# What do the modes of fermion fields look like?

A boson field can be understood as a collection of stationary modes (e.g. plane waves of various polarizations), and for each mode there is a quantum harmonic oscillator. If the QHO for some mode is in an energy eigenstate $|n\rangle$, we interpret this as meaning that the field constains $n$ particles in that mode. Since a QHO has infinitely many energy eigenstates, a boson field can contain arbitrarily many particles in any mode.

Can a fermion field also be understood as a collection of modes? What do the individual modes look like? Because of the Pauli exclusion principle, I presume that each mode is a two-state system with an empty and a filled state. Are these any different from other two-state quantum systems, e.g. a single electron spin?

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The simplest way to show the difference between bosonic and fermionic fields is to consider the fourier expansions of these each.

Following the notation provided in Diagrammatica the most basic bosonic field (a scalar field) has the expression of:

$$\phi(x) = \sum_{\vec{k}} \dfrac{1}{\sqrt{2k_0V}}\{a(\vec{k})e^{i(kx)} + a^{\dagger}(\vec{k})e^{-i(kx)}\}$$ $$k_0 = \sqrt{\vec{k}^2+m^2}$$

where $a(\vec{k})$ is the annihilation operator and $a^{\dagger}(\vec{k})$ is the creation operator.

We enforce locality by stating that two fields $\phi(x)$ and $\phi(y)$ commute if $x$ and $y$ are outside each other's light cones.

When we consider fermionic fields, we have to think in terms of complex fields. Here the field is expanded to have four components. The most basic fermionic field can be written as:

$$\psi(x) = \dfrac{1}{\sqrt{V}} \sum_{\vec{k}} \{a^1(\vec{k})u^1(\vec{k})e^{i(kx)} + a^2(\vec{k})u^2(\vec{k})e^{i(kx)}$$ $$+ b^{3\dagger}(\vec{k})u^3(\vec{k})e^{-i(kx)}+b^{4\dagger}(\vec{k})u^4(\vec{k})e^{-i(kx)}\}$$

and

$$k_0 = \sqrt{\vec{k}^2+m^2}$$

Here

1. $a^1(\vec{k})$ is the annihilation operator of a spin up particle with momentum $\vec{k}$
2. $a^2(\vec{k})$ is the annihilation operator of a spin down particle with momentum $\vec{k}$
3. $b^{3\dagger}(\vec{k})$ is the creation operator of a spin down particle with momentum $\vec{k}$
4. $b^{4\dagger}(\vec{k})$ is the creation operator of a spin up particle with momentum $\vec{k}$

So the application of the annihilation operator to a state with spin up particle with momentum $\vec{k}$ would look like:

$$a^1(\vec{k})|\vec{k},\uparrow \rangle = |0\rangle$$

and the application of the creation operator to the ground state to create a spin down particle with momentum $\vec{k}$ is:

$$b^{3\dagger}(\vec{k})|0 \rangle = |\vec{k},\downarrow\rangle$$

The important addition here is that the commutation rule between the complex fields no longer strictly holds. However, this is circumvented by the definition of the anticommutator.

In the fermionic case, it is the anticommutator that is zero for $\psi_{\alpha}(x)$ and $\psi_{\beta}(y)$ when $x$ and $y$ are outside each other's light cones. The result is that the commutator of the related Hamiltonians $\mathcal{H}(x)$ and $\mathcal{H}(y)$ does equal zero when $x$ and $y$ are outside each other's light cones.

As far as the labeling of kets, the change is that it is no longer possible to simply label one state with $n$ particles $|n\rangle$. One has to keep track of the spins and momentum individually, so if one has some number of particles you would track each one as $|p_1, p_2,\dots,p_n\rangle$

Hope this helps.

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I think you missed an $e$ in the last term of your first equation. – David Z Mar 10 '13 at 4:29
@DavidZaslavsky thanks – user11547 Mar 10 '13 at 4:31