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!


4 Answers 4


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
    Commented Jul 21, 2017 at 14:02
  • 2
    $\begingroup$ @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. $\endgroup$ Commented Jul 21, 2017 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
    Commented Jul 21, 2017 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. $\endgroup$ Commented Jul 21, 2017 at 18:44
  • 2
    $\begingroup$ I would be very interested to see a thought-experiment demonstration of superluminal information transfer involving RQM. This would clarify everything. $\endgroup$
    – Sam Gralla
    Commented Jul 21, 2017 at 19:59

OP has a point. On one hand, P&S on p. 14 argue that in first quantized RQM in the Newton-Wigner (NW) operator formalism the kernel is

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

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 kernel 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! There is a superluminal overlap.

On the other hand, P&S in eq. (2.50) on p. 27 find exactly the same kernel (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 frequency states, so that one cannot dismiss negative frequency states.

  • There are unbounded negative frequency 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 & this related Phys.SE posts.


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


$^1$ Both kernels (A) and (B) satisfy the kernel property

$$ (\Box_f-m^2)\langle {\bf x}_f,\tau_f \mid {\bf x}_i,\tau_i\rangle~=~0, \qquad \Delta\tau~:=~\tau_f-\tau_i~>~0,\tag{D}$$

but only propagators (A) satisfies the normalization

$${}_{\rm NW}\langle {\bf x}_f,\tau_f \mid {\bf x}_i,\tau_i\rangle_{\rm NW} ~\longrightarrow~\delta^3(\Delta {\bf x}) \quad \text{for} \quad \Delta\tau \to 0^+. \tag{E} $$

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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? $\endgroup$
    – user87745
    Commented Dec 23, 2019 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
    Commented Dec 23, 2020 at 0:44
  • $\begingroup$ the second section of the book by Weinberg is called "Relativistic quantum mechanics" and it is claimed that it is a completely consistent theory. Though, he does not discuss the issue of causality, I believe. Still, I think the answer is that one should not compute the transition amplitudes but the commutator. One should just better understand why $\endgroup$
    – Dr.Yoma
    Commented Apr 2, 2023 at 13:41

That is a both long standing and quite subtle issue. I recently published a paper (arXives version here) on the subject, regarding the notion of spatial localization at given time for a massive real Klein-Gordon particle in Minkowski spacetime.

Let us focus on the one-particle states of a quantum field (specifically the massive real KG field), what you call RQT.

There are here two distinct problems with any attempt to define a spatial localization observable (see the paper):

  • One concerns the possibility of superluminal spread of the probability density of detecting a particle (Hegerfeldt's theorem) at different times.
  • The other issue regards a possible causality violation by the post-measurement state after a first detection (Malement's and Malament-like theorems).

The Newton-Wigner relativistic position operator for a single relativistic particle (see my paper for a review on this notion) shows up problems with both issues. It is definitely ruled out by the Hegerfeldt theorem, though it enjoys also physically sound properties, that should be preserved. And also, it is uniquely determined by basic requirments: the very CCR or, more rigorously, the action of the spatial translations on the spectral measure of these selfadjoint operators (from Mackey's imprimitivity theory) as established by Wightman.

A crucial point is that no notion of localization observable can be defined in terms of projection-valued measures (PVMs) i.e., standard selfadjoint observables, because they automatically would produce a superluminal probability detection spread. This is a non-trivial consequence of the fact that the states of single particle (but also all states of the field) have positive energy.

The only possibility compatible with the restrictions arising by the Hegerfeld therem is to define localization observables written in terms of positive operator valued measures (POVMs). Actually the theorem also imposes severe constrains on these type of notions, in particular there cannot exist states where a particle is localized in a bounded region.

However, esploiting POVMs, the first problem can be made harmless as rigorously established in my paper, relying on previous ideas and results by D.Terno and D. Castrigiano. An important point is that this new notion of spatial localization also preserves all good properties of the Newton-Wigner observable. The Heisenberg inequality should be changed though.

Since a POVM does not define a uniquely associated post measurement state procedure, the second issue is still open.


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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