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I'm looking at the spin-squared operator for a two-particle state, and I've been confusing myself about justifying the equality in the posting title. Consider the vector space $V\otimes W$, and two linear operators $A\in\mathcal{L}(V)$ and $B\in\mathcal{L}(W)$. If $(v\otimes w)\in V\otimes W$, then we can define a linear operator $(A\otimes B)$ acting on elements of this space as

$$(A\otimes B)(v\otimes w) \equiv (Av\otimes Bw)$$

With $C\in\mathcal{L}(V)$ and $D\in\mathcal{L}(W)$, this implies $(A\otimes B)(C\otimes D) = (AC\otimes BD)$. Having laid that out, onto my confusion. The spin operator of a two particle state is $\textbf{S} = \textbf{S}_1\otimes\textbf{1}+\textbf{1}\otimes\textbf{S}_2$. Using the above result, the square of this operator would be

$$\textbf{S}^2 = \textbf{S}_1^2\otimes\textbf{1}+\textbf{1}\otimes\textbf{S}_2^2+2\cdot\textbf{S}_1\otimes\textbf{S}_2$$

but I'm failing to see why

$$\textbf{S}_1\otimes\textbf{S}_2 = \sum_{i = x,y,z} S_{1i}\otimes S_{2i}$$

Is this obvious? What am I missing here? Usually one neglects to write out the tensor products, so without formality one directly arrives at $\textbf{S}^2 = (\textbf{S}_1+\textbf{S}_2)^2 = \textbf{S}_1^2+\textbf{S}_2^2 + 2\textbf{S}_1\cdot\textbf{S}_2$, but I'd like to see how this "dot product" pops out of the above tensor product.

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    $\begingroup$ Look in my answers here : Total spin of two spin-1/2 particles, especially equations ( 78 ) , ( 79 ) and ( 80 ) in my F O U R T H___ A N S W E R. $\endgroup$
    – Frobenius
    Commented Dec 22, 2019 at 6:37
  • $\begingroup$ @Frobenius Wow, incredibly thorough answer! Have just skimmed through the four posts; looking forward to going through them in more detail later tonight. Cheers $\endgroup$
    – dsm
    Commented Dec 22, 2019 at 6:56

3 Answers 3

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You're running into a common issue people have with vector operators. That is, the vector operator $\mathbf{S}$ cannot be squared in the usual way an operator can.

Recall that an operator is a linear map from a vector space to itself, $A: V \to V$. By contrast, $\mathbf{S}$ is a vector of operators. It acts on a single ket to produce a vector of kets, $$\mathbf{S}: V \to V \otimes \mathbb{C}^3, \quad \mathbf{S} | \psi \rangle \equiv \begin{pmatrix} S_x |\psi \rangle \\ S_y |\psi \rangle \\ S_z |\psi \rangle \end{pmatrix}.$$ Since the domain and range aren't the same, $\mathbf{S}$ can't be squared by just applying it twice. Instead, the definition of the notation "$S^2$" is in terms of the usual dot product, $$S^2 \equiv S_x^2 + S_y^2 + S_z^2.$$ Because of this, the question is moot. The dot product isn't actually related to the tensor product at all, it's put in by the definition of the square of a vector operator. In fact, it doesn't even make sense to consider the object $\mathbf{S}_1 \otimes \mathbf{S}_2$ for the calculation you're trying to do. That object would not even be a vector operator; it would be a rank 2 tensor operator, being the tensor product of two vectors.

Unfortunately, textbooks generally aren't very clear about this, because it's tempting to just stick to the clean notation and avoid mentioning these details.

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  • $\begingroup$ Perfect, thank you for clearing that up! Makes much more sense $\endgroup$
    – dsm
    Commented Dec 22, 2019 at 6:36
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    $\begingroup$ A bit OT, but you know what I'd like? A textbook of the intermediate-advanced sort, but that spends/wastes a lot of space to continually expand and show the various formulas, operators and variables for "what they are" all the time. It's so easy to get lost in hierarchical notation. Like, an advanced textbook that treats its readers as dummies or just forgetful whimsical readers :) Instead, textbooks seem to revel in making the most condensed, terse, brief texts and notations possible. $\endgroup$
    – BjornW
    Commented Dec 22, 2019 at 11:56
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Here the square of the operator $S^2$ does not mean you apply it twice. Literally, one has by definition $$ {S}^2\equiv S_xS_x+S_yS_y+S_zS_z\, . $$ Having ascertained this, replace $$ S_x=S_{1x}\otimes 1 + 1\otimes S_{2x} $$ etc. Applying $S_x$ twice yields \begin{align} S_xS_x&= (S_{1x}S_{1x})\otimes 1 +2 S_{1x}\otimes S_{2x} + 1\otimes (S_{2x}S_{2x})\, . \end{align} Redoing this for all the components and summing yields $$ S^2= S_{1}^2 + S_{2}^2 +2\sum_{i}S_{1i}S_{2i} $$

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    $\begingroup$ Awesome, thanks! Wanted to accept yours as well, but went with knzhou because the mapping was my main confusion. $\endgroup$
    – dsm
    Commented Dec 22, 2019 at 6:38
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I'm late as usual but here's my two cents

When one writes $\vec S^2$, one actually means the dot product of the Cartesian degrees of freedom of the (Cartesian) vector operator $\vec S$;

In simpler words, $\vec S$ is just a triple of operators that happen to mix as a Cartesian vector if a rotation is applied to the system, but they behave as any other operator for what concerns the action on the states of the Hilbert space;

The dot product, then, is defined regardless of any other tensor structure that $\vec S$ might have, because it only mixes the components of the vector operator as per definition of dot product, simply replacing the multiplication between pairs of components in a real (or complex) vector space, with the composition of operators in the automorphism algebra of the Hilbert space:

$$\vec S^2=S_xS_x +S_yS_y+S_zS_z$$

Where by $S_iS_i$ we mean the operator defined by applying twice $S_i$ (a more than legitimate operation).

Then, aside from that, each of the $S_i$ is also an operator the total Hilbert space defined as $$S_i\equiv S_{i1}\otimes S_{i2}$$ So the composition $S_iS_i$ must also follow the usual rules of the tensor product.

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