# Grassmann numbers as eigenvalues of nilpotent operators?

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. – Beni Bogosel Jun 20 '11 at 14:46
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? – Jyrki Lahtonen Jun 20 '11 at 15:32
@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. – Greg Graviton Jun 20 '11 at 17:07
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 en.wikipedia.org/wiki/Clifford_algebra – Jyrki Lahtonen Jun 20 '11 at 18:53
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? – Alex 'qubeat' Jun 23 '11 at 10:13

Let me just retell a part of the story in slightly different language — maybe it will become less mysterious then.

The story starts with a module $M=\mathbb C[\xi]$ (free supercommutative algebra with one odd generator — aka exterior algebra on one generator) over a Clifford algebra $A=\mathbb C[\xi,\frac\partial{\partial\xi}]$ (“odd differential operators” — i.e. the only relation is $[\frac\partial{\partial\xi},\xi]_+=1$). “Physical notation”: $|0\rangle=1\in M$, $|1\rangle=\xi\in M$, $a=\frac\partial{\partial\xi}\in A$, $a^\dagger=\xi\in A$.

One can extend scalars: instead of an algebra $A$ over (the field of) complex numbers we get an algebra $\tilde A=A[\eta,\bar\eta]=R[\xi,\frac\partial{\partial\xi}]$ over (the ring) $R=\mathbb C[\eta,\bar\eta]$; and module $M$ extends to $A[\eta,\bar\eta]$-module $\tilde M=M[\eta,\bar\eta]=R[\xi]$ (cf. Koszul resolution). Over our base ring, $R$, it's a free module of dimension 2.

Now $\frac{\partial}{\partial\xi}$ has an eigenvector, $\left| \eta\rangle \right.=exp(-\xi\eta)=1-\xi\eta$ with eigienvalue $\eta$.

(Nothing too suprising here: everybody knows that $\exp(\lambda x)$ is an eigenvector of $\frac\partial{\partial x}$. One might ask, why not just use $\exp(\xi)$ then — but an easy calculation show that it's not an eigenvector of $\frac\partial{\partial\xi}$ — because, in fact, we need the argument of exponent have to be even, not odd. That's why we need to extend scalars, I believe — $\xi\eta$ is an even element, and exponentiation works fine.)

But there is still no way to extend it to an eigenbasis. The problem is, we can't divide by $\eta$ (or two exponents — say, $\left|\eta\rangle \right.$ and $\left|\bar\eta\rangle \right.$ — would generate everything). So for any free $\mathbb C[\eta,\bar\eta]$ let's define (Berezin) integral by $\int d\eta d\bar\eta\,(e+f\eta+g\bar\eta+h\eta\bar\eta)=h$. Now $\left|\eta\rangle \right.$ generates our $\tilde M$ if we allow not only summation, but also this “integration”: \begin{align} 1 &=\pm\int d\eta d\bar\eta\,\eta\bar\eta|\eta\rangle;\\ \xi&=\pm\int d\eta d\bar\eta\, \bar\eta|\eta\rangle. \end{align} In other words, \begin{align} \left|0\rangle\langle0\right|&=\pm\int d\eta d\bar\eta\,\bar\eta\eta|\eta\rangle;\\ \left|1\rangle\langle1\right|&=\pm\int d\eta d\bar\eta\,|\eta\rangle\bar\eta\frac{\partial}{\partial\xi}. \end{align} Hence $$\operatorname{Id}= \left|0\rangle\right. \left. \langle0\right|+\left|1\rangle\langle1\right|= \int d\eta d\bar\eta(1\pm\eta\bar\eta)|\eta\rangle\langle\eta|.$$

// Note, in particular, that (unlike $|0\rangle\langle0|$ or $|1\rangle\langle1|$), $|\eta\rangle\langle\eta|$ is not really a projector on anything — (it's unipotent, so) it's an automorphism, actually!

(Oh, I appologize for mixing “mathematical” and “physical” notation freely.)

(So, as for the question in gray from the OP, it's not really a problem: much of linear algebra can be done not only in vector spaces over fields, but also in module over (super)commutative rings — and it's useful to consider not only field extensions but also ring extensions sometimes.)

