Photons, particles and wavepackets

[OOO]

In this way, you will obtain the Schrödinger equation. But how will you recover the probabilistic interpretation of the solutions of the Schrödinger equation? You should not postulate it, but derive from the probabilistic interpretation of QFT. So, how exactly you will do that?

You can't because you've already dropped the information about probabilities in your nonrelativistic approximation. The classical solutions virtually describe wave functionals with infinitely sharp distribution (localized at the classical solution). So I think you'd have to reintroduce the spreading of the wave functional again somehow in order to describe how the time evolution deviates from the classical path.

As to the details I have no clue. But I guess you have.

 

[OOO]

I would very much like to see how to derive the one particle wave function

<br /> \Psi:\mathbb{R}^3\to\mathbb{C},<br />

its equation of motion, and its probability interpretation, by starting from the field wave functional

<br /> \Psi:X^{\mathbb{R}^3}\to\mathbb{C},<br />

its equation motion, and its probability interpretation. (X defines what is the value of the classical field. It can be X=\mathbb{R},\mathbb{C},\mathbb{R}^4,\ldots or something else.)

Well then I'm afraid you'll have to take a look at some QFT textbook.

 

[OOO]

Have you seen this notation a lot already? If so, you wouldn't bother mentioning some sources? I mean, I have not seen this anywhere yet.

Come on, it's a bit OT discussing math 101. I'm sure you know the most stringent definitions of distributions.

 

[jostpuur]

Well then I'm afraid you'll have to take a look at some QFT textbook.

I've already been forced to reinvent the idea of the wave functional on my own, because all that the books tell are the cursed operators and their commutation relations!

 

[jostpuur]

dumb question

Come on, it's a bit OT discussing math 101. I'm sure you know the most stringent definitions of distributions.

What does OT mean?

 

[OOO]

What does OT mean?

off-topic

 

[OOO]

I've already been forced to reinvent the idea of the wave functional on my own, because all that the books tell are the cursed operators and their commutation relations!

Yes, the textbooks have more to say about the Heisenberg picture but it should't be difficult to put it in Schrödinger terms. I wouldn't call this reinvention.

If you don't like operators that much, try out the path integral approach (but of course you'll need some operators to show the equivalence of expectation values in both approaches).

 

[jostpuur]

Yes, the textbooks have more to say about the Heisenberg picture but it should't be difficult to put it in Schrödinger terms. I wouldn't call this reinvention.

It's so relative. There has been incidents where people, who already know QFT in their own opinion, tell me that the wave functional stuff is something that I have come up on my own and that doesn't really belong to the correct QFT. No doubt, because the QFT seems to be operators and Feynman diagrams usually.

 

[Demystifier]

I've already been forced to reinvent the idea of the wave functional on my own, because all that the books tell are the cursed operators and their commutation relations!

Then see
B. Hatfield, Quantum Field Theory of Point Particles and Strings
This pedagogically written textbook on QFT (and string theory, but you don't need to read the part II if you don't like strings) develops the functional Schrödinger representation of QFT in detail.
In addition, it points out some conceptual details on the difference and relation between particles and fields.

 

[Demystifier]

You can't because you've already dropped the information about probabilities in your nonrelativistic approximation. The classical solutions virtually describe wave functionals with infinitely sharp distribution (localized at the classical solution). So I think you'd have to reintroduce the spreading of the wave functional again somehow in order to describe how the time evolution deviates from the classical path.

As to the details I have no clue. But I guess you have.

Yes I do. (Although, it does not really work in the way you sketch above.) But as I already said, first quantization can be deduced from second quantization (QFT) ONLY in the nonrelativistic formulations of both first and second quantizations. This is closely related to the fact that relativistic QM does not have well defined probabilistic interpretation, at least not in the conventional orthodox approach.
It is frequently said that relativistic QFT solves this problem of relativistic QM, but it does not. Instead, it merely sweeps it under the carpet. It is not a problem for most of the practical applications of QFT, but it is a problem as a matter of principle. You cannot just state the usual axioms of RELATIVISTIC QFT and then derive all the rules of nonrelativistic QM as an approximation.

 

[OOO]

You cannot just state the usual axioms of RELATIVISTIC QFT and then derive all the rules of nonrelativistic QM as an approximation.

You say this. But have you tried hard enough ? I can neither confirm nor refute your claim because there would have to be a no-go theorem or a proof of said NR limit for that.

 

[Demystifier]

You say this. But have you tried hard enough ? I can neither confirm nor refute your claim because there would have to be a no-go theorem or a proof of said NR limit for that.

The argument (not a proof) is actually simple. NR QM contains a NR position operator. It should be a NR limit of the relativistic position operator. However, the latter does not seem to exist. I am not sure if there is a rigorous proof that it does not exist, but I know that the most obvious attempts do not really work, for one reason or another.

On the other hand, in my paper I show that the axioms of nonrelativistic Bohmian mechanics CAN be derived as an approximation of the axioms of relativistic Bohmian mechanics (because the axioms of Bohmian mechanics are not based on operators describing observables). In a sense, this makes Bohmian mechanics more powerfull than the orthodox approach.

 

[OOO]

The argument (not a proof) is actually simple. NR QM contains a NR position operator. It should be a NR limit of the relativistic position operator. However, the latter does not seem to exist. I am not sure if there is a rigorous proof that it does not exist, but I know that the most obvious attempts do not really work, for one reason or another.

You're just shifting the problem from the NR limit to the position operator. The fact that you can't show it doesn't mean a proof doesn't exist.

On the other hand, in my paper I show that the axioms of nonrelativistic Bohmian mechanics CAN be derived as an approximation of the axioms of relativistic Bohmian mechanics (because the axioms of Bohmian mechanics are not based on operators describing observables).

What does it help to prove well known physical theory A from speculation B ? Unless you haven't got a Bohmian equivalent to QFT there is no point in doing that.