6
$\begingroup$

A coherent state of the quantum harmonic oscillator is defined as an eigenvector $|\alpha\rangle$ of the annihilation operator $\hat a$ with eigenvalue $\alpha$ or as spatial translations of the ground state of the QHO:$$T_{x_0}|0\rangle = \exp(-\frac i \hbar\hat px_0)|0\rangle:=|\bar x_0\rangle$$the definitions are equivalent when $\alpha$ is real. Coherent states exhibit certain semi-classical properties, such as the following:$$\langle \bar x_0|\hat x_H(t)|\bar x_0\rangle= x_0\cos\omega t,$$where $\omega$ is the angular frequency of the harmonic oscillator, and the $H$ subscript represents a Heisenberg operator. We have also $$\langle \bar x_0|\hat p_H(t)|\bar x_0\rangle = -m\omega x_0 \sin \omega t.$$ So that both the expectation value of position and momentum oscillate with time, in contrast to the energy eigenstates of the harmonic oscillator, which have vanishing expectation values for these operators. So, my question is: why do the coherent states actually behave like oscillators, when the harmonic oscillator energy eigenstates do not? What is a physical intuition for why the expectation values vanish for the QHO energy eigenstates?

$\endgroup$
6
  • 1
    $\begingroup$ Energy eigenstates also behave like those of a classical oscillator, whose energy remains constant in time :) $\endgroup$
    – Roger V.
    Jan 4, 2022 at 8:42
  • $\begingroup$ @RogerVadim But they do not oscillate with time. $\endgroup$ Jan 4, 2022 at 14:33
  • $\begingroup$ oscillator energy does not oscillate with time. In fact, if we start with the time-dependent Schrödinger equation, then there is factor $e^{-i\omega (n+1/2)t}$. $\endgroup$
    – Roger V.
    Jan 4, 2022 at 14:46
  • $\begingroup$ @RogerVadim I am referring to the expectation values of position and momentum. In which case, the factor you mentioned gets cancelled by its inverse. $\endgroup$ Jan 4, 2022 at 14:49
  • $\begingroup$ @DanileWaters I see. I gave an answer - perhaps, it is not complete, but too long for the comments section. $\endgroup$
    – Roger V.
    Jan 4, 2022 at 15:24

2 Answers 2

4
$\begingroup$

The discovery and study of coherent states represents one aspect of one of the biggest problems physicists have faced with the birth and the subsequent development, supported by excellent experimental results, of the quantum mechanics: the search for a correspondence between the new theory, conceived for the analysis of microscopic systems, and classical physics, still fully valid for the description of the macroscopic world.

The history of coherent states begins immediately after the advent of mechanics quantum: their introduction on a conceptual level dates back to a article published in 1926, in which Schrödinger reports the existence of a class of states of the harmonic oscillator that show, in a certain sense, behavior analogous to that of a classic oscillator: for these states it is verified that the energy mean corresponds to the classical value and the position and momentum averages have oscillatory forms in constant phase relation.
Returning to Schrödinger's article, the "almost classical" states from him identified present, in addition to the characteristics already mentioned, an important aspect: being represented by Gaussian wave packets that do not change shape in the time, guarantee the minimization of the product among the uncertainties about position and on the impulse, that is the condition closest to the possibility of measuring simultaneously the aforesaid quantities with arbitrary precision, allowed from classical physics.

So, starting from the following relations: \begin{equation} a^{\dagger}|n\rangle=\sqrt{n+1}|n+1\rangle \quad a|n\rangle=\sqrt{n}|n-1\rangle \end{equation} it is noted that, by virtue of the orthonormality of the states stationary, the diagonal matrix elements of the position and momentum operators are null in the representation of energy, which means that the expectation values ​​of position and momentum on any stationary state are zero instant by instant.

