For a massive particle going at relativistic speeds relative to an observer, would experience a proper time that is different from the proper time of the observer.

Let us now suppose that a massive particle is in a superposition of two very different momemtum eigenstates. What would be the proper time of such a particle? does this time difference cause some interesting physical effects?

I would imagine that it would be equivalent to having a particle in a superposition of two proper times (if that makes sense). After a measurement of the particle's mometum, the system collapses to either momentum $\mathbf{p}_1$ or $\mathbf{p}_2$, and thus to a proper time of either $\tau_1$ or $\tau_2$, respectively. But I am wondering if before the measurement, this mismatch of proper times would have any physical consequence in the observer's description of the evolution of the quantum system.

I know Dirac's equation, but I would not know where to start to verify such a claim. I was unable to find related threads or articles.

Disclaimer, I am not considering the effects of general relativity, just special relativity.

Edit: just thought of something, if I have a particle that decays after time $t$ and the particle is in a superposition of two different states, a state at rest (defined by the observer) and a state with a relativistic speed. Could the particle be in superposition of decayed at rest and not decayed at high speed after time $t$, as measured by the observer?

  • $\begingroup$ Why does proper time matter? Particles don't age, and you can't decay in fixed time, rather it's a fixed probability per unit time. Particles and QM are inherently non-relativistic, so once you bring in relativity you need fields are then all you know are approximate free-particle initial and final states, along with a transition amplitude. $\endgroup$
    – JEB
    Commented Jun 30, 2021 at 14:40
  • $\begingroup$ "What is the proper time of a particle in a superposition?" a superposition of the two proper times? $\endgroup$ Commented Jun 30, 2021 at 16:25
  • $\begingroup$ @JEB if you prefer you can think not of an elementary particle but more like a complex composite quantum system with some internal clock or change $t$ for a half-life time. I sure get that that relativistic QM leads to QFT, which can be much harder, but is proper time never ever treated in quantum mechanics? $\endgroup$
    – Mauricio
    Commented Jun 30, 2021 at 16:43
  • $\begingroup$ @Prof.Legolasov sure, that's what I said in the text above. $\endgroup$
    – Mauricio
    Commented Jun 30, 2021 at 16:44
  • $\begingroup$ arxiv.org/pdf/1710.06504.pdf $\endgroup$
    – JEB
    Commented Jun 30, 2021 at 19:33

1 Answer 1


From comments: OP would be satisfied with a definition of a quantum-mechanical operator that corresponds to the physical observable which is the proper time between coordinate times $t_1$ and $t_2$.

Start with classical special relativity. A particle has a 3-velocity $\vec{v}$ and a 3-momentum $\vec{p}$ which are pointing in the same direction with absolute values related by $$ p = \frac{m v}{\sqrt{1 - \frac{v^2}{c^2}}}. $$

Invert this relation to give $$ v = \frac{c p}{\sqrt{p^2 + m^2 c^2}}. $$

This means that the proper time $T(t_1 \rightarrow t_2)$ can be expressed as a function on the phase space as $$ T(t_1 \rightarrow t_2) = c^{-1} \sqrt{c^2 - v^2} (t_2 - t_1) = c^{-1} \sqrt{ c^2 - \frac{p^2 c^2}{p^2 + m^2 c^2} } (t_2 - t_1) =$$ $$ \frac{m c}{\sqrt{p^2 + m^2 c^2}} (t_2 - t_1). $$

To quantize, note that any well-behaved function $F(\vec{p})$ can be turned into an operator by the following procedure. We fix an eigenbasis of $\hat{p}$ first (these are non-normalizable plane waves $e^{i \vec{p} \vec{x}}$). Then we define the operator $\hat{F}$ to act on the elements of this basis via

$$ \hat{F} e^{i \vec{p} \vec{x}} = F(\vec{p}) \cdot e^{i \vec{p} \vec{x}}. $$

Now making use of the fact that the eigenbasis of $\vec{p}$ is complete, we extend this definition to the whole Hilbert space. Mathematically this means we need the Fourier transform and the inverse Fourier transform:

$$ \hat{F} \Psi(x) = \int \frac{d^3 p}{(2\pi)^3} e^{i p x} F(p) \int d^3y e^{-ipy} \Psi(y) $$

By applying this procedure to $F(p) = T(t_1 \rightarrow t_2)$, you end up with a quantum operator $\hat{T}$ which measures the proper time that passed between coordinate times $t_1$ and $t_2$.

You can build this operator both in Klein-Gordon and Dirac theory, in either case it will be the same differential operator (as the proper time doesn't depend on the spin degrees of freedom).

There is also a manifestly relativistic definition that uses the theory of quantization of constrained systems, and completely agrees with the definition above. I won't give it in this answer, but let me know in the comments if you're curious.

If you want such an operator in QFT rather than QM, then it doesn't exist. In QFT, individual particles are not well-defined classical concepts, as the classical counter part of QFT is a field theory, not a particle theory. Hence proper times of particles are not physical observables. This makes sense: QFT has states where your particle doesn't even exist (like the vacuum for example), so you can't define proper time for a specific particle because you can't fix this specific particle. Nevertheless you can define operators such as "sum of proper times for all involved particles" in free QFT that make sense (but I don't think it is possible to extend this definition to interacting QFT where the concept of a particle is even more obscure).

  • $\begingroup$ How do you get your eqn 3. Proper time depends on the trajectory followed by the particle between the end points. To know the exact trajectory you need to know both position and momentum which is not possible in QM. Further time is intrinsically invariant in standard non relativistic QM. QM cannot handle all this stuff because it is intrinsically non relativistic. $\endgroup$
    – Shashaank
    Commented Jul 3, 2021 at 10:43
  • $\begingroup$ @Shashaank for a free particle, proper time only depends on the momentum due to translation invariance. If you had a particle in a potential, it would also depend on position. In this case you’d have a function on phase space that needed a more sophisticated method to be made into an operator, for example, the Weyl quantization map. $\endgroup$ Commented Jul 3, 2021 at 10:56
  • $\begingroup$ Ok I see what you mean. The particle is only allowed to follow the straight line path. Any paper or book that deals with the stuff you have written above in a bit more detail. Was just curious . $\endgroup$
    – Shashaank
    Commented Jul 3, 2021 at 11:02
  • $\begingroup$ Also after you have constructed such an operator “which corresponds to the………..”, is the particle really in superposition of proper times (whatever that means) $\endgroup$
    – Shashaank
    Commented Jul 3, 2021 at 11:05
  • $\begingroup$ @Shashaank yeah exactly. I don’t have a good reference right now sorry, but if you’re curious about Weyl quantization (or more generic Kontsevich quantization) there are a lot of good literature out there. What I’ve described is just a very special case $\endgroup$ Commented Jul 3, 2021 at 11:24

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