Spin coherent states are often introduced as "the most classical states of a finite-dimensional system", or as the analogous of coherent states of light for finite-dimensional systems. See e.g. (Radcliffe 1971) and (Chryssomalakos et al. 2017).

One way to define them (using a notation similar to Radcliffe 1971) is as the states $$\lvert\mu\rangle=N^{-1/2}\exp(\mu S_-)\lvert S\rangle,$$ where $S_z\lvert S\rangle=S\lvert S\rangle$, $S_-\equiv S_x- i S_y$, and $N$ is a normalisation constant.

While the formal analogy between these states and coherent states of light (a.k.a. Glauber states), $$\lvert\alpha\rangle=\exp(\alpha a^\dagger - \alpha^* a)\lvert0\rangle=e^{-\lvert\alpha\rvert^2/2}\exp(\alpha a^\dagger)\lvert0\rangle,$$ is clear, what I don't find too clear from the references above is why these states should be regarded as "the most classical states", as is stated e.g. in the abstract of (Chryssomalakos 2017). In the optical case, we justify calling $\lvert\alpha\rangle$ classical observing e.g. that it gives Poissonian photon-counting statistics, and that it cannot produce entangled states using only linear operations.

Is there any similar physical justification in the case of spin coherent states?

  • $\begingroup$ I would guess that as $S\to\infty$ (the "classical" limit), those transition into normal coherent states, which are "classical". $\endgroup$ Jul 11, 2020 at 17:36
  • $\begingroup$ @NorbertSchuch that seems to be the case, and is also discussed by Radcliffe around page 5. Still, that doesn't seem like a strong justification for calling these states "the most classical available". The classicality of Glauber states is, afaik, mostly tied to the photon statistics they produce. How would this translate to the spin case? There is no natural notion of subsequent spin states representing different amounts of some underlying excitation (or is there?), so how would one justify the "classicality" here? $\endgroup$
    – glS
    Jul 11, 2020 at 18:19
  • $\begingroup$ To start with, as I said I guess this only makes sense in the large S limit. (After all, for S=1/2 this captures all states?) $\endgroup$ Jul 11, 2020 at 21:32
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    $\begingroup$ One reason why we consider the optical coherent state as classical is that if we plot the expectation value of $x$ as a function of time it oscillates back and forth similar to a classical simple harmonic oscillator. $\endgroup$
    – ChrisM
    Jul 11, 2020 at 23:08
  • $\begingroup$ One additional sense in which coherent states of the SHO are classical is that they are generated by applying a classical drive to a system initialized in the ground state. Perhaps something similar applies to spin coherent states? $\endgroup$
    – Rococo
    Jul 13, 2020 at 14:05

4 Answers 4

  1. They saturate the "displaced" uncertainty relation. If \begin{align} \vert \Omega\rangle=R(\Omega)\vert jj\rangle \end{align} is the coherent state, and \begin{align} J_k^\prime=R(\Omega)J_kR^{-1}(\Omega) \end{align} for any rotation $R(\Omega)$, then \begin{align} \Delta J_x^\prime \Delta J_y^\prime=\frac{1}{2}\vert\langle \Omega \vert J_z^\prime\vert\Omega\rangle\vert \, . \end{align}
  2. Their Wigner function is localized on the sphere, as for instance the WF of this coherent state with $J=9$ rotated about $\hat y$ by $\beta=2\pi/9$: enter image description here
  3. The time-evolution of a coherent state under a Hamiltonian which is linear in the generator of $SU(2)$ is just a continuous rotation on the sphere, without the WF changing its shape.
  • $\begingroup$ Re 3: Doesn't every state have its phase-space distribution on the sphere rotate rigidly under the dynamics of a Hamiltonian linear in the generators of SU(2)? $\endgroup$ May 18, 2021 at 18:29
  • $\begingroup$ @QuantumMechanic actually I shot from the hip here and now I'm not so sure about that now. Let me think a little more. $\endgroup$ May 18, 2021 at 19:04
  • $\begingroup$ I'm fairly certain - this follows because of the completeness of the SU(2)-coherent states, or from the definitions of the quasiprobability distributions (easiest for me to see with Husimi, because rotating the state rotates the arguments of the Husimi function). $\endgroup$ May 18, 2021 at 19:10
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    $\begingroup$ @QuantumMechanic yes that's basically the argument: rotating the state amounts to rotating the arguments. I can't think of a counterexample right now (my head is somewhere else) but that doesn't mean there isn't one. I'm almost sure this is right. $\endgroup$ May 18, 2021 at 19:17
  • $\begingroup$ @QuantumMechanic This follows from covariance of the WF. Unless the function is pathological in some way that is truly bizarre (in which case on can wonder about the physics of such function) the motion will be “rigid” under the action of a group element, i.e under the action of a Hamiltonian linear in the generators. $\endgroup$ May 20, 2021 at 1:38

