# Why does relativistic Quantum mechanics make sense? [duplicate]

During the last few weeks, I've spent quite some time thinking about Relativistic Quantum mechanics and have realized that special Relativity and Quantum mechanics can be woven together into a single theory as shown by Paul Dirac.

• But then one thing that I am confused about is what happens when the wave function collapses. As far as my understanding goes, Quantum mechanics acts deterministic only until observation, the location of a particle when observed is completely random but can be described by probabilities. Special Relativity on the other hand says that the universe is like a 4D block, and everything is already there. Does this imply that the location of a particle is predetermined and it appears random because there is no way for us to send information to the past. But if, in theory we can go beyond 4 dimensions and into 5 Dimensions, can we know exactly where we'll find the particle if we observe it.

• Now based on this, if in theory, we conduct the double slit experiment and we "see" the location of a particle later in time from a 5 Dimensional universe and then come back and perform the experiment, will we get the same results?

• So is the universe a deterministic block or random?

## marked as duplicate by knzhou, Cosmas Zachos, user191954, Kyle Kanos, glSNov 10 '18 at 13:47

• Easy: it does not. The real theory is QFT; relativistic point-particle QM is but a historical accident. – AccidentalFourierTransform Sep 28 '17 at 18:44
• Can you elaborate on how this situation would turn out in QFT and how it gets rid of determinism of Special Relativity if it does. Thanks – Chandrahas Sep 28 '17 at 18:47
• Presumably, you will find some answers in 220470, 343237 and 4212. – AccidentalFourierTransform Sep 28 '17 at 18:55

So mostly the union of quantum mechanics and relativity has avoided nonlinear wavefunction collapse into unified quantum field theories, at which point "particles" emerge as resonance spikes in amplitudes to see an average change over at spacetime position $(ct,~\vec r)$ due to a perturbation at spacetime position $(ct',~\vec r').$ See A. Zee, Quantum Field Theory in a Nutshell, I.4 in order to see a relatively quick construction of this based on "jumping on a mattresss."

The ignoring of wavefunction collapse is not as bad as it sounds, as measurement is a somewhat philosophical addition to quantum theory: the various interpretations are all constrained to give physically the same results so there is no experiment to distinguish them. In particular there is one theory of measurement that totally hides the unitary dynamics, and that is the many-worlds interpretation. The claim is that you get entangled with the result of some experiment and then the only coherent description of your phenomenology must be some classical superposition of observers who see the system in a "collapsed" state, and you happen to be unable to see that whole superposition because that would be a substantially different brain-state than any of the brain-states in the ensemble.

However traditional collapse theories have also been brought into QFT's purview; see e.g. Tumulka (2005), On Spontaneous Wave Function Collapse and Quantum Field Theory.

• So if there are no particles, then what do the fields represent? Doesnt the square of the field amplitude represent the probability of finding a particle like in ordinary qm? – Chandrahas Oct 3 '17 at 3:30
• Let me take a very condensed matter approach. Suppose that we have some one-particle Hamiltonian $\hat H$ with wavefunctions $\varphi_n(x)$ and we want to extend it from one particle to non-interacting fermions, we introduce annihilators for each state $a_n$ such that $\{a_m,a_n^\dagger\}=\delta_{mn}$. The field creation operator would look something like $\psi^\dagger(x)=\sum_n\varphi_n(x)~a^\dagger_n.$ It's an operator, not a wavefunction, but $\psi^\dagger(x) \psi(x)$ is an operator whose expectation is the number density of particles at position $x$. Does that help? – CR Drost Oct 3 '17 at 4:28