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...
\end{equation}
This sets apart the
behaviour behavior of the two operators, the manner they go about extracting the information is different. Perhaps, if we have a system of quanta and set the condition if at any time I SEE the system empty it must be destroyed, then $r$ would correspond to looking into the system and taking half a quantum out, and $s$ to not looking and taking half a quantum out. $r$ carries with it the act of observation. However, if one operates further \begin{equation}
ss|0>=\\
rr|0>=\\
...
s^+r=r^+s, Herm...
\end{equation}
So we have mixed hermitian eigenoperators of state
|n>. $|n>$.
Which means a double application of $r^+r$ is equal to the number operator $N=a^+a$, a double application of $ss^+$ is equivalent to the operator $aa^+$ and it implies that $s^+s \equiv rr^+$. This is interesting, we can measure the number of particles in an oscillator by only taking half of one out and putting it back ($r$ then $r^+$ is the operation $r^+r$), but there are two ways to take the particle out and two ways to put it back and each appear as different actions with different 'compensations'.
These sub-operations work in line with the conjugate operation. if we introduce a basic ladder
opperator operator with no coefficients \begin{equation}
\nu^+|n>=|n+1> \\
\nu|n>=|n-1>
...
\end{equation}
If we assume there is an inverse of
r^+ $r^+$ we can eliminate $\nu$ and come to the identity $s^+s \equiv rr^+$ which was true from definition before. We have also \begin{equation}
s\nu^+s=s^+\nu s^+ \\
r\nu^+r=r^+\nu r^+
...
[s^+_k,s^+_l]=f(n)\delta_{kl}
\end{equation}
where $f(n)$ is some relationship between the
|n> $|n>$ state the commutator is applied to. It would be nicer still if
f(n) $f(n)$ was a constant, however this does not appear to be true.
In some respects, with the correct left and right divisions \begin{equation}
...
H=\sum_{k} \hbar \omega a^+_ka_k + \sum_{k,l} a^+_la_k + \sum_{k,l,m,n} a^+_ma^+_na_ka_l
\end{equation}
Which would correspond to a Feynman diagram with incoming states
k,l $k,l$ and outgoing states
n,m $n,m$ respectively. When we expand the normal self interaction terms in the fractional operators \begin{equation}
H= \sum_{k} \hbar \omega a^+_ka_k \\
H= \sum_{k} \hbar \omega r^+_ks^+_ks_kr_k \\