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 Answer 1


Comments to the question (v2):

  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$^1$ $$\tag{5} f(\theta)~=~f(0)+\theta f^{\prime}(0) ~=~f(0)-(-1)^{|f|}f^{\prime}(0)\theta .$$

  12. For references and more information, see also this and this Phys.SE post.


$^1$ 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.

  • 1
    $\begingroup$ 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 $f(\eta) = \exp(-\eta a^\dagger)$. $\endgroup$ Nov 18, 2013 at 21:14
  • $\begingroup$ Thank you very much for your detailed answer. Yes, I realized that A&S have a lot of typos and errors in their formulae. $\endgroup$ Nov 21, 2013 at 0:00

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