In metrology, spin-coherent states are classical in the sense of not reaching the Heisenberg limit, while spin-squeezed states reach. In terms of the $SU(2)$ Wigner function both sets are "classical", since their Wigner functions are positive, as it happens in the harmonic oscillator case with standard coherent states and squeezed states. Squeezed states, however, are nonclassical concerning the Glauber-Sudarshan $P$-distribution, which singles out coherent states (and convex combinations of coherent states) as the only classical states. This comes from the fact that coherent states are the only quantum states behaving classicaly in respect to optical correlation functions of arbitrary degree - see https://books.google.com.br/books/about/Fundamentals_of_Quantum_Optics.html?id=rbSfWTrKwnAC&redir_esc=y for proofs.

There are $SU(2)$ $P$-distributions as well and those single out spin-coherent states and their convex combinations as the only classical ones - see, for example https://www.cambridge.org/core/books/geometry-of-quantum-states/4BA9DCEED5BB16B222A917EAAAD17028 .

A criteria for nonclassicality of non-spin-coherent states can be found in https://arxiv.org/pdf/1112.0809.pdf . Klyachko and others developed the concept of generalized entanglement - for example, https://arxiv.org/pdf/quant-ph/0512213.pdf - in which generalized entangled $SU(2)$ states are those which are not spin-coherent. This notion is however relative to the set of observables being measured. Further connections can be found in https://arxiv.org/pdf/1505.07393.pdf and https://arxiv.org/pdf/0910.3198.pdf.


One thing is true and you probably already know, coherent states of the quantum harmonic oscilator are called the "most classical" only because they minimize Heisenberg uncertainty principle (HUP). For those states this is valid (notice the equal sign): $$\Delta p\Delta x=\hbar/2$$ Since quantum effects generally are said to be encapsulated in HUP it's clear why people refer to CS as the "most classical but still quantum" states.

But as you know, the formalism of CS was developed for many other systems, like spin systems. With all the experience I've had working specifically with spin coherent states (describing magnons) in the last few years I haven't seen books call specifically the spin CS as the "most classical" of states. Probably because that's not the reason why this formalism is used in the first place.

Though an easy generalization might be to say they must minimize the Generalized HUP (called Robertson–Schrödinger inequality, or relation) given for two operators : $$(\Delta X_1)^2(\Delta X_2)^2\geq \dfrac{1}{4}|\langle[X_1,X_2]\rangle|^2+\left(\dfrac{\langle X_1X_2+X_2X_1\rangle}{2}-\langle X_1\rangle\langle X_2\rangle \right)^2$$

And turns out that's exactly what this article says it minimizes (look for the text above equation 22 where it talks about operators that obey SU(2) algebra, spins). Also the articles goes on explaining only the canonical CS states (those from the harmonic oscillator) minimize the standard HUP, and only they. He also mention that the spin CS are not the only states that minimize the Robertson–Schrödinger inequality, they are only part of a much larger set of states that minimize this inequality.

Given this we can conclude that its only precise to say they are the "most classical" states when talking about the harmonic oscillators, other than that what is minimized is the Robertson–Schrödinger inequality. It's totally valid to say that minimizing this inequality is as classical as you get on those systems, but it's wrong to say they are only one that share the podium.

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    $\begingroup$ The ground state of the harmonic oscillator which is a gaussian, also minimize the heisenbergs uncertainty principle. Going by your logic, it should be the most classical state too! $\endgroup$
    – abir
    Jul 13, 2020 at 10:24
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    $\begingroup$ In optics, squeezed Gaussian states also minimize the HUP. I'd say that the major reason why they are 'classical' is because they are eigenstates of the annihilation operator, meaning that under a coherent ansatz, you can replace the field operators by classical numbers (leading to e.g. Gross-Pitaevskii) $\endgroup$
    – Wouter
    Jul 13, 2020 at 11:23
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    $\begingroup$ @abir The ground state of the harmonic oscillator indeed is a coherent state. It looks like you think that PedroDM is saying that there exists one state that is "the most classical". But he is not. He is saying that there is this family of states, the coherent states, that are the most classical ones (all of them). $\endgroup$
    – DoeJohn
    Jul 13, 2020 at 11:40
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    $\begingroup$ @HicHaecHoc still, squeezed states are still Gaussian and saturate Heisenberg's inequality, but are not considered 'coherent states' $\endgroup$
    – Wouter
    Jul 13, 2020 at 13:33
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    $\begingroup$ @Wouter Yes, you're right. The most general family of 1D states that saturate thr uncertainty principle is given by the squeezed states, with the coherent ones as a subset. $\endgroup$
    – DoeJohn
    Jul 13, 2020 at 14:02

