# Proof that eigenvalues of Fermionic creation/annihilation operators are Grassman numbers

It's stated probably in all textbooks on many-body functional integrals that operators that satisfy $$\hat{a}^\dagger \hat{a} + \hat{a} \hat{a}^\dagger = 1$$ must have eigenvalues that satisfy $$\psi \psi' + \psi' \psi = 0 ,$$ where $\psi$ and $\psi'$ are two eigenvalues of the $\hat{a}$ operator: $$\begin{array}{cc} \hat{a} \vert \psi \rangle = \psi \vert \psi \rangle , & \langle \psi \vert \hat{a}^\dagger = \langle \psi \vert \bar{\psi} . \end{array}$$

Seems intuitive, but how does one rigorously prove this?

• "It's stated probably in all textbooks.." Can you give some particular reference which states this ? – user10001 Nov 27 '13 at 15:41
• From the books at my hand at the moment at least Altland & Simons "Condensed matter field theory" and Peskin & Schröder "Introduction to QFT" do this. – Echows Nov 28 '13 at 9:43

Although these equations are correct, it is not quite correct to treat them as eigenvalue equations.

In quantum theory, both bosonic and fermionic operators act on a Hilbert space, thus have ordinary numerical matrix elements and eigenvalues. One of the possible realizations of the Hilbert space in the case of bosonic operators is as a Hilbert space of functions on some manifold. In this case, the creation and annihilation operators can be realized by means of combinations of multiplication and differential operators:

$$\hat{a}_b = \frac{1}{\sqrt{2}}(\hat{x} + i \hat{p}) =\frac{1}{\sqrt{2}}(x + \frac{d}{d x})$$

The analog in the case of fermions is to construct the Hilbert space as a space of functions on a supermanifold rather than over a manifold. This task is accomplished by:

1) Finding a representation by means of multiplication and differentiation operators on this manifold of the canonical fermionic algebra:

$$\{ a_f, a^{\dagger}_f \} = 1$$

2) Then a choice of an odd complex one dimensional manifold is sufficient for such a representation:

$$a_f = \psi$$

$$a^{\dagger}_f = \frac{d}{d\psi}$$

Where $\psi$ is a complex Grassmann variable.

A function on this manifold of the form

$$f(\psi) = e^{\bar{\psi}\psi}$$

is called a coherent state. It is correct to think of the exponential as a Taylor series. The Taylor series is truncated after the first term because $\psi$ is nilpotent because it is a coordinate of a supermanifold. It is easy to see using the rules of the Grassmann algebra.

$$a_f f(\psi) = \psi f(\psi)$$.

$$a^{\dagger}_f f(\psi) = \bar{\psi} f(\psi)$$

These are just the equations given in the question, but they should be interpreted as the action of the canonical operators on the coherent state.

It should be mentioned that this coherent state Hilbert state is just an ordinary Hilbert space expressed by means of functions over a supermanifold and having the inner product:

$$(g, f) = \int \overline{g(\psi) } f(\psi) e^{-\bar{\psi} \psi} d\bar{\psi} d\psi$$

This Hilbert space is just the 2 dimensional complex vector space generated by the vectors $1, \psi$. (The functions do not depend on $\bar{\psi}$, otherwise, there would have been null vectors and the space would not be Hilbert).

Edit

@user10001 as you noticed $a^{\dagger}_f$ has a well defined action on the Hilbert space spanned by $\{1, \psi \}$ as the differential operator $\frac{d}{d\psi}$.

To your second question I used a slightly abuse notation for the coherent state in order to get the same kind of the formulas as in the question, actually one should write it as: $e^{\zeta^{\dagger}\psi}$, where $\zeta$, is another Grassmann number not $\psi$'s complex conjugate. You can think of $\zeta$ as a Grassmann number coefficient. The only way you get $\psi^{\dagger}$ is in the complex conjugation in the inner product integral.

Still, in the answer's (abuse) notation everything is O.K. in the action of $a^{\dagger}_f$ on the coherent state:

$$a^{\dagger}_f e^{\psi^{\dagger}\psi} = \frac{d}{d\psi}(1+\psi^{\dagger}\psi) =\psi^{\dagger}. 1$$

This means that the answer is proportional to the vector $1$ in the Hilbert space with a Grassmann number coefficient $\psi^{\dagger}$.

• I'm not sure if I understand this explanation, but this is probably just my own ignorance since I don't know anything about supermanifolds. Can you recommend any sources for reading about this stuff? – Echows Nov 28 '13 at 9:39
• "The functions do not depend on $\bar{\psi}$" But since $a^{\dagger}_f 1 = \bar{\psi}$ ; so $\bar{\psi}$ should be linearly dependent upon 1 and $\psi$. However I can't see any way to write $\bar{\psi}$ in terms of 1 and $\psi$ using complex coefficients. – user10001 Nov 28 '13 at 12:37
• sorry, I used the rules you mentioned for nonconstant functions which don't apply to the function 1 :) However now the same problem remains if we use function $exp(\psi \bar{\psi})$ in place of 1. – user10001 Nov 28 '13 at 13:18
• @Echows I recommend to start from the following review arxiv.org/abs/math-ph/0202026v1 by Cartier, DeWitt-Morette, Ihl and Sämann – David Bar Moshe Nov 28 '13 at 13:38
• @user10001 I am answering your questions in a separate edit – David Bar Moshe Nov 28 '13 at 13:56