You are party looking in the wrong place .

I would say non relativistic QM is the limit of relativistic QM for low speeds.

There is plenty of this in books of relativistic QM

Then you have to see how QM fits in with Quantum Field Theory.

The end of section 2.3.2 in Schwartz's QFT book (https://schwartzqft.fas.harvard.edu/) might help.

If you expand the Klein-Gordon equation energy term about the rest mass and ignore the second order time derivative of the slow oscillating factor (relative to the large exp(i mc^2 t) factor) the result is Schrodinger's equation. See "QFT in a Nutshell" by Zee, Chapter III.5 for details.

This limit corresponds to energies that are less than the rest masses of the particles so there is not enough energy for particle creation.

Phillip W. Dennis

I don't see what happens with the increasing and lowering operators when passing from high velocities (QFT) to low velocities (non-relativistic QM).

Juan Weisz

Juan, my friend, you are funny today: look!

"There is plenty of this in books of relativistic QM.

Then you have to see how QM fits in with Quantum Field Theory."

Many thanks. The fact that non-relativistic QM is the limit of the relativistic QM is quite trivial. The relativistic QM does not contain raising and lowering operators. "After that, Sofia, please answer your question by yourself."

Sofia D. Wechsler

the harmonic oscilator solution is often done using these operators as in Landau-Lifshits. That is one contact example.

It is

k ( implicitely hbar k as momentum) when you write a(k) or a dagger(k)

There is momentum operator, not velocity operator in QM

You

can also try to read

"Quantum waves from the exclusion principle"

on my RG site. You wont find exactly what you want, but much needed formalism.

Phillip is right, how you express E(p) indicates in what regime you are.

Phillip W. Dennis and Juan Weisz

You are wiser than me. So, please go slowly. Phil, I lost you here:

"When the Hamiltonian operator is constructed it is the sum over free energies E(p) where p is the three momentum and the Fock space occupation number operators N(p) = â†(p)â(p). i.e. Ĥ = sump ( E(p)N̂(p))."

I don't know what you mean by free energy. This is a concept from the thermodynamics. But I think you mean something else. So, please write its formula.

Then, you say,

"The non-relativistic limit is handled via the c-number coefficient alone."

There is no such coefficient in Ĥ as you expressed it.

"The raising and lowering operators don't enter at this point."

No, no! It seems to me as abracadabra. I want to see how these operators disappear from the Hamiltonian as the velocity decreases. They disappear gradually, not suddenly "at this point".

With thanks to both of you, kind regards, and waiting for explanation,

Sofia

Sofia D. Wechsler

Let me cleanup my terminology. By "free energy" I meant the energy for non-interacting particles. The relativistic formula for a free particle of momentum p is E(p) = sqrt( m2 + p2).

This E(p) is the c-number coefficient of N̂(p) in the Hamiltonian operator.

By "don't enter" I meant don't enter into the argument. The operators cannot disappear and cannot turn into c-numbers in the limit. In the limit, they are still there in the QFT equation but their expectation values don't change in the classical limit. Since that is the case to carry them around in calculations would be unnecessary extra baggage. The main point is that in the non-relativistic limit particle numbers do not change, they are constants of the state. Dropping the occupation number operators for notational convenience places us back in mutli-particle non-relativistic QM.

Phil

it sounds like having spotted a discontinuity.... it is relevant to what occurs at least in one case with the Lorentz Transformations where a discontinuity of behavior appears..

I would look at the question from a different perspective, that is: wherefrom has QFT originated? Many text books on the subject (Bjorken & Drell, last chapters of Messiah, Ryder), derive Klein-Gordon and Dirac field equations in order to satisfy the request of a Lorentz invariant version of Schrödinger equation. Well, this request originates from the idea that wave functions obeying Schrödinger eq. should somehow correspond to some physical porperty of systems, and as such should not propagate in space-time faster than light, whereas wave function just serves the scope of estimating probabilities of events about systems and looking at probabilities as "physical features" it's not really a wise move (you may have a look at "Probability theory: the logic of science" by Edwin T. Jaynes for a very clear account on the issue).

Enforcing this request brings about all kind of difficulties in interpreting quantum fields as the relativistic version of wave functions (one out of many, eigenstates of the Dirac hamiltonian give negative energies for non interacting particles), and a conceptual jump is made to make sense of quantum fields: that is one has to decompose field configurations into eigenmodes of the field equations and "quantize" the amplitudes of such eigenmodes as harmonic oscillator's rising and lowering operators. Then excitations are thought as particles, whose quantum states live in Fock space.

Therefore differently from ordinary QM, which needs just one mathematical space to be formualted, that is a Hilbert space isomorphic to the space of square integrable functions in which states of the systems are, QFT needs two mathematical spaces: the one of field equation solutions and for each eigenmode a Hilbert space for the quantum state of that mode. So a clean low velocity limit of QFT into QM, if it exists, should somehow make the two mathematical spaces collapse into one.

Maybe the less controversial approach to get a low-velocity limit into this context is getting an answer to the problem in QFT one is facing, and then calculate the low velocity limit of the solution, as it is done for instance for some scattering cross sections in QED. Trying to get a clear nonrelativistic limit of QFT looks to me like a no-go.

Sofia D. Wechsler

Nuclear forces aside, the non-relativistic QM follows as the rotating wave approximation in the relativistic QED, where the carrier frequency is the rest mass of the electron (in units hbar = c = 1). The physical requirement here is that all transition frequencies, and hence all optical frequencies, are small compared to the said mass. From the top of my head, this approximation is good up to X-rays. Surely it covers all of optics. Its main problem is that by excluding high frequencies we lose accurate renormalisation, and have to introduce truncations (cf. Bethe's logarithm). In practice, people mostly resort to hand-waving like "the logarithmically divergent term corresponds to uniform all-level shift which cannot be calculated nonrelativistically; we ignore it".

There are two ways of introducing the above approximation. One is "nonrelativistic QED". It is amply covered in various textbooks on quantum optics (I would recommend Cohen-Tannoudjji's' Atoms and Photons). This is more or less the modern way.

The older way is to firstly derive the Dirac equation in an external field, and only then introduce the rotating wave approximation. This results in the Pauli equation with spin; by neglecting spin one gets the Schroedinger equation. This was the standard way in atomic and molecular optics before Hanbury Brown--Twiss. Any standard textbook in atom optics will give you proper directions, but I doubt you should waste time on it. Your interest, to the extent I can judge, is mostly in nonrelativistic correlation measurement. For it, the right framework is "nonrelativistic QED", a.k.a. "quantum optics".

Sofia D. Wechsler

To add, to discuss things like "measurement" and "quantum nonlocality", it is mostly sufficient to stay at the level of structural, or meta-, theory. Namely, one assumes that the EM field and current operators and the quantum averaging (state) in the interaction picture are known, and proceeds as per the Feynman-Dyson perturbative approach (S-matrix, Wick's theorem, etc). Such meta-theory cares little about physical details; things are essentially the same for one mode and full relativistic EM field. The only thing that matters is the field-times-current form of the electromagnetic interaction.

Hi,

You may read a good article on this subject.

Article Renarks on the classical limit of quantum field theories

• by J. -P. Eckmann

Thanks

The Schrödinger equation and Maxwell equations are a nonrelativistic approximation of quantum relativistic field theory (Klein-Gordon equation, relativistic form of Maxwell equations, Dirac equation).

@monakhov

the maxwell equations are certainly not the nonrelativistic limit of anything, because they are relativistic invariant.