Phonon Number Conservation in a Single Mode Oscillation Experiment

In summary, the number of phonons in a classical mode of oscillation of the lattice is not necessarily fixed, but it is definite.
  • #1
dRic2
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Suppose I prepare an experiment where I excite a single mode of oscillation of the lattice, that is something like ##u(x, t) = Ae^{i(kx-\omega t)} ## (in the classical limit). The energy corresponding to that mode should be ##E = \frac 1 2 \rho L^3 A^2 \omega^2 ##. If I equate this equation to ##E(k) = \hbar \omega(k) (N_{k} + \frac 1 2)## can I conclude that the number of phonons is exactly ##N_{k}## ?

I think the answer is no, but I'm not totally sure.

My reasoning is that, if ##N_{k}## is the exact umber of phonons, then the only possible way to describe the state of the lattice is
$$ | \psi > = |0, 0, 0, ..., N_{k}, 0, 0, ..., 0>$$
Isn't this in contradiction with the fact that phonons number is not conserved ? I am a bit confused about this... I can see that the particle number operator N does not commute with the phonon hamiltonian, but I don't know how to interpret this on empirical grounds.
 
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  • #2
dRic2 said:
Suppose I prepare an experiment where I excite a single mode of oscillation of the lattice,
That's a mouthful. All depends on the "experiment". Most excitations are thermal in nature. For example a ideal laser might excite a single mode, but that mode's occupation number would obey a black body distribution for the given oscillation frequency. Making eigenstates of occupation number isn't easy.
 
  • #3
Okay I'm having a little crisis here :sorry:... It's like I'm asking this question for the first time ahahah. As you said, usually one has that the average energy of a singe mode is given by
$$<E(k)> = \hbar \omega(k) ( <n> + 1/2)$$
where ##<n>## is given by BE statistics with ##\mu = 0## (chemical potential) and the overall average energy ##<E>## is just an integral running over all modes (states).

Now my problem comes when we say that phonon number is not conserved: phonons can be created and destroyed. The math is straightforward ##[H,N] \neq 0## so yeah, nothing really to add here, but how can you justify this statement on "empirical" grounds ? How con you visualize this ?

Plus I've been thinking about this: in the thermodynamic limit, gran canonical, canonical and micro canonical description should "merge". But in the canonical and micro canonical description the number of particle is fixed. So in the thermodynamic limit should the phonon number be fixed (aka conserved) ? But that is not possible because ##[H,N] \neq 0##!... Ahhhh I'm loosing it :cry::cry:
 
  • #4
I'm not certain photons (or phonons) not being conserved has anything to do with the question you're asking. In an eigenstate of the occupation number, ##N_k##, has a unique integer value. In an atomic coherent state this is not the case but it's still a pure state, just not an eigenstate of ##N##.

What is ##H## here? Everything I said assumes ##H## is the free field hamiltonian in which case ##[H,N]=0##.
 
  • #5
Thanks for the answer. I'm very tired bacuase I am preparing two exams... I should come back to this maybe tomorrow because I feel like I don't understand a thing now :D
 
  • #6
Ok, so I'm a bit confused right now. I was wrong, [H, N] = 0. But if N commutes with H, shouldn't N be a constant of the motion ? That is, shouldn't the number of phonons be fixed ?
 
  • #7
dRic2 said:
Ok, so I'm a bit confused right now. I was wrong, [H, N] = 0. But if N commutes with H, shouldn't N be a constant of the motion ? That is, shouldn't the number of phonons be fixed ?
Fixed, yes, but definite, not necessarily.

Classical modes are superposition of different phonon-number states.
 
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  • #8
DrClaude said:
Fixed, yes, but definite, not necessarily.

Classical modes are superposition of different phonon-number states.
I think I'm getting there, but I need a bit more help please. I don't quite get the difference between "definite" and "fixed". Let's go back to the example of my fist post for a second. I have this classical mode with fixed energy and momentum (I know it is an extreme example) so the number of phonon is fixed. So I have a constraint on the number of phonons, the total energy and the momentum: how can I not have an unique representation for that state ? The only state that comes to my mind that satisfies those constraint is the obvious ##|0, 0, ..., N_k, ..., 0>##.

PS: An other question now comes to my mind. I always thought that phonons have zero chemical potential because their number is not fixed, but now I have to take into account that this is not the case. (we can address this question later)
 

FAQ: Phonon Number Conservation in a Single Mode Oscillation Experiment

What is phonon number conservation?

Phonon number conservation refers to the principle that the total number of phonons, or quanta of vibrational energy, in a system remains constant over time. In other words, phonons cannot be created or destroyed, but can only be transferred or transformed within the system.

How is phonon number conservation observed in a single mode oscillation experiment?

In a single mode oscillation experiment, the energy of a system is confined to a single vibrational mode. This means that the total number of phonons in the system must remain constant, as any transfer or transformation of energy would result in a change in the number of phonons. By measuring the energy of the system at different points in time, we can observe the conservation of phonon number.

What factors can affect phonon number conservation in a single mode oscillation experiment?

There are several factors that can affect phonon number conservation in a single mode oscillation experiment. These include external forces or perturbations acting on the system, temperature changes, and interactions with other particles or fields. Additionally, the properties of the materials used in the experiment, such as their elasticity and thermal conductivity, can also influence phonon number conservation.

Why is phonon number conservation important in the study of solid state physics?

Phonon number conservation is important in the study of solid state physics because it plays a crucial role in understanding the behavior and properties of materials at the atomic level. By studying how phonons interact and behave in different systems, we can gain insight into phenomena such as thermal conductivity, electrical conductivity, and phase transitions. Phonon number conservation also helps us understand how energy is transferred and dissipated in solid materials.

How does phonon number conservation relate to other conservation laws in physics?

Phonon number conservation is closely related to other conservation laws in physics, such as the conservation of energy and momentum. This is because phonons are a form of energy and their number cannot change without a corresponding change in energy. Additionally, the conservation of phonon number is also linked to the principles of symmetry and time-reversal invariance, as the laws governing phonon interactions must remain the same regardless of the direction of time.

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