1
$\begingroup$

Consider the quantization conditions for a complex Fermi field $\Psi=\Phi_1+i\Phi_2$: $$\{\Psi(x),\Psi(y)\}=\{\Psi^\dagger(x)\Psi^\dagger(y)\}=0,~~~~ \{\Psi^\dagger(x),\Psi(y)\}=\delta(x-y)$$ Compare this to what would happen if we decided to quantize just a real fermonic scalar field $\Phi$: $$\{\Phi(x),\Phi(y)\}=\{\Pi(x),\Pi(y)\}=0,~~~~ \{\Pi(x),\Phi(y)\}=\delta(x-y)$$ This has very bizarre and perhaps unintended consequences: For example, a fermion (arising from a complex scalar field, as in scalar QED) with first-quantized wave function $\psi$ will then be given by $$\left|\psi\right>=\int dx\, \psi(x)\, \Psi^\dagger(x)\left|\Omega\right>\tag{1}$$ Whereas a fermion (arising from a real scalar field $\Phi$) with first-quantized wave function $\psi$ will be given by $$\left|\psi\right>=\int dx\, \psi(x)\, (\Phi(x)-i\Pi(x))\left|\Omega\right>\tag{2}$$ Where $\Pi(x)$ is the momentum field conjugate to $\Phi(x)$. Already, the formulas (1) and (2) do not look even close to each other. The worst thing is, since a complex field can be written as the sum of two real fields, $$\Psi(x)=\Phi_1(x)+i\Phi_2(x)\tag{3}$$ We can actually plug (3) into (1) and thus compare (1) even more closely with (2), to really show that something is really messed up: $$\left|\psi\right>=\int dx\, \psi(x)\, (\Phi_1(x)-i\Phi_2(x)\left|\Omega\right>$$ Now, the formula looks nothing like (2). Can anyone resolve these apparent clashes?

$\endgroup$

Your Answer

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

Browse other questions tagged or ask your own question.