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I want to start out with a real fermionic quantum field theory (in one dimension), with operators $\psi$ satisfying: $$ \{\psi(x), \psi(y) \} = \delta(x-y) $$ I know I won't find a state that's an eigenstate for every point, because that would mean that the $\psi(x)$ at the individual points have to commute. As they also anticommute, that would lead to some contradictions. However, when I pick out just a single $x$, there is still the possibility to have an eigenvector only for this specific $\psi(x)$.

If I naively look at: $$\ \{ \psi(x), \psi(x) \} = 2 \psi(x)\psi(x) = \frac{1}{2} \delta(0) $$ So, if there is any eigenvalue, then it should be $\pm \frac{1}{\sqrt{2}} \sqrt{\delta(0)}$, but I can't really make sense of that, although I still think that it is the right result, if I compare with the discrete case:

In a discrete space with finitely many (evenly numbered) points, and $$ \{\psi(i), \psi(j) \} = \delta_{i, j} $$ I can reproduce this result in the way that the possible eigenvalues for one specific $\psi(i)$ here are $\pm \frac{1}{\sqrt{2}}$, and any other $\psi(j)$ will act on this vector by switching the sign of the eigenvalue. So, up until the infinities that are occurring, this seems like the plausible result.

For the continuous case, one could also define: $$ \psi_i = \int \phi_i(x) \psi(x) \\ \psi_j = \int \phi_j(y) \psi(y) $$ For an appropriately chosen set of $\phi_i$ Then: $$ \int dx \int dy \ \phi_i(x) \phi_j(y) {\psi(x), \psi(y)} = \{\psi_i, \psi_j \} = \int dx \int dy \ \phi_i(x) \phi_j(y) \delta(x-y) \\ = \int dx \ \phi_i(x) \phi_j(x) = \delta_{ij} $$

That means from my uncountable set of $\phi_i(x)$, I can get a countable set of $\psi(x)$ by convoluting with the right test functions, and those will then have nicely behaving eigenvalues. In that sense I also understand this answer, that asserts quantum fields are distributions.

Now my question: I'm aware that those eigenvalues will only hold for one region in space, and don't represent a complete field configuration of a field value at every point - That is, these eivenvalues have very little informational content (which is fine for me, as long as they exist in principle).

However I'm not sure- Am I allowed to talk about measurement outcomes for regions, when I translate the example to spacetime? - I fear this might violate causality. Or is it fine, and simply means that I have to build a measurement device that has a finite size in space time?

In case it violates causality, what other options do I have to make sense of $\sqrt{\delta(0)}$ ?

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  • $\begingroup$ Shouldn't $\{\psi(x), \psi(y) \} = 0$? $\endgroup$ Commented Jul 8, 2022 at 12:17
  • $\begingroup$ @flippiefanus this isn't the usual complex dirac field, but instead a real fermionic field. If you want, you can take $\psi(x)$ to be the real part of one component of the usual dirac field. See for example the article by Jackiw (inis.iaea.org/collection/NCLCollectionStore/_Public/22/056/…, p133, V Fermionic systems) $\endgroup$ Commented Jul 8, 2022 at 12:27

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