I have seen this question and I believe I understand the answer to it. However, AFAIK, only for bosons the causality condition is a vanishing commutator. For fermions we expect the *anticommutator* $[\phi,\phi^\dagger]_+$ to turn zero. The answer given to the question above does not seem to address this.

# [Physics] In QFT, why do fermions have to anticommute in order to insure causality

anticommutatorcausalitycommutatorquantum-field-theoryspecial-relativity

#### Related Solutions

According to Weinberg in his text, the components of most quantum fields are not really measurable in any obvious way, so it's best not to think in those terms.

However, the fields do have to get commuted past each other when you evaluate the S-matrix, and then the Lorentz invariance of the S-matrix depends crucially on the fields commuting at space like separations.

Lots of aspects of the physical interpretation in QFT are at best subtle, and philosophically weak but plausible-sounding heuristic arguments are not uncommon. (You can already see people disagreeing about something so basic as whether $\phi$ creates a particle in the comments!) I found the early chapters in Peskin hard going for this exact reason- it's much better when you get to phenomenology and the physics is less opaque. If you want a book you can't argue with, try Weinberg- but this does come at the price of taking twice as long to cover the material, unfortunately in a rather idiosyncratic notation that makes it hard to dip in and out of.

**General calculation in Schrodinger picture**

We can do a straightforward calculation of your setup in the Schrodinger picture. Let $x,y$ label points in space. Before $t=0$, the state is in the vacuum $|0\rangle$. Just after $t=0$, the state is $e^{-i \epsilon_s \phi(x)} |0\rangle$, for some spatial point $x$ where you place the source at $t=0$. (Really you might want to consider turning on a source smeared over a small neighborhood of $x$, to avoid singularities in the following discussion, but I'll ignore that.) At later time $t>0$, we have state $$|\psi(t)\rangle = e^{-iHt}e^{-i \epsilon_s \phi(x)} |0\rangle =e^{-iHt}e^{-i \epsilon_s \phi(x)} e^{iHt} |0\rangle = e^{-i \epsilon_s \phi(x,t)}|0\rangle$$

Then at time $t$, we make a measurement at spatial point $y$. For a moment I'll ignore your desired detector model and speak generally. Say we make a measurement in a spatial region $Y$. The observables measurable by an observer local to $Y$ are precisely the observables generated by (sums and products of) operators $\phi(y), \pi(y)$ for any $y \in Y$. Choose an observable $A_Y$ of this form, e.g. $A_Y = \phi(y)$ for some $y \in Y$, or $A_Y = \int_{y \in Y}\phi(y)\, dy$. Assume the points $(Y,t)$ are spacelike from the point $(x,t=0)$. Then the expectation of $A_Y$ in $|\psi(t)\rangle$ is $$\langle \psi(t) | A_Y |\psi(t)\rangle = \langle 0 |e^{i \epsilon_s \phi(x,t) } A_Y e^{-i \epsilon_s \phi(x,t)} | 0 \rangle = \langle 0 | A_Y |0 \rangle.$$ So the expectation of $A_Y$ is the same as if you hadn't turned on the source at $x$ at $t=0$. The second equality uses $[\phi(x,t),A_Y]=0$, by assumed spacelike separation.

**Being careful with first-order expansion in $\epsilon$**

Here's a possible point of confusion. Just after $t=0$, and to first order in $\epsilon_s$, we have
$$ |\psi \rangle = |0\rangle - i \epsilon_s \phi(x) |0\rangle + O(\epsilon_s^2).$$
The second term $\phi(x) |0\rangle$ seems to dominate over the higher-order terms in $\epsilon_s$, and this may seem to suggest causality violation: an observable $A_Y$ at spacelike $Y$ still has nonzero expectation value in this state, i.e. $\langle 0 | \phi(x) A_Y \phi(x) |0\rangle \neq 0$. (Incidentally if you choose $A_Y=\phi(y)$ the expectation value is zero by symmetry, but you could choose e.g. $A_Y=\pi(y)$.). However, this observation is entirely unproblematic; the expectation value of $A_Y$ with respect to *just* the first-order term is not directly related to any measurement. If we include both the zero'th order and first-order terms in $|\psi\rangle$, then we find $\langle \psi | A_Y |\psi\rangle = \langle 0 | A_Y |0 \rangle + O(\epsilon^2)$, because the first-order contributions cancel.

