Phonons as a quasiparticles in a quantum LHO

[LagrangeEuler]

I am confused with phonons as a quasiparticles in quantum LHO. When I say $E_n=(n+\frac{1}{2})\hbar \omega$ why $n$ is number of phonons? Why no magnons, excitones...

And one more question. Why number of maxima of $|\psi_n(x)|^2$ is related to number of phonons, so that
number of phonons=number of maxima of $|\psi_n(x)|^2$-1?
Thanks a lot for your answers!

 

[hutchphd]

1. This is a definition and what you "call" them is up to you. The Hamiltonians all have a similar mathematical form: the physical system represented by the variables determines the particular physics being described...mechanical, electrical, magnetic, or a combination
2. The number of zeros in the solution to a Sturm-Liouville system (look it up) counts the eigenvalues. I think this result then holds for any non-positive potential. For the LHO there my be an easier explanation but I don't know it.

 

[LagrangeEuler]

Why in all quantum mechanics books they used phonons? Why no photons? What is the reason for that?

Do you maybe know some book where I can find more about Sturm-Liouville systems?

 

[PeterDonis]

Why in all quantum mechanics books they used phonons?

They don't. How many QM textbooks have you looked at?

 

[LagrangeEuler]

More then 10. Please tell me one where this is not the case.

 

[PeterDonis]

More then 10.

Which ones? What chapters/pages are you seeing the reference to phonons on?

Please tell me one where this is not the case.

Ballentine, Chapter 6, is a treatment of the quantum harmonic oscillator which does not use the term "phonon" at all. In fact, the only places where Ballentine uses the term "phonon" are in Chapter 19, specifically referring to the normal modes of vibration in a crystalline solid.

 

[HomogenousCow]

The gist of it is that you can rewrite the Hamiltonian of a spatial array of coupled oscillators into a sum of harmonic oscillator hamiltonians at different frequencies. These harmonic oscillators are not actually particles tied to springs but they satisfy the same algebra so you can interpret them as particles. Ultimately the reason why people use this formalism is because it helps them solve the system and get the results that they presumably want, the quasi-particle interpretation is just a bunch of words to dress it up.

 

[Dr_Nate]

The gist of it is that you can rewrite the Hamiltonian of a spatial array of coupled oscillators into a sum of harmonic oscillator hamiltonians at different frequencies. These harmonic oscillators are not actually particles tied to springs but they satisfy the same algebra so you can interpret them as particles. Ultimately the reason why people use this formalism is because it helps them solve the system and get the results that they presumably want, the quasi-particle interpretation is just a bunch of words to dress it up.

I am very confused by why you are saying this.

Coupled HO's will give you phonon dispersion relations which have a dependence on crystal momentum $k$. Whenever, I've seen a phonon spectrum modeled or measured there is a cap to the maximum energy.

An array of uncoupled HO's are what we would find in an Einstein solid. These uncoupled HO's are a poor man's modelling of optic phonons. This model was intended to model specific heat capacity. The oscillators have a flat dispersion as there is no $k$ dependence. These levels can continue up in energy indefinitely. However, the ground state is really the only one that contributes to the specific heat.

Looking at the Hamiltonian for an uncoupled array of HO's
$$H_u=\sum_{i=0}^N \left(\frac{p_i^2}{2m} + \frac{m\omega^2}{2} x_i^2 \right) $$
and the Hamiltonian for a coupled array of HO's
$$H_c=\sum_{i=0}^N \left(\frac{p_i^2}{2m} + \frac{m\omega^2}{2} \left(x_i^2 +x_{i+1}^2 -2x_ix_{i+1} \right) \right) ,$$
I don't see how they could be equal.

 

[Dr_Nate]

...the quasi-particle interpretation is just a bunch of words to dress it up.

We use the term quasiparticle because one might intuitively think it is impossible to resolve the behaviour of the individual particles separate from all the things they are correlated with. Surprisingly, we can still model solids as if they had things that behave similar to particles that can exist in vacuum. We call these quasiparticles to remind us that they aren't the same thing as regular particles.

 

[LagrangeEuler]

Ok. I made mistake. There is really no a lot of phonons corespoding to state $|n \rangle$ in books in quantum mechanics. So when I speak about operator $\hat{n}$ it is operator of what?

 

[vanhees71]

Why in all quantum mechanics books they used phonons? Why no photons? What is the reason for that?

Do you maybe know some book where I can find more about Sturm-Liouville systems?

Well, unfortunately many introductory QM textbooks start with a historically motivated chapter trading a misleading picture of photons as a kind of localized "light particles". That's a very misleading old-fashioned idea going back to Einsteins famous article of 1905 on "light quanta as a heuristic aspect". It's indeed very heuristic and completely outdated today. The only way to adequately describe photons is relativistic quantum field theory.

