# Why are relativistic quantum field theories so much more restrictive than non-relativistic ones?

Part of the reason that relativistic QFT is so hard to learn is that there are piles of 'no-go theorems' that rule out simple physical examples and physical intuition. A very common answer to the question "why can't we do X simpler, or think about it this way" is "because of this no-go theorem".

To give a few examples, we have:

Of course all these theorems have additional assumptions I'm leaving out for brevity, but the point is that Lorentz invariance is a crucial assumption for every one.

On the other hand, nonrelativistic QFT, as practiced in condensed matter physics, doesn't have nearly as many restrictions, resulting in much nicer examples. But the only difference appears to be that they work with a rotational symmetry group of $SO(d)$ while particle physicists use the Lorentz group $SO(d-1, 1)$, hardly a big change. Is there a fundamental, intuitive reason that relativistic QFT is so much more restricted?

• I thought it might be linked to the fact that the Lorentz group isn't compact, but then again the salient comparison is actually the Galilean group versus the Poincaré group! Mar 31 '18 at 13:30
• I think almost all of the things you mentioned are just a consequence of unitarity, not necessarily Lorentz invariance, although maybe in my thinking Lorentz invariance is entering in some subtle way. I think things like Reeh-Schlieder though can be stated nonrelativistically. Mar 31 '18 at 17:47
• There is a no-go theorem which states that you cannot intuitively understand why relativistic QFT is so much more restricted. Apr 1 '18 at 17:48
• @Dvij ... which is a corollary of the more general no-go theorem "you cannot intuitively understand anything in quantum mechanics" ;-P Apr 1 '18 at 17:53
• @Dvij I would laugh if it weren't so true! Apr 1 '18 at 22:36

One of the reasons relativistic theories are so restrictive is because of the rigidity of the the symmetry group. Indeed, the (homogeneous part) of the same is simple, as opposed to that of non-relativistic systems, which is not.

The isometry group of Minkowski spacetime is $$\mathrm{Poincar\acute{e}}=\mathrm{ISO}(\mathbb R^{1,d-1})=\mathrm O(1,d-1)\ltimes\mathbb R^d$$ whose homogeneous part is $\mathrm O(1,d-1)$, the so-called Lorentz Group1. This group is simple.

On the other hand, the isometry group of Galilean space+time is2 $$\text{Bargmann}=\mathrm{ISO}(\mathbb R^1\times\mathbb R^{d-1})\times\mathrm U(1)=(\mathrm O(d-1)\ltimes\mathbb R^{d-1})\ltimes(\mathrm U(1)\times\mathbb R^1\times\mathbb R^{d-1})$$ whose homogeneous part is $\mathrm O(d-1)\ltimes\mathbb R^{d-1}$, the so-called (homogeneous) Galilei Group. This group is not semi-simple (it contains a non-trivial normal subgroup, that of boosts).

There is in fact a classification of all physically admissible kinematical symmetry groups (due to Lévy-Leblond), which pretty much singles out Poincaré as the only group with the above properties. There is a single family of such groups, which contains two parameters: the AdS radius $\ell$ and the speed of light $c$ (and all the rotation invariant İnönü-Wigner contractions thereof). As long as $\ell$ is finite, the group is simple. If you take $\ell\to\infty$ you get Poincaré which has a non-trivial normal subgroup, the group of translations (and if you quotient out this group, you get a simple group, Lorentz). If you also take $c\to\infty$ you get Bargmann (or Galilei), which also has a non-trivial normal subgroup (and if you quotient out this group, you do not get a simple group; rather, you get Galilei, which has a non-trivial normal subgroup, that of boosts).

