2
$\begingroup$

It's often noted that Bosonic fields result from quantizing classical field theories defined on a regular numbers, whereas Fermionic fields arise when quantizing a classical field theory defined on Grassmann numbers.

It is no surprise, then, that the expectation values of operators associated with Bosonic fields are normal (non-Grassmann) numbers. For concreteness, we can consider a real scalar field $\phi(x)$ with some associated raising and lowering operators $b^\dagger_q$ and $b_q$. A coherent state of the field $|\alpha> = Ae^{\alpha b^\dagger_q}|0\rangle$ has expectation value $\langle\alpha| b^\dagger_q |\alpha\rangle = \alpha$, clearly a run-of-the-mill number.

Since the expectation value of a lowering operator for a Bosonic field is a regular (non-Grassmann) number, should the expectation value of a Fermionic field operator $\psi(x)$ or a corresponding lowering operator $a_q$ be a Grassmann number? Although I can't understand why this wouldn't be the case (shouldn't the "typical value" of a Grassmann number drawn from some probability distribution be a Grassmann number?), it seems easy to construct a counterexample. The state

$$|\text{confusing}\rangle = \frac{1}{\sqrt{2}}(1 + a^\dagger_q)|0\rangle$$

Leads to $\langle \text{confusing}|a_q|\text{confusing}\rangle = \frac{1}{2}$. Yet it's not clear to me why such an expectation value should generally be a regular number. As a result, I'm curious if there are any clear ways of understanding the origin of this behavior.

Is the expectation value of a Fermi field operator a Grassmann number? Why or why not?

Improve this question
$\endgroup$
3
  • $\begingroup$ Possible duplicate: physics.stackexchange.com/q/269699/2451 $\endgroup$
    – Qmechanic ♦
    Commented Dec 2, 2019 at 3:30
  • $\begingroup$ Forget all about path integrals and Grassmann numbers. In good ol' canonical quantization, you have a standard Hilbert space over the complex numbers. Some of the degrees of freedom in this Hilbert space happen to represent particles with half-integer spin. In no case do you need a Grassmann number, all expectation values are manifestly complex. $\endgroup$
    – knzhou
    Commented Dec 3, 2019 at 21:42
  • $\begingroup$ You could choose to introduce fermionic coherent states involving Grassmann coefficients, in which case all bets are off, but you don't have to. $\endgroup$
    – knzhou
    Commented Dec 3, 2019 at 21:45

1 Answer 1

Reset to default
2
$\begingroup$

Theoretically speaking, the expectation value of a Fermi field operator must be a Grassmann number. How do we observe a Grassmann number-value in experiments? Beats me!

(Added note: Since here we are concerned with field quantization and Fock space rather than Hilbert space, one can NOT escape from Grassmann number given the anti-symmetric requirement for the multi-particle state. )

Fortunately, in reality there is no such case as fermion field acquiring a non-zero vacuum expectation value (VEV). Otherwise it would break some crucial symmetries (Lorentz, particle number, etc.) and surely would open a Pandora's box.

That being said, a pair of Fermi fields CAN acquire a non-zero VEV, such as the cooper pair condensation $<\psi_\uparrow\psi_\downarrow>$ in the celebrated BCS theory of superconductivity. In this case, the VEV is an ordinary number, since the multiplication of two Grassmann-odd number is Grassmann-even.

But hold on! is the multiplication of two Grassmann-odd number indeed a real number? Actually it's a ring instead of a real number (with weird properties like $c^2 = (ab)^2 = 0$, for $a$ and $b$ being Grassmann-odd numbers).

In other words, our cherished canonical/path integral QFT is performing the sleight of hand of transforming rings into real number VEVs. It's a can of worms mathematicians would surely cringe from. But we physicists are the dauntless and shameless bunch, are we?

Improve this answer
$\endgroup$
1
  • $\begingroup$ In case the fermion is ungauged, you could just add a background source, say $\delta L = \bar{\psi} \chi + \text{h.c.}$ for some Grassmann field $\chi$, and then indeed the vev of $\psi$ will be $\propto \chi$. The presence of such a background field indeed breaks various symmetries, but that's fine, the formalism of QFT can deal with it. $\endgroup$
    – Hans Moleman
    Commented Dec 3, 2019 at 21:43

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.