1

The following paragraph is from pag. 55 of Peskin and Schroeder's An Introduction to Quantum Field Theory:

First consider the propagation amplitude $\langle 0|\psi(x) \bar{\psi}(y)|0 \rangle$, which is to represent a positive-energy particle propagating from $y$ to $x$. In this case we want the (Heisenberg) state $\bar{\psi}(y)|0 \rangle$ to be made up of only positive-energy, or negative-frequency components (since a Heisenberg state $\Psi_H=e^{+iHt}\Psi_S$).

Now, my doubts:

(1) This has nothing to do with the above paragraph, but with my understanding of the mathematical construction of the theory: before in the same chapter, to construct an invariant quadratic on the fields, we made the 'replacement' $u^\dagger \to \bar{u}=u^\dagger\gamma^0$, where $u$ is the spinor part in the solution to Dirac's equation. My question is: why is this replacement physically acceptable? That is, doesn't it change the postulated probabilistic quantities, like the propagation amplitude from the part I quoted (since $\psi$ is proportional to $u$)?

(2) Regarding the quoted passage, why do we want the state $\bar{\psi}(y)|0 \rangle$ to be made up of only positive-energy or negative-frequency components? And why are these two requirements 'equivalent' here?

(3) Why the consideration which caused my second question follows from the fact that $\Psi_H=e^{+iHt}\Psi_S$ or, if it doesn't, what the does the author mean with '(since [...])'?

GaloisFan
  • 1,722
  • Answer to (1): You want to construct an invariant quadratic on the fields, that means, an expression which is Lorentz-invariant: $\overline{u}u$ is lorentz-invariant, whereas $u^{\dagger} u$ is not. $u^{\dagger} u$ behaves like zero-component of a four-vector which changes under Lorentz-transformations. In particular $\int |\Psi|^2 d^3x$ is not Lorentz-invariant. – Frederic Thomas Jan 25 '18 at 13:52
  • Short (but correct) answer: Peskin and Schroeder is a bad book. Get a better one. – AccidentalFourierTransform Jan 25 '18 at 16:45
  • @FredericThomas Thanks for the answer! Yes, I understand why we would want to do that, but can't see why we are allowed to (without spoiling the physical predictions). For instance, even if the necessity of Lorentz invariance is a physical requirement because of quantities proportional to $u^{\dagger}u$ that should be invariant, isn't the particular choice of fixing this by multiplying on the right by the gamma matrix arbitrary? – GaloisFan Jan 25 '18 at 20:43
  • @AccidentalFourierTransform Can you indicate a good one for me to try? Thanks a lot! – GaloisFan Jan 25 '18 at 20:45
  • 1
    @GaloisFan sure. Have a look at https://physics.stackexchange.com/a/381048/84967 – AccidentalFourierTransform Jan 25 '18 at 22:17
  • you may want to have a look https://physics.stackexchange.com/q/669457/42982 – ann marie cœur Oct 02 '21 at 22:18

0 Answers0