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@Alex No-no, for even variables $[\frac{\partial}{\partial x},x]=1$, but for odd variables $[\frac{\partial}{\partial \xi},\xi]_+=1$ (I should write it explicilty in the answer, perhaps, thanks). – Grigory M Jun 22 '11 at 9:56
I deleted the comment before read the answer – Alex 'qubeat' Jun 22 '11 at 10:00
@Gigrory M: Ok, I can understand the extension procedure. But could you expand on the structure of $\tilde M$ as an $R[\frac{\partial}{\partial\xi}]$ module? I believe it's not a free $R[\frac{\partial}{\partial\xi}]$ module, otherwise we wouldn't need this fishy "integral". – Greg Graviton Jun 23 '11 at 7:39
@Grigory M: Thanks a lot for the update, now I have a way to talk about this properly. :-) Unfortunately, however, it seems that the structure of $\bar M$ is not that simple. It is not the case that $|\eta\rangle = 1-\eta\xi$ and $|\bar\eta\rangle := 1-\bar\eta\xi$ span $\bar M$ as an $R$-module; it is impossible to find a linear combination $\xi = |1\rangle = r_1|\eta\rangle + r_2|\bar\eta\rangle$. Somehow, the "completeness relation" with the integral is supposed to be a substitute for this, but I don't see how. – Greg Graviton Jun 23 '11 at 13:35
@Grigory M: You seem to have done this before, do you have any references on supercommutative algebras that might help me with this question? – Greg Graviton Jun 23 '11 at 13:37

Let us consider exterior (Grassmann) algebra $\Lambda\mathbb R^{2n}$ with $2n$ generators $v_k$, $k=1, \ldots, 2n$. It is $2^{2n}$-dimensional real algebra. It is possible to introduce operators (1, p.15) $M_k$ and $\delta_k$, $k=1, \ldots, 2n$, where $M_k(1) = v_k$, $M_k(\omega) = v_k \wedge \omega$ and $\delta_k$ is adjoint of $M_k$ (see 1, p.15 for more, except, I am using notation $M_k$ and $\delta_k$ instead of $\delta_v$ and $M_v$ for basic vector $v=v_k$ and Euclidean norm instead of arbitrary $B$).

Now we have representation of Clifford algebra with $2n$ generators $e_k = M_k - \delta_k$. In the (1) $M_k$ and $\delta_k$ were called creation and annihilation operators, but it is better to use here a complexification with operators: $a_k = (e_{2k-1}+i e_{2k})/2$ and $a_k^\dagger = (e_{2k-1}-i e_{2k})/2$ and to introduce variables $\eta_k = v_{2k-1}+i v_{2k}$ and $\bar{\eta}_k = v_{2k-1}-i v_{2k}$, $k=1, \ldots, n$. So we may talk about $\Lambda\mathbb C^n$ now.

An exterior algebra is a linear space. An element of the space were denoted as $| \eta \rangle$. An element of dual space is $\langle \eta |$. Both linear spaces have real dimensions $2^{2n}$.

Equations like $a_k | \eta \rangle = \eta_k | \eta \rangle$, $k=1, \ldots, n$ simply mean that $| \eta \rangle$ may not contain
$\eta_k$, i.e. it is element of $2^n$-dimensional exterior algebra $\bar\Lambda\mathbb C^n$ with generators $\bar{\eta}_k$. Comparison with equation on eigenvalues may be a misleading. I think it is better to use more formal presentation, like (2).

1) J. E. Gilbert and M. A. M. Murray, Clifford algebras and Dirac operators in harmonic analysis, (CUP, 1991).

2) L. D. Faddeev and A. A. Slavnov, Gauge fields: Introduction to quantum theory, (Benjamin, 1980, 1982).

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@Alex 'qubeat'. Wait, I think you're mixing up two things: the Fock space for the operators $a,a^\dagger$ and the new Grassmann numbers $\eta,\bar\eta$, which exist on top of that. So, it's a "Grassmanified Fock space". (For instance, Grigory uses $\xi$ to denote the state $|1\rangle$. We have $|\eta\rangle = 1-\eta\xi$) – Greg Graviton Jun 24 '11 at 18:53
I doubt. But Fock space certainly presents there. Anyway, $a$ and $\eta$ are formally different things in my models. – Alex 'qubeat' Jun 24 '11 at 20:19
@Greg Graviton: An example of the construction: $|\eta\rangle = c_{00} + c_{01}\eta + c_{10}\bar{\eta}+c_{11}\eta\bar{\eta}$, $a|\eta\rangle = c_{00}\eta + c_{01} + c_{10}\eta\bar{\eta}+c_{11}\bar{\eta}$, $\eta|\eta\rangle = c_{00}\eta + c_{10}\eta\bar{\eta}$, so from $a|\eta\rangle = \eta|\eta\rangle$ follows $|\eta\rangle = c_{00} + c_{10}\bar{\eta}$. – Alex 'qubeat' Jun 25 '11 at 12:45
@Alex 'qubeat'. The intention was that $a$ does not act on $\eta$, but on something else, namely $\xi$. There are two anticommuting things here: the Fock state $\xi$ and the coefficients $\eta,\bar\eta$. – Greg Graviton Jun 27 '11 at 6:30
@Greg Graviton: $a$ acts on $|\eta\rangle$, not on $\eta$, but there is standard map $\mathbb C^n \to \Lambda^1 \mathbb C^n$. – Alex 'qubeat' Jun 27 '11 at 9:58