Another reason is that the postulate of causality is trivial in non-relativistic systems (because there is an absolute notion of time), but it imposes strong restrictions on relativistic systems (because there is no absolute notion of time). This postulate is translated into the quantum theory through the axiom of locality, $$[\phi(x),\phi(y)]=0\quad\forall x,y\quad \text{s.t.}\quad (x-y)^2<0$$ where $[\cdot,\cdot]$ denotes a supercommutator. In other words, any two operators whose support are casually disconnected must (super)commute. In non-relativistic systems this axiom is vacuous because all spacetime intervals are timelike, $(x-y)^2>0$, that is, all spacetime points are casually connected. In relativistic systems, this axiom is very strong.

These two remarks can be applied to the theorems you quote:

• Reeh-Schlieder depends on the locality axiom, so it is no surprise it no longer applies to non-relativistic systems.

• Coleman-Mandula (see here for a proof). The rotation group is compact and therefore it admits finite-dimensional unitary representations. On the other hand, the Lorentz group is non-compact and therefore the only finite-dimensional unitary representation is the trivial one. Note that this is used in the step 4 in the proof above; it is here where the proof breaks down.

• Haag also applies to non-relativistic systems, so it is not a good example of OP's point. See this PSE post for more details.

• Weinberg-Witten. To begin with, this theorem is about massless particles, so it is not clear what such particles even mean in non-relativistic systems. From the point of view of irreducible representations they may be meaningful, at least in principle. But they need not correspond to helicity representations (precisely because the little group of the reference momentum is not simple). Therefore, the theorem breaks down (as it depends crucially on helicity representations).

• Spin-statistics. As in Reeh-Schlieder, in non-relativistic systems the locality axiom is vacuous, so it implies no restriction on operators.

• CPT. Idem.

• Coleman-Gross. I'm not familiar with this result so I cannot comment. I don't even know whether it is violated in non-relativistic systems.

1: More generally, the indefinite orthogonal (or pseudo-orthogonal) group $\mathrm O(p,q)$ is defined as the set of $(p+q)$-dimensional matrices, with real coefficients, that leave invariant the metric with signature $(p,q)$: $$\mathrm O(p,q):=\{M\in \mathrm{M}_{p+q}(\mathbb R)\ \mid\ M\eta M^T\equiv \eta\},\qquad \eta:=\mathrm{diag}(\overbrace{-1,\dots,-1}^p,\overbrace{+1,\dots,+1}^q)$$

The special indefinite orthogonal group $\mathrm{SO}(p,q)$ is the subset of $\mathrm O(p,q)$ with unit determinant. If $pq\neq0$, the group $\mathrm{SO}(p,q)$ has two disconnected components. In this answer, "Lorentz group" may refer to the orthogonal group with signature $(1,d-1)$; to its $\det(M)\equiv+1$ component; or to its orthochronus subgroup $M^0{}_0\ge+1$. Only the latter is simply-connected. The topology of the group is mostly irrelevant for this answer, so we shall make no distinction between the three different possible notions of "Lorentz group".

2: One can prove that the inhomogeneous Galilei algebra, and unlike the Poincaré algebra, has a non-trivial second co-homology group. In other words, it admits a non-trivial central extension. The Bargmann group is defined precisely as the centrally extended inhomogeneous Galilei group. Strictly speaking, all we know is that the central extension has the algebra $\mathbb R$; at the group level, it could lead to a factor of $\mathrm U(1)$ as above, or to a factor of $\mathbb R$. In quantum mechanics the first option is more natural, because we may identify this phase with the $\mathrm U(1)$ symmetry of the Schrödinger equation (which has a larger symmetry group, the so-called Schrödinger group). Again, the details of the topology of the group are mostly irrelevant for this answer.

