When one calculate the amplitude for a particle to propagate between two points, the results seems to violate causality. One book that makes some comments about this is Peskin & Schroeder, chapter 2, page 10. They use a square-root Hamiltonian. Even if we use the local Klein-Gordon equation and we start with say a delta function, then the wave function will spread superluminally.

In the case of a relativistic string we can follow Polchinski's book for example and we get the string spectrum (page 23) but there is nothing about causality or superluminal propagation.

So again if we start with a localized wave packet for the string, will this wave packet spread superluminally? If yes, that would mean that string theory is inconsistent?

I am currently a beginer in string theory and qft but I think this should be an important question addressed at the begining. Regarding the calculation for the amplitude for a string to propagate between two spacetime points I have not found any other than Emil Martinec, arXiv:hep-th/9304037, but it seems that he is using String Field Theory. But that paper makes me think that: as in particle theory we have to ensure causality going from first quantization to QFT, we need to address the analogous question in string theory (and perhaps go unavoidably to SFT?).

  • $\begingroup$ "if we start with a localized wave packet for the string" - how does that look? The states of string theory are not states corresponding to a particular "position" of the string, but corresponding to particular "excitations"/particles "on" it, and since string theory is not a quantum field theory, asking for the amplitude for the string to propagate does not really make sense unless you explain how you think that quantity should be defined and computed within the framework of string theory. $\endgroup$
    – ACuriousMind
    Commented Jul 27, 2017 at 9:59
  • $\begingroup$ See for exmple Zwiebach's book page 268, there is table with wavefunctions (but in momentum space, with can transform them to position space anyway). Also in page 269 there are wave equations eq (12.188) , (12.189) and (12.194). This are differential equations so to get a particular solution I need a initial value, I mean a function at t=0. This function is the initial wave packet I am free to choose. I can choose this function to be a gaussian function for example or a similar smooth funtion in small region but vanishing outside that region. Is that clear? $\endgroup$
    – MoYavar
    Commented Jul 27, 2017 at 14:45
  • $\begingroup$ Ah, the thing is, string theory doesn't really say that these states just evolve according to the free Hamiltonian there. If you want to compute amplitudes for anything, you must look towards chapter 25 in Zwiebach. $\endgroup$
    – ACuriousMind
    Commented Jul 27, 2017 at 15:34
  • $\begingroup$ Yes I know off course that the relevant observables are scattering amplitudes. At the end of the day what we want to calculate with strings are say graviton-graviton scattering amplitudes or other physically meaningful, gauge-invariant quantities. But my logic is: Before going to the interacting theory we have to be sure everything is well in the free theory, dont we? $\endgroup$
    – MoYavar
    Commented Jul 27, 2017 at 16:19
  • $\begingroup$ You say string theory doesn't really say that these states just evolve according to the free Hamiltonian there, but I think it should be interacting string theory doesn't really say that these states just evolve according to the free Hamiltonian there but free string theory does, am I right? $\endgroup$
    – MoYavar
    Commented Jul 27, 2017 at 16:21

1 Answer 1


As an experimentalist I have a working view of quantum mechanical calculations, whether from simple potential problems, field theory and even strings if they manage to be predictive.

Probability is the same in classical or quantum mechanics, by construction. It is the calculational method that differs.

Take the probability function for a throw of dice to come up 6. There is no causality. It is just a mathematical function coming out of calculations. If you need space variation take the probability of a child in greece to have an IQ of 120. This is a probability function coming from measurements, but the single throw ( one child) answer comes out immediately, is there, no matter whether Greece, China, or US. nothing superluminal. Quantum mechanics is about probability distributions for observables.

Quantum mechanical postulates are at the root of all wave function calculations, number 5 is relevant here.

  1. For a system described by a given wavefunction, the expectation value of any property q can be found by performing the expectation value integral with respect to that wavefunction.

These postulates are true for all equations, and underlie all predictions.

These are mathematical manipulations leading to predictions of probability distributions.

They are mathematical functions uniquely described on paper. The free particle wavefunction on which field theoretical concepts are built with creation and annihilation operators , is just a mathematical function defined on all four dimensional space. It is an interaction/experiment which will pick up an observable at a specific (x,y,z,t). The accumulation of these observables will give a distribution to be compared to the predicted by field theoretical calculations distribution, but it is fundamentally a probability distribution whether a cross section or angular distribution.

