1
$\begingroup$

In QFT is it the case that the electron matter field and anti-electron matter field (using the electron as a specific example) are truly distinct physical fields versus different excitation modes of a single electron matter field?

$\endgroup$
2
  • 2
    $\begingroup$ What is your definition of "distinct physical field" w.r.t. a quantum field? $\endgroup$
    – ACuriousMind
    Commented Feb 11, 2019 at 15:43
  • $\begingroup$ I guess in the case of matter fields, that it is an independent spinor field with its own set of field values extended through spacetime. In the cases of the electron and anti-electron that there are two separate fields required to model reality. I previously thought that there was a single electron matter field and the electron and anti-electron were just conjugate (loose terminology) modes of excitation, but recent reading seemed to indicate otherwise. If separate fields, it seems odd to two totally separate fields so similar in properties and structure. $\endgroup$
    – CSnowden
    Commented Feb 11, 2019 at 20:40

1 Answer 1

1
$\begingroup$

The answer to your question is kind of. The mode expansion for an electron field is given by

$$ \Psi (x)=\int {\frac {d^{3}p}{(2\pi )^{3}}}{\frac {1}{\sqrt {2E_{p}}}}\sum _{s}\left(a_{\mathbf {p} }^{s}u^{s}(p)e^{-ip\cdot x}+b_{\mathbf {p} }^{s\dagger }v^{s}(p)e^{ip\cdot x}\right).$$

If we act on the vaccum with this, what we get is essentially an electron in a position eigenstate which has spin $s$. Similarly for the conjugate field $\bar{\Psi}$ but now we have a positron.

So are $\Psi$ and $\bar{\Psi}$ "independent"? The only way I see fitting to answer this is by simply just looking at the anti commutation relations. For the same field and its conjugate at two different spacetime points, they are are given by $\{\Psi(x), \bar{\Psi}(y)\}=0$. So these fields anti commute. Would you call this "independent"? I'm not sure, as it seems by the reasoning above that they at first "appear independent but satisfy a constraint", if you will.

$\endgroup$
1
  • $\begingroup$ Thanks for the response - can you clarify about the field and conjugate field being at different positions yet anti-commuting? It seems if an electron is at one position and an anti-electron at another they would not cancel. Also, does this mode expansion applied to a vaccum state give an electron if none exists there - I realize there are virtual particle pairs but didn't think that was the intent here. Thanks! $\endgroup$
    – CSnowden
    Commented Feb 13, 2019 at 3:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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