# I Semiclassical vs classical

1. Oct 7, 2016

### spaghetti3451

A system can only be called semiclassical if there are parts of the system that are described classically and parts decsribed quantum-mechanically. In this paradigm, physical quantites are described in a power series of $\hbar$, with the zero order corresponding to classical physics and higher orders corresponding to quantum corrections.

Given the above, I do not see why the limit $\hbar \rightarrow 0$ called the semiclassical limit and not the classical limit. After all, in the limit that $\hbar \rightarrow 0$, the system becomes classical and there are no quantum-mechanical corrections.

Last edited by a moderator: Oct 7, 2016
2. Oct 7, 2016

### Mentz114

I think that definition is too restrictive. The interaction between light and matter can be modelled as an interaction between atoms ( treated QM wise) and classical light (Jaynes-Cummings model). Now that is a hybrid of classical and quantum modelling.

3. Oct 7, 2016

4. Oct 7, 2016

### Simon Phoenix

That's not quite correct. The JCM is a model of a two-level atom interacting with a single quantized EM field mode in a high Q cavity. It is perhaps the simplest fully-quantized model of the interaction between light and matter. It's nice because in the rotating wave approximation it can be solved exactly. Even this simplest of models leads to some interesting behaviour - so, for example, if we begin with the field mode in a coherent state we see a complicated evolution for the atomic inversion (the so called collapse and revival phenomenon).

The JCM is not a semiclassical model at all.

5. Oct 7, 2016

### Simon Phoenix

But letting h tend to zero does not always give us classical behaviour. In other words classical mechanics is not the small h limit of QM. What is the small h limit of entanglement, for example?

6. Oct 7, 2016

### Mentz114

You are right. I mistook JC for the (semi-classical) rotating wave approximation. It's a long time since I studied this ...

7. Oct 7, 2016

### spaghetti3451

Okay, so back to my question, ....

8. Oct 7, 2016

### Mentz114

The RWA uses a classical EM field and a wave function to get transition probabilities - which have no classical value according to the definition above.

I have seen RWA called semi-classical. Anyhow, it seems to a problem of nomenclature, not physics.

9. Oct 7, 2016

### Simon Phoenix

σ
I've not seen the RWA described as a semi-classical approximation. In the interaction picture the interaction Hamiltonian contains products of atomic and field raising and lowering operators. The products of raising and lowering operators (so terms like σa+) oscillate at low frequencies (basically the detuning between the atom and field) whereas products of raising-raising or lowering-lowering oscillate at approximately twice the mode frequency (for small detuning). The RWA amounts to neglecting these rapidly oscillating terms. I'm not sure why this is considered to be 'semi-classical'.

10. Oct 7, 2016

### Mentz114

OK, you don't give an inch so I had to look something up which I hope will clear up the confusion I've caused.

There exists a model of the interaction between the EM field and an atom, which uses the unquantised EM field. The derivation of the transition probabilities uses the RWA, where ( as you say) some anti-resonant terms are dropped.

Now forget I mentioned JC or RWA. I assert that the model I refer to is semi-classical but does not follow the definition above of S-C.

Furthermore this has no relevance to anything and I'm sorry I mentioned it at all.

Last edited: Oct 7, 2016
11. Oct 7, 2016

### Simon Phoenix

Lol - yeah - sorry for the nit-picking too. In my defence a chunk of my PhD was on the JCM

I think it is relevant though because I don't think a semiclassical model for the JCM will give anything like the JCM/RWA behaviour - and the quantum behaviour (even with the RWA) can't be understood as some perturbative correction to a semi-classical model in powers of Planck's constant.

12. Oct 7, 2016

### Mentz114

You are absolutely right to correct my inaccuracies and I thank you.
I didn't say it is a good model.

13. Oct 7, 2016

### Simon Phoenix

When I started out it was still a fairly hot topic to find behaviours that could not be explained with a semi-classical model, that is where matter was treated quantum mechanically but the EM field classically. The JCM is a really sweet model to investigate because it's simple, it's exactly solvable and it has a rich quantum behaviour that has no classical, or even semi-classical limit. So I think it's a useful example when talking about the efficacy of semiclassical approaches and h tending to zero limits.

14. Oct 8, 2016

### A. Neumaier

No, this is called quantum-classical. Semiclassical refers to a treatment of a true quantum system by methods that make use of the limit $\hbar\to0$, but often keeps low order correction terms. Otherwise it would be classical.

15. Oct 8, 2016

### vanhees71

Well, this is a debate on semantics again. It depends a bit on the subject you are discussing, what you call "semiclassical". In atomic physics and quantum optics usually semiclassical refers to the approximation that the matter (electrons, atoms, molecules, condensed matter) is treated with quantum theory and the electromagnetic field as a classical field. It's amazing, how much can be described with this semiclassical model. Often the quantization of the em. field is only a small correction. An important example is the photoelectric effect, which is not proving em.-field quantization but is well described in the "semiclassical approximation" in this sense.

You can also consider the opposite approximation, i.e., you have a classical charge-current-density distribution and consider the quantized em. field coupled to it. For an oscillating charge distribution you then get coherent states of the quantized electromagnetic field, which shows that classical em. waves (light) is not a naive stream of photons but a highly non-trivial quantum superposition of states with all photon numbers (the coherent state). Itzykson and Zuber call this approximation "hemiclassical" to distinguish it from "semiclassical", but I've never seen any other text using this convention.

