# B forces of nature and QM

1. Jun 29, 2017

### mieral

First, if you want to refer to the Schrodinger Equation only.. is it only called Quantum Mechanics? Or does the word Quantum Mechanics also include Quantum Field Theory? If true. Then what words should you refer to the non-relativistic Schrodinger Equations only?

About the forces of nature. The Hamiltonian contains a term for the forces of nature like strong force, weak force or em force contribution to the potential energy, and we use the Hamiltonian to calculate the wave function.. is this correct?

But do people use the Schrodinger Equation to compute for the strong force or weak force? Can you cite an application where the original non-relativistic Schrodinger Equation is used in the strong or weak force instead of QFT?

2. Jun 29, 2017

### houlahound

I don't think there is an official glossary stating when you call it QM or QFT or anything else.

there is just a bunch of problems to solve, physical laws and invented mathematical tools to solve them - call them what you prefer.

if you get to the task of solving a specific problem and you want to be peer reviewed you will have to state what actual methods you employed precisely.

3. Jun 30, 2017

### mieral

What I was specifically asking was. In atoms we commonly use schrodinger equation only because it is not relativistic and it is connected to the electromagnetic field. How about in the strong and weak force. Do physicists also use schrodinger equation a lot too or mostly QFT?

4. Jul 1, 2017

### bhobba

QM is formally just an extension of probability theory:
http://www.scottaaronson.com/democritus/lec9.html
https://arxiv.org/abs/quant-ph/0101012
https://arxiv.org/abs/1402.6562

Now that can be applied to normal; classical mechanics and you get what we usually call QM. Schrodinger's equation etc believe it or not follow from symmetry considerations ie the probabilities are frame independent ie the same irregardless of where you are, what time it is or how fast you are travelling, It's very intuitive, but really you are applying the Principle of Relativity. It is this that is usually applied to the hydrogen atom etc.

To go to QFT is easy - you simply divide the field into a lot of small blobs and apply the QM above to those blobs then let the blob size to go to zero. Strangely and interestingly when you do some mathematical manipulation you end up with the second quantization formulation of ordinary QM:

In this way particles emerge from fields. It can be applied to atoms as well and you get strange phenomena like spontaneous emission etc, but normal QM is considered good enough for most things. QFT is used for the stuff you mention because it combines fields and particles in the one framework.

Even in string theory many now think it's really QFT in disguise - in modern times it has morphed from what was originally thought:
https://www.theatlantic.com/science/archive/2016/09/the-strange-second-life-of-string-theory/500390/

Thanks
Bill

5. Jul 1, 2017

### vanhees71

Usually, at least in Germany, Quantum Mechanics deals with QT formulated in the "1st-quantization formulation", i.e., it's strictly speaking only applicable to non-relativistic quantum theory since it deals with situations only, where the number of particles doesn't change in interactions. It can be formulated in the configuration-space representation, where the pure states are represented as functions $\psi(t,\vec{x}_1,\ldots,\vec{x}_N)$, where $N$ is the fixed number of particles.

Sometimes one talks about "Relativistic Quantum Mechanics" which labels the old-fashioned approach to relativistic QT that tries to formulate it in terms of the 1st-quantization formulation (as it is, e.g., treated in Bjorken&Drell vol. I). Personally for me relativistic QT should not be introduced in this way but right away as QFT ("2nd quantization formulation"). Of course, there's also non-relativsitic QFT, which is equivalent to QM if the interactions are thus that the particle number is conserved, but it's very flexible to also deal with quasiparticles, which are a very powerful concept.

6. Jul 1, 2017

### mikeyork

I don't understand why you claim that QFT is the only way to consider interactions where particles are not conserved. First of all, there is also S-matrix theory. But in principle we can view particle transitions in the same way that we consider quantum number "addition" using vector-coupling coefficients. In fact this is how, using symmetry groups, that even in QFT we treat quantum numbers other than energy-momentum. But energy-momentum coupling has not yet been fully resolved even in QFT.

7. Jul 1, 2017

### vanhees71

Well, I've no clue, how you want to describe a system with changing particle numbers by a single wave function, with a fixed number of configuration-space variables, as in the 1st-quantization formalism. Of course, you can consider abstract S-matrix theory, but usually we derive the S matrix from QFT.

8. Jul 1, 2017

### mikeyork

If you think of your wave-function as a special case of a scalar product in Hlbert space that describes a change of representation, then you can extend that to include transitions between representations of differing numbers of particles. For example, particle decay is a transition from a single-particle representation to a two- (or more) particle representation -- just like the way we add orbital and spin angular momentum for instance. If you think about it, this is really what you do to describe dynamic coupling in either S-matrix theory or Feynman diagrams. As regards deriving the S-matrix from QFT, I would say the whole point of S-matrix theory was that it was very definitely an alternative to QFT. (Just check out anything from Chew or Stapp or many others in the 60s and 70s.)

