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Consider a nonrelativistic free fermion Hamiltonian given by $$H = \sum_{i} (\epsilon_i - \mu) \psi^\dagger_i \psi_i \ ,$$ where we have explicitly included the chemical potential $\mu$. The ground state is given by the "Fermi surface" $$|\text{GS}\rangle= \prod_{i|\epsilon_i < \mu} \psi^\dagger_i|0\rangle \ ,$$ and it is with respect to this nonempty ground state that correlation functions, e.g. the propagator $$\left<\text{GS} | T \psi(r,t) \psi^\dagger(r',t')|\text{GS}\right>$$ are computed.

Why do we never consider the same thing in relativistic QFT?

For example, a real scalar field Hamiltonian is given by $$H_0 = \int d^d r \left(\frac{m^2}{2}\phi^2 + \frac{1}{2}\pi^2 + \frac{1}{2}(\nabla\phi)^2\right)$$ with $\pi = \partial_t\phi$.

This is diagonalized by setting $$\phi \propto \sum_k a_k e^{i(kx-\omega t)} + a^\dagger_k e^{-i(kx-\omega t)}$$ with $\omega=\sqrt{k^2+m^2}$, and the ground state is given by the vacuum $|\text{GS}\rangle = |0\rangle$.

What if we explicitly add a chemical potential, i.e. $$H = H_0 - \sum_k \mu a^\dagger_k a_k \ .$$ I suppose this breaks Lorentz invariance? If so, how about instead considering a complex scalar field, and setting $$H=H_0 - \sum_k \mu (a_k^\dagger a_k - b_k^\dagger b_k) \ .$$ Is this Lorentz invariant?

Qmechanic
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Abhi Sarma
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4 Answers4

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Any energy cutoff will not be Lorentz-invariant.

Also, note that for fermions you can reinterpret your original annihilation operatprs of particles as creation operator of holes. I.e. $\psi_k=\tilde{\psi}_k^\dagger$. This way the ground state becomes vacuum annihilated by hole annihilation operators.

OON
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You are mistaking the conventional treatment that you are reading, with the reality.

It just so happens that there are few avenues whereby we require both SR and degenerate materials.

In particular, in a condensed matter textbook, they will need to have a ground state that treats the degenerate electrons, but the exposition will be much easier if we ignore the SR contributions. No need to be dealing with Dirac equation, although the corrections are sometimes necessary. Maybe a few orders worth of corrections would be way more than needed.

Whereas in a HEP textbook, the fundamental issue is the definition of the fields themselves become very tenuous, and so in SR QFT, the bulk of the exposition is to deal with renormalisation and other headaches. The important part is that even the definition of interacting particles in free space, vacuum, is already very very difficult, let alone in matter.

But that does not mean that the two cannot be combined. If you have to do precise calculations, then you need both. In particular, SR+degeneracy models are everywhere in astrophysics. You should just look those up.

naturallyInconsistent
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Just want to add two things not already mentioned in the answers above:

  1. In CMP the chemical potential is ultimately due to the background positive charges. It is physically justified.

  2. Scalar chemical potential is the relativistic mass. It is consistent with Lorentz symmetry.

Vokaylop
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TL; DR: Ground state is not a state with no particles, but a state with no excitations. In QFT particles are the excitations.

If we follow the discussion in most condensed matter books, then it deals with excitations/quasiparticles, rather then the actual electrons. Indeed, a conduction electron in solid state is not the same as a free electron - even without interactions it has dispersion determined by the band shape, and then it is further "dressed" by the interactions. Thus, the condensed matter really deals with particle and hole excitations, defined in respect to the ground state. E.g., in non-interacting case, if we define the ground states as the filled Fermi sphere: $$ |GS\rangle=\prod_{i, \epsilon_i \leq \mu}\psi_i|0\rangle $$ then the particle and hole operators are introduced as $$ \psi_i=\begin{cases}p_i,\text{ if } \epsilon_i>\mu\\ h_i^\dagger,\text{ if } \epsilon_i<\mu \end{cases} $$ and the Hamiltonian takes form $$ H=\sum_ip_i^\dagger p_i + \sum_i h_i^\dagger h_i $$ with the ground state defined by $$ p_i|GS\rangle=h_i|GS\rangle=0. $$

This is even more manifest when dealing with bosonic quasiparticles, such as phonons, magnons, plasmons, etc. For discussion of the interacting case see this answer and the books on QFT in many-body physics.

The real QFT treats the free particles themselves as the excitations, so naturally the vacuum states has none of them.

Related: Electrons and holes vs. Electrons and positrons

Roger V.
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