# Wigner functional for fermionic fields (QFT in phase space)

I'm curently studying the Wigner functional formulation of Quantum Field Theory, which is derived from the Schrödinger picture: the operators which act on the states of the Fock space are functions of the time independent fields $$\phi(\textbf{x})$$. In the position representation, each state $$|\Psi\rangle$$ is represented by a functional of the fields $$\Psi[\phi] = \langle\phi|\Psi\rangle$$, where $$|\phi\rangle$$ are the eigenstates of the field operators. These are realized as the kernels $$\hat{\phi}(\textbf{x})\rightarrow\delta[\overline{\phi}-\phi]\phi(\textbf{x})$$, and their conjugate momenta as $$\hat{\pi}(\textbf{x})\rightarrow \delta[\overline{\phi}-\phi]\frac{\delta}{\delta\phi(\textbf{x})}$$. With that in mind, the states are calculated from the time dependent Schrödinger equation:

$$i\frac{\partial}{\partial t}|\Psi(t)\rangle = H|\Psi(t)\rangle$$

For the phase space formulation of this theory, all the operators are written as functionals of the fields and their conjugate momenta $$\mathcal{O}[\phi,\pi]$$, and the state is reprsented b the Wigner functional $$W[\phi,\pi]$$, which obeys the Moyal equation:

$$\{\{W,H\}\} = W[\phi,\pi]\star H[\phi,\pi] - H[\phi,\pi]\star W[\phi,\pi] = 0$$

Where the Moyal star product is defined as:

$$A[\phi,\pi]\star B[\phi,\pi] = A[\phi,\pi] \exp\left(\frac{i\hbar}{2}\int d^{3}\textbf{x}\ \frac{\overleftarrow{\delta}}{\delta\phi(\textbf{x})}\frac{\overrightarrow{\delta}}{\delta\pi(\textbf{x})} - \frac{\overleftarrow{\delta}}{\delta\pi(\textbf{x})}\frac{\overrightarrow{\delta}}{\delta\phi(\textbf{x})} \right) B[\phi,\pi]$$

The Wigner functional can be obtained from the Shrödinger one by means of the Wigner transform:

$$W[\phi,\pi] = \int \mathcal{D}\eta \ \Psi^{*}\left[\phi-\frac{\hbar}{2}\eta\right]e^{-i\int d^3\textbf{x}\ \eta(\textbf{x})\pi(\textbf{x})}\Psi\left[\phi+\frac{\hbar}{2}\eta\right]$$

Now, everything works well for the scalar and elctromagnetic fields. Nevertheless, I've encountered several difficutes when I try to apply this formulation to the fermionic field. Firstly, fermionic field operators and their momenta anticommute, which means that their eigenvalues also have to: $$\{ \psi(\textbf{x}),\psi(\textbf{y}) \} = 0$$. This can be solved, however, by expressing the field functions $$\psi(\textbf{x})$$ a Grassmann variables and performing all the calculations taking that into account. My main problem is that the conjugate momentum of the fermionic field is its adjoint, $$\pi_{\psi}(\textbf{x}) = i\psi^{\dagger}(\textbf{x})$$. Therefore, the value of the field $$\psi$$ uniquely determines that of the momentum $$i\psi^{\dagger}$$. Thus, wouldn't a Wigner functional $$W[\psi,\psi^{\dagger}]$$ be only effectively dependent on the field $$\psi$$? How can you construct a phase space if he conjugate momentum is related to the field? Recall that this doesn't happen for the scalar or the electromagnetic field since, although the conjugate momentum is defined as $$\pi = \partial_0\phi$$ in the Heisenberg picture, the time independence of the operators in the Schrödinger picture eliminates this relation.

• Once $\psi(x)$ is no longer a field operator, but a Grassmann-valued function, it no longer holds that $i \psi(x)^\dagger = \pi(x)$ is related to $\psi(x)$ by complex conjugation of some sort. – Lorenz Mayer Mar 17 at 14:16
• I don't really understand. For example, in the case of Majorana fermions, $\hat{\pi}(x) = i\hat{\psi}(x)$. How would $\psi(x)$ being Grassmann variables make them independent? – Marcosko Mar 17 at 14:22

## 1 Answer

I think the confusion arises out of a misunderstanding of "complex conjugation" in the Grassmann formalism.

Grassmann "variables" are simply elements of an exterior algebra. If $$V$$ is a vector space, and $$\xi_1,...,\xi_n$$ some vectors, then $$\xi_1 \cdots \xi_n$$ is simply a way of writing the exterior product $$\xi_1 \wedge \cdots \wedge \xi_n$$.

What people call the "complex conjugate" of a Grassmann variable is usually the following: They take two sets of Grassmann variables, which they call $$\xi_1,...,\xi_n$$ and $$\overline{\xi}_1,...,\overline{\xi}_n$$. In terms of linear algebra, they have two vector spaces, which are called $$V$$, and $$\overline{V}$$, which have the same dimension $$n$$, and they choose a basis of both. Then define an anti-linear map $$J : V \rightarrow \overline{V}$$ as:

$$J(\xi_i) =\overline{\xi}_i \ .$$

Then one obtains what is called a real structure on $$V \oplus \overline{V}$$ as

$$\gamma = \begin{pmatrix} 0 & J^{-1} \\ J & 0 \end{pmatrix} \ .$$

Then one hides $$\gamma$$ by denoting its action by a bar: $$\gamma(v)=:\overline{v}$$.

This procedure is, to my knowledge, barely useful, except in making certain fermionic formulas look more like their bosonic analogues. In particular, the difference between Majorana and complex fermions, at the level of their Grassmann-variable formulation, isn't that one uses "real" and the other uses "complex" Grassmann variables. It is simply that for Majorana fermions one has one set of Grassmann variables while for complex fermions one has two - Better: one has one set which has a peculiar structure.

• I think I understand. So, you are saying that the complex conjugate map not being completely defined for Grassmann variables is what makes the eigenvalues of $\hat{\psi}$ and $\hat{\psi}^{\dagger}$ independent, right? – Marcosko Mar 18 at 21:36