9. Jul 1, 2017

### vanhees71

What you describe above, however, in fact should be equivalent to QFT, right? It's just most convenient to use QFT and not the cumbersome formalism as Dirac's sea formulation of QED. S-matrix theory has become out of fashion. I don't know whether anybody is considering it anymore today. The reason is the proof of the renormalizability of non-Abelian gauge theories by 't Hooft and Veltman in the early 70s.

10. Jul 1, 2017

### mikeyork

In most circumstance, yes. But I think it is QM itself that is the fundamental underlying theory and QFT merely one incomplete development.
Well, S-matrix phenomenology was the starting point for string theory. I don't consider fashion to be a strong physical argument. In any case, it is the idea of representation transition by vector coupling that I consider to be the appropriate starting point for all dynamical theories. So my point is that I don't think your characterization of QM not providing interaction dynamics is correct.

11. Jul 1, 2017

### vanhees71

I don't what you have in mind with this "vector-coupling" thing. Do you have a reference?

Well, I call the overall concept quantum theory (QT). QM is non-relativistic QT in the first-quantization formalism, i.e., a proper subset of QT. It's just a question of definition.

12. Jul 1, 2017

### mikeyork

A vector coupling is just a scalar product between representations. It is just more often used to describe couplings of pairs of eigenstates to a single eigenstate such as $<JM|l,m,s,\nu>$ tells us how to couple angular momentum. So Clebsch-Gordon coefficients are just another name for vector-coupling coefficients in the specific context of SU(2). So when I say that dynamics come from vector coupling, what I mean is extending C-G coefficients to include all quantum numbers without factorizing so that one explicitly includes the energy-momentum dependencies. My suggestion is that QFT boils down to a particular way of calculating them. Is that incorrect?

13. Jul 1, 2017

### mikeyork

Oops! I just realized that most people probably think of the word "vector" in vector coupling to refer to Cartesian vectors such as classical ideas of angular momentum, whereas I have always thought of it as coupling state vectors from different representations! Could be a possible source of confusion.

14. Jul 1, 2017

### Staff: Mentor

All of these are questions about words, not physics. There is no "right" answer to them.

15. Jul 1, 2017

### Staff: Mentor

As far as I know, no. In commonly studied cases such as the structure of atoms and molecules, the effects of the weak force are negligible and the effects of the strong force are limited to the nucleus, which is just treated as a point charge at $r = 0$, with no internal structure.

When modeling the internal structure of the nucleus, in order to compute things like the expected rest masses of various nuclei (which involves the strong force) or the expected half-life of unstable nuclei (which involves both the strong and weak forces), there are various phenomenological models (such as the "liquid drop" model), but I don't think those are properly understood as using the Schrodinger Equation to compute bound states the way we do for atoms and molecules. A key part of the problem here is that we don't know the correct form of the potential $V$ for the strong or weak forces; in fact it's not even clear that there is such a potential for those interactions. So we can't even write down a Schrodinger Equation anyway.

16. Jul 1, 2017

### mieral

But even when solving for QFT.. we also need to know the potential.. is it not?
Or is there alternative form of computation that doesn't use the potential in QFT for the strong or weak force.. and what method is that? I heard about S-matrix.. doesn't it use the potential?

17. Jul 1, 2017

### Staff: Mentor

No, we don't. We need to know the Lagrangian. They're not the same.

QFT never uses the potential. It uses the Lagrangian. The various formulations of QFT (such as S-matrix) just do different things with the Lagrangian. If you want to find out more, consult a QFT textbook; a discussion of this would be far too much for a PF thread.

18. Jul 1, 2017

### mikeyork

Only a hard-bitten field theorist or someone who graduated after about 1975 would imagine that the S-matrix was a formulation of QFT or had anything to do with a Lagrangian. Leaving aside the issue of taking the unphysical asymptotic limit of a time-dependent operator -- actual S-matrix theory itself is a fundamentally time-independent theory -- instead of Lagrangians, it is founded on unitarity and analyticity and, in most versions, the "particle bootstrap" (AKA "particle democracy"). And it was the foundational basis on which string theory was built. The wikipedia entry gives a simple and reasonably accurate summary.

19. Jul 1, 2017

### Staff: Mentor

Well, my degrees are from 1987 and 1988, so I guess I qualify.

I think it would still be valid to state that S matrix theory does not use a potential in the sense of the Schrodinger Equation, though.

20. Jul 1, 2017

### mikeyork

Yes, that is certainly correct.

Edit: On reflection, I think dispersion relations for elastic scattering might have originally been treated using potential theory. But it is obviously essential to go beyond potential theory in discussing inelastic interactions.

Last edited: Jul 1, 2017