Weyl Spinors Transformation, QFT1, Peskin, Chapter 3

[Pouramat]

Homework Statement

The transformation law for Weyl spinors is as following (3.37); these transformation laws are connected by complex conjugation; using the identity (3.38)

Relevant Equations

(3.37) and (3.38) peskin

\begin{align}
\psi_L \rightarrow (1-i \vec{\theta} . \frac{{\vec\sigma}}{2} - \vec\beta . \frac{\vec\sigma}{2}) \psi_L \\
\psi_R \rightarrow (1-i \vec{\theta} . \frac{{\vec\sigma}}{2} + \vec\beta . \frac{\vec\sigma}{2}) \psi_R
\end{align}
I really cannot evaluate these from boost and rotation generator which was introduced in (3.26) and (3.27) Peskin.
Although the main porblem is the identity introduced below:
$$
\sigma^2 \vec\sigma^* = -\vec \sigma \sigma^2
$$
My attempt:
$$
\sigma^1 =
\begin{pmatrix}
0 & 1 \\
1 & 0 \\
\end{pmatrix};
\sigma^2 =
\begin{pmatrix}
0 & -i \\
i & 0 \\
\end{pmatrix};
\sigma^3 =
\begin{pmatrix}
1 & 0 \\
0 & 1 \\
\end{pmatrix}
$$
the h.c. of the Pauli Sigma matrices is as following:
$$
(\sigma^1)^* =
\begin{pmatrix}
0 & 1 \\
1 & 0 \\
\end{pmatrix};
(\sigma^2)^* =
\begin{pmatrix}
0 & i \\
-i & 0 \\
\end{pmatrix};
(\sigma^3)^* =
\begin{pmatrix}
1 & 0 \\
0 & 1 \\
\end{pmatrix}
$$
so $\sigma^2 = diag(1,1)$,right?
I think my problem is that I cannot write the identity in components form.

 

[vanhees71]

Of course $\sigma^3=\mathrm{diag}(1,-1)$. I'd say the most simple way is just to prove the equation by doing the matrix multiplications for the three matrices :-).

 

[Pouramat]

Of course $\sigma^3=\mathrm{diag}(1,-1)$. I'd say the most simple way is just to prove the equation by doing the matrix multiplications for the three matrices :-).

Ohhh, Yes, it was a typo :)
the LHS:
$$
\large \begin{pmatrix}
0 & 1 \\
1 & 0 \\
\end{pmatrix}
+
\begin{pmatrix}
0 & i \\
-i & 0 \\
\end{pmatrix}
+
\begin{pmatrix}
1 & 0 \\
0 & -1 \\
\end{pmatrix} =
\begin{pmatrix}
1 & 1+i \\
1-i & -1 \\
\end{pmatrix}
$$
which should be equal to RHS:
$$
-\large( \begin{pmatrix}
0 & 1 \\
1 & 0 \\
\end{pmatrix}
+
\begin{pmatrix}
0 & -i \\
i & 0 \\
\end{pmatrix}
+
\begin{pmatrix}
1 & 0 \\
0 & -1 \\
\end{pmatrix})
=
\begin{pmatrix}
-1 & 1+i \\
1-i & 1 \\
\end{pmatrix}
$$
but it is not!

 

[vanhees71]

I have no idea what you are doing here. You are supposed to show that (I'll give one example)
$$\sigma^2 \sigma^{1*}=-\sigma^1 \sigma^2.$$
Take the right-hand side and use the anti-commutation relations
$$\{\sigma^j,\sigma^k\}=2 \delta^{jk}$$
to get
$$-\sigma^1 \sigma^2 =+ \sigma^2 \sigma^1.$$
Since $\sigma^1$ is real this hows that
$$-\sigma^1 \sigma^2=\sigma^2 \sigma^{1*}.$$
Now you can prove the other two identities!

 

[Pouramat]

I have no idea what you are doing here. You are supposed to show that (I'll give one example)
$$\sigma^2 \sigma^{1*}=-\sigma^1 \sigma^2.$$
Take the right-hand side and use the anti-commutation relations
$$\{\sigma^j,\sigma^k\}=2 \delta^{jk}$$
to get
$$-\sigma^1 \sigma^2 =+ \sigma^2 \sigma^1.$$
Since $\sigma^1$ is real this hows that
$$-\sigma^1 \sigma^2=\sigma^2 \sigma^{1*}.$$
Now you can prove the other two identities!

thank you @vanhees71. Yes you are right.

