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While studying the representation theory, I came up with the following example, but it seems hard for me to solve.

For an integer or half-integer $j$, let $V_j$ be a $(2j+1)$-dimensional complex vector space with basis $$\{{|j, j\rangle, |j, j-1\rangle, \cdots, |j,-j\rangle}\}$$ which transforms as an irreducible representation of $SU(2)$. Consider the tensor product $V_j\otimes V_j$. Generally this space is reducible, and it is well known that $$V_j\otimes V_j \approx V_0 \oplus V_1 \oplus \cdots \oplus V_{2j}$$ as a direct sum of irreducible representation. Then, what is the explicit element in $V_0$ on RHS in terms of the tensor product state?

  • This question is motivated from my pure interest. This is not the homework-like question.
Laplacian
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  • There exists one only product state with $0$ angular momemtum, basis of the 1-dimensional Hilbert $V_0$ in all respects identical to $\mathbb C$. – Frobenius Aug 12 '20 at 12:53
  • @Frobenius Yes, and my question is that what is the state in the "tensor product space" $V_j \otimes V_j$ that maps the element in $V_0$ according to the isomorphism. – Laplacian Aug 12 '20 at 13:15
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    P.S. I am frustrated that this question is considered off-topic. I came up with this question myself, and this is from my pure curiosity. – Laplacian Aug 12 '20 at 13:16
  • The homework policy holds regardless of the origin of the question, and it's based on a consideration of the text of the post -- which here reads precisely like a homework question. If this comes only from your own curiosity, then phrase it that way. – Emilio Pisanty Aug 12 '20 at 13:29
  • Moreover, one thing that makes this read like homework is that the text is explicitly aware of the $j\otimes j = 0\oplus \cdots\oplus 2j$ rule, but somehow misses all of the relevant context around that formula. The answer is a straightforward calculation in terms of Clebsch-Gordan coefficients, and the current text reads like it's a set-piece demanding precisely that calculation. If you do know how to work with CG coefficients, then you should explain how much you know and where you get stuck in that calculation. If you don't know what CG coefficients are -- that doesn't make any sense. – Emilio Pisanty Aug 12 '20 at 13:33
  • @EmilioPisanty For given (small) value of $j$, one can explicitly calculate this by CG coefficient. However, what I am asking is the general state for arbitrary $j$, and this is where I stuck. Also, I edited the question to not look like a homework question. – Laplacian Aug 12 '20 at 13:35
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    This question is absolutely suitable for this site. For your answer, I'd suggest looking at https://en.m.wikipedia.org/wiki/Clebsch%E2%80%93Gordan_coefficients . Under the special cases section, the first case given is what you need. – fewfew4 Aug 12 '20 at 13:36
  • @LucashWindowWasher Thanks very much! I resolved my question, and your link is the precise answer to my question. I am sad that I cannot accept this answer. – Laplacian Aug 12 '20 at 13:38
  • @eigenvalue I (personally) don't see how the edits help take the question out of a homework format; I would suggest a much more thorough restructuring, but that's just me. The question is currently on the Reopen review queue, and it's up to the reviewers there to determine whether your edits thus far are sufficient. – Emilio Pisanty Aug 12 '20 at 14:14
  • Other approaches to @Lucash's "look in Wikipedia" pointer are to use the DLMF or simply the Mathematica function ClebschGordanCoefficient. At some point in the past, calculating CG coefficients was indeed a tedious task in drudgery, but (particularly for simple cases like this) that's not really the case these days. – Emilio Pisanty Aug 12 '20 at 14:19
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    \begin{equation} \boldsymbol{|}0,0 \boldsymbol{\rangle} \boldsymbol{=}\dfrac{1}{\sqrt{2j+1}}\sum\limits_{m\boldsymbol{=-}j}^{m\boldsymbol{=+}j}\left(\boldsymbol{-}1\right)^{j\boldsymbol{-}m}\boldsymbol{|}j,m \boldsymbol{\rangle\otimes |}j,-m\boldsymbol{\rangle} \tag{01}\label{01}
    \end{equation}     – Frobenius Aug 12 '20 at 23:05
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    @Frobenius Thanks. I found the same form in the wiki link above. – Laplacian Aug 13 '20 at 01:50
  • May be my answers there : Τotal spin of two spin-1/2 particles help you a little in other cases. – Frobenius Aug 13 '20 at 04:32

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