# Query about Grassmann Variables in Lewis Ryder's book

1. Mar 26, 2012

### maverick280857

Query about Grassmann Variables a + b\eta

Hi,

What does

$$a + b\eta$$

where a and b are c-numbers and $\eta$ is a Grassmann number, really mean?

At first sight, this does not seem like a valid thing to do, as this can't be interpreted as a regular composition and sum in Grassmann space -- there is no concept of an identity which could be 'multiplied' with a in the first term (and so the first term is a c-number, whereas the second is a scalar multiple of a Grassmann number).

And yet we do this all the time!

Last edited: Mar 26, 2012
2. Mar 26, 2012

### Ilmrak

From the mathematical point of view you simply defined the operation of sum and product of a Grassman number with C-numbers.

In physics that number can be interpred as the number resulting by the application of a sum of a bosonic and a fermionic destruction operators on the product of a bosonic and a fermionic coherent state:

$\{a,a^\dagger \}=1 \, , \; [b,b ^\dagger ]=1 \, , \\ |\eta \rangle = e^{-\eta a^\dagger} |0\rangle\, , \; |\phi \rangle = e^{ \phi b^\dagger} |0\rangle \, , \\ \Rightarrow (a+ b)|\eta;\phi \rangle = (\phi + \eta)|\eta;\phi \rangle$

I know this was a stupid answer... but that's the best I can do now

Ilm

3. Mar 26, 2012

### tom.stoer

It's not more difficult than z=a+ib with z² = (a² - b²) + 2iab; replacing i by η and i²=-1 by η²=0 you arrive at Grassmann numbers

4. Mar 26, 2012

### maverick280857

I see your point. However, in writing z = a + ib, you are really writing

$$z = a \times \mathbb{1} + b \times i$$

where $\mathbb{1}$ is an identity element of the space of complex numbers. Put another way, the first term $a$ is as much of a complex number as the second term $ib$, so adding them is like adding two elements of the same space.

However, with Grassmann numbers, you are effectively adding a c-number to a c-number multiple of a Grassmann number. When you do that, what is the entity you generate? (I realize that my question really points to the definition of '+'.) Is it a Grassmann number, or is it something else?

In other words, if $\eta$ is a Grassmann number (so $\eta \in \mathcal{G}$), and I define

$$f(\eta) = a + b\eta$$

where $a, b \in \mathbb{C}$ (i.e. a and b are c-numbers), is it correct to say that

$$f: \mathcal{G} \rightarrow \mathcal{G}$$

?

5. Mar 27, 2012

### tom.stoer

I don't see you point; why not doing exatly the same with a*1 + b*η and 1²=1, η²=0? then 1 and η span a two-dim. vector space with an additional operation (a*1 + b*η)(c*1 + d*η) = ac*1 + (ad+bc)*η; it's purely formal but it works

In addition there is a matrix representation of Grassmann algebras http://en.wikipedia.org/wiki/Grassmann_number so it shouldn't really bother you

Last edited: Mar 27, 2012