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 Shroder 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!

  • $\begingroup$ I'll have to look up those statements as they are not made explicit by the OP. Nevertheless the issue is interesting enough to do this extra work. $\endgroup$ – my2cts Dec 23 '20 at 0:47

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.

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    $\begingroup$ All true, but it doesn't answer the question. Is there a problem with rqm, or is it merely apparent superluminal propagation, as you suggest? $\endgroup$ – Sam Gralla Jul 21 '17 at 14:02
  • $\begingroup$ @SamGralla I believe it is the most straightforward answer there could be. Yes, RQM is flawed, because it has a different set of observables ($x(t)$ instead of $\phi(x, t)$). If you do the math, you can show that this leads to paradoxes which are absent in QFT. $\endgroup$ – Prof. Legolasov Jul 21 '17 at 18:32
  • $\begingroup$ By your logic non-relativistic quantum mechanics is flawed too, since it has the set of observables x(t) (among others). $\endgroup$ – Sam Gralla Jul 21 '17 at 18:39
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    $\begingroup$ @SamGralla no, because in nonrelativistic QM superluminal information transfer is ok. There is no Special Relativity requiring this to be absent for consistency. The origin of the "fundamental flaw" is that having $x(t)$ as observables means superluminal information transfer is possible. This is incompatible with SR. $\endgroup$ – Prof. Legolasov Jul 21 '17 at 18:44
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    $\begingroup$ I would be very interested to see a thought-experiment demonstration of superluminal information transfer involving RQM. This would clarify everything. $\endgroup$ – Sam Gralla Jul 21 '17 at 19:59

OP has a point. On one hand, P&S on p. 14 argue that in first quantized RQM 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 flawed and not Lorentz covariant. A more careful Lorentz covariant analysis reveals that the RQM propagator is

$$\begin{align}&\langle {\bf x}_f,\tau_f \mid {\bf x}_i,\tau_i\rangle\cr &~=~ \int_{\mathbb{R}^3} \!\frac{\mathrm{d}^3{\bf p}}{(2\pi\hbar)^3}\color{red}{ \frac{1}{2\sqrt{{\bf p}^2+m^2}} }\exp\left[\frac{i}{\hbar}\left( {\bf p}\cdot \Delta {\bf x} - \Delta \tau \underbrace{\sqrt{{\bf p}^2+m^2}}_{\text{Hamiltonian}}\right)\right], \end{align}\tag{B} $$

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 real 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!


  1. M.E. Peskin & D.V. Schroeder, An Intro to QFT; p. 14 + p. 27.
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    $\begingroup$ Thanks for this clarifying answer--I never tried to get the RQM propagator in a covariant form and always implicitly assumed that the two propagators are just different. Regardless, I realized that both the propagators have support outside of the lightcone but the QFT formulation doesn't violate causality because causal influence is related to the commutator in QFT whereas, in RQM, the non-vanishing propagator is an issue because the propagator itself is the indicator as to whether there can be causal influence between the said events. Is that correct? $\endgroup$ – Dvij D.C. Dec 23 '19 at 0:15
  • $\begingroup$ The term negative energy states is a misnomer. This is about negative frequency while energy is $E=\hbar |\omega |$. The fact that negative frequency states cannot be dismissed is not an issue, negative frequency states are not unbounded, the charge-current density does not need to be of definite sign. $\endgroup$ – my2cts Dec 23 '20 at 0:44

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.


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