# Relativity and Quantum Mechanics

I have been thinking about the problem of relativistic path integrals and I encountered several difficulties. Let's assume we have a particle initially a position $$x_i$$ at $$t_i$$ in a certain reference frame. In another inertial reference frame the positions are $$x_i'$$ at $$t_i'$$. Feynman's path integral method allows us to calculate the conditional problem to observe the particle at $$x_f$$ at $$t_f$$ given that it was initially at position $$x_i$$ at $$t_i$$: $$P(x_f,t_f|x_i, t_i)$$ In another frame of reference the probability is $$P'(x_f',t_f'|x_i', t_i')$$ Common sense would tell us that $$P(x_f,t_f|x_i, t_i)=P'(x_f',t_f'|x_i', t_i')$$ Since we are observing the same event. However now the problem comes: $$\int_{-\infty}^{+\infty} dx_f P(x_f,t_f|x_i, t_i) = 1$$ Which basically means that the probabilities have to add up to 1. In the different frame of reference the same must be true, but according to special relativity the simultaneity hypersurface is different, but still the probabilities along that hypersurface have to add up to 1. But these probabilities are in general not the same as the probabilities in a different reference frame. If my arguemtn is correct then it seems to me that we are left with three choices: 1. The probabilities are different in each frame of reference. This is absurd. 2. Not all the paths are allowed such that this condition is indeed satisfied. This is also absurd. 3. There is no free will, meaning that an observer can't arbitrarily choose his velocity.

Is my argument correct?

• I don't see where the contradiction is. Why can't they still add up to 1? – Eric David Kramer Mar 25 at 14:44
• Perhaps I should have shown a calculation. But a priori this should not be the case. – user139383 Mar 25 at 14:55
• Notice the the QM formulation of the path-integral isn't necessarily relativistic invariant (because QM isn't). In the general formulation both the action and the path integral measure are invariant under Lorentz transformation, so the whole integral is, but this isn't necessarily true for the QM particle states. – gented Mar 25 at 16:10
• Also, in QM it is unclear how position eigenstates $|x(t)\rangle$ transform under Lorentz transformations: the position $|x\rangle$ is an element of the Hilbert space whereas $t$ is the parameter of the Hamiltonian flow: they are treated differently. – gented Mar 25 at 19:02
• @gented Yes I guess both your comments are right. However I thought about it in more detail and I think if the way I argue is correct, then no standard probabilistic theory would be compatible with special relativity. Because the arguments I gave are not confined to QM. – user139383 Mar 25 at 20:41

## 1 Answer

I think that if you impose that the probabilities are equal, you can actually derive the relativistic scalar propagator. Please forgive me that I don't have time to work out the details here. But it has to work. That is, under a boost,

$$| x , t \rangle \to \int\!dy' K(x'-y',t') |y'\rangle$$

where $$K$$ is some kind of Fourier transform of $$\sqrt{E'/E}$$. (Recall that $$\int d^3p/E \sim \int d^4p \delta(p^2-m^2)$$ is Lorentz invariant.)

Imposing some kind of composition law

$$\int\!dx_1 P(x_2 ,t_2|x_1,t_1)P(x_1,t_1|x_0,t_0) \sim P(x_2,t_2|x_1,t_1)$$

which I think includes the integral you wrote in your question should then give you a consistency condition which should allow you to solve for $$P(x_f,t_f|x_i,t_i)$$.

In short, you are looking for unitary scalar representations of the Lorentz group. It has to work. What you have shown in your question is that this cannot be done in an arbitrary way. The propagator is uniquely fixed by Quantum Mechanics and Special Relativity.

P.S. You don't have to deduce from this that not all paths in the path integral are allowed. I think it should be sufficient that paths that violate causality give large action whose phases will cancel in the Path Integral, leaving the causal path as the dominant one.