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Consider a system of two 1/2-spins. Under some conditions the Hilbert space can be decomposed into the direct sum of spin-0 and spin-1 representations: $\frac12\otimes\frac12=0\oplus1$.

When I write this formula on the board, I immediately get an objection that $1/4$ is not equal to 1 ! My question is as follows, how to explain this equation to the audience of physicists. Preferably in one or two sentences, concise, mathematically correct, but without going into much mathematical details.

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  • $\begingroup$ What do you mean by "audience of physicists?" Are these students, professors, or what? $\endgroup$ – Chris Jan 29 '18 at 9:44
  • $\begingroup$ @Chris Actually, I have heard this question from a professor. But I imagine, that students would also have this idea. $\endgroup$ – yarchik Jan 29 '18 at 9:45
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    $\begingroup$ What do you understand when you write $\frac12\otimes\frac12=0\oplus1$ on the board? It's better to write $2\otimes 2=3\oplus 1$. $\endgroup$ – SRS Jan 29 '18 at 10:24
  • $\begingroup$ @SRS That at least dimensionality (but not only) is the same $(2*1/2+1)*(2*1/2+1)=4=(2*0+1)+(2*1+1)$. $\endgroup$ – yarchik Jan 29 '18 at 10:29
  • $\begingroup$ This sounds like a question about teaching, not physics. I recommend writing $\mathbf{R}_{1/2}$ instead of $1/2$, or something like that, the first couple times. If you actually wrote down the equation $1/2 \times 1/2 = 0 + 1$ without any explanation whatsoever of what the symbols mean, it's not the audience's fault for not understanding. It's your fault for being unclear. $\endgroup$ – knzhou Jan 29 '18 at 10:47
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Each spin-1/2 particle is associated with a $(2\times\frac{1}{2}+1)$=2-dimensional vector space $\mathbb{V}$ as far as its spin degree of freedom is concerned. A composite system of two spin-1/2 particles is associated with a 4-dimensional vector space which is a direct product $\mathbb{V}_1\otimes \mathbb{V}_2$ of two 2-dimensional vector spaces $\mathbb{V}_1$ and $\mathbb{V}_2$. Under a similarity transformation a $4\times 4$ matrix representing an element of $SU(2)$ that acts on the space $\mathbb{V}_1\otimes \mathbb{V}_2$, can be reduced to a block-diagonal form consisting of block matrices of dimensions $3\times 3$ and $1\times 1$ acting on invariant subspaces of dimensions 3 and 1 respectively.

In technical terms, it means that the 4-dimensional representation is reducible into a 3-dimensional and 1-dimensional irreducible representations, and symbolically written as $2\otimes 2=3\oplus 1$ which respectively corresponds to three triplet states of spin-1 and one singlet state of spin-0 of the composite system.

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  • $\begingroup$ Do you have some reference, preferably well-established book, that your notation in the second paragraph is actually used in this form? I have doubts that this is common. $\endgroup$ – yarchik Jan 29 '18 at 11:00
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    $\begingroup$ It's a standard notation in group theory, and I guess it's universal. You should find it in Sakurai's quantum mechanics book. @yarchik $\endgroup$ – SRS Jan 29 '18 at 11:05
  • $\begingroup$ I am sorry for being picky, but I just scrolled through the Sakurai, Modern Quantum Mechanics book, 1994 year edition and I could not find such notations. What edition are you using, could you please be more specific, Eq. number, etc.? I am quite convinced that to name representations by their dimensions is not common, viz. representations of the point groups. In any case, I am thankful for your textual explanations. It can be a good start. $\endgroup$ – yarchik Jan 29 '18 at 12:41
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    $\begingroup$ @yarchik Since your question has the term 'Hilbert space', I implicitly assumed that you're familiar with the notion of finite-dimensional vector spaces (I'm reluctant to use Hilbert space for finite dimensional vector spaces) associated with the spin state of a quantum system. It's just the Hilbert space if you like but only associated with spin state. $\endgroup$ – SRS Jan 29 '18 at 13:36
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    $\begingroup$ @yarchik This is standard notation. For an on-site explainer, try the question What does "the ${\bf N}$ of a group" mean?. $\endgroup$ – Emilio Pisanty Feb 1 '18 at 10:00

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