Why do we not use the Schrödinger equation/standard QM to describe phonons (or other quasi particles)? 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?
 A: 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.).
A: 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.
