# Why can we quantize macro(meso)scopic harmonic oscillator?

+ 4 like - 0 dislike
745 views

It is well known that we have got many kinds of quantized macro(meso)scopic harmonic oscillators or so in tiny mechanical systems. People are talking about cavity cooling and so on.

However, it is since the first time of learning quantum mechanics that I've reckoned that quantum mechanics is a theory merely for small particles with few degrees of freedom. At least once you use it as an exact starting point for any large problem, you will apply it to small particles, e.g., writing down individual terms in a Hamiltonian.

Somehow the quantum mechanical macro-harmonic oscillator just looks unfathomable to me. How to understand this kind of macroscopic quantum state? Is it just a system whose vast many constituents are in the same conventional small-particle harmonic oscillator quantum state (kinda reminiscent of BEC)? Or anything else?

This post imported from StackExchange Physics at 2014-08-22 05:06 (UCT), posted by SE-user huotuichang
+1. Nice question. I am sure you are familiar with (1) 2nd quantization from microscopic harmonic oscillators (so nothing really is macro(meso)scopic harmonic oscillator there); and (2) emergent degree of freedoms and topological orders. I provide some thoughts below following this two comments.

This post imported from StackExchange Physics at 2014-08-22 05:06 (UCT), posted by SE-user Idear

+ 1 like - 0 dislike

(1) Oppose your question- in the macro or mesoscopic states, I thought that we usually still consider (the collection of many) microscopic harmonic oscillators, such as the cavity examples, or BEC or superfluids?

A well-known example is a 1+1D superfluid-insulator transition. (Are you familiar with this model?) Given a microscopic Lattice Hamiltonian: $$H=-t \sum_{\langle i,j \rangle} (\psi_i^\dagger \psi_j +\psi_j^\dagger \psi_i) + U \sum_i (\hat{N}_i - \langle\bar{N}\rangle)^2$$ With $\psi_j$ of some boson operator and boson number operator is $N_i=\psi_j^\dagger \psi_j$. You can show that the (2nd) quantization, with $\psi_j = \sqrt{N_j} e^{i \theta_j}$ of U(1) phase $\theta$, with commutators $[ \psi_i, \psi_j^\dagger]= \delta_{i,j}$. You can derive $$\boxed{ [ \theta_i, \hat{N}_j]=-i \delta_{i,j}}.$$ Continuum field limit is free Klein-Gordon equation. Above all have linear dispersion $\boxed{\omega \propto k}$. This is superfluid mode when U(1) symmetry is broken, and

The derivation here for this commutator $[ \theta_i, \hat{N}_j]=-i \delta_{i,j}$ gives something you may refer to a macro or mesoscopic harmonic oscillator (in disguise, analogue to $[x,p]=i \hbar$ for a single site harmonic oscillator), but it is NOTHING mysterious but an overall effect of a collection of microscopic phenomena. The degree of freedom and quantization are from the microscopic creation/annihilation operators on each site. So they are just a phenomena from a collection of many microscopic harmonic oscillators.

(2) Support your question- there are examples of condensed matter system, one consider emergent degrees of freedom, where quasiparticles (such as 2+1D anyons) are indeed quite different from the fundamental constituents. See an example of emergent topological Chern-Simons theory-where one can derive a phenomena of a macro or mesoscopic harmonic oscillator in your own language (by doing a quantization on the intrinsic emergent gauge fields(anyons) ), and many other examples such as in toric code or in the string-net model.

This post imported from StackExchange Physics at 2014-08-22 05:06 (UCT), posted by SE-user Idear
answered Jan 26, 2014 by (1,455 points)
Thanks! But sorry, I'm not familiar with that 1+1D superfluid-insulator transition model. I can't see why it shows sth. like a big harmonic oscillator.

This post imported from StackExchange Physics at 2014-08-22 05:06 (UCT), posted by SE-user huotuichang
@ huotuichang, yes, let me clarify and reorganize the answer soon, get back to you certainly.

This post imported from StackExchange Physics at 2014-08-22 05:06 (UCT), posted by SE-user Idear
Thanks! The new commutator in the square box, why could I refer to it with "macro"? It looks like site index-dependent commutator, not kinda collective or macro.

This post imported from StackExchange Physics at 2014-08-22 05:06 (UCT), posted by SE-user huotuichang
@ huotuichang, yes, so if you know it is actually microscopic harmonic oscillators(SHO), then didn't you just spell the answer? So my first comment is to oppose your thinking on macro SHO. My comment is that it is actually micro SHO instead of macro.

This post imported from StackExchange Physics at 2014-08-22 05:06 (UCT), posted by SE-user Idear
@ huotuichang, the 2nd comment is that in the topological phases (or topological QFT), you can reduce the many (or infinite) degrees of freedom of QFT problem to a finite dimensional problem (like quantum mechanics) because of finite number of ground states at lowest energy. In that case, there are indeed "macro" SHO, where you consider the finite emergent degrees of freedom (such as anyons). The 2nd comment provides alternate thoughts, supporting your "macro" SHO thinking.

This post imported from StackExchange Physics at 2014-08-22 05:06 (UCT), posted by SE-user Idear
+ 0 like - 0 dislike

To start, the macroscopic system is subject to the same laws as the microscopic one, although it is harder to isolate it from it's environment. In any case, your harmonic oscillator can be considered to be made out of many particles each with their individual Hamiltonian operator, along with interactions between each particle, so the total Hamiltonian is just the sum of these, and acts on the wavefunction for the entire system. Now, you can always choose different variables to describe the system, and a convenient variable change turns out to be $x_\mathrm{com}$, the center of mass of the system, and $x_i$, the position of the $i^\mathrm{th}$ particle relative to the center of mass. You will mostly find that the variable $x_\mathrm{com}$ doesn't enter any of the interaction terms, due to translation invariance of the problem, except where it enters the potential function $V(x_\mathrm{com})$ and the kinetic energy. Therefore you can write your wavefunction as $\Psi(x_\mathrm{com}, x_1,x_2,\dots) = \Psi_\mathrm{com}(x_\mathrm{com}) \times \psi(x_1,x_2,\dots)$, or, at least, a superposition of such states. Once you have a solution that satisfies $$(T_\mathrm{com} + V_\mathrm{com}) \Psi_\mathrm{com}(x_\mathrm{com}) = E_\mathrm{com} \Psi_\mathrm{com}(x_\mathrm{com})$$ you can substitute that solution into the wavefunction for the entire system and solve for the motion of the other particles. In other words, the center of motion factors out and it's dynamics can be considered separately.

Having said that, the normal solutions you obtain for harmonic oscillators aren't much good for macroscopic systems, because they have such large uncertainty, and don't look at all like classical behaviour, so you would then consider coherent states instead.

This post imported from StackExchange Physics at 2014-08-22 05:06 (UCT), posted by SE-user lionelbrits
answered Jan 25, 2014 by (110 points)
+ 0 like - 0 dislike

When you excite an composite oscillator, all of the microscopic bits are oscillating in phase with each other.

Quantumly speaking, it's not just that all the microscopic bits are each separately excited, but rather it really is crucial that their oscillatory phases are all lined up in some sense.

This post imported from StackExchange Physics at 2014-08-22 05:06 (UCT), posted by SE-user Nanite
answered Jan 26, 2014 by (50 points)

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOverf$\varnothing$owThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.