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?

  • $\begingroup$ 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


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?

  • $\begingroup$ 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. $\endgroup$ Commented Dec 3, 2019 at 21:43

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