}$ particle is an element of a two-dimensional Hilbert space, let's call it $\mathcal H_\frac{1}{2}$. The spin state of a system consisting of two spin-$\frac{1}{2}$ particles is $\mathcal H_\frac{1}{2}\otimes\mathcal H_\frac{1}{2}$ called the tensor product of $\mathcal H_\frac{1}{2}$ with itself.

If $|+\rangle$ and $|-\rangle$ are the usual spin up and spin down basis elements for $\mathcal H_\frac{1}{2}$, namely eigenvectors of $\sigma_z$ with eigenvalues $\pm\frac{\hbar}{2}$ $$ \sigma_z|+\rangle = \frac{\hbar}{2}|+\rangle, \qquad \sigma_z|-\rangle = -\frac{\hbar}{2}|-\rangle $$ then a basis for the tensor product $\mathcal H_\frac{1}{2}\otimes\mathcal H_\frac{1}{2}$ consists of all tensor products of basis elements of $\mathcal H_\frac{1}{2}$ of which there are $4$ in this case; $$ |+\rangle\otimes|+\rangle, \qquad |+\rangle \otimes|-\rangle, \qquad |-\rangle \otimes |+\rangle, \qquad |-\rangle \otimes|-\rangle $$ In particular, note that the dimension of the vector space of the two-particle spin $\frac{1}{2}$ system is twice the dimension of the single particle spin system.

This far, we have set the stage for defining what $\sigma^1_z$ and $\sigma^2_z$ are. The superscripts simply mean that the operator in question only acts on either the first or the second factor in a tensor product state depending on whether the superscript is a $1$ or a $2$. For example $$ \sigma_z^1(|+\rangle\otimes|-\rangle) = (\sigma_z |+\rangle)\otimes|-\rangle = \frac{\hbar}{2}|+\rangle\otimes|-\rangle $$ while $$ \sigma_z^2(|+\rangle\otimes|-\rangle) = |+\rangle\otimes(\sigma_z|-\rangle) = -\frac{\hbar}{2}|+\rangle\otimes|-\rangle $$ and in fact one writes $$ \sigma^1_z = \sigma_z\otimes I, \qquad \sigma_z^2 = I\otimes \sigma_z $$ which indicates that for example $\sigma_z^1$ acts like $\sigma_z$ in the first factor of the tensor product and like the identity matrix in the second factor, and vice versa for $\sigma_z^2$. Similarly, if we write down $$ \vec\sigma^1\cdot\vec\sigma^2 = \sigma_x^1\sigma_x^2 + \sigma_y^1\sigma_y^2+\sigma_z^1\sigma_z^2 $$ then every operator with a $1$ superscript only acts non-trivially on the first factor in a tensor product state while every operator with a superscript $2$ only acts non-trivially on the second factor in a tensor product state.

  • Upvote. I think your answer is clearer than mine :-) – Siva Apr 8 '13 at 7:00
  • Thanks. Though I am not familiar with the math but I did search the tensor product online, I think I have some idea now. May I ask that what's the main reason to perform tensor product of two Pauli matrices? I am thinking in this way: if we have two particles of no interaction inbetween (so the particles are independent). So if I combine the two Pauli matrices on the diagonal elements and the so the off-diagonal elements are ZERO. But if you use the tensor product, the resulting matrix has the off-diagonal blocks are not zero, so can I say Tensor product takes the interaction into account? – user1285419 Apr 8 '13 at 18:01
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    You might find this physics.stackexchange.com/questions/54896/… useful. I think there are other physics.SE posts regarding what the tensor product does for us in quantum. I'm not sure I'd go so far as to say that the reason we use the tensor products for composite systems is to take the interaction into account, but as you point out, it certainly does allow us to correctly take interactions between subsystems into account as an important characteristic of the construction. – joshphysics Apr 8 '13 at 19:06
  • it is great, thanks. – user1285419 Apr 9 '13 at 14:00

This is often confusing to people getting acquainted to QM and you need to stare at it for a while and convince yourself about how it works.

Firstly $\sigma^{x,y,z}$ are the Pauli spin matrices and $\vec{\sigma}_1 \cdot \vec{\sigma}_2 \equiv \sigma_1^x \otimes \sigma_2^x + \sigma_1^y \otimes \sigma_2^y + \sigma_1^z \otimes \sigma_2^z$

The $\sigma_z$ operator for each particle might have the same matrix form, but remember for each particle it acts on a different Hilbert space. $\sigma_1^z$ acts on the Hilbert space of particle 1 and $\sigma_2^z$ acts on the Hilbert space of particle 2.

