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...

  • $\begingroup$ 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/… . $\endgroup$
    – Ron Maimon
    Commented Apr 5, 2012 at 19:04

1 Answer 1


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 idiosyncrasy of $SU(2)$, for $SU(3)$ and higher you have to distinguish between the representations and their complex conjugate. People have a 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 of $\varepsilon$ 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.

  • $\begingroup$ 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. $\endgroup$ Commented Apr 5, 2012 at 18:16
  • $\begingroup$ 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 $1^*/2$. $\endgroup$
    – Lagerbaer
    Commented Apr 5, 2012 at 22:10
  • $\begingroup$ 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). $\endgroup$
    – Lagerbaer
    Commented Apr 5, 2012 at 22:37
  • $\begingroup$ 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 $\endgroup$
    – Lagerbaer
    Commented Apr 5, 2012 at 23:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.