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This question is an extension of this question, I asked previously.

Let us denote the unique irreducible unitary representations of $\mathrm{SU}(2)$ by $V_{j}$, where $\mathrm{dim}(V_{j})=2j+1$. It is a general fact that the $j$-spin representation can be viewed as a submodule of the tensor product $V_{1/2}^{\otimes 2j}$. This can be seen by applying the Clebsch-Gordan decomposition theorem recursively. Now, in some lecture, I have seen the claim that any pure tensor of the form

$$\vert z\rangle^{\otimes 2j}\in V_{1/2}^{\otimes 2j}$$

for some element $\vert z\rangle\in V_{1/2}$ can be viewed as an element of $V_{j}$. My question is, why is this true? So, why do we know that in the decomposition $$V^{\otimes 2j}_{1/2}=V_{j}\oplus\mathrm{something}$$

the state $\vert z\rangle^{\otimes 2j}$ lives purely in $V_{j}$. For the vector $\vert\uparrow\rangle$ this was explained in the answer of Mike Stone in the previous post, since he argued that $\vert\uparrow\rangle^{\otimes 2j}=\vert j,j\rangle$. If $\vert z\rangle$ is an arbitrary element of $V_{1/2}$, then I can write it as $$\vert z\rangle=c_{1}\vert\uparrow\rangle+c_{2}\vert\downarrow\rangle$$ for some coefficients $c_{1},c_{2}\in\mathbb{C}$. Is it then maybe possible to express $\vert z\rangle^{\otimes 2j}$ in term of the basis $$\{\vert j,m\rangle\}_{-j\leq m\leq j}$$ of $V_{j}$?

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  • $\begingroup$ Going through the calculations, it seems to be true for $j=1$, at least. If it's true, I suspect it will have something to do with the symmetry of the $V_j$ subspace under the exchange of the individual $V_{1/2}$ subspaces. $\endgroup$ Nov 29, 2021 at 19:18
  • $\begingroup$ You may always rotate your axes to one set where the z axis north pole goes to the north pole of z', where your state $|z\rangle= |\uparrow\rangle$. Thus, this reduces to the answered question. $\endgroup$ Nov 29, 2021 at 19:27

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If it's true for $|\uparrow\rangle$, it has to be true for all states $|z\rangle$ by rotational invariance of the total momentum. In other words, we know the total angular momentum of $|\uparrow\rangle^{2j}$ is given by $\hbar^2 j(j+1)$, so the total angular momentum of $|z\rangle^{2j}$ is also $\hbar^2 j(j+1)$, and thus made of a linear combination of states of the form $|m,j\rangle$.

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  • $\begingroup$ Okay, I see, this makes sense. But, is there also a way to explicitely write $\vert z\rangle^{\otimes 2j}$ in terms of the basis $\vert j,m\rangle$ of $V_{j}$? $\endgroup$
    – B.Hueber
    Nov 29, 2021 at 19:29
  • $\begingroup$ @Udalricus.S. You could just act on the state $|j,j\rangle$ with the appropriate rotation operator. You know the canonical map $SO(3)\mapsto SU(2)$, so you just need to figure out how an appropriate rotation taking $|\uparrow\rangle$ to $|z\rangle$ acts on the state $|j,j\rangle$ using this mapping. $\endgroup$ Nov 29, 2021 at 19:34
  • $\begingroup$ But I think this just determines the state up to a phase, right? Because the rotation which I choose is only unique up to an initial orientation around the $\vert\uparrow\rangle$ direction...To be precise, I need this discussion only for the state $\vert z\rangle:=\vert +\rangle:=\frac{1}{\sqrt{2}}(\vert\uparrow\rangle+\vert\downarrow\rangle)$. $\endgroup$
    – B.Hueber
    Nov 29, 2021 at 19:46
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    $\begingroup$ @Udalricus your particular state is the positive eigenvalue eigenstate of $\sigma_x$, so you only need rotate your z-axis to the x-axis, which can be done without superfluous phases. $\endgroup$ Nov 29, 2021 at 19:53
  • $\begingroup$ @Udalricus.S. How could you ever determine a state not "up to a phase"? $\endgroup$ Nov 29, 2021 at 20:14
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It is true because $V_j$ is the only part of this tensor product that is fully symmetric.