There is no propagation in the mathematics of the wave function, they span the whole (x,y,z,t) phase space for the given boundary conditions. So the question of supeluminal , in my opinion, does not enter. If a mathematical solution, wavefunction, assumed for a measurable observable, as for example the position x for a particle, gives a superluminal probability, in my opinion, it is the wrong mathematical model for the particle, i.e. wrong boundary conditions have been assumed in the modeling.

In the example given in the comments (page 14) the analysis leads to the need of new modeling, consistent with my statement above.

"Relativistic causality is inconsistent with a single particle theory. The real world evades the conflict through pair production. This strongly suggests that the next thing we should do is develop a multi-particle theory "

AFAIK real elementary particles are modeled with wavepackets, not by a plane wave solution. The plane wave solutions of relativistic equations are used in order to define the quantum field on which creation and annihilation operators act, and the particle is a wavepacket moving on this field.

One should have a clear distinction between possible mathematical solutions, and the specific quantum mechanical model for given systems and boundary conditions. It is nature one needs to model, and mathematics is not a generator of nature ( the platonic view), but a tool used as needed to model nature.

If a mathematical solution proposed for modelling a particle, as the plane wave solution was, comes into contradictions with data, ( in this case no superluminal data exist ) the modeling needs modification, as is logically suggested in Coleman's lecture notes.

The question:

Do string-wave functions always spread superluminally?

Should be answered by, wavefunctions (strings or ...) modeling real physical systems cannot give a superluminal probability, and thus will have to be used appropriately in the model so as to give a correct physical behavior, any supeluminal part mathematically cancelled out.

  • $\begingroup$ Hi again. So at the end of the day, Does your answer imply that Relativistic Quantum Mechanics is perfectly causal? $\endgroup$
    – MoYavar
    Commented Jul 28, 2017 at 22:45
  • $\begingroup$ My answer says that causal is an experimental observation. Relativistic quantum mechanics is a mathematical system, functions. They are used in modeling interactions, and the postulates that connect the abstract mathematics to experiments pick up that subset of solutions which display probabilies, and yes the predictions are causal FOR THE PROBABILITY distributions. There is zero probability that the incoming lines are detected/modeled after the outgoing lines in a feynman diagram $\endgroup$
    – anna v
    Commented Jul 29, 2017 at 5:28
  • $\begingroup$ Could you please add to your answer what is your view on the arguments that are usually found at the beginning of textbooks like "It is possible to measure a particle’s position in our theory, x is an observable, and this leads to a conflict with causality" Coleman page 14. These arguments are bsed on the propagation of the wave function. I think this is a good reason to say that The Quantum Theory of a Relativistic Point Particle has a problem. Do you think they are missleading something when they say that the particle can travel faster than light? $\endgroup$
    – MoYavar
    Commented Jul 29, 2017 at 16:17
  • $\begingroup$ " "It is possible to measure a particle’s position in our theory, x is an observable, and this leads to a conflict with causality" Coleman page 14. I think it is a mistaken assumption that the wavefunction is physical. The wavefunction is a mathematical construct whose complex conjugate squared gives the probability of a measurement giving a specific value. In the same way that the caluclated trajectory of a ball does not mean that the ball is instanteniously at the beginning and at the end of the trajectory ( superluminal) probabilities are there, and measurements follow the probabilities $\endgroup$
    – anna v
    Commented Jul 29, 2017 at 17:57
  • $\begingroup$ Coleman conclude in this chapter "Relativistic causality is inconsistent with a single particle theory. The real world evades the conflict through pair production. This strongly suggests tha tthe next thing we should do is develop a multi-particle theory " and is clearly stating "in this theory",. I am not in a position to check his mathematics. My experimentalist's conclusion is that "it is the wrong wavefunction for the boundary conditions" giving inconsistent probability distributions, hence the need for field theory. The free particle solution cannot represent a particle, particles are $\endgroup$
    – anna v
    Commented Jul 29, 2017 at 18:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.