 

[Pouramat]

dear @vanhees71
And 1 more question, do you have any idea to explicitly show that $\sigma^2 \psi^*_L$ transforms like a right-handed spinor? using the identity.

 

[vanhees71]

Take the conjugate complex of the transformation law for $\psi_L$ and multiply with $\sigma^2$. Then use the (anti-)commutation relation for the complex conjugated Pauli matrices just proven.

 

[Preyn]

Take the conjugate complex of the transformation law for $\psi_L$ and multiply with $\sigma^2$. Then use the (anti-)commutation relation for the complex conjugated Pauli matrices just proven.

I have a question to that. Peskin asks us to show that $\sigma^2\psi_L$ transforms like a right-handed spinor. So why can we just consider the transformation of $\psi_L$ alone to then multiply by $\sigma^2$, and not the transformation of whatever the object $\sigma^2\psi_L$ is?

Edit: I think that my question is why we don't need to take the $\sigma^2$ in front into account when doing the transformation. After all, $\sigma^2\psi_L\ne\psi_L$, so it should be taken into account when transforming, right?

 

[vanhees71]

You now the transformation representing Lorentz transformations. The infinitesimal version is already sufficient, i.e., the generators for left- and right-handed Weyl spinors. Then it's just Pauli-matrix gymnastics to prove that $\sigma^2 \psi_L$ transforms like $\psi_R$. Mathematically it's of course not the special pauli matrix $\sigma^2$ which is behind this but the antisymmetric product of two spinors given by $\epsilon_{ij}$ (with $\epsilon_{12}=-\epsilon_{21}=1$) which is relevant here.

That it's an antisymmetric product instead of a symmetric hints at the fact that one should rather argue with Grassmann-number valued fields for fermions, which is what comes out also using the path-integral formulation.

 

[Preyn]

You now the transformation representing Lorentz transformations. The infinitesimal version is already sufficient, i.e., the generators for left- and right-handed Weyl spinors. Then it's just Pauli-matrix gymnastics to prove that $\sigma^2 \psi_L$ transforms like $\psi_R$. Mathematically it's of course not the special pauli matrix $\sigma^2$ which is behind this but the antisymmetric product of two spinors given by $\epsilon_{ij}$ (with $\epsilon_{12}=-\epsilon_{21}=1$) which is relevant here.

That it's an antisymmetric product instead of a symmetric hints at the fact that one should rather argue with Grassmann-number valued fields for fermions, which is what comes out also using the path-integral formulation.

I’m sorry, but I fail to see how this answers my question. What doesn’t make sense for me is the following:
We are asked to show that the object $\sigma^2\psi_L^*$ (in my previous post I forgot the complex conjugate, sorry) transforms as a right-handed spinor. For me, this means that we need to consider $$\sigma^2\psi_L^*\to(\sigma^2\psi_L^*)’.$$However, you are saying that we should consider $$\psi_L^*\to(\psi_L’)^*,$$to then just multiply by $\sigma^2$. I can see that this gives the correct result, but I fail to see why we are allowed to do it like this, considering that, in general, these are two different quantities. I hope that this clarifies my confusion.

 

[vanhees71]

But then you solved the problem. I don't understand, what's still unclear.

 

[Preyn]

But then you solved the problem. I don't understand, what's still unclear.

I wouldn't consider it solved, since I don't know why I am allowed to do the transformation like this. The goal for me is to not blindly solve it and move on, but to understand why it works like this. What I want to know is why I can consider the transformation $\psi_L^*\to(\psi_L')^*$, even though one is asked to consider the transformation $\sigma^2\psi_L^*\to(\sigma^2\psi_L)'$, since we are asked to show that $\sigma^2\psi_L^*$ transforms as a right-handed spinor. I fail to see what confuses you in my question...

 

[vanhees71]

Why? If you know how $\psi_L$ transforms, you can just take the conjugate complex of the equation to know how $\psi_L^*$ transforms. Then you multiply by $\sigma^2$ to find out our $\sigma_2 \psi_L^*$ transforms and compare this with the transformation of $\psi_R$.