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I haven't been taught tensor product in class but they have taught us addition of spin. I looked up online in this link->http://homepage.univie.ac.at/reinhold.bertlmann/pdfs/T2_Skript_Ch_7.pdf#page=10 (pg 148, pg 10 in the pdf) and found an explanation. I think I understand most of it except this step:

$$ \vec{S}^{(A)} \otimes \vec{S}^{(B)} = \frac{\hbar^2}{4}(\sigma_x \otimes \sigma_x + \sigma_y \otimes\sigma_y + \sigma_z \otimes \sigma_z). $$

I know that:

$$ S^{2} = \frac{\hbar^2}{4}((\sigma_x)^2 + (\sigma_y)^2 + (\sigma_z)^2), $$

but I don't see the connection.

Physics_maths
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drewdles
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3 Answers3

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I don't think the author should use the tensor product $\otimes$ in $$\vec{S}^{(A)} \otimes \vec{S}^{(B)} = \frac{\hbar^2}{4}(\sigma_x \otimes \sigma_x + \sigma_y \otimes\sigma_y + \sigma_z \otimes \sigma_z)$$ because he really doesn't mean tensor product. Rather, $\vec S$ is a vector operator, that is, its components transform like the components of a vector, only they are operators, not numbers. Then the inner product $\vec S^A \cdot \vec S^B$ makes sense, $$\vec S^A \cdot \vec S^B = S_x^A \circ S_x^B + S_y^A \circ S_y^B + S_z^A \circ S_z ^B$$ where $\circ$ is composition as operators (matrix multiplication, if you prefer). Note that this isn't necessarily symmetric! If $\vec S = S^A \otimes 1 + 1 \otimes S^B$, then we obtain $$\vec S \cdot \vec S = \vec S^A \cdot \vec S^A \otimes 1 + 2 (\vec S^A \otimes 1) \cdot (1 \otimes \vec S^B) + 1 \otimes \vec S^B \cdot \vec S^B .$$

To find the expression you had trouble with, note that $\vec S^A_ x \otimes 1 = \frac{\hbar}{2} \sigma_x \otimes 1$, $ 1 \otimes \vec S^B_x = \frac{\hbar}{2} 1 \otimes \sigma_x$. Then $$S^A_x \circ S^B_x = \frac{\hbar^2}{4}\sigma_x \otimes \sigma_x$$ and naturally it's the same for $y$ and $z$.

Robin Ekman
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The tensor product makes perfect sense! It is the inner product that does not! These vectors "live" in different Hilbert spaces, you can't make an inner product out of those, it does not make sense.

What the equation means is simply the statement of 2 different particles with 2 different Hilbert spaces. Particle A has its state vector in Hilbert space A and same for B, so $\vec{S}_A$ acts on the states of A and same for $\vec{S}_B$ on B.

Since $\vec{S}_A=(S_x,S_y,S_z)_A=\frac{\hbar}{2}(\sigma_x,\sigma_y,\sigma_z)_A$ and $\vec{S}_B=(S_x,S_y,S_z)_B=\frac{\hbar}{2}(\sigma_x,\sigma_y,\sigma_z)_B$, the tensor notation is the formal way to state that the first operator (S_A) will act on the A state vector and S_B on the B state vector. One simple example would be one component of the tensor product acting on an entangled state of A and B in the z-direction basis of these. So when operating on a state like:

$\left|\frac{1}{2} \frac{1}{2}\right>_A\otimes\left|\frac{1}{2}-\frac{1}{2}\right>_B+\left|\frac{1}{2} -\frac{1}{2}\right>_A\otimes\left|\frac{1}{2}\frac{1}{2}\right>_B$

each operator only acts on the corresponding state.

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

The expression $$ \frac12 \left(\sigma_x\otimes\sigma_x+\sigma_y\otimes\sigma_y+\sigma_z\otimes\sigma_z\right) $$ 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 $\otimes$ but one "understands" 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 $(\vec\sigma,\vec\sigma)$, but he gives no physical justification 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.

Perhaps someone has other "ancient" references.

Kyle Kanos
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Zeno
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