Thinking in terms of Schur-Weyl duality, the "$\oplus$ something" part basically contains all the other Young diagrams, which are associated with other $J$ values less than $j$ and indeed will occur more than once.

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If your question were: "How does the state $|+\rangle=(\uparrow +\downarrow )/\sqrt{2}$ tensor with itself 2j times in this (suboptimal, z-eigen-) basis", (where I've skipped the Dirac ket qualifiers for simplicity of writing), the answer is straightforward. Your state $|+\rangle$ is the positive-eigenvalue eigenstate of $\sigma_x$, to which you may rotate $\sigma_z$, by a minus-right-angle rotation around the y-axis. (Can write this as a half-angle Pauli exponential, if you really need it.)

Consequently, in the x-basis, the problem is reduced to the answered question referred to. The peculiarity of this state is that the tensor product is fully symmetric in all 2j tensor multiplications.

If however, you wish to see how this simple state presents in the original, suboptimal, z-basis, it is probably sufficient to cover the j=2 case, $$ ((\uparrow +\downarrow )/\sqrt{2} )^{\otimes 4}= \frac{1}{2^2}( \uparrow\uparrow\uparrow\uparrow\\ +\downarrow \uparrow\uparrow\uparrow +\uparrow\downarrow \uparrow\uparrow+ \uparrow\uparrow\downarrow \uparrow+ \ \uparrow \uparrow\uparrow\downarrow \\ + \downarrow \downarrow \uparrow\uparrow + \downarrow \uparrow\downarrow \uparrow + \downarrow \uparrow\uparrow \downarrow + \uparrow \downarrow \downarrow \uparrow + \uparrow\downarrow \uparrow \downarrow +\uparrow\uparrow \downarrow \downarrow \\ + \downarrow\uparrow\uparrow\uparrow + \uparrow\downarrow\uparrow\uparrow+ \uparrow\uparrow\downarrow\uparrow+ \uparrow\uparrow\uparrow\downarrow\\ +\downarrow \downarrow \downarrow \downarrow ), $$ with 1,4,6,4,1 combinations in each line, specified by the Pascal triangle. 16 ($=2^4$) states in all. The combinatorics is self-evident, and it would be up to you what compact formalism you'd wish to adopt. In the coupled orthonormal basis, this normalized state amounts to the normalized $\tfrac{1}{4}(|2,2\rangle+\sqrt{4}|2,1\rangle+\sqrt{6}|2,0\rangle+\sqrt{4}|2,-1\rangle+|2,2\rangle)$.


Post @Jahan Claes comment edit

Consequently, in terms of the "inefficient" $|j, m\rangle$ basis, the state is but $$ \frac{1}{2^j}\sum^j_{m=-j} \sqrt{\binom{2j}{j-m} } |j,m\rangle ~. $$

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    $\begingroup$ My interpretation is that OP wants the formula in the $|m,j\rangle$ basis $\endgroup$ Nov 29, 2021 at 23:26
  • $\begingroup$ @Jahan Claes Thanks. If so, I added an edit... $\endgroup$ Nov 29, 2021 at 23:50
  • $\begingroup$ Sorry to bother you again, but I am not sure if your formula below with the sum is correct. Using your spelled-out formula above, I get $\vert +\rangle^{\otimes 4}=\frac{1}{4}(\vert 2,2\rangle+2\vert 2,1\rangle+8\vert 2,0\rangle+2\vert 2,-1\rangle+\vert 2,-2\rangle)$, but this is different from your formula below with the sum, which gives $\vert +\rangle^{\otimes 4}=\frac{1}{4}(\vert 2,2\rangle+6\vert 2,1\rangle+\vert 2,0\rangle+6\vert 2,-1\rangle+\vert 2,-2\rangle)$... $\endgroup$
    – B.Hueber
    Nov 30, 2021 at 20:35
  • $\begingroup$ Apologies, I was half asleep when I changed variables. Fixed it now. The binomial coeffs 1,4,6,4,1 follow from $\binom{4}{0}, \binom{4}{1},\binom{4}{2},\binom{4}{3},\binom{4}{4}$. $\endgroup$ Nov 30, 2021 at 21:11
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    $\begingroup$ You are right, yet again! I neglected to normalize states in the coupled basis, living proof that Clebsching in one's head is treacherous. Will fix, and hit me again, if there are further snags! $\endgroup$ Dec 1, 2021 at 14:19

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