# Why is $\overline{\Lambda_s \psi} = \bar{\psi} \Lambda_s^{-1}$?

In my text Schwartz writes: since $$\bar{\psi} \gamma^\mu \psi$$ transforms like a 4-vector, we can deduce that

$$\Lambda_s^{-1} \gamma^\mu \Lambda_s = (\Lambda_V)^{\mu \nu} \gamma^\nu, \tag{10.78}$$

where $$\Lambda_s$$ and $$\Lambda_V$$ are the Lorentz transformation acting on Dirac spinors and 4-vectors, respectively.

In this argument, it seems that the equality

$$\overline{\Lambda_s \psi} = \bar{\psi} \Lambda_s^{-1}$$

is implied. I am having trouble seeing why this is true. Expanding this out, we need

$$\psi^\dagger \Lambda_s^\dagger \gamma^0 = \psi^\dagger \gamma^0 \Lambda_s^{-1}$$

so this is equivalent to showing that

$$\Lambda_s^\dagger \gamma^0 \Lambda_s = \gamma^0.$$

In my attempt I used the Weyl representation, although I assume it must be true in general too. In the Weyl representation we have

$$\Lambda_s = \exp(i \theta_{\mu \nu} S^{\mu \nu}), \tag{10.71}$$ $$S^{ij} = \frac{1}{2} \epsilon_{ijk} \begin{bmatrix}\sigma_k & 0 \\ 0 & \sigma_k\end{bmatrix}, \; S^{0i} = -\frac{i}{2} \begin{bmatrix}\sigma_i & 0 \\ 0 & -\sigma_i\end{bmatrix}. \tag{10.72}$$

Now I can verify the equation for certain $$\Lambda_s$$, namely those generated by a single $$S^{\mu \nu}$$. For example with $$S^{01}$$,

$$\Lambda_s = \exp(2i \beta S^{01}) = \exp \begin{bmatrix} 0 & \beta & 0 & 0 \\ \beta & 0 & 0 & 0 \\ 0 & 0 & 0 & -\beta \\ 0 & 0 & -\beta & 0 \end{bmatrix} = \begin{bmatrix} \cosh\beta & \sinh\beta & 0 & 0 \\ \sinh\beta & \cosh\beta & 0 & 0 \\ 0 & 0 & \cosh\beta & -\sinh\beta \\ 0 & 0 & -\sinh\beta & \cosh\beta \end{bmatrix}$$

and the desired $$\Lambda_s^\dagger \gamma^0 \Lambda_s = \gamma^0$$ can be easily verified. Similarly for any other specific $$S^{\mu \nu}$$. But this isn't enough to show it in general. The form of $$\exp(i \theta_{\mu \nu} S^{\mu \nu})$$ gets complicated for general $$\theta_{\mu \nu}$$. And we can't decompose it like $$\exp(i \theta_{01} S^{01}) \exp(i \theta_{02} S^{02}) \cdots$$ since the $$S^{\mu \nu}$$ components don't commute in general. So I'm stuck with trying to figure out why $$\Lambda_s^\dagger \gamma^0 \Lambda_s = \gamma^0$$ for general $$\Lambda_s$$.

Why $$\Lambda_s^\dagger \gamma^0 \Lambda_s = \gamma^0$$ for general $$\Lambda_s$$?

Expanding infinitesimal form of $$\Lambda_s = \exp{\left(i \theta_{\mu\nu}S^{\mu\nu}\right)}$$ and keeping only first order terms gives this:

\begin{align} \Lambda_s^\dagger \gamma^0 \Lambda_s &= \left(1-i \theta_{\mu\nu}S^{\mu\nu}\right) \gamma^0 \left(1+i \theta_{\mu\nu}S^{\mu\nu}\right) \\ & \approx \gamma^0 - {i \gamma^0 \theta_{\mu\nu}S^{\mu\nu}} + i \gamma^ 0 \theta_{\mu\nu}S^{\mu\nu} \\ & = \gamma^0 \end{align}

• This works for infinitesimal Lorentz transformations but not in general, right?
– jcai
May 25, 2020 at 8:51

I found that part of Schwartz's derivations on the next page explains this retroactively. As preliminaries, note that $$(\gamma^0)^2 = \mathbf{1}$$ and $$\gamma^{0\dagger} = \gamma^0$$, as well as

$$\gamma^0 (S^{\mu\nu})^\dagger \gamma^0 = S^{\mu \nu}.$$

These can be derived from the Dirac algebra or checked explicitly with the Weyl representation. Then we can derive,

$$(\gamma^0 \Lambda_s \gamma^0)^\dagger = \gamma^0 \exp(i\theta_{\mu \nu} S^{\mu \nu})^\dagger \gamma^0 = \exp(-i\theta_{\mu \nu}\gamma^0 S^{\mu\nu\dagger}\gamma^0) = \exp(-i\theta_{\mu\nu}S^{\mu\nu}) = \Lambda_s^{-1}.$$

The only nontrivial step is the use of the identity

$$\gamma^0 \exp(M) \gamma^0 = \exp(\gamma^0 M \gamma^0)$$

for any matrix $$M$$. We can check this by Taylor expanding:

$$\exp(\gamma^0 M \gamma^0) = \sum_{n=0}^\infty \frac{1}{n!} (\gamma^0 M \gamma^0)^n = \sum_{n=0}^\infty \frac{1}{n!} \gamma^0 M^n \gamma^0 = \gamma^0 \exp(M) \gamma^0,$$

where we got $$(\gamma^0 M \gamma^0)^n = \gamma^0 M^n \gamma^0$$ by repeatedly contracting $$\gamma^0 \gamma^0 = \mathbb{1}$$.

To nudge the result to the desired form, we have,

$$\gamma^0 \Lambda_s^\dagger \gamma^0 = (\gamma^0 \Lambda_s \gamma^0)^\dagger = \Lambda_s^{-1}$$

$$\Lambda_s^\dagger \gamma^0 \Lambda_s = (\gamma^0)^{-1} = \gamma^0,$$

as desired.

In general, $$\overline{AB}=\bar B \bar A$$. Hence $$\overline{\Lambda_{s}\psi}=\bar \psi \Lambda ^{-1}_{s}$$ since $$\Lambda_{s}$$ is real and $$\Lambda^{t}_{s}=\Lambda^{-1}_{s}.$$

• I don't understand what you mean. The $\bar{\psi} \equiv \psi^\dagger \gamma^0$ notation only applies to Dirac spinors. Writing $\overline{\Lambda_s \psi} = \bar{\psi} \bar{\Lambda}_s$ would be nonsensical because $\bar{\Lambda}_s$ isn't well-defined. Even if you did define $\bar{\Lambda}_s = \Lambda_s^\dagger \gamma^0$, the equation $\overline{\Lambda_s \psi} = \bar{\psi} \bar{\Lambda}_s$ would be incorrect.
– jcai
May 25, 2020 at 3:50
• Also $\Lambda_s$ isn't always real.
– jcai
May 25, 2020 at 3:52
• $\overline{\Lambda_s \psi} = \bar{\psi} \Lambda_s^{-1}$ was in the title of the post - and you asked it again in the body. You're only entitled to one question per post and checking your calculations is off topic. And $\Lambda_s$ is the Lorentz transformation which is real - the exponential form is a complex exponential but it's not denoted by $\Lambda_s$. May 25, 2020 at 19:59