Spin coherent state for general spin $S$

In Wen's book on Many-body QFT, he claimed that the coherent state for a spin-$S$ particle can be written as a tensor product of $2S$ spin-1/2 coherent states:

$$|\hat{n}\rangle=|z\rangle\otimes|z\rangle...\otimes|z\rangle$$ where each $|z\rangle$ is a spin-1/2 spinor.

If we look at the degrees of freedom of both sides of the above equation, the spin-$S$ coherent state has $2S+1$ free parameters, whereas the $2S$ spin-1/2 have $4S$ free parameters in total. How do we understand this?

By definition, a spin coherent state $\vert \theta,\phi\rangle_S$ is just a rotation of the $\vert S,S\rangle$ state.

What makes the state "coherent" is that all the individual spin states point in the same $(\theta,\phi)$ direction (are "coherently aligned"), thus \begin{align} \vert \theta,\phi\rangle_S &:= \vert \theta,\phi\rangle_1\otimes\ldots\otimes \vert \theta,\phi\rangle_{2S} \tag{1} \\ &=R_z(\phi)R_y(\theta)\vert S,S\rangle \\ & \; = [R_z(\phi)R_y(\theta)\vert +\rangle_1]\otimes \ldots \otimes [R_z(\phi)R_y(\theta)\vert +\rangle _{2S}] \end{align} $\vert \theta,\phi\rangle_S$ clearly depends on only 2 parameters, as it should, and indeed there is no freedom in the orientation of any of the particle spins (except the first) are they must all be aligned in the same $(\theta,\phi)$ direction as the first. Thus, the right hand side of (1) depends on only two angles also.

The state in Eq.(1) is clearly symmetric under permutation of the particle indices, so taking $\theta=\phi=0$ gives
$$\vert S,S\rangle = \vert +\rangle_1\otimes \vert +\rangle_2\otimes\ldots \otimes \vert +\rangle_{2S} \tag{2}$$ This state is an eigenstate of $L_z$ $$L_z= L_z^{(1)}+L_z^{(2)}+\ldots +L_z^{(2S)}$$ with eigenvalue $M=S$, and is killed by the raising operator $$L_+= L_+^{(1)}+L_+^{(2)}+\ldots +L_+^{(2S)}$$ where $L_+^{(k)}$ acts on the state of particle $k$ alone.

• but why is the decomposition valid in the first place? – M. Zeng Apr 18 '18 at 3:50

A coherent spin state, when expressed in terms of composite spin 1/2 particles, is the tensor product of all spin 1/2 particles where each spinor is in the same state. In your expression, each $|z\rangle$ is identical, so there are in fact only two degrees of freedom on the right hand side (simply the degrees of freedom of a single spin half particle).

On the left hand side, the coherent spin state is not just any state with total spin S, but a highly constrained combination (as you see from the righthand side decomposition into spin half particles). There are again only two degrees of freedom.

The usual picture is that the coherent spin state lives on a collective Bloch sphere of radius S. The position on the collective Bloch sphere, parametrized by angles $\theta, \phi$, similarly describe the Bloch sphere position of each spin half particle in the decomposition.

Mathematically: a single spin half can be parametrized by angles $\theta, \phi$ as: $$|\psi(\theta, \phi)\rangle = \cos (\theta/2) |\downarrow\rangle + e^{i\phi} \sin(\theta/2) |\uparrow\rangle$$

Then the collective spin state is $$|\Psi(\theta, \phi)\rangle = |\psi(\theta,\phi)\rangle^{\otimes N}$$

• could you elaborate a bit on why the decomposition is possible? – M. Zeng Apr 18 '18 at 3:51
• Normally I actually think of the picture where you start with an ensemble of spin half particles, each in the same state, and you want to describe their collective spin dynamics - and if each spin is in the same state, then their collective state is a 'coherent spin state', as described above. The decomposition is possible based on the very specific definition for what a coherent spin state is. – Harry Levine Apr 18 '18 at 21:12

You need to look only at the symmetric combinations (under particle index permutation) of spin 1/2. So for instance, with $N=2$, the familiar triplet state has the same multiplicity as spin 1 vector boson. For $N=3$, you find:

$$|+\frac 3 2\rangle = |\uparrow\uparrow\uparrow \rangle$$

$$|+\frac 1 2\rangle = (|\downarrow\uparrow\uparrow \rangle + |\uparrow\downarrow\uparrow \rangle +|\uparrow\uparrow\downarrow \rangle)/\sqrt 3$$

$$|-\frac 1 2\rangle = (|\downarrow\downarrow\uparrow \rangle + |\uparrow\downarrow\downarrow \rangle +|\downarrow\uparrow\downarrow \rangle)/\sqrt 3$$

$$|-\frac 3 2\rangle = |\downarrow\downarrow\downarrow \rangle$$

and so on. There are many more combos of 3 spin 1/2 particles, but only those four are symmetric under interchange.

As you increase $N=2S$, you'll aways find $N+1$ states that are totally symmetric and they are in a 1-to-1 correspondence with the $2S+1$ states of a spin $S$ particle.

• thank you for your answer. For a particular $S_z$ eigenstate on the LHS, why do we only keep the symmetric combinations? – M. Zeng Apr 18 '18 at 19:22
• For any number N of spinors (or vectors or tensors), every integer partition of N gives a symmetry class. N=N is totally symmetric, N=1+1+1...+1 is totally antisymmetric. N = (N-1) +1 and (N-2) +2 and (N-2) +1+1 and so on are mixed symmetries--all permutation symmetries are closed under rotation (Schur-Weyl duality + Robinson Schensted correspondence). Since the total symmetric one contains all spins totally aligned: it is the maximal total S--which is what you're interested in. In summary: N get partitioned, Each partition is a Young Diagram. Each standard Tableau for a diagram is a.... – JEB Apr 18 '18 at 19:56
• ...representation of the permutation group. Apply that permutation to the particle index gives a closed-under-rotations sub-space of the total space spanned by all the spinors. E.g, for 2 spinors: 2 =2 --> triplet state. 2 = 1+1 --> singlet state. With 3 spins, 3 = 3 is the symmetric combos, 3 = 2 +1 give 2 different mixed symmetry sub spaces, and 3 = 1 + 1 +1 is antisymmetric, but has dimension 0--you can't get S=0 from 3 spins. The Hook Length Formula for Young Diagrams tells you the dimensions of the sub-spaces. – JEB Apr 18 '18 at 20:03