In Shankar, QFT and Condensed Matter, p73, it says

$$\langle S,S|\vec S|SS\rangle=\vec kS,\tag{6.3}$$ $$\langle\Omega|\vec S|\Omega\rangle=S(\vec i \sin\theta \cos\phi+\vec j \sin\theta \sin\phi+\vec k \cos\theta),\tag{6.4}$$

where $|\Omega \rangle\equiv |\theta \phi\rangle$ is the spin coherent state.

I have problem understanding how those equations comes from. - It seems to me that they are only valid for $S_z$ operator, because i can only understand (6.3) when $\vec S =\vec{S_z}, k=\hat{z}$. What do i miss?

  • Even if (6.3) is valid for spin operators in all direction, i can't derive (6.4). Consider $$\langle S,S|U^{\dagger}\vec S U|SS\rangle=\langle\Omega|\vec S |\Omega\rangle,$$ but how dose r.h.s of (6.3) change?

I also see other books that describes the topic in deferent ways, but they makes me further confuse and i struggled and failed to connect them in a clear way. I list them as following.

In Altland & Simons, Condensed matter field theory, p138,

In view of the fact that the states |g ̃(φ, θ, ψ)⟩ cover the entire Hilbert space $H_S$,we are led to suspect that the latter bears similarity with a sphere. To substantiate this view, let us compute the expectation values $$n_i\equiv \langle \tilde{g}(\phi,\theta,\psi|S_i|\tilde{g}(\phi,\theta,\psi\rangle, i=1,2,3.\tag{3.49}$$ To this end, we first derive an auxiliary identity which will spare us much of the trouble that will arise in expanding the exponentials appearing in the definition of $\tilde{g}\rangle$. By making use of the identity $(i\not = j)$ $$ e^{−iφ\hat{S_i}}\hat{S_j}e^{iφ[\hat{S_i},\ ]} = e^{iφ[\hat{S_i},\ ]} \hat{S_j} = \hat{S_j}\cos\phi+\epsilon_{ijk}\hat{S_k}\sin\phi,\tag{3.50}$$ where the last equality follows from the fact that $cos x (sin x)$ contain $x$ in even (odd) orders and$[\hat{S_j},]^2 \hat{S_i} =\hat{S_i}$, it is straightforward to obtain(exercise)$\vec n =S(\sinθ\cosφ,\sinθ\sinφ,\cosθ)$, i.e. $\vec n$ is the product of $S$ and a unit vector parameterized in terms of spherical coordinates.

  • I have no idea what does (3.50) mean, and how to derive $\vec n$.

In Fradkin, Field Theories of Condensed Matter Physics,p193, it just claims

the diagonal matrix elements of the $SU(2)$ generators $\vec S$, $$\langle \vec n|\vec S|\vec n\rangle=S\vec n.\tag{7.13}$$


1 Answer 1

  1. (6.4) is meant to be understood component by component, v.g. $\langle S_x\rangle = S\sin\theta\cos\phi$. I've never seen (6.3) with $\vec k$ but I've seen it with $\vec n$ as you have later. The simplest way to derive (6.4) is to start with $S_z\vert SS\rangle$ and then $\langle SS\vert U^\dagger U S_z U^\dagger U\vert SS\rangle.$
  2. As to your last, (7.13) is again to be understood as an equation valid component by component, i.e. $\langle \vec n\vert S_x\vert \vec n\rangle= S\sin\theta\cos\phi$, with $n_x=\sin\theta\cos\phi$.
  3. I've never seen the Altland&Simons notation.

The simple way to derive this is to remember that \begin{align} \vert SS\rangle =\vert +\rangle_1\otimes\ldots \otimes \vert +\rangle_N \end{align} with $S=N/2$. Basically, the top state $\vert SS\rangle$ is the $N$-fold product of the single spin up states.

Since \begin{align} \hat S_i = \hat S_i^{(1)}+ \ldots + \hat S_i^{(N)} \end{align} it follows that \begin{align} \langle SS \vert \hat S_i\vert SS\rangle = {_1\langle} +\vert \hat S_i^{(1)}\vert +\rangle_1+\ldots + {_N\langle} +\vert \hat S_i^{(N)}\vert +\rangle_N \end{align} so it's enough to compute one expectation value and, since the remaining $N-1$ ones will be identical, multiply the result for one expectation value by $N$. Thus \begin{align} \langle SS \vert \hat S_i\vert SS\rangle = N\times {_1\langle} +\vert \hat S_i^{(1)}\vert +\rangle_1\, . \end{align}


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