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In page 42 of Peskin and Schroeder, for

$$\left( \mathcal{J}^{\mu\nu} \right)_{\alpha\beta} = i \left( {\delta^\mu}_{\alpha} {\delta^\nu}_{\beta} - {\delta^\mu}_{\beta} {\delta^\nu}_{\alpha} \right)$$

$$S^{\mu\nu} = \frac{i}{4} \left[ \gamma^\mu , \gamma^\nu \right]$$

and infinitesimal $\omega_{\mu\nu}$, they state that

$$\left[ \gamma^\mu , S^{\rho\sigma} \right] = {\left( \mathcal{J}^{\rho \sigma} \right)^\mu}_\nu \gamma^\nu$$

is equivalent to

$$\left( 1 + \frac{i}{2} \omega_{\rho\sigma} S^{\rho\sigma} \right) \gamma^\mu \left( 1 - \frac{i}{2} \omega_{\rho\sigma} S^{\rho\sigma} \right) = {\left( 1 - \frac{i}{2} \omega_{\rho\sigma} \mathcal{J}^{\rho\sigma} \right)^\mu}_\nu \gamma^\nu$$

which I believe.

However, they then immediately generalise this to the non infinitesimal case and say that

$$\Lambda_{\frac{1}{2}}^{-1}\gamma^{\mu}\Lambda_{\frac{1}{2}}={\Lambda^{\mu}}_{\nu}\gamma^{\nu}$$

where

$$\Lambda_{\frac{1}{2}}=\exp{\left(-\frac{i}{2}\omega_{\mu\nu} S^{\mu\nu}\right)}.$$

How is this allowed? I understand that generally Lorentz transformations can be built up by compounding infinitesimal transformations, but this is for a single transformation by itself, not something like the identity above.

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  • $\begingroup$ oh well if you taylor expand $\Lambda_{\frac{1}{2}}$ first 2 terms will be that transformation you write at 3rd line. and if omega infitesaml you can ignore higher order terms. $\endgroup$
    – physshyp
    Commented Dec 20, 2021 at 2:50
  • $\begingroup$ @physshyp Please post answers as answers, not as comments $\endgroup$
    – BioPhysicist
    Commented Dec 20, 2021 at 3:02
  • $\begingroup$ @physshyp This still doesn't show that the non infinitesimal version satisfies the relation. $\endgroup$
    – Ghorbalchov
    Commented Dec 20, 2021 at 13:10
  • $\begingroup$ you can just prove it by applying the rule you wrote in the second line. which is very straightforward. $\endgroup$
    – physshyp
    Commented Dec 20, 2021 at 23:40
  • $\begingroup$ This is not proving it but rather showing that the finite version implies the infinitesimal version which is already obvious. $\endgroup$
    – Ghorbalchov
    Commented Dec 20, 2021 at 23:45

2 Answers 2

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Actually the relation $$\Lambda_{\frac{1}{2}}^{-1}\gamma^{\mu}\Lambda_{\frac{1}{2}}={\Lambda^{\mu}}_{\nu}\gamma^{\nu}$$ can be deduced from the requirement of relativistic covariance of the Dirac equation. That is the form of Dirac equation remains unchanged under a Lorentz tranformation. Then we can expand $\Lambda_{\frac{1}{2}}$ and $\Lambda^{\mu}_{\nu}$ on both sides of the above relation to obtain the explicit form of $\Lambda_{\frac{1}{2}}$. I think this is more natural than that in Peskin's book.

For more details, you may look at Quantum field theory Chapter 2.5.3 by Eduardo Fradkin

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  • $\begingroup$ This has the same problem though: once you have found $\Lambda_{\frac{1}{2}}$ for an infinitesimal transformation, how do you know that the non infinitesimal version still satisfies the relation? $\endgroup$
    – Ghorbalchov
    Commented Dec 20, 2021 at 13:07
  • $\begingroup$ My understanding is that firstly the Lorentz transformation induces a linear transformation $\Lambda_{\frac{1}{2}}$ in the spinor space(i.e. $\Lambda_{\frac{1}{2}}$ acts on spinor indices). Secondly, to make sure the Dirac equation is covariant under Lorentz transformation, we can find out the relation$ \Lambda_{\frac{1}{2}}^{-1}\gamma^{\mu}\Lambda_{\frac{1}{2}}={\Lambda^{\mu}}_{\nu}\gamma^{\nu} $must hold. Now the matter is how to find out the explicit form(which is just a representation) of $\Lambda_{\frac{1}{2}}$. $\endgroup$
    – Not even trivial
    Commented Dec 20, 2021 at 16:33
  • $\begingroup$ Since $\Lambda_{\frac{1}{2}}$ itself forms a Lie group, we know Lie groups are usually generated by some generators and be expressed in some exponential form $\Lambda_{\frac{1}{2}}=\exp{\left(-\frac{i}{2}\omega_{\mu\nu} S^{\mu\nu}\right)}.$ where $S^{\mu\nu}$ here is the generators. So that relation holds for finite transformation. $\endgroup$
    – Not even trivial
    Commented Dec 20, 2021 at 16:33
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The key thing I was missing is closure. The product of two infinitesimal transformations must also satisfy the relation, which is easy to show in this case. The exponential is then just formed as a product of infinitely many infinitesimal transformations, but as you have already shown that it is closed under multiplication, the exponential will automatically satisfy it.

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