9
$\begingroup$

Coherent states are quantum states that are said to act "as classically as possible". You can define coherent states for the harmonic oscillator, or more generally for any collection of harmonic oscillators, such as a free quantum field. It is often said that a quantum field in a coherent state behaves like a classical field, e.g. that this is how classical electromagnetism emerges from QED.

I've only been taught about coherent states in optics, where you need to use special procedures and expensive equipment to create them. But if the world looks like it's made of classical fields, it must be easy to make a coherent state, so easy that it's inevitable.

How are coherent states of quantum fields made in the real world?

$\endgroup$
  • 1
    $\begingroup$ Do you have a link on your statement "quantum states that act as classically as possible". Laser set ups are par excellence coherent and certainly are not classical, i.e describable with classical equations. One needs quantum mechanics to understand what is going on. $\endgroup$ – anna v May 26 '17 at 4:07
  • 3
    $\begingroup$ @annav When I've heard people say that, they usually just meant that they have the minimum $\Delta x \Delta p$ allowed by the uncertainty principle. $\endgroup$ – knzhou May 26 '17 at 4:15
  • $\begingroup$ "incredibly contrived" is a strong descriptor for a laser. It's pretty easy to make coherent states of light - though it might take some work to conclusively distinguish them from other possibilities. $\endgroup$ – Emilio Pisanty May 26 '17 at 7:14
  • 3
    $\begingroup$ I'm not convinced the implied leap from "Coherent states are the most classical states" to "The world looks classical so it must be full of coherent states" is justified. Remember that $\hbar$ is tiny on our scales - what does it matter if the uncertainty of the quantum states is minimized or not when even those with large uncertainty still have approximately zero uncertainty in the classical $\hbar\to 0$ limit? If you want to make an argument that the world is full of coherent states you find need to understand and present the classical limit of QFT in detail. $\endgroup$ – ACuriousMind May 26 '17 at 12:50
  • 1
    $\begingroup$ What's the status on this post, i.e. what more information is needed before we can select an accepted answer? $\endgroup$ – DanielSank Jul 27 '17 at 18:30
5
$\begingroup$

Coherent states arise naturally in many quantum systems besides the free electromagnetic field. Many of them are mentioned in the following review by WM Zhang, which is actually dedicated to the same question asked here.

I should emphasize that the examples given in this answer refer to the usual Heisenberg-Weyl coherent states as well as generalized coherent states including squeezed states coherent states on Lie groups etc. Please see the following book by Perelomov.

There are many examples of coherent states besides the electromagnetic field; may be the most known examples are:

  1. Ground states of Bose-Einstein condensates can be approximated by coherent states of the condensate fields.
  2. The BCS ground state of superconductor can be approximated by a coherent state of the Cooper pairs.
  3. Ground states of a superfluid can be approximated by a coherent state.
  4. Ground states of certain phases of ferromagnets and anti- ferromagnets can be approximated by coherent states, please see the following lecture note by Christopher Henley .

The above examples correspond to large systems with spontaneously broken symmetry. This fact stems from the general rule that in spontaneously broken theories, the condensate state can be approximated by a coherent state, which can be heuristically explained as follows, suppose that the number of condensate atoms is $N$. The action of the annihilation operator per unit volume on the condensate state is: $$\frac{a}{\sqrt{V}}|N\rangle = \sqrt{\frac{N}{V}}|N-1\rangle$$ When $N\rightarrow\infty$, the states on both sides of the equation have the same limit and we obtain: $$ \lim_{N, V, \rightarrow \infty} \frac{a}{\sqrt{V}}|\mathrm{Condensate}\rangle = \sqrt{\rho}|\mathrm{Condensate}\rangle$$

Where $\rho$ is the condensate density. This shows that the condensate state is a coherent state.

In systems of interacting of atoms with radiation fields, the atoms can be approximated by two state systems. An atomic gas in this approximation, has an SU(2) (angular momentum) symmetry of the two state systems. In this case also when this symmetry is broken, the atomic gas will reside in a coherent state of the angular momentum. This is a generalized coherent state of Perelomov- Gilmore type and still the above rule applies.

