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?

  • $\begingroup$ 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
    Commented Jul 12, 2019 at 20:04

1 Answer 1


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.

  • $\begingroup$ yes, sorry it's +. Can the fermionic annihilation and creation operators be represented somehow? With differential operators or something? $\endgroup$ Commented Jul 12, 2019 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
    Commented Jul 12, 2019 at 20:04
  • $\begingroup$ @MadMax how would I guess they square to zero? From what does it follow? $\endgroup$ Commented Jul 12, 2019 at 20:13
  • $\begingroup$ @AlexandreGurchumelia, you might want to brush up on your Grassmann calculus. $\endgroup$
    – MadMax
    Commented Jul 12, 2019 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
    Commented Jul 15, 2019 at 12:33

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