Phonon Number Conservation in a Single Mode Oscillation Experiment

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SUMMARY

The discussion centers on the conservation of phonon numbers in a single mode oscillation experiment, specifically examining the relationship between energy equations and phonon number states. The participants clarify that while the number of phonons, denoted as ##N_{k}##, can be fixed in certain states, it is not conserved due to the non-commutation of the number operator ##N## with the Hamiltonian ##H##, leading to the creation and destruction of phonons. The conversation also touches on the implications of thermal excitations and the statistical mechanics governing phonon behavior, emphasizing the distinction between fixed and definite phonon numbers.

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  • Familiarity with Hamiltonian mechanics and operator algebra
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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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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.
 
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:
 
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##.
 
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
 
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 ?
 
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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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)
 

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