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We were just taught about phonons in solid state physics class. Last year we did QM and now we are starting QFT as well.

Phonons are excitations of a condensed matter field. I thought that we would be using the Schrödinger equation (SE), but no. It is also nowhere in the syllabus. I read questions on this site about using the Schrödinger equation for quantum fields and many answers say that the SE describes particles without any spin, but the phonon is spin zero. Why can we not use the SE or normal (no relativity) QM and for phonons and if I'm wrong and you can, then how?

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    $\begingroup$ The SE for the zillions of coupled oscillators involved is monstrously unwieldy. Instead, the manifestly superior Dirac creation-annihilated solution is applied. $\endgroup$
    – Cosmas Zachos
    Commented Mar 2, 2022 at 3:38
  • $\begingroup$ @CosmasZachos Answers should be posted as answers. It seems like you have important information here $\endgroup$
    – BioPhysicist
    Commented Mar 2, 2022 at 13:30
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    $\begingroup$ We could, but it is not very practical. A good excercise is to solve the SE for a chain of atoms connected by springs. $\endgroup$
    – Roger V.
    Commented Mar 2, 2022 at 14:41
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    $\begingroup$ @BioPhysicist . I don't know that it is important. I am merely reminding the OP, ostensibly taking QFT, why he was definitely told the Fock space formulation is being used instead of the impossible infinity of SEs, a nightmare to repackage and decouple into normal modes. If he only read the drill, he'd instantly get the point, no? He claims he took elementary QM. Ideally, I'm trying to see what he failed to see, and how, or retract the question. $\endgroup$
    – Cosmas Zachos
    Commented Mar 2, 2022 at 15:01
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    $\begingroup$ ...ideally, they've done a 2D oscillator, and two coupled oscillators in his QFT intro, in which case the point is self-explanatory... $\endgroup$
    – Cosmas Zachos
    Commented Mar 2, 2022 at 15:07

2 Answers 2

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In condensed matter physics, or any field theory where particle numbers are not conserved (like particle physics), the Schrodinger equation does not work. The Schrodinger equation needs the condition that particle numbers are constant.

In standard quantum mechanics, where we do use the Schrodinger equation, in it we have the wave function $\psi(x)$ and we encounter the term $\rho=\psi^*(x)\psi(x)$ which represents the probability (density) of finding the particle at $x$ and when we integrate $\rho$ over all space, we require it to equal unity, a constant. That is, $$\int \psi^*(x)\psi(x)dx=1$$ which is called the normalization condition and is the result of requiring the particle to be located somewhere in the space, which is obviously a fair assumption. That is, in QM and by extension the Schrodinger equation, we require this condition.

In we then go to a theory where particles numbers are not conserved, or where particles are annihilated or created, the requirement in the above equation no longer makes sense. For example, if a particle were to suddenly annihilate or "disappear" the probability to find it just before it disappears will be one, and zero thereafter. So we cannot use the Schrodinger equation to describe phonons, or indeed any field theory where particles are created and destroyed.

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    $\begingroup$ well… technically the probability density $\rho=\psi(x,t)^*\psi(x,t)$ can vary with $t$ is $\psi(x,t)$ is not an eigenstate of $H$, so the fact that $\rho$ varies in $t$ does imply conservation or non-conservation.. The integral of $\rho$ over space (rather than $\rho$ itself) is a conserved quantity. $\endgroup$
    – ZeroTheHero
    Commented Mar 2, 2022 at 4:02
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    $\begingroup$ @ZeroTheHero You're absolutely right. Edited! Thanks. $\endgroup$
    – joseph h
    Commented Mar 2, 2022 at 4:06
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    $\begingroup$ I might be missing something, but normalization of the wave function and the SE are two different things. Perhaps this answer could better link the two beyond just saying we encounter them together. $\endgroup$
    – BioPhysicist
    Commented Mar 2, 2022 at 13:28
  • $\begingroup$ Thanks joseph h. Is there a way to formulate the schrodinger equation to describe phonons with just relativity but not include changing particle numbers? $\endgroup$
    – gemmima
    Commented Mar 4, 2022 at 1:22
  • $\begingroup$ I think davidhigh's answer and Cosmos Zachos's comment are correct, and this answer appears to be wrong. The $d^3x$ integral isn't correct even for systems of fixed particle count unless the particle count is $1$ and it has no other properties like spin. The integral is over the phase space, whatever shape that happens to be. It can be a Fock space. $\endgroup$
    – benrg
    Commented Mar 4, 2022 at 1:54
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I'm arguing here in a different way than the answer by @josephh. In my opinion, you can definitely apply the Schrödinger equation -- resp. a probably differently named version of it -- to phonons as well, but you have to extend the framework.

In the standard setting you're basing the whole formalism on a Hilbert space $\mathcal H_N$ with a definite particle number $N$, and by solving the SE you're searching for a solution $|\Psi_N\rangle \in \mathcal H_N$.

In the case of phonons, instead you have to use the Fock space $\mathcal F$, which is the direct sum of all Hilbert spaces with varying particle number $N$. In this space, you can define the physical processes of creation and annihilation in the sense that a wavefunction goes from subspace $\mathcal H_N$ to $\mathcal H_{N\pm1}$.

In practice, however, this detailed description is usually to complex, and so, as mentioned in the other answer, the density matrix is commonly used. It doesn't care about the microstate of your system, but rather gives the probability (density) of finding the particle at a point in space $x$.

By the way: In the same manner I'd also argue that you can apply the Schrödinger equation to a system with spin $\neq 0$ (even if it's called differently). The idea is similar: e.g. for electrons, instead of using the Hilbert space $\mathcal H$, one uses an extended (Hilbert) space $\mathcal H \times \{\uparrow,\downarrow\}$, where $\uparrow,\downarrow$ are spin-states of a spin-1/2 particle, and applies the (form of the) Schrödinger equation again, with a specific Hamiltonian. The particular form of the Hamiltonian then basically determines the kind of equation and its name (e.g. Pauli equation, etc.).

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    $\begingroup$ Yes, this is a sensible proposition, the SE can definitely be used, just requires some non-run-of-the-mill workout. $\endgroup$
    – Mahir Lokvancic
    Commented Mar 2, 2022 at 19:29
  • $\begingroup$ Once E. Fermy answered that QFT is QM in terms of occupation numbers. $\endgroup$
    – Vladimir Kalitvianski
    Commented Mar 3, 2022 at 9:29
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    $\begingroup$ Thanks but that would no longer be standard Schrodinger equation/qm which was what I was asking about. $\endgroup$
    – gemmima
    Commented Mar 4, 2022 at 1:19
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    $\begingroup$ @gemmima I know the answer says "you have to extend the framework", but this is standard QM. The wave function is defined over a phase space, which generally doesn't have the shape $\mathbb R^n$, even in undergraduate QM. $\endgroup$
    – benrg
    Commented Mar 4, 2022 at 2:15
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    $\begingroup$ @benrg: yup, that's standard QM, thanks for pointing. "Extend the framework" was meant relatively to the context of the OP. $\endgroup$
    – davidhigh
    Commented Mar 4, 2022 at 7:40

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