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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.

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  • $\begingroup$ How did you compute $P_2|2,0\rangle_c$? Are you sure there are no typos in what you wrote? Shouldn't the $|1,0\rangle$ be $|1,-1\rangle$? Also, did you use $\vec{s}_1\cdot \vec{s}_2 = \frac12 ( S^2 -s_1^2 - s_2^2)$? $\endgroup$ – MannyC Mar 6 at 5:18
  • $\begingroup$ Yes, there was a typo. Thanks for pointing out. $\endgroup$ – git-able Mar 6 at 6:04
  • $\begingroup$ What I did was write $\vec{s}_1 \cdot \vec{s}_2 = s_{1x}\otimes s_{2x} + s_{1y} \otimes s_{1y} + s_{1z} \otimes s_{2z}$ and act it on the states. I calculated for the explicit $9 \times 9$ matrix $P_2$ and act it on the states (their corresponding column matrices) using a computing software for a quick check. $\endgroup$ – git-able Mar 6 at 6:16
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    $\begingroup$ As of now, this is a "check-my-work" question which is unlikely to be of value to future visitors. $\endgroup$ – Norbert Schuch Mar 6 at 8:42
  • $\begingroup$ Ultimately, what I would like to ask is if the properties, $P^2=P$ and $\text{Eigenvalues}(P)=\{1,0\}$, of a projector $P$, are only true in a certain representation (I edited the first paragraph to start with this general question). I just constructed my post this way because I encountered this problem when doing a quick check on subspace projectors of two-spin system. $\endgroup$ – git-able Mar 8 at 2:44
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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

$\{0\}$

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  • $\begingroup$ I actually derived $P_2$ almost similar to what you discussed here (I actually followed the linked post). This is just the same way as writing $P_2$ as $P_2 = \sum_{M^2_Z} |2,M^2_Z \rangle \langle 2, M^2_Z |$, and therefore $P^2 = P$ and $\text{Eigenvalues}(P_2)=\{0,1\}$. So I know this, that in the "single-spin" representation (the two spins treated as a single spin), the said properties of projector are true. However, what I would like to understand is why it seems not to be true (as my example showed) in the two-spin representation. $\endgroup$ – git-able Mar 8 at 2:59
  • $\begingroup$ I read later the answer to that post, otherwise I would have avoided the repetition. What you really wanted to know is why does this not work when you compute explicitly the tensor product and plug the explicit values of the matrices? That is never the most convenient way to do it so I didn't want to give an answer where I do it explicitly. But my guess is that you messed up either the Mathematica code or the definition of the tensor product. And if you didn't use Mathematica it's very easy to miss things in a $9 \times 9$ matrix multiplication. $\endgroup$ – MannyC Mar 8 at 3:15
  • $\begingroup$ I am using Mathematica. But I will check my code again. $\endgroup$ – git-able Mar 8 at 3:27
  • $\begingroup$ I tried it with Mathematica and it works. You can check if the matrix you get is the same (look at the edit). $\endgroup$ – MannyC Mar 8 at 5:01
  • $\begingroup$ Well, that is embarrassing, but thank you for doing a consistency check. It turns out $S^2 \neq S.S$ in matrix multiplication in Mathematica, and I have been using $S^2$ for the last term of $P_2$. When I used $S.S$, we have the same $P_2$ and everything works. $\endgroup$ – git-able Mar 8 at 5:43

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