In atomic, molecular and condensed-matter physics you get very far with non-relativistic quantum physics and the approximation of the electromagnetic field as being a classical field. Except for the phenomenon of spontaneous emission all very fundamental properties can be well approximated in this semiclassical approximation. That's why in most QM textbooks they talk about photons only in this unfortunate misleading heuristic way and then they don't mention photons anymore, because for that you have to read on in a QFT book.

 

[hutchphd]

I am very confused by why you are saying this.

I am confused by your confusion. There are optical phonons and acoustic phonons. You seem to limit yourself only to the optical variety.

 

[hutchphd]

Do you maybe know some book where I can find more about Sturm-Liouville systems?

https://en.wikipedia.org/wiki/Sturm–Liouville_theory

And any diff. eq. text should cover it.

 

[HomogenousCow]

I am very confused by why you are saying this.

Coupled HO's will give you phonon dispersion relations which have a dependence on crystal momentum $k$. Whenever, I've seen a phonon spectrum modeled or measured there is a cap to the maximum energy.

An array of uncoupled HO's are what we would find in an Einstein solid. These uncoupled HO's are a poor man's modelling of optic phonons. This model was intended to model specific heat capacity. The oscillators have a flat dispersion as there is no $k$ dependence. These levels can continue up in energy indefinitely. However, the ground state is really the only one that contributes to the specific heat.

Looking at the Hamiltonian for an uncoupled array of HO's
$$H_u=\sum_{i=0}^N \left(\frac{p_i^2}{2m} + \frac{m\omega^2}{2} x_i^2 \right) $$
and the Hamiltonian for a coupled array of HO's
$$H_c=\sum_{i=0}^N \left(\frac{p_i^2}{2m} + \frac{m\omega^2}{2} \left(x_i^2 +x_{i+1}^2 -2x_ix_{i+1} \right) \right) ,$$
I don't see how they could be equal.

They are equal by a relabelling of variables, this is a standard manipulation that is shown in many solid state textbooks.

 

[Dr_Nate]

I am confused by your confusion. There are optical phonons and acoustic phonons. You seem to limit yourself only to the optical variety.

Well now I seem to be confused by your confusion of my confusion.

The uncoupled Hamiltonian will give the rather unphysical Einstein mode, which is like I said a poor man's model of optical phonons.

The coupled Hamiltonian I've given is for a monatomic chain. It will give only acoustic phonons. To add a set of optic phonons, I would need to split the Brillouin zone in half by either using a diatomic basis or two different coupling constants.

 

[HomogenousCow]

The coupled Hamiltonian I've given is for a monatomic chain. It will give only acoustic phonons.

The concept of phonons here precisely emerges because you can write the Hamiltonian of coupled oscillators into a sum of uncoupled oscillators, the latter uncoupled oscillators are not analogous to the former ones but they satisfy the annihilation/creation operator commutation relations.

 

[Dr_Nate]

The concept of phonons here precisely emerges because you can write the Hamiltonian of coupled oscillators into a sum of uncoupled oscillators, the latter uncoupled oscillators are not analogous to the former ones but they satisfy the annihilation/creation operator commutation relations.

This sounds like an array of coupled oscillators are not physically equivalent to a sum of uncoupled oscillators. Am I understanding you correctly?

 

[hutchphd]

The coupled Hamiltonian I've given is for a monatomic chain. It will give only acoustic phonons. To add a set of optic phonons, I would need to split the Brillouin zone in half by either using a diatomic basis or two different coupling constants.

OK.
The array of coupled oscillators gives rise to eigensolutions at various frequencies determined by the details of the potential. Each eigensolution looks like an harmonic oscillator and the n can be thought of as an occupation number This is basic Solid State (tranverse, longitudinal acoustic optical ; phonons, magnons excitons plasmons ) .

So I still truly do not understand your overall confusion.

 

[PeterDonis]

when I speak about operator $\hat{n}$ it is operator of what?

It's the number operator for the quantum harmonic oscillator, which is a general model that can describe many different things. Which specific thing it is describing depends on the context.

 

[vanhees71]

First look at the classical model. You have a set of points connected by springs all with the same spring constant. It's intuitively clear that the dynamics leads to oscillations of the masses relative to each other around their equilibrium positions.

Then there are some special configurations of vibrations, where all the particles oscillate with the same frequency. These are the eigenmodes of the oscillating system. Each eigenmode has a frequency $\omega_k$ and $k$ labels the eigenmodes.

All possible oscillations can be described as superposistions of these eigenmodes.

If you write down the Hamiltonian in terms of the eigenmodes, you see that in this way the system is described as a sum over simple harmonic oscillators which don't interact with each other. That's why using these special coordinates and momenta makes it very easy to quantize the system: It's just a set of independent harmonic oscillators, which can be described with annihilation operators $\hat{a}_k$ (where $k$ still labels the eigenmodes) and creation operators that are simply the Hermitean conjugate of the annihilation operators $\hat{a}_k^{\dagger}$.