• Can you really construct a counterexample to Reeh-Schlieder? Apr 1 '18 at 6:03
• @RyanThorngren The concept of non-relativistic Reeh-Schlieder is meaningless (because the concept of local algebras is also meaningless: if $c\to\infty$, all points are causally connected). Therefore, strictly speaking there are no counter-examples, because the theorem is incompatible with the situation. But in OP's sense, which states "... forbids position operators in relativistic QFT", then the counter-example is straightforward: in non-relativistic systems, the position operator $\hat X$ is well-defined, Galilei covariant, and local. No such operator exists in the relativistic regime. Apr 1 '18 at 14:15
• I would state Reeh-Schlieder that $\langle O^\dagger(x) O(0) \rangle \ge 0$, with 0 only for the identity operator. This follows from reflection positivity. I don't know how to construct the position operator in many-body quantum mechanics. Apr 1 '18 at 17:37
• @RyanThorngren 1) The version of R-S I had in mind is that of wikipedia, i.e., that the vacuum is a cyclic vector (for algebras over open sets in Minkowski). I don't know whether this is equivalent to your version. 2) In principle, you just take $\hat X=\bigoplus_i \hat X_i$, where $\hat X_i=1\otimes\cdots 1\otimes \hat X\otimes 1\cdots 1$, where the $\hat X$ is at the position $i$. Apr 1 '18 at 17:46
• I think they are equivalent, or at least the version on wikipedia follows from the statement I made. Local observables can just mean ones whose support on Hilbert space is bounded in real space. I don't think the X operator you write down is so bounded in, say, a hopping Hamiltonian model. Apr 1 '18 at 19:30

Lorentz invariance is also an indirect contributor to the restrictions that renormalizability places on the theory. The logic goes something likes this:

1. The action must be Lorentz invariant, thus the number of spatial derivatives must equal the number of time derivatives in the action.
2. We want the Hamiltonian to play the role of energy (has lower bound and provides stability), therefore the action can have no more than two time derivatives (for more, see: this previous answer on renormalization).
3. By 1 and 2, the propagator $\Delta(x,y)$ will, necessarily, diverge like $\left([x^\mu-y^\mu][x_\mu-y_\mu]\right)^{-1}$ as $x\rightarrow y$ (equivalently, the momentum space form looks like $(p^\mu p_\mu)^{-1}$ for $p\rightarrow \infty$).

It is that third step that leads to the divergences that require renormalization, and renormalizibility is very restrictive on what terms the action is allowed to contain. Without Lorentz invariance, we could add more spatial derivatives without time derivatives, produce a well behaved finite propagator, and work with a much broader class of theories.

Granted, as discussed in the linked answer, you could relax 2, some, but that doesn't allow any theory, just more.

Here's one (incomplete) perspective, mostly about the infrared:

For a field with given charges under Lorentz and all other symmetries, there is essentially only one theory with quadratic action to first order in derivatives. For integer spins it's $\partial_\mu \phi \partial^\mu \phi + m^2 \phi^2$ and for half integer spins it's $\bar \psi \gamma^\mu D_\mu \psi + m \bar \psi \psi$. This is a fact of representation theory, and the fact that all you have to contract spacetime indices are $g_{\mu \nu}$, $\gamma^\mu$, and things with more spacetime indices. Note that the form of these actions determine the bare propagators to be the relativistic ones $1/(p^2 - m^2)$ and $1/(p-m)$, respectively.

On the other hand, if you were to break Lorentz symmetry, say by choosing a vector field $v^\mu$, then you could write terms like $\phi v^\mu \partial_\mu \phi$, which would change the dispersion relation for $\phi$ to be linear in $p$ for momenta parallel to $v$. Note that for timelike $v$ this breaks the Lorentz group $SO(1,d)$ to $SO(d)$.

For an applied magnetic field $F_{ij}$ on fermions we could add a term $\bar \psi F_{ij} \gamma^i \gamma^j \psi$ which can mess up spin-statistics.

I think these new Gaussian fixed points" cause lots of things to go wonky (in the IR) when you do perturbation theory around them.

On the other hand, there aren't that many terms that can lead these ones, and because of that most of the theories we end up studying in condensed matter have emergent Lorentz invariance in the IR. Some significant exceptions are theories with singular fermi surface or others with UV/IR" mixing that cause the field theory to see the lattice.