1
$\begingroup$

The paper "Quantized Grassmann variables and unified theories" says given creation and annihilation operators $b$ and $b^\dagger$ one can represent quaternionic imaginary units $q_1$, $q_2$ and $q_3$ in a following way (I expressed $q$-s in terms of $b$-s)

$$\begin{align} q_1&=b-b^\dagger, \\ q_2&=-i(b+b^\dagger), \\ q_3&=i\left[b^\dagger,b\right]. \end{align}$$

Paper doesn't give further details about this. Now from what I know quaternionic units should satisfy Hamilton's famous formula $q_1^2=q_2^2=q_3^2=q_1q_2q_3=-1$ (unless they're in some other basis). Also as I understand it $b$-s can be written as

$$\begin{align} b &= \frac{1}{\sqrt{2}}(x + ip), \\ b^\dagger &= \frac{1}{\sqrt{2}}(x - ip), \end{align}$$

where $p=-i\partial/\partial x$ so we have $\left[x,p\right]=i$, and also $\left[b,b^\dagger\right]=1$. So upon these substitutions $q_3$ becomes $-i$ which of course has the property that squaring it gives you $-1$. As for the first two units I calculated and got

$$\begin{align} q_1^2 = -2p^2 \\ q_2^2 = -2x^2 \end{align}$$

and I don't see how they represent quaternions. $q_1$ and $q_2$ definitely do not square to $-1$, nor do they have proper commutation relations with $q_3$, they commute with $q_3=-i$ but they shouldn't. What am I missing?

$\endgroup$
  • $\begingroup$ The multiplication operation that you do on quaternions is not represented by multiplying (or composing) the operators. The multiplication is mapped to the (anti)commutator. So for example $q_1^2= 1$ would actually mean $\{q_1,q_1\} = 1$. $\endgroup$ – MannyC Jul 12 at 20:04
2
$\begingroup$

The operators are presumably fermion annihilation and creation operators obeying $b^2=(b^\dagger)^2$=0 and $\{ b, b^\dagger\}=1$.

I think you want ${\bf j}$ (your $q_2$) to be ${\bf j}=-i(b+b^\dagger)$ though.

$\endgroup$
  • $\begingroup$ yes, sorry it's +. Can the fermionic annihilation and creation operators be represented somehow? With differential operators or something? $\endgroup$ – Alexandre Gurchumelia Jul 12 at 19:56
  • $\begingroup$ @AlexandreGurchumelia, of course you can have a representation just like the bosonic case: p=−i∂/∂x, only that x here is a super/grassmann dimension. That being said, why are you so bent on a representation? The communication relationships of the annihilation and creation operators would suffice for any calculation. $\endgroup$ – MadMax Jul 12 at 20:04
  • $\begingroup$ @MadMax how would I guess they square to zero? From what does it follow? $\endgroup$ – Alexandre Gurchumelia Jul 12 at 20:13
  • $\begingroup$ @AlexandreGurchumelia, you might want to brush up on your Grassmann calculus. $\endgroup$ – MadMax Jul 12 at 20:21
  • $\begingroup$ You can use Pauli sigma matrices: ${\bf i}\to i\sigma_x$, ${\bf j}\to i\sigma_y$, ${\bf k}\to i\sigma_z$ and then $a^\dagger \to \sigma_+\equiv (\sigma_x+i\sigma_y)/2$, $a \to \sigma_-= (\sigma_x-i\sigma_y)/2$. $\endgroup$ – mike stone Jul 15 at 12:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.