# Different quantum fields for different particles?

I started my studies about quantum field theory but I have some problem to understand the whole concept. I have the following questions:

1. The Lagrangian contains a dirac term for fermions and a maxwell term for gauge bosons. But does this mean that there is only one dirac field for all fermions? I guess this is not the case but I don't understand why we do not sum over the masses and other quantum numbers in the Lagrangian to distinguish the fields?

2. We interpret particles as excitations of the field. Should we assume that there is one "field" for each particle or do we create for example three electrons into a common field? But what about the pauli exclusion principle in this case? Is it impossible to have two fermions of the same kind (and same quantum numbers) in the whole universe (the occupation number of fermions is 0 or 1)?

• Right now you're probably just studying QED, with only the photon and the electron. In general, there's a quantum field for every type of particle, and the full Standard Model Lagrangian has a ton of terms. – knzhou Apr 27 '17 at 19:53
• You can have two electrons with the same quantum numbers in different places, and also an electron and a muon with the exact same everything. What you can't have is two electrons in the same place with the same quantum numbers. – knzhou Apr 27 '17 at 19:54
• But if I describe the electron in the momentum space with precise momentum then it is delocalized in the position space, so the momentum of two electrons can not be the same? But why do we not distinguish between electrons and muons in our Lagrange density? – Core2016 Apr 27 '17 at 19:58

## 1 Answer

1. In case you have different types of elementary particles, you must add different fields in the Lagrangian. You will have multiple Dirac terms like this: $$\mathcal{L} = \bar{\psi}_1 (i \gamma^{\mu} \partial_{\mu} - m_1) \psi_1 + \bar{\psi}_2 (i \gamma^{\mu} \partial_{\mu} - m_2) \psi_2 + \dots.$$

2. This only is true for different types of elementary particles (e.g. electron, muon, tau). But the theory of the quantum electron field is capable of describing multiple electrons on its own, no extra fields are needed.

3. Look up "Fock space". For bosons, it consists of symmetrized tensor products of 1-particle states: $$\mathcal{H} = \mathcal{H}_0 \oplus \mathcal{H}_1 \oplus \mathcal{H}_2 \oplus \dots,$$ where $$\mathcal{H}_n = \text{Sym}\,\left[ \mathcal{H}_1 \otimes \dots \otimes \mathcal{H}_1 \right].$$ Let me explain the notation. $\mathcal{H}_1$ stands for the 1-particle Hilbert space (space of particle wavefunctions, etc), whereas $\text{Sym}$ is symmetrized tensor product. Symmetrization is a direct consequence of particle interchangeability: there's no "first" and "second" particle, both are equivalent.

4. For fermions (e.g. electrons) we use antisymmetrization instead: $$\mathcal{H}_n = \text{Assym} \, \left[ \mathcal{H}_1 \otimes \dots \otimes \mathcal{H}_1 \right].$$ Interchange between two fermions changes the sign of the wavefunction. This is, of course, unobservable directly (only projective classes of wavefunctions matter in QM), so fermions are also interchangeable. But it reflects the Fermi-Dirac statistics.

5. Note also that the Pauli principle follows directly from fermionic Fock space. E.g. for two fermions, $$\mathcal{H}_2 = \text{Assym}\, \left[ \mathcal{H}_1 \otimes \mathcal{H}_1 \right].$$ Its elements are wavefunctions $$\Psi(x_1, x_2)$$ of two arguments (generalized arguments denote both continuous momenta quantum numbers and discrete internal spin quantum numbers) satisfying $$\Psi(x_1, x_2) = - \Psi(x_2, x_1).$$ This means that $$\Psi(x_1, x_2) = 0 \quad \text{when} \quad x_1 = x_2,$$ which is precisely the exclusion principle in the mathematical form.

• Thanks for your answer. It clearified most of my problems. Should I think of two interacting electrons (or also free electrons) with the same quantum number at the same position that one occupies the state n and the other occupies the state m in this fock space? – Core2016 Apr 28 '17 at 16:07
• @Core2016 $\mathcal{H}_n$ part of $\mathcal{H}$ corresponds to the subspace of states of the field with $n$ particles in it. So if you have 4 electrons, then your state lies in $\mathcal{H}_4$. If, on the other hand, your state lies in $\mathcal{H}_3 \oplus \mathcal{H}_4$, then your field is in superposition of having 3 and 4 particles. – Prof. Legolasov Apr 29 '17 at 0:23
• Ok. But then it is impossible to have two or more electrons with the same momentum (in the same quantum state) all over the universe? This sounds ab bit strange to me. – Core2016 Apr 29 '17 at 11:35
• @Core2016 yes indeed, it is the Pauli exclusion principle and a direct consequence of Fermi quantization. This shouldn't seem too strange. Remember that in the real world particles can never have definite momenta, they are always smeared in the momentum space. We can loosely say that collisions occur with probability zero. However in the atom, for example, momentum states are discrete, and exclusion principle is directly observable. – Prof. Legolasov Apr 29 '17 at 16:17
• Thanks that helped me a lot. What do you mean with "that collisions occur with probability zero"? With this in mind could we say that the electrons in an atom are not part of the quantum field since the levels are discret and for all atmos of the same kind the same? – Core2016 Apr 29 '17 at 17:24