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I am having trouble grasping the projection operators in the context of composite spins system, e.g. with two spin-1. First off, a projector $P$ is said to be an operator that squares to itself, $P^2=P$. With that, its eigenvalue is either 0 or 1. Are these properties, $P^2=P$ and $\text{Eigenvalues}(P)=\{1,0\}$, only true in a certain representation?

Edit: The answer to the above question is no and that indeed those properties hold for both of the representations. What I did below has a mistake. Thanks to Mane.andrea for doing a consistency check also in Mathematica. It turns out I had it wrong when I wrote $M^2$, instead of $M.M$, where $M$ is the matrix form of $(\vec{s}_1 \cdot \vec{s}_2)$.

Consider a two-spin-1 system, the projector onto the subspace whose total spin is equal to 2 is:

$$P_2 = \frac{1}{3} + \frac{1}{2} (\vec{s}_1\cdot \vec{s}_2 ) + \frac{1}{6} (\vec{s}_1 \cdot \vec{s}_2)^2$$

where $\vec{s}_1$ be the spin operator for the first spin-1 and $\vec{s}_2$ for the second. Hence, when $P_2$ acts on a state with a total spin of 2, the eigenvalue is 1, otherwise, the eigenvalue is 0. (A post on detailed derivation)

Let us write the states in the composite system as $|J,M_z^J \rangle_c$, i.e. given the total spin operator $\vec{S} = \vec{s}_1 + \vec{s}_2$, we have (set $\hbar=1$)

$$S^2 |J,M^J_z \rangle_c = J(J+1)|J,M^J_z \rangle_c,$$ $$S_z|J,M^J_z \rangle_c = M^J_z |J,M^J_z \rangle_c.$$

Meanwhile, the spin-1 states simply as $|m_z \rangle$, such that

$$s^2|m_z \rangle = 2|m_z \rangle, \qquad s_z|m_z\rangle = m_z|m_z \rangle.$$

Let me list the states with a total spin $J=2$:

$$|2,2 \rangle_c = |1,1 \rangle,$$ $$|2,1 \rangle_c = \frac{1}{\sqrt{2}} \big( |0,1 \rangle + |1,0 \rangle \big),$$ $$|2,0 \rangle_c = \frac{1}{\sqrt{6}} \big( |-1,1 \rangle + 2|0,0 \rangle + |1,-1 \rangle \big),$$ $$|2,-1 \rangle_c = \frac{1}{\sqrt{2}} \big( |0,-1 \rangle + |-1,0 \rangle \big),$$ $$|2,-2 \rangle_c = |-1,-1 \rangle.$$

I expect then that the action of $P_2$ on these states will be $P_2|2,M^2_z\rangle_c = |2,M^2_z\rangle_c$, i.e. eigenvalue is 1. This is true except for $|2,0\rangle_c:$

$ P_2 |2,0\rangle_c = \bigg( \frac{1}{3} + \frac{1}{2} (\vec{s}_1\cdot \vec{s}_2 ) + \frac{1}{6} (\vec{s}_1 \cdot \vec{s}_2)^2 \bigg)\bigg(\frac{1}{\sqrt{6}} \big( |-1,1 \rangle + 2|0,0 \rangle + |1,-1 \rangle \big) \bigg) $ $ \qquad \qquad = (\frac{2}{3})^{1/2} \big( \frac{2}{3} |1,-1\rangle + |0,0\rangle + \frac{2}{3} |-1,1 \rangle \big) \neq |2,0 \rangle_c $

Moreover, $P_2$ does not square to itself. I will not show it here, but $P_2$ does not have an eigenvalue of 0 when it acts on $|1,1\rangle_c, |1,-1 \rangle_c, |0,0\rangle_c$.

Are the statements $(P_2)^2=P_2$ and Eigenvalues($P_2)=\{1,0\}$ true only on a certain representation? This is what I can potentially conclude from my endeavor above, but I would really appreciate additional insights and suggested readings because what I have written above is how deep my understanding of this subject goes.

Added: So I know that I can just write $P_2$ as

$$P_2 = \sum_{M_Z=-2}^2 |2,M_Z \rangle_c \langle 2, M_Z|_c, $$

so it is straightforward that $P_2^2 = P_2$ and $\text{Eigenvalues}(P_2)=\{1,0\}$. This is in a single-spin representation (treating the two spin-1 as a single spin; with subscipt "c" above), and the properties is true. In the two-spin representation, $P_2$ is ($\frac{1}{3} + \frac{1}{2} (\vec{s}_1\cdot \vec{s}_2 ) + \frac{1}{6} (\vec{s}_1 \cdot \vec{s}_2)^2)$ while the states with total spin of 2 are what I listed above (the right-hand-side terms).

The problem is in the two-spin representation, the eigenvalue of $P_2$ is not 1 when acted on state $|2,0\rangle_c$ and $P_2$ does not annihilate all states orthogonal to states of total spin equals 2.

git-able
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1 Answers1

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In order to differentiate between eigenvalues and operators I'll use a hat. By completing the square we can say $$ \vec{\hat s}_1 \cdot \vec{\hat s}_2 = \frac12({\hat S}^2-{\hat s}_1^2-{\hat s}_2^2) = \frac12 S(S+1)-2\,. $$ Thus the projector can be rewritten as $$ P_2 = \frac13 + \frac12\left( \frac12 S(S+1)-2\right) + \frac16\left( \frac12 S(S+1)-2\right)^2\,. $$ After some straightforward algebra this can be factorized as $$ P_2 = \frac{1}{24} (S-1)S(S+1)(S+2)\,. $$ Thus $P_2$ is manifestly zero on states with $S=0,1$ and it's equal to $1$ on states with $S=2$.

