# Probability Postulate in Relativistic QM?

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1. Jan 24, 2015

### referframe

In non-relativistic QM, the probability postulate works very well in position and momentum space. But, I have read that in relativistic QM (Dirac or QFT), the probability postulate in position space does not work because the corresponding probability density function is not Lorentz covariant. But, what about probability in momentum space in relativistic QM? Is a probability interpretation also not possible there - is it also not Lorentz covariant?

2. Jan 24, 2015

### atyy

In non-relativistic QM, the probability postulate is the Born rule, which is given in various forms of increasing generality in http://arxiv.org/abs/1110.6815. Even in non-relativistic QM, it does not refer specifically to position or momentum. The Born rule also holds in relativistic quantum theory, except that the observables and the states are different. The general rules from non-relativistic QM do carry over to relativistic quantum theory, and yield Lorentz invariant probabilities, eg. http://arxiv.org/abs/1007.3977.

3. Jan 24, 2015

### referframe

atyy:

Ok, but I can find no references to "Born rule" or "probability" in either of my QFT books or in other online sources for QFT. What are the "different" "observables" and "states" that you refer to that have a probabilistic interpretation in relativistic QM?

4. Jan 24, 2015

### atyy

In some cases the field itself is an observable in QFT. Roughly, in QM (including QFT) the order in which non-commuting observables are measured matters. If A and B do not commute, measuring A before B changes the distribution of B, compared to when B is measured before A. Thus, in relativistic QFT, if measurements are taken at spacelike separation, the observables must commute, otherwise there might be faster than light transmission of classical information. The requirement in relativistic QFT that spacelike observables commute is found in eg. http://www.damtp.cam.ac.uk/user/tong/qft.html (Eq. 2.86). That the field itself is an observable is not so obvious in QFT books, because there is a long chain from correlators of fields to the LSZ formula to scattering cross sections etc to get a match to how experiments are done.

The Born rule is rarely stated in QFT books because all the postulates of QM are assumed, but one can find the Born rule stated eg. in the QFT texts of Weinberg https://www.amazon.com/The-Quantum-Theory-Fields-Volume/dp/0521670535 (Eq 2.1.7) and Haag https://www.amazon.com/Local-Quantum-Physics-Theoretical-Mathematical/dp/3540610499 (Eq I.3.1).

Last edited by a moderator: May 7, 2017
5. Jan 24, 2015

### bhobba

I thought it was implicit in the propagator, <x|x'> which is used extensively in QFT, |<x|x'>|^2 gives the probability of being at position x then later being at position x'.

Or am I missing something?

Thanks
Bill

6. Jan 24, 2015

### atyy

7. Jan 24, 2015

### bhobba

Agreed that is a big issue. There is also the issue the state resides in a Fock space.

Thanks
Bill

8. Jan 25, 2015

### vanhees71

First of all one should mention that the very fundamental principles of quantum theory are the same in non-relativistic and relativistic theory. The Born rule is a the heart of the application of the Hilbert-space formalism to describe real-world experiments. Any good QFT book (my favorites are the three volumes by Weinberg, Quantum Theory of Fields) mentions it in the beginning (perhaps not under the name Born's rule).

In short it's as follows: A (pure) state can be represented by a ray in Hilbert space, which itself can be represented equivalently by the statistical operator $\hat{\rho}=|\psi \rangle \langle \psi|$, where $|\psi \rangle$ is any normalized vector in the ray.

An observable (like position, momentum, angular momentum, etc.) is represented by a self-adjoint operator on the Hilbert space. Possible values are the eigenvalues of the operators. Suppose you measure an observable $A$ (represented by the self-adjoint operator $\hat{A}$) on a system prepared in the state represented by $\hat{\rho}$. Let $a$ by an eigenvalue of $\hat{A}$ and $|a,\beta \rangle$ a complete set of orthonormal eigenvectors to the eigenvalue $a$. Then Born's rule states that
$$P_{\psi}(a)=\sum_{\beta} \langle a,\beta|\hat \rho a,\beta \rangle.$$

This formalism shows you that to define the probabilities correctly first of all you need an observable. The observable algebra built by the operators representing the observables is determined from a modern form of the correspondence principle: You look for representations of the symmetry group of the physical system. The most important symmetry underlying all physics is the symmetry structure of space-time. In the case of non-relativistic quantum theory it's the Galilei group for special-relativistic theory it's the Poincare group.

As it turns out for massive particles (the only case making sense in non-relativistic QT as this also comes out of the symmetry analysis based on the Galilei group) of any spin (also spin comes out of the symmetry analysis in a quite natural way) there can be defined a position observable in both non-relatistic and relativistic theory.

In relativistic QT there's also the possibility of massless quanta like the photon. It turns out that you cannot define a position observable for massless particles with spin $s \geq 1$.

It's also pretty intuitive, why the localization of a relativistic particles makes problems: To localize a particle well you must confine it somehow to a small volume in space, which e.g., can be done with various kinds of traps. Now according to the Heisenberg-Robertson uncertainty relation, $\Delta x \Delta p \geq \hbar/2$ (which is valid in both non-relativistic and relativistic theory, provided there exists a position observable in the strict sense). This implies the better you confine the particle to a small area, i.e., the smaller the standard deviation of position, $\Delta x$, becomes, the broader the spread $\Delta p$ in momentum space. This means with quite some probability the particles can have large momentum. Hitting the "wall" (in reality some kind of potential well provided by the trap) with high momentum however means that according to the relativistic QFT there's always some probability to create new particles destroying the original particle you wanted to localize. For massless particles it's very easy to create very soft ones from hitting a potential well with a harder one (like for photons, you need charged particles to confine it and hitting with a photon a charged particle it will wiggle around and emit more radiation, i.e., photons), because there's no energy threshold for creation of a massless particle.

9. Jan 25, 2015

### Demystifier

It is true that it is not Lorentz covariant, but it does not necessarily need to be a problem. According to the operational view of QM, the aim of QM is not to give the probability that the particle is at certain position. Instead, its aim is to give the probability that the particle will be found at certain position. In the latter case, the probability only has a contextual meaning, i.e. it only makes sense when there is a measuring apparatus that will measure an observable. Each apparatus defines a distinguished Lorentz frame, the one in which the apparatus is at rest. Therefore, in a purely operational QM, physical observables do not need to be Lorentz covariant.

By the way, something similar occurs even in classical (non-quantum) relativity. Due to Lorentz contraction, the length of a rod transforms from one Lorentz frame to another in a non-covariant way. The length of a rod is neither a Lorentz scalar, nor a Lorentz vector or tensor. Yet, there is an experimental procedure which (in principle) can measure Lorentz contraction.

There is also an example in general relativity. The gravitational force, being related to Christoffel connections, is also not a tensor and hence not a covariant object. Yet, the gravitational force can certainly be measured, probably more easily than any other "observable" in general relativity.