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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.

yarchik
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  • What do you mean by "audience of physicists?" Are these students, professors, or what? – Chris Jan 29 '18 at 09:44
  • @Chris Actually, I have heard this question from a professor. But I imagine, that students would also have this idea. – yarchik Jan 29 '18 at 09:45
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    What do you understand when you write $\frac12\otimes\frac12=0\oplus1$ on the board? It's better to write $2\otimes 2=3\oplus 1$. – SRS Jan 29 '18 at 10:24
  • @SRS That at least dimensionality (but not only) is the same $(21/2+1)(21/2+1)=4=(20+1)+(2*1+1)$. – yarchik Jan 29 '18 at 10:29
  • 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. – knzhou Jan 29 '18 at 10:47
  • @knzhou Completely agree with you, of course it is not the audience fault. Therefore the question here. – yarchik Jan 29 '18 at 10:56
  • @yarchik Why have you deleted your question about v-representability? It seemed like a reasonable question, I just tried to ask for some clarifications. I could not fully read your last comment (just saw the preview), but my intention certainly wasn't to "heavily criticize" you, just to request some clarifications about what exactly you were asking. – Norbert Schuch Aug 20 '18 at 20:45

1 Answers1

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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.

SRS
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  • 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. – yarchik Jan 29 '18 at 11:00
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    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 – SRS Jan 29 '18 at 11:05
  • 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. – yarchik Jan 29 '18 at 12:41
  • I also wanted to point out that you are using some kind of jargon in your first paragraph: "Each particle is associated with space" ... – yarchik Jan 29 '18 at 12:44
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    @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. – SRS Jan 29 '18 at 13:36
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    @yarchik This is standard notation. For an on-site explainer, try the question What does "the ${\bf N}$ of a group" mean?. – Emilio Pisanty Feb 01 '18 at 10:00