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I have some heuristic idea of how to think about state preparation in quantum mechanics. It may be revolving around the idea of using filters, cooling/heating, Stern-Gerlach type setup, etc. However, how does one prepare a ground state of a quantum field (at least approximately, not necessarily global vacuum)? For non-interacting theory, is it as simple as "making a vacuum chamber"? I found that to be weird if that's the case, because I will not be able to tell if I am in EM vacuum, interacting vacuum, or in fact classical vacuum at all.

I cannot seem to find this answer anywhere, eventhough I thought vacuum state is one of the most important states in QFT. One possible guess I had was that vacuum state was mere conceptual thing that should exist but not needed empirically, since one's measurement processes may not involve the vacuum (e.g. particle physics mostly cares about scattering). But I honestly do not know.

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  • $\begingroup$ I am not sure of the answer, but could you probably rephrase (or explain in more detail) what state you have in mind here? What I mean is, if the ground state is the actual ground state then you don't have a field (i.e. particle); this is different to having a particle and preparing it in a specific quantum state as you've described. $\endgroup$ – Helen - down with PCorrectness Jun 21 '19 at 21:18
  • $\begingroup$ Here is an example of differentiation between EM vacuum and an excited field (single photon) in a microwave photon cavity: nature.com/articles/nature05589 $\endgroup$ – wcc Jun 21 '19 at 21:36
  • $\begingroup$ @Helen ground state of an EM field should be one in, in particle language, which there is no "photon", i.e. vacuum state. I am assuming non-interacting fields. This is ground state of the theory. In QM it's slightly different; ground state of hydrogen atom does not mean I have no hydrogen atom. $\endgroup$ – Everiana Jun 21 '19 at 22:45
  • $\begingroup$ @wcc I will have to read it up. Thanks for the reference! $\endgroup$ – Everiana Jun 21 '19 at 22:46
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I can only provide an answer from an experimental quantum optics angle (i.e. low energy limit of QED):

Once you have defined a particular mode of the electromagnetic field (for example in a cavity, or any kind of spatio-temporal mode) and applied the usual "canonical quantisation" procedure, you are looking at a more or less large Hilbert space (formally equivalent to a more or less large collection of harmonic oscillators). The "vacuum state" is then just the state that has no excitation, i.e. in the number basis you would just write $|0\rangle$, in case of a single mode.

Experimentally, once you are in a dark room, the vacuum state is a very good approximation to the "true quantum state" of almost any modes that can be defined in the optical regime, as the vast majority of modes contain no photons (as you can even guess from your visual impression). This is even true in a bright room :D, there is just such a ginormous number of ways the electromagnetic field can oscillate.

If you go to lower energies like microwave regime (or hotter environments), the approximation is not so good anymore. In thermal equilibrium the field would be best described by a thermal state, where you as the experimenter have minimal information about the quantum state, you just know one parameter of the radiation field, and that is its temperature (determining the mean number of quanta in a particular mode).

So if you wanted to experimentally "prepare" the radiation field (or a single mode of it) in the vacuum state $|0\rangle$, you don't have to take crazy measures at optical frequencies, you get the vacuum state for free (which is a pure quantum state! This is indeed very beneficial for all of quantum optics). But at next order your main enemy is basically black body radiation of the environment. So an option to improve things would be to literally cool the cavity walls with a fridge.

So it has almost nothing to do with somebody evacuating a chamber of air.

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  • $\begingroup$ That's an answer of the kind I am looking for, but I think I missed a few jumps. (1) why is optical regime a good approximation for vacuum? I get the large Hilbert space part, but if I first read QFT, I would think that dark room is very different from bright room. Is there a way to make this concrete? (2) Are you using single mode approximation (i.e. vacuum = vacuum at that mode)? $\endgroup$ – Everiana Jun 21 '19 at 22:51
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    $\begingroup$ (1) At optical frequencies (some $5 \cdot 10^{14}$ Hz) the probability that a mode is thermally excited is just very low. This can be obtained from the Bose-Einstein distribution, but you can get a feel how improbable this is by comparing the energy of a quantum of light ($\hbar \omega$ ~ 2 eV) to the thermal energy per degree of freedom ($k_B T$ ~ 1/40 eV). (2) Essentially to the same quality of approximation we are in a global vacuum state, i.e. $|0\rangle \otimes |0\rangle \otimes ...$. There are just so few photons (even if your eye sees quite a few) compared to the vast number of modes... $\endgroup$ – Georg E. Jun 21 '19 at 23:00
  • $\begingroup$ As an exercise you could try to compute the rate of photons impinging on 1 m² of the surface of the earth from bright sunlight (assuming the sun is a perfect black body radiator at 5800 K) in a 10 MHz window around the HeNe laser wavelength of 633nm, and compare it to the number of photons a HeNe laser with a few mW power emits. $\endgroup$ – Georg E. Jun 21 '19 at 23:10
  • $\begingroup$ Thanks for the guides, I might try at some point. One last thing: when you say "take crazy measures at optical frequency", you mean I do experiment using optical cavity? And then if I want to do it in microwave regime, I need to use microwave cavity? (I am not quantum optician, but I am guessing some resonance is used to filter all other irrelevant frequencies) Sorry, being a theorist I really want a better experimental intuition. $\endgroup$ – Everiana Jun 21 '19 at 23:17
  • $\begingroup$ @Everiana, yes you use optical cavity (pair of mirrors) for optical photons, and microwave cavity for microwave photons. Because of the huge difference in wavelength, obviously microwave cavity is typically larger than optical cavity. In quantum optics experiments, microwave cavities are made of superconductors (for very high Q factor), and I guess the low working temperature has the added benefit of reducing blackbody radiation leaking into the mode volume. $\endgroup$ – wcc Jun 21 '19 at 23:24

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