At the end of section 9 on page 49 of Dirac's 1966 "Lectures on Quantum Field Theory" he says that if we quantize a real scalar field according to Fermi statistics [i.e., if we impose Canonical Anticommutation Relations (CAR)], the quantum Hamiltonian is no longer any good because it gives the wrong variation of the creation operator $\hat{\eta_{k}}$ with time. Unfortunately, I can't make anything go wrong, so would someone show my mistake, or explain what calculation I should do to understand Dirac's remark. Here's my calculation.

The quantum Hamiltonian is, $$ \hat{H}=\int d^{3}k |k|\hat{\eta_{k}}\hat{\eta_{k}}^{\dagger} $$ and the Heisenberg equation of motion is, $$ \frac{d\eta_{k}}{dt}=-i[\eta_{k},H]_{-}=-i\int d^{3}k'|k'|(\eta_{k}\eta_{k'}\eta_{k'}^{\dagger}-\eta_{k'}\eta_{k'}^{\dagger}\eta_{k}) $$ where the hats to indicate operators have been left out and $[A,B]_{-}$ is a commutator. Now assume that the $\eta's$ obey Fermi statistics, $$ [\eta_{k}^{\dagger},\eta_{k'}]_{+}=\eta_{k}^{\dagger}\eta_{k'}+\eta_{k'}\eta_{k}^{\dagger}=\delta(k-k') $$ and use this in the last term of the Heisenberg equation, $$ \frac{d\eta_{k}}{dt}=-i\int d^{3}k'|k'|(\eta_{k}\eta_{k'}\eta_{k'}^{\dagger}+\eta_{k'}\eta_{k}\eta_{k'}^{\dagger}-\eta_{k'}\delta(k-k'))=i|k|\eta_{k} $$ In the above equation, the first two terms in the integral vanish because of the anticommutator $[\eta_{k},\eta_{k'}]_{+}=0$ and the result on the right is the same time variation of $\eta_{k}$ that one gets quantizing using Bose statistics: nothing seems to have gone wrong.

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    $\begingroup$ Since you are using anticommutator for quantization so causality would require that the fields anticommute for all spacelike separated points. Since form of the field (as got from time evolution of eta's) will be same as in bosonic case so the fields will fail to anticommute for spacelike separated points. Hence time evolution of eta's is wrong since it is not in accordance with causality. $\endgroup$
    – user10001
    Commented May 25, 2013 at 4:53
  • $\begingroup$ related: physics.stackexchange.com/q/63957 $\endgroup$
    – user4552
    Commented Jul 28, 2013 at 3:45
  • $\begingroup$ For the analogue Phys.SE question with Grassmann-odd (instead of Grassmann-even) fields, see physics.stackexchange.com/q/17893/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Jul 29, 2013 at 17:19

1 Answer 1


I will firstly point out some apparent misconceptions in the question and subsequently I will explain what goes wrong when quantizing a theory of integer spin fields or particles with anticommutators, and vice versa.

First, if we quantize a real Klein-Gordon field using anticommutators, the Hamiltonian vanishes (or is a field-independent constant). At the level of fields, the Hamiltonian for this field is a sum of squares $H=\sum_i A_i^2 (x)$ (one $A_i$ is, for example, $\nabla\phi$). Since $\{A_i(x),A_i(y)\}=0$ ($\{\phi(x),\phi(y)\}=0$), $A_i^2=0$ for every $i$, and therefore $H=0$. At the level of creation and annihilation operators $H\sim \int_p\,a_p^{\dagger}a_p+a_pa_p^{\dagger}\sim\int_p\,\{a_p,a^{\dagger}_p\}$. As $\{a_p,a^{\dagger}_q\}\sim\delta^3 (p-q)$, the Hamiltonian is an operator-independent constant. Let's see what happens when considering a complex scalar Klein-Gordon field, a more interesting case.