There are many great answers here so I will summarize before adding more:

  1. Spin-coherent states are displacements of some fiducial state (whose classical properties must still be determined), just like canonical coherent states (CCS).
  2. They saturate the uncertainty relation $\Delta S_x \Delta S_y\geq\frac{1}{2}|\langle S_z\rangle|$ for any three angular momentum operators satisfying the usual commutation relations, similar to CCS saturating the Heisenberg uncertainty principle.
  3. They have localized Wigner functions on the sphere, just like CCS being localized in phase space.
  4. Their phase space distributions rotate rigidly under Hamiltonians of the form $H=\mathbf{r}\cdot\mathbf{S}$ for some real vector $\mathbf{r}$ (dubious because this is true for all states). This is also true for all continuous-variable states under a displacement operation there, so again this is not a good way of choosing coherent states.
  5. They achieve the standard quantum limit for estimating a single phase, equivalent to a rotation around an axis perpendicular to the direction of their spin. CCS achieve the same limit in precision for phase estimation.
  6. They are eigenstates of some spin operator $\mathbf{r}\cdot\mathbf{S}$, just like CCS are eigenstates of the annihilation operator.

More answers:

  1. The fiducial state being displaced is a minimum uncertainty state, is a maximally-localized state in phase space, and is annihilated by one of the ladder operators. For SU(2) we know that $S_-|S,-S\rangle=0$ similar to $a|0\rangle=0$ with continuous variables.
  2. These states uniquely minimize the total variance $\Delta^2 S_x+\Delta^2 S_y+\Delta^2 S_z$, just like CCS minimize $\Delta^2 x+\Delta^2 p$ (the latter is sometimes known as total variance, quadrature coherence scale, mean quadrature variance, total noise, etc.), again for any set of orthogonal angular momentum operators (ie it doesn't matter if you displace the phase space before calculating these quantities).
  3. They satisfy $\langle \Omega|\mathbf{S}|\Omega\rangle=S \mathbf{n}$, where $\Omega$ encodes the angular coordinates $(\theta,\phi)$, the unit vector $\mathbf{n}=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)$ points in that direction, and $S$ is the total spin. For some calculations this lets us replace the spin operators with their expectation values. For example, if the spins about which we are talking are representing the optical polarization of $N=2S$ photons, the classical Stokes parameters are $S \mathbf{n}$ and so we can replace the Stokes operators with scalar values. This means that we can consider only the classical Poincaré sphere when trying to understand the optical polarization properties of these states, which is not true of other quantum states.
  4. They minimize other variance inequalities. For example, if we define $\mathbf{n}_{||}=\langle \mathbf{S}\rangle/|\langle \mathbf{S}\rangle|$ and two other mutally orthogonal unit vectors $\mathbf{n}_{\perp 1}$ and $\mathbf{n}_{\perp 2}$, the spin-coherent states are the only ones that simultaneously saturate the three inequalities \begin{aligned} \Delta^2 S_{\perp 1}\Delta^2 S_{\perp 2}&\geq|\langle S_{||}\rangle|\\ \Delta^2 S_{\perp 1}\Delta^2 S_{||}&\geq 0\\ \Delta^2 S_{||}\Delta^2 S_{\perp 2}&\geq 0. \end{aligned}
  5. They form an overcomplete basis for the state space of the system $\int |\Omega\rangle\langle\Omega|d\Omega\propto\mathbb{I}$, just like CCS.
  6. Their Husimi $Q$-functions are the most localized in phase space. This is seen in many ways, one of which is the relatively recent proof of the Lieb conjecture, stating that the spin coherent states minimize the Wehrl entropy of the phase space distribution. CCS also uniquely minimize the Wehrl entropy in their phase space.
  7. If the Husimi function is expanded in terms of spherical harmonics, the sum of the squared absolute values of the expansion coefficients is maximized by the spin-coherent states. Edit: this means that the states mostly have a dipole moment, less so a quadrapole moment, and so on to all orders. Other, more quantum states hide their information in higher-order moments that require higher-order correlation measurements to be unearthed.
  8. They have the largest inverse participation ratios in terms of Husimi functions, again meaning they are the most localized in phase space, just like CCS are the most localized.
  9. One can create a Majorana constellation with $N=2S$ points for any state with total spin $S$, and the spin-coherent states have all $N$ points degenerate to the same location on the sphere.

(Mostly taken from this recent article and refs therein.)

Edit: more answers

  1. If the total spin $S$ is made from $N=2S$ qubits in a symmetric superposition, the spin-coherent states are the only pure states that are completely separable in a first-quantized description. This means they are the only states for which we can unequivocally say what each qubit is doing, independent of the others, which is something that we can typically say about classical states but not always something we can say about quantum ones.

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