# Grassmann numbers

I am reading Zee's Quantum Field Theory in a Nutshell and am having some questions about Grassmann numbers. Let $$x,y$$ be Grassmann numbers. I think I have two relations:

$$e^{x+y}=1+x+y\tag{1}$$ and $$e^{x+y}=e^xe^y=(1+x)(1+y)=1+x+y+xy,\tag{2}$$

Which is wrong and why? In fact, my question originates from the derivation of the formula:

$$\int dx\int dy\,e^{yAx}=\det A\tag{3}$$

where $$x=(x_1,x_2,\dots,x_N),y=(y_1,y_2,\dots,y_N)$$ both are $$N$$ Grassmann numbers and $$A$$ is an antisymmetric $$N$$ by $$N$$ matrix. For, if $$(1)$$ holds, it seems that $$\int dx\int dy\,e^{yAx}=\int dx\int dy\exp\left(\sum_{ij}y_iA_{ij}x_j\right)=\int dx\int dy\left(1+\sum_{ij}y_iA_{ij}x_j\right)=0\,,\tag{4}$$ which is, of course, wrong. So, where is the problem?

• Does the first equality in (2) hold when $x$ and $y$ don't commute? Do they commute here? Dec 14, 2022 at 10:20
• They are Grassmann numbers and therefore anticommute. Dec 14, 2022 at 10:21
• @rioiong yeah he's just saying $e^{x+y} \neq e^xe^y$ if x and y don't commute Dec 14, 2022 at 11:25
• Yes, he is right and I have found the problem in the derivation of (4). Dec 14, 2022 at 11:40

The power series is the definition of exponentiation.

$$e^{x+y}=e^xe^y$$ is a property of that power series that can be derived for the special case when $$x$$ and $$y$$ commute.

• For the generalized version of $e^{x+y}=e^xe^y$ see the Baker-Campbell-Hausdorff formula en.wikipedia.org/wiki/… Dec 14, 2022 at 10:34
• Watch Leonard Susskind's youtube lecture on "Supersymmetry and grand unification", end of lecture 4, where he discovers after a comment from the audience that with Grassmann numbers, $e^{x+y}=e^xe^y$ doesn't work. His surprised reaction is priceless. But he rightly notes that the reason why he never noticed this is that it's not really important. Aug 22 at 7:17
• @NadavHar'El Thanks for this! But I think this fact about exponentials is important (so you dont end up computing the wrong thing). It is true for non commuting QM operators too. Susskind probably just had a momentary brain-fart. There's no way he had never used this fact about exponentials. Aug 22 at 17:08

It seems that the problem is here. For Grassmann numbers $$x,y,a,b,c,d$$, we have $$e^{x+y}=1+x+y$$but$$e^{ab+cd}=1+ab+cd+\frac{1}{2}(2abcd)\neq 1+ab+cd$$ Therefore, $$(1)$$ is correct but the derivation in $$(4)$$ is wrong.

• I am struggling with the same question as OP, but I don't understand how you obtained the correct expression for $e^{ab + cd}$. Would you be able to elaborate or point me in a direction to learn more? Thank you :) Feb 24 at 17:13
• I think I realized the answer just after commenting; it follows just from the power series definition of the exponential function. I was thinking that the term $(ab + cd)^2$ had to vanish on account of the constituents being Grassmann numbers... but I suppose $ab$ and $cd$ aren't, actually, Grassmann numbers? Feb 24 at 17:16
• No, $ab$ and $cd$ are not Grassmann numbers. Feb 25 at 9:36
• Wow, I didn't realize this (that if a and b are grassman numbers, ab isn't). It turns out that although individual grassman numbers anti-commute, products of an even number of grassman numbers - like ab and cd actually commute as can be seen by moving elements around from $abcd$ to $cdab$ and watching the sign flip an even number of times. Aug 22 at 7:32