# Irreducible tensor representations with “covariant” indices

As a follow-up of my question on the "most general" $\mathrm{SU}(2)$-symmetric interaction of two spin 1/2 particles, I ponder the following question:

Consider an operator acting just on one particle of spin 1/2, and just on the spin part. In second quantization, it can be written as $$\sum_{\alpha\beta} \psi^\dagger_\alpha V_{\alpha \beta} \psi_\beta$$

Now, under $SU(2)$, the operators will transform as $$\psi^\dagger_\alpha = \sum_{\alpha'} \psi^\dagger_{\alpha'} D_{\alpha\alpha'}$$ with $D \in SU(2)$.

I can move this transformation to the operator and conclude that it transforms like $$V_{\alpha\beta} \rightarrow \sum_{\alpha'\beta'} D_{\alpha\alpha'} V_{\alpha'\beta'} D_{\beta\beta'}^\dagger$$.

(I might have gotten some of the daggers and indices backwards, but the important point is that one matrix is the adjunct of the other).

Now if I understand correctly, this allows me to conclude that $V_{\alpha\beta}$ transforms "reducibly like a tensor product of two spin 1/2 representations", i.e., like $\frac{1}{2} \otimes \frac{1}{2}$. From that, I conclude that $V$ should decompose into one component that transforms like a spin $0$ particle and one component that transforms like a spin $1$ particle.

Indeed, as $V$ is a $2\times 2$ matrix, I can write it as a linear combination of the unit matrix and the pauli matrices. The former transforms like a scalar, i.e., like spin $0$, whereas the latter transform like vectors. However, I have trouble relating this to what I know about combining spin $1/2$ particles using Clebsh-Gordan coefficients, where I have, e.g., for the singlet $$|0 0\rangle = \frac{1}{\sqrt{2}} \left( \uparrow \downarrow - \downarrow \uparrow\right)$$

Because of this conceptual problem, I also have trouble generalizing this to the case of two interacting spin 1/2 particles, which then should lead to an interaction that transforms reducibly as $1/2 \otimes 1/2 \otimes 1/2 \otimes$ and gives rise to two different singlets.

I'd appreciate it if someone could disentangle my misconceptions...

EDIT: I forgot to specify what I mean with "covariant" in the title: I think an important thing to notice is that the matrix elements $V_{\alpha\beta}$ are not the 4 elements of a "cartesian" tensor of rank 2. The entire operator $V$ might be an element of a tensor of higher rank, or even the sum of elements of tensors of different rank. A simple example for such a thing would be an operator "1 + x", which is the sum of a rank zero tensor (scalar) and the element of a cartesian tensor of rank 1.

Now, what I mean with covariant is that one of the indices of $V_{\alpha\beta}$ transforms with matrix $D$ and the other with matrix $D^\dagger$.

Also important is that the components of the operator transform with the actual $SU(2)$-matrices and not with some rotation matrix $R \in SO(3)$. I guess this is why I have trouble translating the standard literature on tensor operators to my situation...

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You are right about the Pauli/trace thing. This is explained (perhaps in too much generality, since it deals mostly with large representations) in my answer to this question: physics.stackexchange.com/questions/10403/… . – Ron Maimon Apr 5 '12 at 19:04

The issue I think you are having (although it is not entirely clear from what you write) is that

• you have to distinguish between representations and their complex conjugate representations
• specific to $SU(2)$, representations are isomorphic to their complex conjugate representations
• the isomorphism that connects the $\frac{1}{2}$ representations and it's complex conjugate is multiplication by $\varepsilon_{\alpha\beta}$, the completely anti-symmetric two dimensional matrix.

Your spinor $\psi_\beta$ transforms in the $\frac{1}{2}$ representation. And $\psi^\dagger_\beta$ transforms in the complex conjugate representation $\frac{1}{2}^*$. If you look at your formula for the transformation of $\psi$ and $\psi^\dagger$ they are not the same. However if you look at the transformation properties of $\psi^t_{\alpha}\varepsilon_{\alpha\beta}$ you will see that this transforms exactly the same as $\psi^\dagger$ because $\varepsilon \sigma^t \varepsilon= \sigma$ for all of the Pauli matrics $\sigma$. The operator you decomposed (correctly) transforms like $\psi^\dagger\otimes \psi$, which has a singlet part $\psi^\dagger_\alpha\delta_{\alpha\beta}\psi_\beta$. However the two state system which you were originally taught Clebsch-Gordon composition transforms like $\psi^\dagger \otimes \psi^\dagger$, (since you get it from creating two $\frac{1}{2}$ particles). To get it to look the same we insert a copy of the $\varepsilon$ matrix (using $\epsilon^2 = -1$. So the singlet part of $\psi^\dagger\otimes \psi^\dagger$ is $\psi^\dagger_\alpha\epsilon_{\alpha\beta}\psi^\dagger_\beta$, which is what you wrote in ket notation in your answer. The triplet part is $\psi^\dagger\varepsilon\sigma\psi^\dagger$ which you could check is the same as what you know from QM.

To deal with higher dimensional you just have to use an $\varepsilon$ for each index you want to change to the complex conjugate. Just to reiterate, this is an idiosyncracy of $SU(2)$, for $SU(3)$ and higher you have to distinguish between the representations and their complex conjugate. People have careful notation for this stuff, which is laid out clearly in Srednicki IIRC, and I think Ron has an answer explaining it somewhere.

By the by, in many places you will see that people use $i\sigma_y$ instead $\varepsilon$ since they have the same components in the usual convention. This is a little confusing since it makes it look like there's something special about the $y$ direction which is of course not true. It's just a notational coincidence, and I would recommend avoiding using it (especially since strictly speaking as linear maps $\varepsilon$ and $\sigma_y$ don't even operate on the same spaces).

Hope that helps.

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I posted this as you posted your edit. It's pretty unclear what you are trying to say, but I hope this answer will still cover it. – BebopButUnsteady Apr 5 '12 at 18:16
I haven't fully digested your answer yet, but I feel like it is exactly what I was looking for. What I cumbersomely called "covariant" index really relates to what you say about $1/2$ and $1^*/2$. – Lagerbaer Apr 5 '12 at 22:10
One last question... I try to see how $\epsilon_{\alpha\beta} \psi^\dagger_{\beta}$ transforms the same as $\psi$, but I get a minus sign in there, basically from inserting $\epsilon$ twice. Is that a mistake or does that not affect the end result? Is the gist just that for every matrix $D \in SU(2)$, we know that $-D$ is also in $SU(2)$? (Which makes sense as $(-D)(-D)^\dagger = DD^\dagger = 1$ and $\det(-D) = \det(D) = 1$ for matrices of even dimension). – Lagerbaer Apr 5 '12 at 22:37
Ah, I forgot that of course the matrix $D$ is not just a pauli matrix, but can actually be parametrized as $\cos \phi/2 + i \vec{n} \cdot \vec{\sigma} \sin \phi/2$ and if I then do the magic with the $\epsilon$ matrix I get $\epsilon D^* \epsilon = -D$, which cancels the $-1$ from inserting $\epsilon^2$. Thank you very much, your answer was a great help – Lagerbaer Apr 5 '12 at 23:24