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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 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 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 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?

This paper 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)

$q_1=b-b^\dagger$,

$q_2=-i(b-b^\dagger)$,

$q_3=i\left[b^\dagger,b\right]$.

Paper doesn't give further details about this. Now from what I know quaternionic units should satisfy Hamiltons 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

$b = \frac{1}{\sqrt{2}}(x + ip)$,

$b^\dagger = \frac{1}{\sqrt{2}}(x - ip)$,

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

$q_1^2 = -2p^2$

$q_2^2 = -2x^2$

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 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?