# A few simple questions about Grassmann numbers: commutation relations and derivatives

I'm trying to learn about Grassmann numbers from the book "Condensed Matter Field Theory" by Altland and Simons, but I am currently encountering some difficulties. I have several smaller questions which are connected.

• Does a Grassmann number $\eta_i$ and a creation operator $a^\dagger_j$ commute or anti-commute? Altland and Simons only defined $[\eta_i, a_j]_+ =0$. I would suggest to take the conjugate of this and arrive at an equation for the creation operator, but the authors refused to define a conjugation for Grassmann numbers as it can lead to some difficulties.

• The authors defined functions of Grassmann numbers by its Taylor expansion: $f(\eta) = f(0)+f'(0)\eta$. But how do I get the equation $e^{-\eta a^\dagger} = 1-\eta a^\dagger$ which I encountered in several notes? The derivative of $e^{-x a^\dagger}$ is $-a^\dagger e^{-x a^\dagger}$. So I would guess that $e^{-\eta a^\dagger} = 1 - a^\dagger \eta$. One solution would be that a creation operator commutes with a Grassmann number, but I am not sure if this holds.

• Can anyone recommend a good introduction to Grassmann numbers and their use in field theory which does not delve too deep into the mathematics but concentrates on performing calculations?

1. When working with superobjects (both supernumbers and superoperators), we normally assume that they have definite Grassmann-parity.

2. The Grassmann-parity $|\hat{A}|$ of a Grassmann-even (Grassmann-odd) superoperator is 0 (1) modulo 2, respectively.

3. A supernumber $z$ can be viewed as special case of a superoperator $\hat{A}$ in the same way an ordinary number can be viewed as a operator.

4. The supercommutator of two superoperators $\hat{A}$ and $\hat{B}$ is defined as $$[\hat{A},\hat{B}]~:=~\hat{A}\hat{B}-(-1)^{|\hat{A}||\hat{B}|}\hat{B}\hat{A} .$$

5. We say that two superoperators $\hat{A}$ and $\hat{B}$ supercommute if their supercommutator vanishes $[\hat{A},\hat{B}]=0$.

6. A supernumber $z$ supercommutes with any superoperator $\hat{A}$ that doesn't depend on $z$.

7. Point 6 implies that a Grassmann-odd supernumber $\theta$ and a Grassmann-odd superoperator $\hat{A}$ (that doesn't depend on $\theta$) have a truncated Taylor expansion: $$\tag{1} \exp(\theta\hat{A})= 1 + \theta\hat{A}+ \frac{1}{2}(\theta\hat{A})^2+ \ldots = 1 +\theta\hat{A}.$$

8. A superfunction $$\tag{2} f(\theta)~=~\theta a + b~=~ (-1)^{|a|} a\theta + b$$ in one Grassmann-odd supernumber $\theta$ must necessarily be an affine function.

9. Left differentiation is defined by $$\tag{3} f^{\prime}(\theta)~:=~\frac{d}{d\theta}f(\theta)~:=~a~=~f^{\prime}(0).$$ The word left here means that the differentiation operator $\frac{d}{d\theta}$ acts on the function (2) from the left.

10. One can also define a corresponding right differentiation $$\tag{4}\frac{d^{R}f(\theta)}{d\theta}~:=~(-1)^{|a|} a ~=~-(-1)^{|f|} f^{\prime}(\theta),$$ but the sign factor is less natural to work with in practical calculations.

11. Thus we may write equation (2) as a Taylor expansion$^1$ $$\tag{5} f(\theta)~=~f(0)+\theta f^{\prime}(0) ~=~f(0)-(-1)^{|f|}f^{\prime}(0)\theta .$$

$^1$ Note that Altland and Simons (A&S), Condensed Matter Field Theory, (2010) p.162, has a crucial sign mistake in their analogue of formula (5). It is clear from a formula in the text between eqs. (4.14-15), that A&S are using left (as opposed to right) differentiation.
• Ok, so I guess that Grassmann numbers $\eta$ are supernumbers with parity 1 and (fermionic) creation/annihilation operators are superoperators with parity 1. So, according to 5) a Grassmann number and a creation operator supercommute (in this case, anticommute). I can see how to arrive at the equation $\exp(-\eta a^\dagger) = 1- \eta a^\dagger$ with your Taylor expansion (because $-\eta$ is also a Grassmann number), but I would like to use the definition from Altland & Simons: $f(\eta) = f(0) + f'(0) \eta$ with $f(\eta) = \exp(-\eta a^\dagger)$. – hauntergeist Nov 18 '13 at 21:14