37
$\begingroup$

In classical physics, particles and fields are completely different stuff. However, when a field is quantized, the particles appear as its excitations (e.g. photon appears as a field excitation in the quantization of electromagnetic field). In fact, for all the elementary particles, there is a corresponding field.

I am interested whether this is also true for any composite particles. Could we define, for any given composite particle, a field for which, upon quantization, that composite particle appears as its excitation? Is there, for example, anything like "hydrogen atom field"?

$\endgroup$
22
$\begingroup$

It depends on the exact circumstance whether or not such an idea is a good approximation for the physics you want to describe.

For the hydrogen atom, you're usually not interesting in it's "scattering behaviour", you're interested in its internal energy states, how it behaves in external electromagnetic fields, etc. Such internal energy states are not well-modelled by QFT. In particular, you'll usually want to consider the proton as "fixed" and the electron as able to jump between its different energy levels. Considering the "hydrogen atom" as an indivisible (or atomic, as it were...) object is not particularly useful.

But there are composite particles where associating a field is perfectly sensible, for example the pion, whose effective field theory describes the nuclear force between hadrons - and the hadrons are also composite particles that are treated with a single field here, for instance by means of chiral perturbation theory.

There are, besides an interest in scattering behaviour (which you also might legitimately have for the hydrogen atom or other atoms, I'm not implying you should never treat the hydrogen atom this way), other reasons to model certain objects as the particles of a field:

Modern many body physics as in condensed matter theory is essentially quantum field theory, too, and it is very frequent there to have fields for composite particles, or even pseudo-particles like phonons. For instance, a simple but powerful model for superconductivity, the Landau model, just treats a conductor as a bunch of charged bosonic particles, thought of as the quanta of a field, coupled to the electromagnetic field, and superconductivity is then another instance of the Higgs mechanism of quantum field theory.

$\endgroup$
  • 1
    $\begingroup$ So, in principle, there are corresponding fields for all the composite particles, but they aren't useful in calculations if you are not interested in the scattering processes of that particles? $\endgroup$ – Annera Aug 20 '16 at 13:12
  • 2
    $\begingroup$ "there are" is the wrong verb. "There can be defined/modeled" corresponding fields for situation where creation and annihilation operators on a basic field wave function can be useful Back in 1961 I sat through a nuclear physics course where field theory was used , probably something like this tphys.physik.uni-tuebingen.de/faessler/Fuchs/Skript/Chap_8.pdf . It is too long ago to remember $\endgroup$ – anna v Aug 20 '16 at 14:00
  • 4
    $\begingroup$ @Annera While I agree with the first sentence of this post, the rest is surely off the mark. "Scattering behavior" is not the reason for defining fields, as this post seems to imply. This is just a narrow high energy physicist's POV of fields. Fields are indispensable for the quantum statistical description of ensemble of particles including atoms. Because of their extended nature, an accurate description would be non-local interactions, and internal energy states of individual atoms would have to be accommodated by adding additional fields. The correct answer to your question is "yes". $\endgroup$ – QuantumDot Aug 20 '16 at 18:35
  • 2
    $\begingroup$ I should clarify: I disagree with the thesis of your answer "it is not useful." Your reason for supporting that thesis in the middle paragraph are sound, and the counterexample of chiral perturbation theory is ok. But I feel your post is much too narrow (from the HEP point of view) to provide an honest answer to the question. $\endgroup$ – QuantumDot Aug 20 '16 at 19:05
  • 4
    $\begingroup$ I'm not sure I agree with that "usually" - it depends on what you're doing, and if you've got two hydrogen atoms scattering off of each other (as you do in cold atom experiments, obviously, but also in many astrophysical contexts, for example) then you mostly care about the elastic scattering and you can treat each atom as a single unit. I'm not saying you should change your answer, but a nod to the fact that you sometimes do want to look at those regimes (like Mark does in his work all the time) would be good (particularly considering HNQ status). $\endgroup$ – Emilio Pisanty Aug 21 '16 at 2:31
31
+200
$\begingroup$

$\def\rr{{\bf r}}\def\ii{{\rm i}}\newcommand{\ket}[1]{\lvert#1\rangle}$I do mostly atomic physics (where, for example, the scattering of atoms is very pertinent) so I have a slightly different take on what counts as "useful" in QFT. It is often very useful indeed to define an "atomic field operator" $\Psi(\rr)$. The excitations of this field are entire atoms, which may be either bosonic or fermionic, in which case the fields satisfy appropriate (anti-)commutation relations $$\Psi(\rr)\Psi^\dagger(\rr') \pm \Psi^\dagger(\rr')\Psi(\rr) = \delta(\rr-\rr'),$$ where the minus (plus) sign is for bosons (fermions). Thus, if $\ket{0}$ is the vacuum state containing no particles, then a general $N$-particle state corresponds to $$\ket{\phi} = \int\prod_{j=1}^N\mathrm{d}\rr_j\; \phi(\rr_1,\rr_2,\ldots,\rr_N) \Psi^\dagger(\rr_1)\Psi^\dagger(\rr_2)\cdots \Psi^\dagger(\rr_N)\ket{0},$$ where $\phi(\rr_1,\ldots,\rr_N)$ is the $N$-body wave function in coordinate space.