**Locality of your detector model**

What about your detector model? There were a few problems. First, I wouldn't actually call your detector localized to the point $y$. The observables measurable at $y$ are just algebraic combinations of $\phi(y), \pi(y)$. Or again more generally, you could take a small spatial region $Y$ and consider observables $A_Y$ local to $Y$, given by algebraic combinations of $\phi(y), \pi(y)$ for $y \in Y$. If you want to imagine an external system $S$ like your atom coupled locally to $Y$, the coupling Hamiltonian for the detector should be like $$H_{det} = \sum_i O^i_S A^i_Y$$ where $O^i_S$ are some operators on the coupled system $S$, and $A^i_Y$ are operators in the QFT local to $Y$.

Your desired Hamiltonian $H_{det}$ may look like it takes this form, but your operators $\phi_{\pm}(x)$, by which I assume you mean something like
$$\phi_{-}(x) \equiv \int \frac{d^3 p}{(2\pi)^3} \frac{1}{\sqrt{E_p}} a_p e^{ipx}$$
$$\phi_{+}(x) \equiv \int \frac{d^3 p}{(2\pi)^3} \frac{1}{\sqrt{E_p}} a_p^\dagger e^{-ipx},$$
are *not* strictly local to $y$. One way to see this is that they have nonzero commutator with any $\phi(x)$. (Incidentally there's some discussion about this in Section 6 and Eq. 81 here.). The operators $\phi_+(y)$ and $\phi_-(y)$ may *look* like they are local to $y$, by the way they are written, but if you actually re-write $\phi_{\pm}(x)$ in terms of the genuinely local $\phi(x)$ and $\pi(x)$ operators, you will find the $\phi_{\pm}(x)$ are not local.

Moreover, regardless of your detector model, I think you're a bit too quick when you say "the amplitude for the detector to be in the excited state$\dots$." You should actually think about what a measurement of the *detector* subsystem would yield. The analysis will then go similarly to the general discussion at the beginning of the answer.

Finally, what if we insist on using your particular detector $H_{det}$, using $\phi_{\pm}(y)$ couplings? First we must admit it's not truly local to $y$, and that it's really only approximately "localized" to a region of radius $\approx \frac{1}{m}$ around $y$ (for a massive theory). We must further admit that it's not even *strictly* localized to that neighborhood, or any finite region: it's really only "local" to a neighborhood of radius $r$ around $y$ with error $e^{-\frac{r}{m}}$, due to the nonzero commutators $[\phi_{\pm}(y), \phi(x)]$, or the expression of $\phi_{\pm}(y)$ in terms of genuinely local field operators $\phi(x), \pi(x)$. So you shouldn't be surprised if the detector has a $e^{-\frac{r}{m}}$ probability of registering a causality-violating signal in this model.

## Best Answer

Here is the formal answer on your question based on particular result of Pauli theorem. Calculations are rather cumbersome, but they are general.

Arbitrary fermionic field (with invariance under discrete transformations of the Lorentz group)It can be shown that each PoincarĂ©-covariant fermion field with half-integer spin $s = n + \frac{1}{2}$ and mass $m$ can be represented as $$ \tag 1 \hat{\varphi}_{a} = \sum_{\sigma = -s}^{s}\int \frac{d^{3}\mathbf p}{\sqrt{(2 \pi )^{3}2E_{\mathbf p}}}\left( u_{a}^{\sigma }(\mathbf p)\hat{a}_{\sigma}(\mathbf p)e^{-ipx} + v_{a}^{\sigma } (\mathbf p)\hat{b}^{\dagger}_{\sigma}(\mathbf p)e^{ipx}\right), $$ where $$ \tag 2 \hat{\varphi}_{a} = \hat{\varphi}_{\mu_{0}...\mu_{n}} = \begin{pmatrix}\hat{\psi}_{\mu_{0}...\mu_{n}b} \\ \hat{\kappa}_{\ \mu_{0}...\mu_{n}}^{\dot {b}} \end{pmatrix} \in \left( \frac{n + 1}{2}, \frac{n}{2}\right) \oplus \left( \frac{n}{2}, \frac{n + 1}{2}\right), $$ $b, \dot{b}$ are spinor indices, and $$ \tag 3 \gamma^{\mu_{j}}\hat{\varphi}_{\mu_{0}...\mu_{j}...\mu_{n}} = 0, \quad (i\gamma^{\mu}\partial_{\mu} - m)\hat{\varphi}_{\mu_{0}...\mu_{n}} = 0, $$ $$ \tag 4 \quad \partial^{\mu_{j}} \hat{\varphi}_{\ \ \mu_{0}...\mu_{j}...\mu_{n}} = 0, \quad u^{\sigma}_{a}(\mathbf p ) = (-1)^{s + \sigma}\gamma_{5}v^{-\sigma }_{a}(\mathbf p). $$

As you can see, for $n = 0$ eqs. $(1)-(4)$ give the Dirac field.

