# Basic question on a vector superfield

Is the following expression Hermitian:

$$\theta \xi(x)+\theta^*\xi^{\dagger}(x)+\theta\sigma^{\mu}\theta^*\nu_{\mu}(x)$$

$$\theta$$ is a left-handed spinor coordinate [(1/2, 0) representation of SO(4)], $$\xi$$ is a left-handed spinor field, and $$\nu_\mu$$ is a real vector field.

Normally:

$$(\theta \xi)^{\dagger}=\xi^{\dagger} \theta^{\dagger}$$

However, since theta is a coordinate and not a field, it just gets complex conjugated instead of daggered and there is also no order change, so that the first two terms added together in the very top expression are Hermitian and transform into each other under the adjoint operation.

However, if this is the case, then the third term is not Hermitian but anti-Hermitian. The third term is only Hermitian is if you treat the anticommutating variable $$\theta$$ as a field (and hence changes order).

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Avodyne
These expressions look like they're from Srednicki's book. He defines complex conjugation as reversing the order; see page 611, after eq.(95.9).

nrqed
Homework Helper
Gold Member
Is the following expression Hermitian:

$$\theta \xi(x)+\theta^*\xi^{\dagger}(x)+\theta\sigma^{\mu}\theta^*\nu_{\mu}(x)$$

$$\theta$$ is a left-handed spinor coordinate [(1/2, 0) representation of SO(4)], $$\xi$$ is a left-handed spinor field, and $$\nu_\mu$$ is a real vector field.

Normally:

$$(\theta \xi)^{\dagger}=\xi^{\dagger} \theta^{\dagger}$$

However, since theta is a coordinate and not a field, it just gets complex conjugated instead of daggered and there is also no order change, so that the first two terms added together in the very top expression are Hermitian and transform into each other under the adjoint operation.
Theta is a two-component spinor. It's not a quantum field but is still a spinor. The application of the dagger does change the order and the third term is indeed hermitian.

There is also an order change in the first two terms. However, for these terms, there is an identity that says that the order in these expressions does not matter, i.e.

$$\theta^a \xi_a = \xi^a \theta_a$$

or, in terms of everything with lower indices,

$$\theta_a (-i \sigma_2)^{ab} \xi_b = \xi_b (-i \sigma_2)^{ba} \theta_a$$

What happens is that there are two changes of signs: one from moving a spinor component through the other one, and a minus sign from the antisymmetry of the matrix $\sigma_2$

What is confusing is that when they write $\theta \xi [/tex], this is not just the product of the components of the spinors, there is also the [itex] \sigma_2$ matrix in between.

Thanks everyone. It was just confusing because normally complex conjugation involves no reordering since you perform it on ordinary numbers which commute. I also totally forgot about the convention of switching upper and lowers to allow the product of two anticommuting variables to commute.

I'm more used to using the 2-dimensional Levi-Civita instead of $$-i\sigma_2$$ but it's interesting to see that convention is used too.