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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.

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    $\begingroup$ The transformation law $A \rightarrow A = U(\Lambda) A U^{-1}(\Lambda)$ is for operators, but $u_s(p)$ is not an operator, rather, it is the coefficient in front of creation or annihilation operators. $\endgroup$ – Jia Yiyang Nov 23 '13 at 1:19
  • $\begingroup$ @JiaYiyang thanks for correcting me! I've now edited the post; hopefully it is better now. $\endgroup$ – Hunter Nov 23 '13 at 1:42
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    $\begingroup$ Thanks for your reply. Could you please explain to me why $u^s(p)$ is just a number? I thought it was a Dirac spinor (4 component vector) which is also dependend on $p$ (and hence I would think that a Lorentz transformation would influence it). Working under the assumption that $u^s(p)$ is just a number, then I see how I can get to: $\endgroup$ – Hunter Nov 23 '13 at 3:52
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    $\begingroup$ $U(\Lambda) \psi U^{-1}(\Lambda) = \int \frac{d^3 \tilde{p}}{(2\pi)^3} \frac{1}{\sqrt{2 E_{\tilde{p}}}} \sum_s a^s_{\tilde{p}} u^s(\Lambda^{-1} \tilde{p}) e^{-i\tilde{p}\cdot \Lambda x}$. However, I do not understand why $u^s(\Lambda^{-1} \tilde{p}) = \Lambda_{1/2}^{-1}u^s(\tilde{p})$. Have they derived this somewhere in their book? Or can I derive it by looking at infinitesimal transformations of $\Lambda_{1/2}^{-1}$? Or is there something else I am missing? $\endgroup$ – Hunter Nov 23 '13 at 3:53
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    $\begingroup$ No you are totally right; I was being sloppy, and so are Peskin and Shroeder with their notation. They should really write a spinor index on $\psi$ so that the transformation reads $U(\Lambda)\psi^\alpha(x)U^{-1}(\Lambda) = (\Lambda_{\frac{1}{2}}^{-1})^\alpha_{\phantom\alpha\beta}\psi^\beta(\Lambda x)$. What they wrote is just a shorthand for that full expression. This means that in the integral, you will have $(u^s(p))^\alpha$ which are numbers, and then the computation goes through as desired. $\endgroup$ – joshphysics Nov 23 '13 at 18:28
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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.

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  • $\begingroup$ I a sorry but your answer is wrong. The mistake comes from the transformation of the creation operator which is missing the Wigner rotation from the little group for particles with spins. It's the whole point about spins. $\endgroup$ – TwoBs Oct 29 '16 at 21:25
  • $\begingroup$ @TwoBs The Wigner rotation doesn't play a role when we boost in the direction of the spin or rotate about the spin axis, which is the case Peskin & Schroeder are considering. $\endgroup$ – Okazaki Oct 30 '16 at 11:39
  • $\begingroup$ ok, I see, you restricted yourself the trivial case. It wasn't clear from your answer. Could you please edit your answer and add that you are restricting to such a case. Especially in the conclusion about the spinor transformation law which is not the general one. (Or even better, add the general one). After your edits I will also be able to remove the downvoted $\endgroup$ – TwoBs Oct 30 '16 at 17:36

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