# Is Probability conserved in RQM?

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1. Aug 27, 2015

### referframe

Is probability (or probability current) conserved in RQM/QFT? I have never seen a simple answer to this question.

2. Aug 27, 2015

### stevendaryl

Staff Emeritus
I think it's usually considered to be a sign of an inconsistency in a theory if probabilities don't add up to one at all times. Or it's a sign that the quantity that you thought of as probability isn't really a probability (which was the case with the current derived from the Klein-Gordon equation)

3. Aug 27, 2015

### DEvens

Is probability of what conserved?

QFT conserves lots of things. Momentum, energy, angular momentum, various charges.

Other things are not so straightforward.

https://en.wikipedia.org/wiki/Lepton_number
https://en.wikipedia.org/wiki/Weak_hypercharge

Some things simply are not conserved. You can pretty much produce as many photons as you are prepared to expend the effort to produce. So the probability of observing a photon is not conserved.

4. Aug 27, 2015

### referframe

I should have said "Is the position probability conserved?".

5. Aug 27, 2015

### stevendaryl

Staff Emeritus
In quantum field theory, there is not a definite number of particles. And particles of the same type are indistinguishable. So in QFT you would not ask "What is the probability that this particle is at location $x$?" Instead, you would ask: What is the probability of finding a particle at location $x$, with the understanding that finding a particle at one location doesn't preclude the possibility of finding an identical particle at another location at the same time.

6. Aug 27, 2015

### Staff: Mentor

The problems with probability in relativistic QM were all resolved when going to Quantum Field Theory. Doing so also showed how they could be resolved in things like the Klein Gordon equation. Since QFT requires antiparticles negative probabilities in a current density are positive probabilities of the antiparticle.

Thanks
Bill

7. Aug 28, 2015

### Demystifier

There is no simple answer because there is no simple answer to an even more elementary question: What is the position operator for RQM/QFT?

8. Aug 28, 2015

### DEvens

Um... No it's not. The probability of finding a particle at a position isn't conserved. Why would you ever think it was? Particles move around.

I think you are struggling to ask a different question. Maybe you could ask what it is you are really trying to ask.

9. Aug 28, 2015

### referframe

Obviously, since it includes SR, QFT/RQM is considered physically more accurate than the original non-relativistic QM. So does that mean that the Born Rule is a nice approximation of reality but not fundamentally true?

10. Aug 28, 2015

### DEvens

The first sentence has nothing to do with the second sentence. So it clearly does not mean that. Whatever you may mean by "fundamentally true" as opposed to true.

https://en.wikipedia.org/wiki/Born_rule

Depending on how you look at it, the Born rule is either a postulate of, or a derivable consequence of the rest of, quantum mechanics.

Again, it seems like you are struggling to ask some other question. Could you please ask that other question?

11. Aug 28, 2015

### Staff: Mentor

Just on that you might like to look into Gleasons Theorem.

Thanks
Bill

12. Aug 28, 2015

### stevendaryl

Staff Emeritus
The Born rule works as well in QFT as in QM. In both cases, we have states, and we use the states to compute probabilities for measurement results, using the Born rule. The difference is that in single-particle, nonrelativistic QM, the "state" is described by a wave function giving a probability amplitude for finding that particle at a particular position, while in QFT, the "state" describes an indefinite number of particles.

13. Aug 28, 2015

### referframe

I am referring to the "conservation law for probability in quantum mechanics" as referred to in the link:
https://en.wikipedia.org/wiki/Probability_current#Continuity_equation_for_quantum_mechanics

14. Aug 28, 2015