Of course, you can also ask, in how far a quantum system can be approximated by classical physics. E.g., a quantum particle like an electron interacting with a (classical) electromagnetic field in non-relativistic QM can be described by the Schrödinger equation, and this equation can be solved using "singular perturbation theory", which is formally an expansion in powers of $\hbar$ (starting with the singular term $\propto 1/\hbar$, which corresponds to the classical limit, leadking to the Hamilton-Jacobi PDE to describe the motion of classical point particles). This is known as the Wentzel-Kramers-Brillouin (WKB) method and is completely analogous to the mathematically identical treatment of em. waves in classical electromagnetism in powers of the wave length (eikonal approximation), leading to the approximation of wave optics by ray optics. The real fun with this are the classical return points (in QM) or equivalently the boundary of shadows (in classical EM). By the way, this was the way Schrödinger used ingeniously when "deriving" his equation: He thought of classical physics as the eikonal approximation of an equation describing de Broglie's matter waves. Thinking about the Hamilton-Jacobi PDE he realized that the corresponding equation should be his Schrödinger equation. Interestingly he started with the relativistic case and got what's now called the Klein-Gordon equation for scalar bosons. He calculated the hydrogen spectrum and found the wrong finestructure. Since this was known to be a relativistic effect he then went to the non-relativistic case, getting the equation now named after him the Schrödinger equation. There the hydrogen spectrum came out right up to the finestructure which, however, was expected since now it was a non-relativistic theory. The reason for the failure of the Klein-Gordon equation to describe the hydrogen spectrum was of course just that electrons obey the Dirac equation, and doing the calcululation with the Dirac equation indeed gives the correct finestructure.

Ironically in the Bohr-Sommerfeld quantization the naive relativistic treatment of the hydrogen atom, Sommerfeld got the correct finestructure although not knowing about spin at the time at all. For me it's an enigma, why a wrong model with an incomplete treatment of the electron as a spinless particle yields the correct finestructure ;-)).

16. Oct 8, 2016

### A. Neumaier

The early quantum physicists were not dumb; they made models to fit Nature, as we do it today. Since fine structure appears in Nature, they looked for how to account for it in their models....

There is a nice book called The Old Quantum Theory by D. ter Haar,
http://www.sciencedirect.com/science/book/9780080121024
From the preface:
Although it is well known that for a proper discussion of atomic
properties one needs wave mechanics and that the old quantum
theory developed by Bohr, Sommerfeld, Kramers and many others
between 1913 and 1926 is not a proper basis, there are many
atomic phenomena which receive at least a qualitative explanation
in the old quantum theory. An account of this theory seems
therefore to be of some interest

17. Oct 9, 2016

### vanhees71

I didn't imply that the early quantum physicists were dumb, but it's really a funny coincidence that a model, that we know today to be totally flawed, gives the right result. You get the finestructure right, assuming the electron to be a classical point particle (implying no spin) and treat the relativistic Kepler problem and use the Bohr-Sommerfeld quantization condition. Of course, it was a great achievement by Sommerfeld in the time to do this calculation and find the correct finestructure. In fact, I consider Sommerfeld as one of the most brillant physicists and teachers ever (in my opinion his theory textbooks are even better than the Feynman lectures).

18. Oct 9, 2016

### rubi

I wouldn't say it's totally flawed. The Bohr-Sommerfeld quantization condition is still part of one of our best quantization techniques: Geometric quantization. The standard Schrödinger treatment of the hydrogen atom also doesn't include backreaction with the electromagnetic field, so it doesn't really explain the stability of matter either. One needs a full field theoretical treatment for this.

19. Oct 9, 2016

### vanhees71

In which sense doesn't the non-relativistic Schrödinger treatment explain the stability of the atom? Even all excited energy eigenstates are stable. When you quantize the em. field they become unstable, and only the ground state is stable. The most prominent feature you add by quantizing the em. field is the possibility of spontaneous emission!

20. Oct 9, 2016

### rubi

Well, unless you add backreaction, the Kepler orbits are already stable in classical mechanics. The standard treatment of the hydrogen atom includes backreaction neither in the classical nor in the quantum case, so the quantum mechanical treatment doesn't explain the stability of matter more than the classical treatment or the Bohr model do.

Yes, as I said, you need a field theoretical treatment to explain the stability of matter. But neither the Bohr model nor the standard quantum mechanical treatment of hydrogen include the backreaction with fields, so they shouldn't be expected to explain the stability. Of course, the quantum mechanics still has other merits.

21. Oct 9, 2016

### vanhees71

Ok, in this sense, I understand what you mean. The point disinguishing the non-relativistic Kepler-like atom from the classical Kepler-like atom is that in the former case you have a static charge configuration, namely the electron in one of its energy eigenstates, which implies that its state doesn't change with time, while in the latter you have the picture of an accelerated classical charge which should radiate but is just forbidden to radiate ad hoc by Bohr. Nature, however doesn't care about Bohr and thus it's not very convincing compared to the modern QT model. Of course you are right, the full picture (as far as we know it) is to implement the full QED description, leading to the most accurate description including Lamb shift of the energy levels and the spontaneous photon emission if the electron is in an excited state, so that the only really stable state is the hydrogen atom in the ground state.