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The following question is motivated by the construction of the fermionic path/field integral, as done for example in Altland & Simons "Condensed Matter Field Theory".

Consider the vector space $\mathbb C^2$ with two standard basis vectors named $|0\rangle$ and $|1\rangle$. Furthermore, consider the linear operator $a$ defined by

$$ a |0\rangle = 0 \text{ and } a |1\rangle = |0\rangle .$$

(This is the "annihilation" operator for a single fermion). Its hermitian conjugate $a^\dagger$ (the "creation" operator) is given by

$$ a^\dagger |0\rangle = |1\rangle \text{ and } a^\dagger |1\rangle = 0 .$$

Clearly, both operators $a$ and $a^\dagger$ are nilpotent and have the following Jordan normal form

$$ a, a^\dagger \simeq \begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}$$

In particular, these operators are not completely determined by their eigenvalues. (They don't commute or anticommute, though, we have $aa^\dagger + a^\dagger a = 1$.)


However, when constructing the fermionic field integral, physicists treat these operators as if they had useful eigenvalues! Namely, the eigenvalues are taken to be Grassmann-numbers, i.e. two "numbers" $\eta$ and $\bar\eta$ that anticommute with each other, $\eta \bar\eta = - \bar\eta \eta$, and that also anticommute with the operators $a$ and $a^\dagger$. Then, physicists construct the so-called "coherent state"

$$ |\eta\rangle := e^{-\eta a^\dagger} |0\rangle = (1 -\eta a^\dagger) |0\rangle $$

which behaves like an eigenvector for the annihilation operator

$$ a |\eta\rangle = \eta |\eta\rangle .$$

Together with the dual vector,

$$ \langle\eta| = \langle 0| e^{-a\bar\eta} $$

we can write the projection onto the corresponding "eigenspace" as $|\eta\rangle \langle\eta|$. These projections form a "complete set", as can be seen by summing/integrating over the Grassmann variables

$$ \int d\bar\eta d\eta\ e^{-\bar\eta \eta} |\eta\rangle \langle\eta| = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.$$


Now my question: how on earth does this make sense? The basic idea is to expand the available supply of "numbers" to obtain eigenvalues. This is a common idea and can be used to construct many familiar field extensions like $\mathbb R \subseteq \mathbb C$. After all, the imaginary unit $i$ is the eigenvalue of a 90° rotation in two dimensions. The problem here is Grassmann algebras can't be fields and we're no longer dealing with vector spaces.

How can elements of a Grassmann algebra be interpreted as eigenvalues of nilpotent operators?

I guess I'm looking for representations of the matrix algebra $\mathbb C^{n\times n}$ on Grassmann modules or something like that. Probably the "natural" representation on the tensor product $\mathbb C^n \otimes \Lambda \mathbb C^n$.

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    Maybe you should try and post this question on MathOverflow, since no one answered it here for two days.2011-06-20
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    The presence of several anti-commuting elements suggests to me that a suitable Clifford algebra (matching a quadratic form of your choice) might provide a useful framework. Have you tried that?2011-06-20
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    @Jyrki Lahtonen: I'm not sure, how would a Clifford algebra help me? The "problem" I see is that the new "numbers" are no longer a field and we are no longer dealing with vector spaces. You're saying that Clifford modules are nicer than modules for the exterior algebra? I'm not familiar with either one, I would be grateful for any references.2011-06-20
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    I'm not sure whether Clifford algebras help or not. They are a generalization of the exterior algebra. The difference coming from the fact in a Clifford algebra $x^2=Q(x)$, for a quadratic form $Q$. You get the exterior algebra when $Q(x)=0$ for all $x$. I thought that there may be some hope, because quite a bit is known about representation theory of Clifford algebras. But certainly many physicists are more conversant with Clifford algebras than I am, so may be there's nothing there. Start with http://en.wikipedia.org/wiki/Clifford_algebra2011-06-20
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    It seems to me, that description of that stuff, e.g., in "Gauge Fields, Introduction to Quantum Theory, Faddeev and Slavnov" (Translated 1980) is more "comfortable" in comparison with cited book. But, anyway, why "eigenvalues" anticommute with its own operators?2011-06-23

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