So far we have considered the field operator in the Schroedinger picture. In the Heisenberg picture, the time evolution of the field is generated by the Heisenberg equation $$ \ii\hbar\dot{\Psi}(\rr,t) = -\frac{\hbar^2 \nabla^2}{2m} \Psi(\rr,t) + V[\Psi;\rr]\Psi(\rr,t),$$ where the (non-linear) potential reads as $$ V[\Psi;\rr] = V_1(\rr) + \int{\rm d}\rr'\; V_2(\rr-\rr') \Psi^\dagger(\rr',t)\Psi(\rr',t).$$ The first term $V_1(\rr)$ describes a one-particle external potential, the second term $V_2(\rr)$ describes a two-body interaction potential between the atoms. Higher-order (i.e. $n$-body potential) terms are possible as well. It is straightforward to generalise the field $\Psi(\rr)$ to a spinor having a component for each possible internal state of the atom. In this case the potential terms generalise to matrices or higher order tensors coupling together different internal states, while derivative terms in the potential could appear due to artificial gauge fields, for example.

As always in QFT, this is a low-energy effective theory. The non-relativistic approximation for the kinetic energy assumes that the centre-of-mass motion of the atoms is much slower than $c$. The field description of the atoms breaks down at scales where their internal structure becomes important, e.g. length scales comparable to the Bohr radius or energies comparable to the Rydberg energy.

If you want, you can view the above as the quantisation of a complex classical "Schroedinger" field $\Psi(\rr)$ (replace $\Psi^\dagger(\rr) \to \Psi^*(\rr)$). This classical Schroedinger field obeys a (non-linear) Schroedinger equation, with an associated Hamiltonian and Lagrangian etc. We usually do not do this, because the classical field does not describe anything particularly interesting. This is because 1) Nature is quantum and 2) no quantum states exist for which the classical field description is even close to a good approximation. The situation is different in, e.g., electrodynamics, where the observable dynamics of coherent states of the electromagnetic field can be well approximated by the classical Maxwell equations in many cases. However, in atomic field theories, the fundamental excitations are composed of fermions and their number is strictly conserved. This means that coherent states cannot be prepared* and therefore the classical field theory is largely useless (same reasoning applies, e.g. to the classical Dirac equation).


*In Bose-Einstein condensates, the off-diagonal long-range order makes a coherent state description a good approximation for many properties of interest. In this case the classical field theory is very useful; it goes by the name of the Gross-Pitaevskii equation.

$\endgroup$
  • $\begingroup$ That's an interesting viewpoint. What are the key differences between $\Psi(\mathbf r)$ and the wavefunction of one atom? I see that it obeys almost the same Schrödinger equation, except that the potential is now also a functional depending on the entire $\Psi$. $\endgroup$ – Annera Aug 20 '16 at 14:59
  • 2
    $\begingroup$ @Annera For the sake of pedagogy, I would say that $\Psi({\bf r})$ and the wave function have nothing to do with each other. Any similarities are only formal; the physical meaning of the two mathematical objects is deeply different. However, the matrix elements of the operator $\Psi(\bf r)$ are, by construction, coordinate wave functions $\phi_m({\bf r})= \langle \mathbf{r}|\phi_m\rangle$ of 1-particle states, i.e. $\langle\phi_m|\Psi^\dagger({{\bf r}})|0\rangle = \phi^*_m(\bf r)$. Therefore, matrix elements of the Heisenberg equation yield a Schrodinger-like equation for the wave functions $\endgroup$ – Mark Mitchison Aug 20 '16 at 15:16
  • $\begingroup$ Okay, but isn't a system described by a "non-relativistic, number-conserving field theory" equivalent to, or at least also well-described by, standard many-body quantum mechanics? It is certainly true that this can formally be described in a field-theoretic way, which I guess is all this question strictly needs, but it isn't clear from your answer why this might actually be useful (except in the case of BEC). $\endgroup$ – Rococo Feb 9 '17 at 4:41
  • 1
    $\begingroup$ @Rococo This is just the occupation number representation ("second-quantised", if you must) of standard many-body QM in coordinate space. The interpretation of the variables in terms of "fields" is largely historical I guess. The utility of this formalism is almost certainly well documented elsewhere on this site. For example, it takes care of (anti)symmetrisation automatically, and massively simplifies the description of low-energy processes where just a few particles change quantum numbers above an N-body ground state. Otherwise one has to keep track of N redundant parameters explicitly. $\endgroup$ – Mark Mitchison Feb 9 '17 at 9:25
4
$\begingroup$

Every bound state (in particular every molecule and therefore the hydrogen atom) has an associated field, whose density figures in the statistical thermodynamics of chemical equilibrium and chemical reactions. Chemical reaction rates are computed microscopically in terms of the scattering theory of these fields or the corresponding asymptotic particles.

In nonrelativistic quantum mechanics, the corresponding field operators were constructed in Section 3 of the following paper:

W. Sandhas, Definition and existence of multichannel scattering states, Comm. Math. Phys. 3 (1966), 358-374. http://projecteuclid.org/download/pdf_1/euclid.cmp/1103839514

In the relativistic case, there is a corresponding nonperturbative construction in Haag-Ruelle scattering theory, valid at all energies (assuming the validity of the Wightman axioms). https://arxiv.org/abs/math-ph/0509047

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.