A convenient complete set of states are the eigenstates of the Hamiltonian, and these are given by the common eigenstates of the number operators $\hat{n}_k=\hat{a}_k^{\dagger} \hat{a}_k$. The Hamiltonian (after subtracting the constant zero-point energies) simply is
$$\hat{H}=\sum_k \hbar \omega_k \hat{n}_k.$$
The eigenvalues of the $\hat{n}_k$ are $n_k \in \{0,1,2,3\ldots \}$.

What's counted are the excitations of eigenmodes, i.e., the collective vibration patterns that lead to the collective oscillation of each particle with the same corresponding eigenfrequency $\omega_k$.

This entire formalism looks very much like particles in the 2nd-quantization or quantum-field theoretical description of bosons, but indeed it's not particles but the eigenmodes of collective lattice vibrations. Mathematically there's of course no difference between the description of how the quantized lattice vibrations are excited or how many non-interacting particles occupy a given energy state (confined in a finite box to make the analogy complete). That's why one calls these quantized collective modes of lattice vibrations quasi-particles, and since the lattice vibrations are also describing sound waves in the solid body these quasi-particles are named phonons ("sound particles").

That's of course also analogous to calling the excitations of the quantized free electromagnetic field "phonons", because the vibrations of the free em. field are em. waves and particularly light (though in this case the particle nature of these "light particles" is pretty misleading, because the em. field is a massless spin-1 field, but that's another story).

 

[Dr_Nate]

So I still truly do not understand your overall confusion.

I was confused by his first sentence. It's not correct.

The gist of it is that you can rewrite the Hamiltonian of a spatial array of coupled oscillators into a sum of harmonic oscillator hamiltonians at different frequencies.

Acoustic phonons are waves that can transport heat. Uncoupled oscillators can't do that. They are different physical systems.

 

[HomogenousCow]

Acoustic phonons are waves that can transport heat. Uncoupled oscillators can't do that. They are different physical systems.

Have you read a serious treatment of acoustic phonons? As in from a QFT or advanced QM textbook, this is standard stuff. Recasting many body systems into uncoupled oscillators is the foundational idea behind QFT.

 

[hutchphd]

I don't see how they could be equal.

Look in any solid state textbook (Ashcroft/Mermin, Kittel are my favorites) . This is an absolutely fundamental technique.

 

[Dr_Nate]

I think we have more points of agreement than disagreement. I'm not denying the formalism of second quantization. The two systems of coupled and independent oscillators are similar. The Hamiltonians that are typically shown in second quantization sure seem to be similar.

I'm trying to say that when you recast it, you won't have independent oscillators. Instead, you will have independent phonons. You can excite one without exciting another, just like with an array of independent QHOs.

In the both Hamiltonians, we will have $\omega_k$. However, in the case of coupled oscillators it is actually a dispersion relation: $\omega(k)$. This is what shows us that we are dealing with waves and not independent oscillators.

 

[vanhees71]

If you have a Lagrangian quadratic in the field operators, you have a set of independent oscillators. The phonons are the eigenmodes of this set of oscillators.

 

[hilbert2]

And the small-deformation oscillations of even a continuum matter object can be described as phonons that act like harmonic oscillators, so it's not something of very limited applicability.

 

[Dr_Nate]

I believe earlier it was implied that there is consensus on this use of the term independent harmonic oscillators in second quantization. Going through many online course notes, I find that this isn't the case. I find that there is a spectrum of opinions that run from i) yes, it is a harmonic oscillator, to ii) being silent on the issue, to iii) being mostly silent on the issue with some suggestive statements, to iv) implying that it only looks similar to harmonic oscillators.

 

[vanhees71]

Free quantum fields are equivalent to a set of independent harmonic oscillators. That's a mathematical fact and nothing is ambiguous here.

Take as the most simple example non-interacting non-relativistic spin-0 bosons. Start with the single-particle case. Its hamiltonian is
$$\hat{H}=\frac{1}{2m} \hat{\vec{p}}^2.$$
There's a complete set of (generalized) momentum eigenstates which are also energy eigenstates.

To make things mathematically unproblematic start first with a regularization by introducing a finite cubic volume $[0,L]^3$ and consider wave functions with periodic boundary conditions, which still allows for self-adjoint momentum operators and thus the infinite-volume limit can be taken in the final results of physical significance.