When you consider each of the three terms in $\vec{\sigma}_1 \cdot \vec{\sigma}_2$ remember that $\sigma_1^i$ and $\sigma_2^i$ act on different Hilbert spaces. The Hilbert space of each spin $s$ particle is $2s+1$ dimensional, which for a spin $1/2$ particle is $2$ dimensional. So the combined Hilbert space is $2 \times 2 = 4$ dimensional. This total Hilbert space is what $\vec{\sigma}_1 \cdot \vec{\sigma}_2$ acts on.

Thus, the system has 4 states in the total Hilbert space. You should construct the Hamiltonian as a $4 \times 4$ matrix acting on this total Hilbert space. Then, you can diagonalize it to find the spectrum (eigenvalues) and eigenstates (eigenvectors).

Note that my notation is slightly different from yours. The operators in your notation (let's call them $\Sigma$), in terms of the operators in my notation ($\sigma$) will be: $\Sigma_1 = \sigma_1 \otimes \mathbf{1}$ and $\Sigma_2 = \mathbf{1} \otimes \sigma_2$. Now, since both $\Sigma_1$ and $\Sigma_2$ act on the same Hilbert space $\mathcal{H}_{tot}$, we can add them. The Hamiltonian will then be given by $$H = \alpha [\Sigma_1 + \Sigma_2] + \gamma \vec{\Sigma}_1 \cdot \vec{\Sigma}_2$$

Understanding this system

Note that this Hamiltonian gives energy $y$ if the two particle spins are aligned in parallel and $\alpha$ for whatever is aligned along the z-direction (usually that's an external magnetic field in the z-direction).


  • A convenient basis for the total system should be $|--\rangle$, $|-+\rangle$, $|+-\rangle$, $|++\rangle$. Write down the $4 \times 4$ matrix for the exchange operator in this basis and find the eigenstates.
  • Note that this Hamiltonian is symmetric under the exchange of the two particles. So if you construct the exchange operator on the space of states, that is a symmetry of the Hamiltonian... which means that the eigenstates of the exchange operator and the Hamiltonian will be the same! So you would have found the eigenvectors. Then finding the eigenvalues should be simple.
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    Saying that the operators $\sigma^1_z$ and $\sigma^2_z$ act on different Hilbert spaces, although perhaps giving useful intuition, is not mathematically correct. They both act on the same tensor product Hilbert space, but each effects only one factor in the tensor product. – joshphysics Apr 8 '13 at 6:28
  • Why not? $\mathcal{H}_{tot} = \mathcal{H}_1 \otimes \mathcal{H}_2$. $\vec{\sigma_1}$ acts on $\mathcal{H}_1$ and $\vec{\sigma_2}$ acts on $\mathcal{H}_2$. We can upgrade the two operators to $\vec{\sigma_1} \otimes \mathbb{1}$ and $\mathbb{1} \otimes \vec{\sigma_2}$ respectively, which act on the total Hilbert space. So the statement is mathematically perfectly fine. – Siva Apr 8 '13 at 6:35
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    Not exactly. $\sigma^1_z = \sigma_z\otimes I$ acts on $\mathcal H_1\otimes\mathcal H_2$. Also, If one writes a sum like $O_1 + O_2$, then provided one means $O_1 = A\otimes I$ and $O_2 = I\otimes B$ for some operators $A,B$ defined on $\mathcal H_1$ and $\mathcal H_2$, the sum makes complete sense whereas you write "$\sigma_1 + \sigma_2$ makes no sense as you can't assign a Hilbert (vector) space which that operator acts on." Lastly, in the case of two spin $1/2$ copies, $\mathcal H_1 = \mathcal H_2$; the two Hilbert space factors are equal; they are not "different Hilbert spaces." – joshphysics Apr 8 '13 at 6:44
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    As for the two Hilbert space factors: they might be mathematically equivalent as spaces but they represent two different physical objects and their states correspond to different physical details... so physically they are different Hilbert spaces. – Siva Apr 8 '13 at 6:51
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    Yeah, the idea of a tensor product takes some getting used to. – Siva Apr 8 '13 at 20:45

The dot notation means to write a sum of three terms by matching x, y, z matrices for the two particles, as if $\vec{\sigma}^1$ were a vector $(\sigma_x^1,\sigma_y^1,\sigma_z^1)$ and likewise for particle #2. $$ \vec{\sigma}^1\cdot\vec{\sigma}^2 = \sigma_x^1\sigma_x^2 + \sigma_y^1\sigma_y^2 + \sigma_z^1\sigma_z^2 $$

Note that $\sigma_i^1$ and $\sigma_i^2$ act separately as matrices; they don't multiply with each other.

