Does relativistic quantum mechanics (RQM) really violate causality? The Hamiltonian $$H=\sqrt{p^2+m^2}$$ defines a one-particle quantum mechanics in the usual way.  Let us call this theory RQM for short.  Peskin and Schroeder claim that RQM violates causality because the (quantum-mechanical) propagator has support outside the light cone (Sec. 2.1).  I do not believe there can be any causality problem with RQM because

*

*a) Quantum field theory (QFT) of a free Klein-Gordon (KG) field has no problems


*b) The one-particle states of free KG QFT obey the RQM Schrodinger equation.
Put more simply, free QFT is valid, and RQM appears as a limit (restricting to one-particle states).  So how can there possibly be a causality problem with RQM?
Are Peskin and Schroeder wrong/sloppy?  Or is there really a causality problem in RQM?  If the latter, somebody should be able to construct a thought experiment with a grandfather paradox or some other disaster.  Please enlighten!
 A: The operator that measures whether a particle is at a particular position $x$ is the projection operator $\mathcal{O}_x=|x\rangle\langle x|$. Suppose we have another projection operator $\mathcal{O}_y=|y\rangle \langle y|$. In the Heisenberg picture, $\mathcal{O}_y(t)=U^\dagger(t)\mathcal{O}_yU(t)$. Therefore,
$$[\mathcal{O}_x(t=0),\mathcal{O}_y(t)]=|x\rangle\langle x|U^\dagger|y\rangle \langle y|U-U^\dagger|y\rangle \langle y|U|x\rangle \langle x|.$$
Now $\langle y|U|x\rangle$, and its complex conjugate $\langle x|U^\dagger|y\rangle$, are non-zero, as shown in Peskin and Schroeder, even with the covariant integral discussed in Qmechanic's answer. Since $|x\rangle\langle y|U$ and $U^\dagger|y\rangle\langle x|$ are not proportional to each other, the two terms cannot cancel. Therefore, the commutator is not zero, and a measurement performed at $x$ can effect a measurement performed outside of $x$'s light cone.
A: The actual difference is in how these approaches treat measurements.
In the single particle theory, your observable is the particle coordinates $x^i(t)$. Measuring them at $t_1$ and $t_2$ can lead to apparent superluminal propagation.
In QFT, your observable is $\phi(x) = \phi(x^{i}, t)$. (I am ignoring the fact that these are operator-valued distributions). Measuring two of these separated by a space-like interval can not lead to superluminal propagation, as Peskin and Schroder show later when they evaluate the commutator of the fields. No grandfather paradoxes here.
A: OP has a point. On one hand, P&S on p. 14 argue that in first quantized RQM in the operator formalism the propagator is
$$\begin{align}&\langle {\bf x}_f,\tau_f \mid {\bf x}_i,\tau_i\rangle\cr
&~=~ \int_{\mathbb{R}^3} \!\frac{\mathrm{d}^3{\bf p}}{(2\pi\hbar)^3}  \exp\left[\frac{i}{\hbar}\left( {\bf p}\cdot \Delta {\bf x} - \Delta \tau \underbrace{\sqrt{{\bf p}^2+m^2}}_{\text{Hamiltonian}}\right)\right]. \end{align}\tag{A}
$$
P&S write on p. 14:

This integral can be evaluated explicitly in terms of Bessel functions. [...] the propagation amplitude is small but nonzero outside the light-cone, and causality is violated.

See also this Phys.SE answer. However, P&S's normalization of the integrand (A) is not Lorentz covariant, and hence not suitable for RQM. A more careful Lorentz covariant analysis of the path integral formalism reveals that the RQM propagator is$^1$
$$\begin{align}&\langle {\bf x}_f,\tau_f \mid {\bf x}_i,\tau_i\rangle\cr
&~=~ \int_{\mathbb{R}^3} \!\frac{\mathrm{d}^3{\bf p}}{(2\pi\hbar)^3}\color{red}{ \frac{\hbar}{2\sqrt{{\bf p}^2+m^2}} }\exp\left[\frac{i}{\hbar}\left( {\bf p}\cdot \Delta {\bf x} - \Delta \tau \underbrace{\sqrt{{\bf p}^2+m^2}}_{\text{Hamiltonian}}\right)\right], \end{align}\tag{B}
$$
cf. my Phys.SE answer here. Remarkably, P&S's above quote essentially still applies!
On the other hand, P&S in eq. (2.50) on p. 27 find exactly the same propagator (B) in second quantized KG QFT. So OP is correct that RQM appears in the one-particle sector of free scalar QFT.
P&S write on p. 28:

So again we find that outside the light-cone, the propagator amplitude is exponentially vanishing but nonzero. To really discuss causality, however, we  should ask not whether particles can propagate over spacelike intervals, but whether a measurement performed at one point can affect measurement at another point whose separation from the first is spacelike.

And P&S then go on to show that the commutator $[\phi(x),\phi(y)]=0$ vanishes outside the lightcone, so that real KG QFT is causal.
One problem for first quantized RQM (which OP seems well aware of) is that it does not describe particle creation and annihilation per se.
Also the usual objections to first quantized RQM still apply, such as, e.g.:

*

*Local interactions couple to both negative & positive energy states,
so that one cannot dismiss negative energy states.


*There are unbounded negative energy states.


*The relativistic probability density
$$ \rho ~=~\frac{i\hbar}{2mc^2}\left(\psi^{\ast} \partial_t \psi - \psi \partial_t \psi^{\ast}\right) \tag{C} $$
may be negative!


*See also this related Phys.SE post.
References:

*

*M.E. Peskin & D.V. Schroeder, An Intro to QFT; p. 14 + p. 27.

--
$^1$ Both propagators (A) and (B) satisfy
$$ (\Box_f-m^2)\langle {\bf x}_f,\tau_f \mid {\bf x}_i,\tau_i\rangle~=~0, \qquad  \Delta\tau~:=~\tau_f-\tau_i~>~0,\tag{D}$$
but only propagators (A) satisfies the normalization
$$\langle {\bf x}_f,\tau_f \mid {\bf x}_i,\tau_i\rangle ~\longrightarrow~\delta^3(\Delta {\bf x}) \quad \text{for} \quad  \Delta\tau \to 0^+. \tag{E} $$
