Unitary Lorentz transformation on quantized Dirac spinor

Question

I am stuck again on page 59 of Peskin and Schroeder. In particular, I do not know how they get equation (3.110). Let me first give some background in the way that I understand it (but I might be completely wrong).

A unitary operator $U(\Lambda)$ acts on states as follows: \begin{equation} |p,s\rangle \rightarrow U(\Lambda)|p,s\rangle \end{equation} and therefore any operator, such as a Dirac field, transforms as: \begin{equation} \psi'(x) = U(\Lambda)\psi(x)U^{-1}(\Lambda) \end{equation} Now, from equation (3.109): \begin{equation} U(\Lambda) a_p^s U^{-1}(\Lambda) = \sqrt{\frac{E_{\Lambda p}}{E_p}}a^s_{\Lambda p} \end{equation} we can find the transformation of the positive frequency solution of $\psi$: \begin{equation} U(\Lambda) \psi U^{-1}(\Lambda) = U(\Lambda) \int \frac{d^3 p}{(2\pi)^3} \frac{1}{\sqrt{2 E_p}} \sum_s a_p^s u^s(p) e^{-ip\cdot x} U^{-1}(\Lambda) \end{equation} \begin{equation} \Rightarrow U(\Lambda) \psi U^{-1}(\Lambda) = \int \frac{d^3 p}{(2\pi)^3} \frac{1}{\sqrt{2 E_p}} \sum_s U(\Lambda) a_p^s U^{-1} (\Lambda) U(\Lambda) u^s(p)U^{-1}(\Lambda) e^{-ip\cdot x} \end{equation} and using equation (3.109) this becomes: \begin{equation} \Rightarrow U(\Lambda) \psi U^{-1}(\Lambda) = \int \frac{d^3 p}{(2\pi)^3} \frac{1}{2 E_p}\sqrt{2 E_{\Lambda p}} \sum_s a^s_{\Lambda p} U(\Lambda) u^s(p)U^{-1}(\Lambda) e^{-ip\cdot x} \end{equation} and from this point I have no idea how to get to equation (3.110). If anybody could push me in the right direction, then this will be greatly appreciated. (I am aware that the integration measure is Lorentz invariant.)

Another question: does anybody have any other references/notes/books where they discuss how the quantized Dirac operator field transforms? I find P&S explanation thoroughly confusing (as may have become clear from the questions I have been asking recently on this forum :) ), but I cannot find any other book that treats this stuff.

Answer 1

You can find the transformation law for $u^s(p)$ by demanding that the spinor field transform as

$\psi(x)\rightarrow \psi'(x')=U^{-1}(\Lambda)\psi(x')U(\Lambda)=\Lambda_{1/2}\psi(x)$.

You already know how the creation / annihilation operators transform from the condition that the 1-particle states transform correctly and you can then find the correct transformation law for $u^s(p)$. Then, armed with this transformation law you can do the transformation in the opposite direction (which is what Peskin and Schroeder do) and you get their result.

In particular, we have

$U^{-1}(\Lambda)\psi(x')U(\Lambda)=\displaystyle\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}}U^{-1}(\Lambda)a^{s\dagger}_pU(\Lambda)u^s(p)e^{ip.\Lambda x} + $similar terms

where I've ignored the summation and the other operator since its analogous to this.

Changing the dummmy variable $p$ to $\Lambda p$ we get

$U^{-1}(\Lambda)\psi(x')U(\Lambda)=\displaystyle\int\frac{d^3\Lambda p}{(2\pi)^3}\frac{1}{\sqrt{2E_\Lambda p}}U^{-1}(\Lambda)a^{s\dagger}_{\Lambda p}U(\Lambda)u^s(\Lambda p)e^{ip. x}$

since $(\Lambda p)(\Lambda x)$=$px$

We also have $U^{-1}(\Lambda)a^{s\dagger}_{\Lambda p}U(\Lambda)=\sqrt{\frac{2E_p}{2E_{\Lambda p}}}a^{s\dagger}_{p}$ giving us

$U^{-1}(\Lambda)\psi(x')U(\Lambda)=\displaystyle\int\frac{d^3\Lambda p}{(2\pi)^3}\frac{\sqrt{2E_p}}{2E_\Lambda p}a^{s\dagger}_{ p}u^s(\Lambda p)e^{ip. x}$

The measure is Lorentz invariant so we can rewrite it as

$U^{-1}(\Lambda)\psi(x')U(\Lambda)=\displaystyle\int\frac{d^3 p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}}a^{s\dagger}_{ p}u^s(\Lambda p)e^{ip. x}$

Now we demand that this equals

$\Lambda_{1/2}\psi(x)=\Lambda_{1/2}\displaystyle\int\frac{d^3 p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}}a^{s\dagger}_{ p}u^s(p)e^{ip. x}$

and we immediately see that we must have

$u^s(\Lambda p)=\Lambda_{1/2}u^s(p)$.

Now you can apply the inverse transformation, $\psi(x)\rightarrow U(\Lambda)\psi(x)U^{-1}(\Lambda)$ to get the result Peskin & Schroeder have.