Then you have a set of momentum-eigenfunctions (I use natural units with $\hbar=1$),
$$u_{\vec{p}}(\vec{x})=\frac{1}{L^{3/2}} \exp(\mathrm{i} \vec{p} \cdot \vec{x}), \quad \vec{p} \in \frac{2 \pi}{L} \mathbb{Z}^3.$$
The general solution of the Schrödinger equation is given by
$$\psi(t,\vec{x})=\sum_{\vec{p}} a(\vec{p}) u_{\vec{p}}(\vec{x}) \exp(-\mathrm{i} E(\vec{p}) t).$$
Now you want to quantize the field to describe a many-body system of non-interacting bosons.

Via the Lagrangian for the Schrödinger equation
$$\mathcal{L}=\mathrm{i} \psi^* \partial_t \psi + \frac{1}{2m} (\vec{\nabla} \psi^*) \cdot (\vec{\nabla} \psi)$$
you get the canonical field momentum to be $\psi^*$, and thus the bosonic equal-time field-commutation relations for the field operators (in the Heisenberg picture of time evolution)
$$[\hat{\psi}(t,\vec{x}_1),\hat{\psi}^{\dagger}(t,\vec{x}_2)]=\mathrm{i} \delta^{(3)}(\vec{x}_1-\vec{x}_2), \quad [\hat{\psi}(t,\vec{x}_1),\hat{\psi}^{\dagger}(t,\vec{x}_2)]=0.$$
The decomposition of the field operator in terms of momentum eigenmodes leads to
$$\hat{\psi}(t,\vec{x})=\sum_{\vec{p}} \hat{a}(\vec{p}) u_{\vec{p}}(\vec{x}) \exp(-\mathrm{i} E(\vec{p}) t).$$
The equal-time commutation relations lead to the commutation relations for the annihilation and creation operators
$$[\hat{a}(\vec{p}_1),\hat{a}(\vec{p}_2)]=0, \quad [\hat{a}(\vec{p}_1),\hat{a}^{\dagger}(\vec{p}_2)]=\delta_{\vec{p}_1,\vec{p}_2},$$
where this $\delta$ is a nice Kronecker-$\delta$ which makes life easier instead of the Dirac-$\delta$ you get for the infinite-volume limit.

This shows that the free QFT is equivalent to an infinite set of independent harmonic oscillators. You can also go on with the analysis of the Lagrangian to get the Hamiltonian for the many-body system (as well as momentum and angular momentum). The Hamiltonian reads (in terms of the annihilation and creation operators)
$$\hat{H}=\sum_{\vec{p}} E(\vec{p}) \hat{N}(\vec{p}), \quad \hat{N}=\hat{a}^{\dagger}(\vec{p}) \hat{a}(\vec{p}).$$
The Hilbert space of the many-particle system is given by the common eigenstates of the number operators $\hat{N}(\vec{p})$. There's a unique ground state, which is the eigenstate with eigenvalue $0$ for all $\hat{N}(\vec{p})$, called "the vacuum", $|\Omega \rangle$. All the harmonic oscillators are in their ground state, i.e., $\hat{a}(\vec{p}) |\Omega \rangle=0$ for all $\vec{p}$.

The general bosonic number-basis states are given by
$$|\{N(\vec{p}) \}_{\vec{p}} \rangle = \prod_{\vec{p}} \frac{1}{\sqrt{N(\vec{p})!}} \left [\hat{a}^{\dagger}(\vec{p}) \right]^{N(\vec{p})} |\Omega \rangle.$$
All such states are allowed with
$$N=\sum_{\vec{p}} N(\vec{p}) \quad \text{finite.}$$
This is an elegant formulation of the basis of bosonic Fock states, which are cumbersome to write in the "1st-quantization formalism" for a fixed number of particles since you'd have to take care of the symmetrization of the corresponding product states to take the indistinguishability of the bosons correctly into account. This is achieved in the QFT ("2nd quantization") formalism automatically due to the commutation relations of the creation operators.

The very same holds true for fermions (despite the fact that they have in addition a (half-integer) spin). The only difference is to assume canonical equal-time anticommutators leading also to anticommutators for the creation and annihilation operators in the mode decomposition of the quantum field.

 

[Dr_Nate]

Take as the most simple example non-interacting non-relativistic spin-0 bosons...

It doesn't seem like you included the coupling. But, even if you did, I know you could show similar looking equations. These equations would probably not explicitly show that there is a dispersion relation.

I think we diverge on what an uncoupled HO is defined to be. For me, it is something like: a single body in a harmonic potential with no other interactions. Having $N$ of these comes with some very specific notions of specific heat and no heat conduction. Thus, my beef is that if we expand this definition of uncoupled HO, rather than using the qualifier similar to, we are talking about very different physical objects when we use the same words.

 

[vanhees71]

For interacting fields the oscillators by definition are coupled, and you can't define creation and annihilation operators in the Heisenberg picture anymore. Except in special cases in low space-time dimensions there's no closed solution for the Heisenberg equations of motion anymore and you rely on the methods of perturbation theory in the interaction picture.