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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.

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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 $z$.

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    $\begingroup$ Can you explain in what sense you write spin $S$ as vector? Because Pauli matrices are basis in su(2) algebra, they can't be coordinates of the vector, do they? $\endgroup$ – xxxxx May 14 '14 at 17:15
  • $\begingroup$ I mean, is it just a notation? $\endgroup$ – xxxxx May 14 '14 at 17:19
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    $\begingroup$ It depends on how you understand vector. Yes, the components of a world-vector are always numbers, never Pauli matrices. However $S$ is not a world-vector. It is a covector taking values in a space of operators that forms a representation of the $\mathfrak{su}(2)$ Lie algebra. You put in a world-vector $\vec v$, and out comes the operator corresponding to the observable "the spin along $\vec v$". In index notation we would write $S_i{}^a{}_b$ where $i$ is a world-vector index and $a$ and $b$ are the spinor (Pauli matrix) indices. $\endgroup$ – Robin Ekman May 14 '14 at 17:21
  • $\begingroup$ Often the spinor indices are omitted since it is implicit in $S$ being the spin operator that it carries such indices. $\endgroup$ – Robin Ekman May 14 '14 at 17:26
  • $\begingroup$ I would agree that the tensor product symbol is redundant, since the separate spaces are signaled by the A and B labels, but the tensor product helps to emphasize that. Bertlman was complacent in skipping the dot product on top of the tensor product, namely the correct expression is $\vec{S}^{(A)}\cdot \otimes \vec{S}^{(B)} = \frac{\hbar^2}{4}(\sigma_x \otimes \sigma_x + \sigma_y \otimes\sigma_y + \sigma_z \otimes \sigma_z)$ interpreted just the way you did. Note the inventor thereof, Dirac, in his book, skips the tensor product and uses his own notation for the dot product. $\endgroup$ – Cosmas Zachos May 10 '16 at 0:10
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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.

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  • $\begingroup$ Sorry but I still dont see why: $$ \vec{S}^{(A)} \otimes \vec{S}^{(B)} = \frac{\hbar^2}{4}(\sigma_x \otimes \sigma_x + \sigma_y \otimes\sigma_y + \sigma_z \otimes \sigma_z). $$ $\endgroup$ – drewdles May 14 '14 at 17:34
  • $\begingroup$ One operator is $\vec{S}\otimes 1 = (\sigma^x\otimes 1, \sigma^y\otimes 1, \sigma^z\otimes 1)^T$, the other is $1\otimes \vec{S}$. They both act on the same Hilbert space, and there's nothing strange about the inner product. The use of the tensor product quoted by the OP is quite unorthodox, however. $\endgroup$ – Mark Mitchison May 14 '14 at 18:49
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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.

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