I am reading Peskin's and Schroeder (P&S), "An introduction to Quantum Field Theory", specifically the first paragraph on page 499 in section 15.4 "Basic Facts about Lie Algebras". At some point, the authors claim that the invariant combination of two spinors is:


and I would like to ask what is meant by the above-mentioned inner product? Does the author (secretly) imply that one of the two spinors ($\eta_{\alpha}$ or $\xi_{\beta}$) is actually a complex conjugate of another spinor? Or is the complex conjugate form of one of the two spinors given by contracting the one spinor with the Levi-Civita tensor, i.e.:


or something like that? And if so, why?

Any help will be appreciated!


2 Answers 2


P&S are talking about the spinor/defining/fundamental representation $$\eta,\xi~\in~ V~\cong~ \mathbb{C}^2$$ of $SU(2)$.

  1. The expression $\epsilon^{\alpha\beta}\eta_{\alpha}\xi_{\beta}$ is $SU(2)$-invariant because the determinant of an $SU(2)$-matrix is 1.

  2. One can use this to show that the complex conjugate spinor representation $\bar{V}$ with complex conjugate matrix $\bar{U}=\epsilon U \epsilon^{-1}$ is equivalent to the spinor representation $V$.

  3. In fact the spinor representation $V\cong\mathbb{H}$ is a pseudoreal/quaternionic representation of $SU(2)\cong U(1,\mathbb{H})$.

  • $\begingroup$ So, given that $\eta_{\alpha}\rightarrow U_{\alpha\beta}\eta_{\beta}$ and that $\xi_{\alpha}\rightarrow U_{\alpha\beta}\xi{\delta}$, then the expression $\epsilon^{\alpha\beta}\eta_{\alpha}\xi_{\beta}\rightarrow \epsilon^{\alpha\beta}U_{\alpha\gamma}\eta_{\gamma}U_{\beta\delta}\xi_{\delta}$ is invariant because of the determinant being one right?? It checks out I think because the terms proportional to $\epsilon^{\alpha\beta}U_{\alpha0}U_{\beta0}\eta_0\xi_0$ and $\epsilon^{\alpha\beta}U_{\alpha1}U_{\beta1}\eta_1\xi_1$ vanish $\endgroup$
    – schris38
    Jun 10 at 13:18
  • $\begingroup$ And how can we use this to show that the complex conjugate spinor representation is equivalent to the original one? If this is true, does this explain why we do not form invariant products from combining one spinor from one representation and one spinor from the other (complex conjugate), and rather we use one from either? $\endgroup$
    – schris38
    Jun 10 at 13:21
  • $\begingroup$ I mean it doesn't matter from which representation we take our spinors to form invariant quantities, since the two representations are equivalent... $\endgroup$
    – schris38
    Jun 10 at 13:24
  • 1
    $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Jun 10 at 13:35
  • $\begingroup$ Okay thank you so much. $\endgroup$
    – schris38
    Jun 10 at 13:45

I would like add to Qmechanic's excellent answer a bit of context from the practical point of view.

Apparently the question arises in classical Quantum mechanics not accounting for relativity. So then we can assume that spinors in 3D transform under the effect of rotations like representations of SU(2) which is the universal covering group of SO(3), the 3D rotation group. The "S" (special) means that the determinant of the unitary matrices of SU(2) is 1.

A bilinear object like $\epsilon_{\alpha\beta}\eta^\alpha \xi^\beta$ with $\eta^\alpha$ and $\xi^\alpha$ as components of non-relativistic spinors in 3D space transforms like

$$\eta'^1 = a\eta^1+b\eta^2\quad\text{and} \quad \eta'^2 = c\eta^1 + d\eta^2$$

if $$U = \left(\begin{array}{cc} a & b\\ c & d\end{array}\right)$$

is a transformation matrix $\in SU(2)$.

If $\xi$ transforms in the same way we find that

$$\epsilon_{\alpha\beta}\eta^\alpha\xi^\beta=\eta'^1\xi'^2 - \eta'^2\xi'^1 = (ad-bc)( \eta^1\xi^2 - \eta^2\xi^1) = \eta^1\xi^2 - \eta^2\xi^1$$

due to the unimodularity of the matrices of SU(2) ($det U=1$).

This means that under SU(2) $$\epsilon_{\alpha\beta}\eta^\alpha\xi^\beta$$ and in particular $$\epsilon_{\alpha\beta}\eta^\alpha\eta^\beta$$ transforms as a scalar.

On the other hand we expect the bilinear

$$\eta^1 \eta^{\ast 1} + \eta^2 \eta^{\ast 2}$$

to be invariant under unitary transformations -- it transforms like a scalar too. We now can identify both bilinear products with each other which leads to the identification of $(\eta^{\ast 1}, \eta^{\ast 2})$ with $(\eta^2, -\eta^1)$.

In other words it means that in 3D spinors and complex-conjugated spinors transform in a very similar way, technically speaking both respresentations are equivalent. This is only true in 3D-space. In Minkowski space, however, the transformation group of spinors is SL(2,C) and the representations of spinors and their complex-conjugated counterpart are no longer equivalent.

It is useful to add that if apart from contravariant spinors $\eta^\alpha$ also covariant spinors $\xi_\alpha$ can be introduced and defined:

$$\eta_1 = \eta^2 \quad \text{and}\quad \eta_2 = -\eta^1$$

We can shortcut this as:

$$\epsilon_{\alpha \beta}\eta^\beta =\eta_\alpha \equiv \eta^{\ast \alpha}$$

and then we can write:

$$\sum_{\alpha=1,2} \eta^\alpha\eta^{\ast \alpha} = \sum_{\alpha=1,2} \eta^\alpha \eta_\alpha$$

  • $\begingroup$ Okay! This is most certainly useful! Thank you so much! $\endgroup$
    – schris38
    Jun 10 at 15:58

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