1
$\begingroup$

I am taking a course on Quantum Mechanics and last class we saw the coherent states of the harmonic oscillator, namely states $ | \lambda \rangle $ such that $$ a | \lambda \rangle = \lambda | \lambda \rangle $$ and we saw how they represent the classical situation when $ \lambda \to \infty $.

The professor also told us that there are coherent states for the hydrogen atom as well, and that these represent (in some limit I guess) states in which the electron describes classical orbits around the nucleus (circles and ellipses).

Now I am wondering: does all quantum systems give rise to some "coherent states" that can be interpreted as the classical situation? Or these two examples belong to some special type of systems that can be interpreted in classical ways?

For example I cannot think of a coherent state for a Stern-Gerlach experiment...

|cite|improve this question
$\endgroup$
1
$\begingroup$

Coherent states are eigenstates of annihilation operator. If you can reduce/transform a particular problem to a simple harmonic oscillator problem, then by construction itself there will be coherent states.

In the context of the hydrogen atom problem, hereby we consider only the angular momentum part of the problem. If we look at the Schwinger's oscillator model of angular momentum [1], we see that the Hilbert space can be characterized by using the number states of two independent harmonic oscillators. The operators $J_+$, $J_-$ and $J_z$ can be written in terms of the creation and the annihilation operators of these oscillators.

In this situation we see that coherent states are naturally arising in the problem. Since every problem cannot be mapped to a simple harmonic oscillator problem, we don't have coherent states arising in an arbitrary problem.

[1] Chapter 3, Section 3.8, Schwinger's oscillator model of angular momentum, Modern Quantum Mechanics, J.J. Sakurai.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ This is not correct. The definition of coherent states as eigenstates of a lowering operator holds only for harmonic oscillator states. This definition cannot hold in finite dimensional spaces. The proper generalization is due to Perelomov, and a coherent state is defined as the group translate of a fiducial state, usually taken to be the lowest or highest weight state. Thus, spin coherent states exists in finite dimensions. The construction has nothing to do with mapping to harmonic oscillator states either. $\endgroup$ – ZeroTheHero Apr 29 '18 at 23:33
  • $\begingroup$ The Perelomov construction was shown by Onofri to naturally inherit a symplectic structure, meaning one can definite a proper Poisson bracket using as variables the parameters of a coherent state (and their conjugate). The work of Gazeau and collaborators (see for instance Coherent states, wavelets and their generalizations. Springer Science & Business Media, 2012.) generalizes coherent states in a variety of other contexts. See also the recent work in arxiv.org/abs/1803.01318 for connections to a class of orthogonal polynomials $\endgroup$ – ZeroTheHero Apr 29 '18 at 23:37
  • $\begingroup$ Finally, a good (but somewhat old) review on coherent states is the work Zhang, Wei-Min, and Robert Gilmore. "Coherent states: theory and some applications." Reviews of Modern Physics 62.4 (1990): 867. $\endgroup$ – ZeroTheHero Apr 29 '18 at 23:47
  • $\begingroup$ I see your objection with finite dimensional spaces, thanks. I didn't completely understand your definition though. By "a coherent state is defined as the group translate of a fiducial state, usually taken to be the lowest or highest weight state", do you mean to say that either J+/J- play the role of the annihilation operator here, since they annihilate the highest/lowest weight state? $\endgroup$ – Bruce Lee Apr 30 '18 at 0:17
  • 1
    $\begingroup$ @ZeroTheHero Also I think you should put up your comments as an answer, it will be nice. $\endgroup$ – Bruce Lee Apr 30 '18 at 0:24
0
$\begingroup$

Today I went back with my professor and asked this same question. Here goes my version of his answer. Apparently the answer is yes. Every quantum system has some set of coherent states that reproduce what we would wait in the classical limit.

In the case of the harmonic oscillator these are the eigenstates of the annihilation operator. For this particular case which has a quadratic potential the coherent states are stable in time, meaning that they will be bouncing back and forth indefinitely.

For the electron in the hydrogen atom there are "semi-classical" states in which the electron seems as orbiting the nucleus, although these states are not stable due to dispersion (in contrast to the coherent states of the harmonic oscillator).

For a free particle we can build a gaussian profile and it is a semi-classical state representing a point like particle that moves with some speed. Again, due to dispersion, this will be true for some finite time.

For the Stern-Gerlach experiment there are also semi-classical states (or what I have previously called, maybe wrong, coherent states). A semi-classical state would be a particle with a total angular momentum $\vec{J}(=\vec{L}+\vec{S})$ consisting of a superposition (maybe gaussian, I don't know) of many different values. Thus if we repeat the experiment many times with the particle identically prepared each time, we would see a classical distribution on the screen.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Well, for the hydrogen atom you have semi-classical circular states which are actually stationary states of the Hamiltonian. They become unstable due to environmental coupling, but that's also true for coherent states of the harmonic oscillator. $\endgroup$ – Mark Mitchison May 11 '18 at 3:03
0
$\begingroup$

Mathematically speaking a coherent state can be defined as a state that saturates an uncertainty relation (be it Heisenberg or Schrödinger-Robertson uncertain relation ...)

The formal development leading to the determination of coherent states doesn't need to be particularised to any system, so theoretically you can have coherent states for every system (not only with the Heisenberg algebra).

|cite|improve this answer
$\endgroup$

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.