# Representing quaternionic algebra with creation and annihilation operators?

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?

• 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 $\{q_1,q_1\} = 1$. Commented Jul 12, 2019 at 20:04

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.
• 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 $a^\dagger \to \sigma_+\equiv (\sigma_x+i\sigma_y)/2$, $a \to \sigma_-= (\sigma_x-i\sigma_y)/2$. Commented Jul 15, 2019 at 12:33