Complex scalar (spin = 0) field quantized with anticommutators

Here, it is micro-causality what fails. Consider a free complex scalar field, and a bilinear local observable $\hat O(x)=\phi^{\dagger}(x)o(x)\phi(x)$, with $o(x)$ a real c-number function. Then, causality tells us that the commutator of two of these operators separated by a space-like distance is to vanish. One can check that: $$[\hat O(x),\hat O(y)]=o(x)o(y)[\phi^{\dagger}(x)\phi(x), \phi^{\dagger}(y)\phi(y)]\\ =o(x)o(y)\left(\phi^{\dagger}(x)\phi(y)-\phi^{\dagger}(y)\phi(x)\right)\,\{\phi(x),\,\phi^{\dagger}(y) \}$$

And using the expression of a complex, free Klein-Gordon field in terms of creation and annihilation operators, we can compute the anticommutator by making use of the assumed canonical anticommutation relations between creation and annihilation operators. The result is (you should check all this)

$$\{\phi(x),\,\phi^{\dagger}(y) \}=2\int d^3\tilde {\bf p}\, \cos(p(x-y))$$

where $d^3\tilde {\bf p}$ is a standard notation for the Lorentz-invariant measure. Using the Lorentz invariance of the previous expression and the fact that it doesn't vanish for $x_0=y_0$, we can conclude that $\{\phi(x),\,\phi^{\dagger}(y) \}$ and, as a consequence, $[\hat O(x),\hat O(y)]$ don't vanish for space-like separations, which violates causality.

Therefore, both real and complex scalar fields refuse to be quantized with anticommutators.

Spin $1/2$ field quantized with commutators

Starting with the Dirac Hamiltonian, one gets $$H\sim \int\, a^{\dagger}a-bb^{\dagger}$$

Then, in order to have a minimum-energy vacuum state, we need a Hamiltonian that is bounded from below. The $b$-modes have a negative sign in the Hamiltonian so that there are two alternatives:

  • Exchange the standard action of the $b$ operators on the Hilbert space. That is, $b^{\dagger}$ is going to annihilate quanta and $b$ is going to create them, so that $$H|p\rangle_b\sim H\,b|0\rangle_b \sim [H,b]|0\rangle_b \sim \sqrt{m^2+p^2}|p\rangle_b$$ where we have made use of $[b,b^{\dagger}]\sim \delta^{3}$. However, doing this we end with states of negative norm $$_b\langle p|p'\rangle_b=\langle 0|b^{\dagger}_p\,b_{p'}|0\rangle=-2\,\left|{\sqrt{m^2+p^2}} \, \right|\,\delta^3(p-p')\, \langle 0|0\rangle \; , $$ which prevents from a probabilistic interpretation (negative probabilities are nonsensical).
  • The alternative is to use anticommutators (i.e., fermi-statistics), which reverse the sign in the Hamiltonian. This is the choice that works.

These obstacles are a consequence of Pauli's spin-statistics connection theorem.

  • $\begingroup$ This comment is about the part dealing with the real scalar field. When one goes from the classical Hamiltonian to the quantum Hamiltonian, there are two possibilities; $H \approx \int a a^{\dagger}$ and $H \approx \int a^{\dagger}a+a a^{\dagger}$. I used the first one in my question and could not get a contradiction when using anticommutators whilst you used the second one and showed the quantum Hamiltonian is zero. Upon reading Dirac's lectures more carefully, I notice that he only claims a contradiction can be found using the second Hamiltonian. Is the first Hamiltonian bad? $\endgroup$
    – user7154
    Commented Jul 30, 2013 at 21:50
  • $\begingroup$ Hello @StephenBlake In the first possibility, one must make use of the so-called 'normal ordering' prescription, which says that creation operators go on the left, ignoring the commutator. This prescription has advantages and drawbacks for bosonic fields, but it is not justified whatsoever for fermionic fields. $\endgroup$ Commented Jul 30, 2013 at 23:23
  • $\begingroup$ How did you get the anticommutator $\{\phi(x),\phi^{\dagger}(y)\}$? In order to calculate the commutator $[\phi(x),\phi^{\dagger}(y)]$, I would work in the classical field theory and calculate the Poisson bracket and then get the quantum commutator using Dirac's quantization rule. This route doesn't work for the anti-commutator because there is no classical theory and no analogue of the Poisson bracket. $\endgroup$
    – user7154
    Commented Aug 1, 2013 at 21:06
  • $\begingroup$ Hello @StephenBlake Use the expression of a complex, free Klein-Gordon field in terms of creation and annihilation operators, and then compute the anticommutator making use of the assumed canonical anticommutation relations between creation and annihilation operators. $\endgroup$ Commented Aug 1, 2013 at 21:58

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