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Consider having $N$ spins $1/2$, so the overall state of $N$ particles can be described by the total spin value $S=0 \ldots N/2$ (let us set $N$ to be even for simplicity), and the projection of the total spin on the $z$-axis $S_z=-S\ldots S$. In total there are $2^N$ states for this system, however, I am interested in states, where only $2$ spins out of $N$ are flipped, so there are $N(N-1)/2$ of such states.

My question is - is that possible for an arbitrary value of $N$ write down the expansion of a total state of the system $|S,S_z\rangle$ from this Hilbert subspace in the following form $$|S=N/2,S_z=-(N-2)/2\rangle = \sum\limits_{i\ne j} C_{i,j} |\uparrow_i \uparrow_j\rangle \otimes|\downarrow \:\rangle^{\otimes (N-2)}~ ?$$

$\textbf{Updated:}$ As an illustrative example, consider just $2$ spins $1/2$. Then we have $4$ states in total described in a picture of a total spin: $|S=1,S_z=-1\rangle, |S=1,S_z=0\rangle, |S=1,S_z=+1\rangle, |S=0,S_z=0\rangle$.

As the Clebsch-Gordan coefficients suggest, these states has the following representations in terms of individual spins:

$$ |S=1,S_z=-1\rangle = |\downarrow \downarrow \rangle \\ |S=1,S_z=0\rangle = \frac{1}{\sqrt{2}} \left( |\downarrow \uparrow \rangle + |\uparrow \downarrow \rangle \right) \\ |S=1,S_z=+1\rangle = |\uparrow \uparrow \rangle \\ |S=0,S_z=0\rangle = \frac{1}{\sqrt{2}} \left( |\downarrow \uparrow \rangle - |\uparrow \downarrow \rangle \right) $$

(In the last equation the overall sign might be different, I don't remember the convention).

The question is, can I do similar for a particular state of $N$ spins?

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  • $\begingroup$ Sure. Why not? Can you clarify the question? $\endgroup$ Mar 16 at 20:12
  • $\begingroup$ Hello, Norberb Schuch, I have edited the question slightly: I would like to expand the state $S=-N/2,S_z=-(N-2)/2$ (the total spin momentum magnitude is maximal, but the projection is $S_z=-(N-2)/2$ as the $2$ spins out of $N$ are up). $\endgroup$
    – Sl0wp0k3
    Mar 16 at 22:35
  • $\begingroup$ Do you want to know if it is possible (then the answer is just "yes"), or what the expansion is? $\endgroup$ Mar 17 at 10:48
  • $\begingroup$ Dear Norbert Schuch, basically, I want to understand the strategy. The "stupid" way to do this is to use the angular momenta summation technique, but for $N$ spins it is... problematic $\endgroup$
    – Sl0wp0k3
    Mar 17 at 11:58
  • $\begingroup$ As my answer says, for the maximum spin state is must be the permutationally symmetric state. You can also understand it by applying the spin lowering operator twice to the maximum Sz state. $\endgroup$ Mar 17 at 18:36
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Yes, it is possible.

Clearly, $S_z$ is the total $z$ spin, i.e. the sum of all spin values. Thus, the state is a superposition of states with two spins flipped, as you write.

Then, all states with maximal spin are fully symmetric. This means that all $C_{i,j}$ must be equal: $C_{i,j}=k$ for all $i<j$. For normalization reasons, $k=1/\sqrt{N(N-1)/2}$ (if you only consider $i<j$).

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  • $\begingroup$ will those be in an $S$-irrep or spread over all irreps with $S\ge 1$? $\endgroup$ Mar 20 at 23:23
  • $\begingroup$ @ZeroTheHero Which "those"? And no, this is not an irrep, since it is one state and the only 1D irrep is S=0. $\endgroup$ Mar 20 at 23:24
  • $\begingroup$ actually I see it now. Interesting. $\endgroup$ Mar 21 at 1:07
  • $\begingroup$ Thank you Norbert, I marked this as an answer, but this particular state is the simplest one among those I was interested in. I kinda wanted to fill the whole space of two flipped spins 1/2 (of dimension N*(N-1)/2) with states that are eigenstates of a total S_z. $\endgroup$
    – Sl0wp0k3
    Apr 8 at 21:51

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