Condensed Matter – Why Schrödinger Equation Isn’t Used for Phonons

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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?

Best Answer

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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