0
$\begingroup$

I'm reading chapter 3 in Peskin on the Dirac equation. First of all, they say since Dirac satisfies Klein Gordon it can be written as a linear combination of plane waves. This is fine. So a general solution can be written as a superposition of plane waves of the form $\psi(x)=u(p)e^{-ip.x}$. Plugging this into the Klein-Gordon equation gives us $p^2=m^2$, so we have $p_0$ constrained by $p_0=\sqrt{\textbf{p}^2+m^2}=E_\textbf{p}$. Given this constraint, would it be more accurate to in fact just write $u(p)=u(\textbf{p})$? My reason for asking is as follows;

Given $\psi(x)$ and fourier transforming the spatial components we have $\psi(x)=\displaystyle\int\frac{d^3p}{(2\pi)^3} \tilde{\psi}(\textbf{p},t)e^{i\textbf{p}.\textbf{x}}$

Putting this into the Klein-Gordon equation gives us $\left(\frac{\partial^2}{\partial t^2}+\textbf{p}^2+m^2\right)\tilde{\psi}(\textbf{p},t)=\left(\frac{\partial^2}{\partial t^2}+E_\textbf{p}^2\right)\tilde{\psi}(\textbf{p},t)=0$

The most general solution of this PDE is then of course

$\tilde{\psi}(\textbf{p},t)=a_\textbf{p}u(\textbf{p})e^{-iE_\textbf{p}t}+b^*_{-\textbf{p}}v(-\textbf{p})e^{iE_\textbf{p}t}$ where $u(\textbf{p})$ and $v(-\textbf{p})$ are four component spinors and $a_\textbf{p}$ and $b^*_{-\textbf{p}}$ are arbitrary coefficients depending on $\textbf{p}$.

Putting this back into the expression for $\psi(x)$ and switching $\textbf{p}\rightarrow-\textbf{p}$ in the second term gives us $\psi(x)=\displaystyle\int\frac{d^3p}{(2\pi)^3} (a_\textbf{p}u(\textbf{p})e^{-ip.x}+b^*_{\textbf{p}}v(\textbf{p})e^{ip.x})$

Each term is separately a solution so putting the first term into the dirac equation gives us $(\gamma^\mu p_\mu-m)u(\textbf{p})=0$

In the rest frame $\textbf{p}=0$ so $p^0=m$ and we find $u(\textbf{0})=\sqrt{m}\left( \begin{array}{c} \xi\\ \xi\\ \end{array} \right)$ with $\sqrt{m}$ included for convenience.

Anyway Peskin does the same thing except they find $u(p_0)=\sqrt{m}\left( \begin{array}{c} \xi\\ \xi\\ \end{array} \right)$ and then boost to find $u(p)$. Then when they write the quantized field in the Schrodinger picture they use $u(\textbf{p})$ and in the Heisenberg picture its back to $u(p)$.

I'm just curious if what I have and what they have are actually the same thing. All books have the same notation as Peskin, but notes by people like Tong and Beisert use my notation, but don't find solutions in different frames like we do so I can't really compare?

Long story short; am I right and is Peskin just using awkward notation?

$\endgroup$
  • $\begingroup$ Each term is separately a solution? A solution to Klein-Gordon, sure, but you then put it into the Dirac equation. Why? $\endgroup$ – Timaeus Jul 30 '15 at 19:27
1
$\begingroup$

The method you used to solve the equation is correct. The reason for which they take a dependence of $u$ from $p^0$ is that they want to write a generic solution for every mass term in the Dirac Equation. In fact, if you change the mass of the particle, you cannot expect $u(\textbf{p})$ to be the same (otherwise, you can label this solution as $u_{m}(\textbf{p})$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.