We have $$ \sigma_z^1 = \left(\begin{matrix}1&0\\0&-1\end{matrix}\right) $$ but acting only on the particle #1 factor of the wavefunction, and $$ \sigma_z^2 = \left(\begin{matrix}1&0\\0&-1\end{matrix}\right) $$ action only on the particle #2 factor.

For example, if you have some wavefunction (not necessarily an eigenstate of anything) being particle #1 spin up and particle #2 down, $$ \psi =\psi^1\psi^2 = {\left(\begin{matrix}1\\0\end{matrix}\right)}^1{\left(\begin{matrix}0\\1\end{matrix}\right)}^2 $$ and some arbitrary matrix acting on particle #1, $$ U^1 = \left(\begin{matrix}a&b\\c&d\end{matrix}\right) $$ and another arbitrary matrix for particle #2, $$ V^2 = \left(\begin{matrix}e&f\\g&h\end{matrix}\right) $$ then $$ U^1 \psi = (U^1\psi^1 )\psi^2 = {\left(\begin{matrix}a\\c\end{matrix}\right)}^1{\left(\begin{matrix}0\\1\end{matrix}\right)}^2 $$ and $$ V^2 \psi = \psi^1(V^2\psi^2 ) = {\left(\begin{matrix}0\\1\end{matrix}\right)}^1{\left(\begin{matrix}f\\h\end{matrix}\right)}^2 $$

Note that tacking superscripts onto column spinors like that to indicate particle identity isn't normal practice outside of textbooks or places where things need to be explained in such detail. A common way to deal with this is to form a four dimensional space, the product of the two spin spaces of the particles. The basis vectors would be $\uparrow^1\uparrow^2, \uparrow^1\downarrow^2, \downarrow^1\uparrow^2, \downarrow^1\downarrow^2$ or some nice linear combination.

Think about it: we need an index to count along the spinor components (the 2-dimensional Hilbert space of complex numbers), an index to count spatial dimensions (x,y,z), and an index to count particles. We're dealing with a three-dimensional entity, a matrix with rows, columns and "another". Stick with theoretical physics, and we'll add chromodynamic charge, more spatial dimensions to deal with curved spacetime, and sprinkle in a dash of supersymmetry or technicolor or whatever is hot with the kids these days. And you'll refer to the Pauli matrices as "Clebsch-Gordan coefficients" but at least they're the simplest nontrivial case. Have fun!

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    Hi DarenW, thanks for your reply. I am reading all replies carefully and I also check it up in the textbook. It seems that $\vec{\sigma}^1\cdot\vec{\sigma}^2$ should be 4x4 matrix. I don't know if it is correct but it looks like that I should correct that to be $$\vec{\sigma}^1\cdot\vec{\sigma}^2 = \sigma_x^1\otimes\sigma_x^2 + \sigma_y^1\otimes\sigma_y^2 + \sigma_z^1\otimes\sigma_z^2$$ – user1285419 Apr 8 '13 at 18:06

Actually I "believe" that this topic is not really "settled"...

The expression 1/2 (σx⊗σx +σy⊗σy +σz⊗σz) is introduced by P. Dirac in his famous book "The Principles of Quantum Mechanics" IV ed. chap IX p. 221 where he uses this expression (actually he does not write explicitly the tensor or Kronecker product sign ⊗ but one "undersdands" that it corresponds to it). He uses this expression in order to obtain a permutation operator for spin 1/2 particles (electrons). He then "identifies" it to the expression (σ⃗,σ⃗), but he gives no physicaljustification for the notation.... He uses these expressions in order to calculate the exchange energy which is a fundamental topic in Physics (Magnetism) and is linked to the Pauli symmetry principle.

Perahps someone has other "ancient" references.

Zeno Toffano



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