Recent content by diracsgrandgrandson

Yes, it is part of a larger exercise. It's Peskin&Schroeder 3.4 (e). One has to construct the diagonalized Hamiltonian. I have arrived at the following expression: $$H = \frac{1}{2}\int d^3p\xi^{rT}p\sigma\xi^{s}(a^{r\dagger}_pa^s_p+a^{s\dagger}_pa^r_p),$$ where ##p\sigma =...

But the anticommutator that is independent of ##p## is not the same expression as ##A(p)##, and therefore it being independent does not say anything about ##A(p)##. Am I missing something?

We use the invariance of the measure under ##p\rightarrow -p## to get $$-\int d^3p\xi^{rT}\mathbf{p}\mathbf{\sigma}\xi^s(a^{r\dagger}_{-p}a^s_{-p}+a^{s\dagger}_{-p}a^r_{-p}) = -\int d^3p\xi^{rT}\mathbf{p}\mathbf{\sigma}\xi^sA(-p).$$ If this pesky ##A(-p)## can be shown to be equal to ##A(p)## or...

Ahh I see, that clarifies it, thank you. I find the problem is formulated in a very confusing way, at least for a beginner like me :)

Doesn't that give zero? $$ \begin{align*} \chi^\dagger \sigma^2 \chi^* &= \begin{pmatrix} \chi_1^* & \chi_2^* \end{pmatrix} \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \begin{pmatrix} \chi_1^* \\ \chi_2^* \end{pmatrix} \\ &= \begin{pmatrix} \chi_1^* & \chi_2^* \end{pmatrix} \begin{pmatrix} -...

I am stuck at the final part where one is supposed to show that the derivative of the second term of the action gives the mass term in the Majorana equation. For $$\chi^T\sigma^2\chi = -(\chi^\dagger\sigma^2\chi^*)^*$$ we get $$\frac{\delta}{\delta\chi^\dagger}(\chi^\dagger\sigma^2\chi^*)^*$$...