As a bonus, with this observation we can easily build a projector for any pair of spins to any value of their composite spin. First evaluate this product

$$ P_S = \frac{1}{\mathcal N}\prod_{\substack{S' =|s_1-s_2|\\S'\neq S}}^{s_1+s_2}\left({\hat S}^2-S'(S'+1)\right)\,,\qquad \mathcal{N} = P_S\big|_{\hat{S}^2 \to S(S+1)}\,. $$ where $\mathcal N$ is a normalization that makes $P_S^2 = P_S$. The final result will be a polinomial in ${\hat S}^2$ which can be written in the other form by replacing $$ {\hat S}^2 \to {\hat s}_1^2+{\hat s}_2^2 + 2 \,\vec{\hat s}_1 \cdot \vec{\hat s}_2\,. $$ Recall that, since we fix the spins $s_1$ and $s_2$, then ${\hat s}_{1,2}^2$ are just numbers.

EDIT:

Since OP wanted to verify the alternative way of doing the computation as a consistency check. I'll post here the $9 \times 9$ matrix for comparison. The basis I chose is the following: in the $|s_{1z},s_{2z}\rangle$ notation $$ \begin{aligned} &|1,1\rangle &&= (1,0,0,\ldots ,0)^T\,,\\ &|1,0\rangle &&= (0,1,0,\ldots ,0)^T\,,\\ &|1,-1\rangle &&= (0,0,1,\ldots ,0)^T\,,\\ &|0,1\rangle && = \cdots\\ &|0,0\rangle && = \cdots\\ &|0,-1\rangle && = \cdots\\ &|-1,1\rangle && = \cdots\\ &|-1,0\rangle && = (0,\ldots,1,0)^T\,,\\ &|-1,-1\rangle && = (0,\ldots,0,1)^T\,.\\ \end{aligned} $$ Then I computed the action of the generators in the spin 1 representation for each factor and expanded the result in this basis. As an example the generator $s_{1x}\otimes s_{2x}$ reads

$$ s_{1x}\otimes s_{2x} = \left( \begin{array}{ccccccccc} 0 & 0 & 0 & 0 & \frac{1}{2} & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & \frac{1}{2} & 0 & 0 & 0 & 0 \\ 0 & \frac{1}{2} & 0 & 0 & 0 & 0 & 0 & \frac{1}{2} & 0 \\ \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 & 0 & \frac{1}{2} & 0 & \frac{1}{2} \\ 0 & \frac{1}{2} & 0 & 0 & 0 & 0 & 0 & \frac{1}{2} & 0 \\ 0 & 0 & 0 & 0 & \frac{1}{2} & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & \frac{1}{2} & 0 & 0 & 0 & 0 \\ \end{array} \right)\,. $$

The final result for $P_2$ instead reads

$$ P_2 = \left( \begin{array}{ccccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & \frac{1}{6} & 0 & \frac{1}{3} & 0 & \frac{1}{6} & 0 & 0 \\ 0 & \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & \frac{1}{3} & 0 & \frac{2}{3} & 0 & \frac{1}{3} & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & \frac{1}{2} & 0 & \frac{1}{2} & 0 \\ 0 & 0 & \frac{1}{6} & 0 & \frac{1}{3} & 0 & \frac{1}{6} & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & \frac{1}{2} & 0 & \frac{1}{2} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \end{array} \right)\,. $$

I'll merely copy-paste my code. With v[i,j] I indicate the vectors of the basis $|i,j\rangle$ exactly as listed above. With s[i] the generators in the spin 1 $\mathrm{SU}(2)$ representation (I don't report here the code with their definitions since it's obvious).

(*Expands into my basis. V is just a name.*)
toBasis = {a_, b_, c_} :> a V[1] + b V[0] + c V[-1];

s[i_, j_] := Flatten[
Table[
  CircleTimes[
  (*Compute the action of the spin 1 generators*)
    s[i].v[m] /. toBasis,
    s[j].v[n] /. toBasis
  ] //. {
  (*Uses bilinearity*)
    CircleTimes[c___, a_ V[b_], d___] :> a CircleTimes[c, V[b], d], 
    CircleTimes[a___, b_ + c_, d___] :> 
    CircleTimes[a, b, d] + CircleTimes[a, c, d]
  } /. {
 (*Writes V as an explicit vector from my basis, and 0 as the zero vector*)
   CircleTimes[V[x_], V[y_]] :> v[x, y], 
   CircleTimes[a___, 0, b___] :> 0 v[1, 1]},
 (*Iterates to find all s[i,j]*)
 {m, 1, -1, -1}, {n, 1, -1, -1}],1]

And then define

s1dots2 = s[1, 1] + s[2, 2] + s[3, 3];

P2 = 1/3 IdentityMatrix[9] + 1/2 s1dots2 + 1/6 s1dots2.s1dots2;

Result:

P2.v[1, 1] - v[1, 1]
P2.(v[0, 1] + v[1, 0]) - (v[0, 1] + v[1, 0])
P2.(v[-1, 1] + 2 v[0, 0] + v[1, -1]) - (v[-1, 1] + 2 v[0, 0] + v[1, -1])
P2.(v[0, -1] + v[-1, 0]) - (v[0, -1] + v[-1, 0])
P2.v[-1, -1] - v[-1, -1]

$\{0, 0, 0, 0, 0, 0, 0, 0, 0\}$

I didn't check if it was zero on the states with lower $S$ because I didn't feel like digging up my Clebsch-Gordan table. Obviously it's also easy to see

P2.P2 - P2 // Flatten // DeleteDuplicates

$\{0\}$

MannyC
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