The stationary states just analyzed are characterized by distributions of constant probabilities with respect to the position over time; the wait values ​​of the position e of the impulse are null at all times: this aspect is a fundamental one difference with the states of the classic oscillator, for which, once the energy is defined (as long as different from zero), the observables position and momentum evolve over time according to sinusoidal functions and are always in phase quadrature with each other. Also, if yes calculate the uncertainties on position and momentum for a steady state with n photons, yes gets the uncertainty relation $\Delta x \Delta p=(n+1/2)\hbar$.
it is therefore possible to obtain the minimization of the product of the uncertainties on impulse and position, which represents the maximum similarity with classical mechanics.

A state that is as similar as possible to the classical case must therefore have the following characteristics:

1): The evolution over time of the position and momentum expectation values ​​must be of a simple periodic type, with a constant phase ratio between position and impulse.

2): The wave functions must be as narrow as possible around the value average of the position, so that the probability distribution with respect to the position may tend, by varying appropriate parameters, to a delta function of Dirac;

3): The product of the uncertainties on the position and on the impulse must be minimal.

So you can see this "classical" behavior, that admits these particular states, as an intrinsic property of QHO.
Furthermore I have say that: also the harmonic oscillator energy eigenstates actually behave like oscillators.

Maybe this answer is a bit too long but I hope it can help you.

$\endgroup$
2
  • $\begingroup$ I had a thought regarding the oscillatory behavior of the stationary states. Considering them as translated ground states, can we think about the oscillation of $\langle \bar x_0|\hat x_H(t)|\bar x_0\rangle$ in time as if we had "pulled" the spring by a distance $x_0$, as in classical mechanics? $\endgroup$ Jan 4, 2022 at 14:44
  • $\begingroup$ You can think as if you had "pulled" the spring but only if you think $\langle x \rangle_t$ as a statistical average of several measures of the same experiment prepared in the same state. And also if the distribution of values is narrow i.e. the state of the particle is localized in regions much more smaller compared to the typical length in the experiment. In this limit you can use average value as a good approximation. And so you can think as you said. This is the "limit bridge" beteween CP and QM: consequence you can see starting from the Heisenberg rapresentaion. $\endgroup$ Jan 4, 2022 at 15:28
4
$\begingroup$

The energy states of a harmonic oscillator can be thought of as averaged over the initial phase of the oscillations.

Indeed, it is instructive to look at how the correspondence principle works for a harmonic oscillator at high energies:

  • For a quantim oscillator the probability of finding the oscillator in the interval $(x,x+dx)$ is given by $$p(x)dx=|\psi_n(x)|^2dx$$
  • For classical oscillator the same probability density can be calculated as $$p(x)=\int_0^{2\pi}\frac{d\phi}{2\pi}\delta\left(x-A\cos(\omega t + \phi)\right),$$ where $A$ is the amplitude of the oscillations, and $\phi$ is the initial phase of the oscillator, averaged upon. After a bit of mathematical manipulations we arrive at $$ p(x)=\frac{1}{\pi}\frac{1}{\sqrt{A^2-x^2}}.$$

The quantum and classical cases then compare as in the figure linked here: enter image description here More detailed discussions are available from many sources and can be found by googling harmonic oscillator classical limit or harmonic oscillator correspondence principle.

Remark: This could also be viewed as the expression of the phase-number uncertainty, as the coherent states have well-defined phase, $\phi$, whereas the eigenstates have well-defined number of quanta $n$.)

$\endgroup$
2
  • $\begingroup$ So I see that you were speaking of the semi-classical nature of QHO energy eigenstates. Thanks for clarifying that they do behave somewhat classically. $\endgroup$ Jan 4, 2022 at 15:33
  • $\begingroup$ Indeed, one could call the HO eigenstates incoherent states, since they average over phase... in fact, all this might be tied to the phase-number uncertainty (coherent states have well-defined phase, while the eigenstates have well-defined $n$.) $\endgroup$
    – Roger V.
    Jan 4, 2022 at 15:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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