The massless case is the same, but $\sigma $ may take only following sets of values: $\{-s\}$, $\{s\}$, or $\{-s,s\}$.

You can read about general aspects of irreducible PoincarĂ© representations which is realised as a sum of fields of creators and annihilators in Weinberg QFT Vol. 1 (chapter about general causal fields). Equations $(3)-(4)$ may be given as the requirement that field $(1)$ transforms under an irreducible representations of the PoincarĂ© group with spin $s$ and mass $m$.

Causality for fermionic theories and anticommutatorFrom the causality principle we must have $$ \tag 5 [\hat{\varphi}_{a}(x), \hat{\bar{\varphi }}_{b}(y)]_{\pm} = 0, \quad (x - y)^{2} < 0 , \quad g_{00} = 1. $$ In $3 + 1$-dimensional spacetime and for indistinguishable particles, the first homotopy group of the configuration space with 3 spatial dimensions is the permutation group $S_N$. This means that the clockwise and anticlockwise exchange of two particles is equal and hence there are only two possible statistics, Bose-Einstein or Fermi-Dirac, $$ \tag 6 [\hat{a}_{\sigma}(\mathbf p ), \hat{a}^{\dagger}_{\sigma {'}}(\mathbf k) ]_{\pm} = \delta (\mathbf p - \mathbf k ) \delta_{\sigma \sigma {'}}. $$

Let's use $(1)-(4)$ and $(6)$ for clarifying the statistic which is obeyed by field $(1)$.

We have by using these equations and the identity $[\gamma_{\mu} , \gamma_{5}]_{+} = 0$ that $$ \tag 7 [\hat{\varphi}_{a}(x), \hat{\bar{\varphi }}_{b}(y)]_{\pm} = \sum_{\sigma}\int \frac{d^{3}\mathbf p}{(2 \pi )^{3}2E_{\mathbf p}}\left( u^{\sigma}_{a}(\mathbf p)\bar{u}^{\sigma}_{b}(\mathbf p) e^{-ipX} \mp \gamma_{5}u^{\sigma}_{a}(\mathbf p)\bar{u}_{b}^{\sigma}(\mathbf p)e^{ipX}\right), $$ where $X = x - y$ and $\hat{\bar {\varphi}} = \hat{\varphi}^{\dagger}\gamma_{0}$.

By using the second identity of $(3)$ and the requirement of Lorentz covariance, it can be shown that $\sum_{\sigma}u^{\sigma}_{a}(\mathbf p)\bar{u}_{b}^{\sigma}(\mathbf p) = R_{ab}(p)$ can be represented in a form $$ R_{ab}(p) = (\gamma^{\mu}p_{\mu} + m)P_{ab}(p), $$ where $P_{ab}(p)$ is constructed from sum of summands with even number of momentums and even number of gamma-matrices and from summands with odd numbers of momentums and odd number of gamma-matrices. So by using $[\gamma_{\mu} , \gamma_{5}]_{+} = 0$ again, we may get $$ [\hat{\varphi}_{a}(x), \hat{\bar{\varphi }}_{b}(y)]_{\pm} = \int \frac{d^{3}\mathbf p}{(2 \pi )^{3}2E_{\mathbf p}}\left( R_{ab}(p) e^{-ipX} \mp R_{ab}(-p)e^{ipX}\right) = $$ $$ \tag 8 = R_{ab}\left( i\partial_{x}\right)(D(X) \mp D(-X)), \quad D(X) = \int \frac{d^{3}\mathbf p }{(2 \pi )^{3}2E_{\mathbf p}}e^{-ipX}. $$ It can be shown that for $X^{2} < 0$, the function $D(X)$ satisfies the relation $D(X) = D(-X)$, so $(8)$ vanishes if and only if fermionic fields have Fermi-Dirac statistics.