What goes wrong when one tries to quantize a scalar field with Fermi statistics?

Question

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.

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.