Coherent states are the solution of the Hartree-Fock approximations in atomic and molecular physics. The slater determinant is actually a coherent state over a Grassmann manifold. Please see the following review by Chiumiento, and Melgaard.

In quantum field theories with massless particles, it was found that using coherent state bases of asymptotic states instead of the particle number basis leads to a faster cancelation of the infrared divergences, please, see for example the following work by More and Misra and the following thesis by Catana. This is because the asymptotic charged fermion states are surrounded by a coherent state cloud of soft photons.

It is also worthwhile also to mention that electrons restricted to the lowest Landau level are described by coherent states. Please see the following lecture note by Christopher Henley.

Although not requested in the question, I have to mention that coherent states have many applications in mathematics. Their importance lies in the fact that they generate coherent state representations of groups, please see for example the following work by Berceanu and Schlichenmaier. Their importance lies in the fact that their construction philosophy can be extended to other mathematical objects such as quantum groups, groupoids which have not found extensive application in physics yet.

Coherent states (in systems other than the free electromagnetic field) can also be manufactured, rather than naturally obtained. This is actually a problem in quantum control and one of the principles that can be applied is the following:

In contrast orthonormal bases in a Hilbert space, which are parametrized by a discrete variable, coherent states are parametrized by a phase space. For example a harmonic oscillator coherent state $|z\rangle$ is parametrized by $z\in \mathbb{C}$ the phase space of a single degree of freedom particle. There is a family of Hamiltonians whose action on the coherent states can be implemented as a classical action on the phase space variable: $$ e^{i\hat{H}t} |z\rangle = |\phi_t \cdot z\rangle$$ Where $\phi_t \cdot z$ is the Hamiltonian $H$ flow of the point $z$, namely the solution of the Hamilton equation of motion. The relation between the quantum Hamiltonian $\hat{H}$ and its classical symbol $H$ is given be certain quantization rules, in our case they can be chosen to be the Wick or anti-Wick ordering. In the case of the harmonic oscillator coherent states, the distinguished family of Hamiltonians are the linear Hamiltonians: $$\hat{H} = \lambda(t) a + \bar{\lambda(t)} a^{\dagger}$$ Thus these Hamiltonians make the system evolve within the coherent state manifold. In order to bring the system to a desired coherent state, we must start from an easily prepared coherent state (in the Harmonic oscillator the ground state) and act on it by this type of Hamiltonian. The system will evolve to a general coherent state controlled by our choice of the Hamiltonian. For example, the Harmonic oscillator whose total Hamiltonian $$ H_T = \omega a^{\dagger} a + \lambda(t) a + \bar{\lambda(t)} a^{\dagger}$$

evolves to a coherent state $|\Lambda(t)\rangle$, where $$\Lambda(t) = \int_{0}^{t} e^{i \omega\tau} \lambda(\tau) d\tau$$ This method is used for example to generate phonon coherent states.

There is another principle allowing the manufacture of coherent states: There are cases where coherent states appear as steady states of open system quantum dynamics, approximated by Lindblad equation. For example those implemented in cooling processes in which Bose-Einstein condensates are prepared. See for example the following work by Albert, Bradlyn, Fraas, and Jiang

$\endgroup$
  • $\begingroup$ I have added some references and some additional applications. Actually, this subject is so vast that I am sure that there are some other important areas not covered by the answer. $\endgroup$ – David Bar Moshe Aug 1 '17 at 13:34
4
$\begingroup$

Coherent states appear in nature because they're what you get when you drive a harmonic oscillator through a dipole interaction. Suppose we push on a harmonic oscillator with a time dependent force $F(t)$. The Hamiltonian associated with that push is $$\hat{H}_x = -\hat{x} F(t) \equiv -(\hat{a} + \hat{a}^\dagger) f_x(t)$$ where

  • we've used $\hat{x} = x_\text{zpf}(\hat{a}+\hat{a}^\dagger)$ where $x_\text{zpf}$ is the zero point fluctuation of the oscillator, i.e. $ x_\text{zpf}^2 = \langle 0 | \hat{x}^2 | 0 \rangle$.

  • we've defined $f_x(t) \equiv F(t) x_\text{zpf}$.

A drive proportional to $\hat{x}$ like this is very common in Nature because, for example, it's common to couple to one of the degrees of freedom of an oscillator and push on it.

From now on, we'll drop the operator hats. Let's also add in a $y$ drive term so that the total drive is \begin{align} H_\text{drive} &= -(a + a^\dagger) f_x (t) - (-i)(a - a^\dagger)f_y(t) \\ &= a(-f_x + i f_y) + a^\dagger(-f_x - i f_y) \, . \\ \end{align} Define $f(t)\equiv -f_x(t)+i f_y(t)$ so that $$H_\text{drive} = af(t) + a^\dagger f(t)^* \, .$$ The full Hamiltonian in the Heisenberg picture is then \begin{align} H(t) &= H_\text{oscillator}(t) + H_\text{drive}(t) \\ &= \hbar \Omega \left(a^\dagger(t) a(t) + \frac{1}{2} \right) + (a(t) f(t) + \text{h.c.}) \end{align} where $\Omega$ is the frequency of the oscillator. Heisenberg's equation of motion works out to $$\left( \frac{d}{dt} + i \Omega \right) a(t) = -\frac{i}{\hbar}f(t)^* \, \text{Id}$$ where $\text{Id}$ means the identity operator. It's reasonably intuitive here (but you should check with explicit calculation) that $a(t)$ picks up only phase factors and scalar offsets. In particular, the solution is $$a(t) = a(0)e^{-i \Omega t} - \frac{i}{\hbar} \int_0^t e^{i \Omega t'} f(t')^* \, dt' \, . $$ Therefore, if we start in a coherent state and drive the system, the new state is still a coherent state: \begin{align} a(t) | \alpha \rangle &= \left( a(0)e^{-i \Omega t} - \underbrace{\frac{i}{\hbar} \int_0^t e^{i \Omega t'} f(t')^* \, dt'}_\text{some complex number $c$} \right) | \alpha \rangle \\ &= \underbrace{\left( \alpha e^{-i \Omega t} - c \right)}_{\alpha'} |\alpha \rangle \\ &= \alpha' |\alpha \rangle \, . \end{align}

Therefore, we still have a coherent state, it's just a different one than before the driving. The ground state is a coherent state, so we've shown that if you start in the ground state, and then drive the system with a dipole interaction, you always get a coherent state.

$\endgroup$
  • $\begingroup$ I read the post as asking about coherent states of fields that are not the EM field, i.e. the electron field or, say, some Boson icons field for child atoms in a tap undergoing Bose-Einstein condensation. $\endgroup$ – Emilio Pisanty Aug 1 '17 at 13:43
  • $\begingroup$ Reading this again, I really like this answer! But can you explain where exactly the $y$ drive term comes from? Why is it $a - a^\dagger$? $\endgroup$ – knzhou Jan 24 '18 at 10:19
  • $\begingroup$ @knzhou The $y$ drive is a drive coupled to the variable conjugate to $x$. In a mechanical system, $y$ would be momentum, in which case the $f_y$ term is kind of unphysical and would usually be zero. Of course, there are electromagnetic oscillators etc where driving both conjugate variables is quite physical. The $y$ coupling is $a-a^\dagger$ because that's the operator canonically conjugate to $a+a^\dagger$, as you can check by working out the commutators. If you do that, you must take $x_\text{zpf}$ and $y_\text{zpf}$ into account to get the $\hbar$ in the commutator. $\endgroup$ – DanielSank Jan 24 '18 at 21:15
  • $\begingroup$ @knzhou For a complete discussion of $x_\text{zpf}$, $y_\text{zpf}$ and the relations between canonically conjugate (i.e. $\hat{x}$ and $\hat{p}$) operators and the raising/lowering operators, see this other post, but note that what's called $x$ here is called $u$ in the other post. $\endgroup$ – DanielSank Jan 24 '18 at 21:16
  • $\begingroup$ @DanielSank Ah, right, I figured it was the conjugate variable but I didn't know why it'd be called $y$. I guess because it's the $y$ axis in phase space? $\endgroup$ – knzhou Jan 24 '18 at 21:17

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.