Abstract

# Abstract

02/06/2014, Benedict Irwin.

We define fractional creation and annihilation operators and explore the consequences and sub algebras. It is found that the number operator presents itself as an interaction term. The fractional operators are shown to be linked through an Abelian operator.

# Introduction

Take the bosonic creation an annihilation operators $a|n>=\sqrt{n}|n-1> \\ a^+|n>=\sqrt{n+1}|n+1>$

suppose that $$a=sr$$ and $$a^+=r^+s^+$$, then one could define $r|n>=\sqrt[4]{n}|n-\frac{1}{2}> \\ s|n>=\sqrt[4]{n+\frac{1}{2}}|n-\frac{1}{2}> \\ r^+|n>=\sqrt[4]{n+\frac{1}{2}}|n+\frac{1}{2}>\\ s^+|n>=\sqrt[4]{n+1}|n+\frac{1}{2}>$

and then $sr|n>=\sqrt{n}|n-1>=a|n> \\ r^+s^+|n>=\sqrt{n+1}|n+1>=a^+|n+1>$

This appears fine “mathematically”. But what of the interpretation. Does there exist quantities like $$s^+s$$ and what do the mean? In some sense, if the operators $$a$$ and $$a^+$$ apply to an isolated harmonic oscillator, measuring the number of quanta in that field is done by taking one out to observe and putting one back making a compensation for that action. For fractional changes, we take half a quantum out, then another half, then put them back one half at a time.

Another notable feature of this is that one can make a half exchange with the vacuum under the right conditions! $s|0>=\sqrt[4]{\frac{1}{2}}|\frac{-1}{2}> \\ r|0>=\sqrt[4]{0}|\frac{-1}{2}>$

This sets apart the 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 $ss|0>=\\ rr|0>=\\ sr|0>=\\ rs|0>=$

It can be seen that the $$rs$$ operation bypasses the vacuum and enters negative states. This is an issue.

There exist eigenoperators for small changes $s^+s|n>=\sqrt{n+\frac{1}{2}}|n> \\ r^+r|n>=\sqrt{n}|n> \\ ss^+|n>=\sqrt{n+1}|n> \\ rr^+|n>=\sqrt{n+\frac{1}{2}}|n>\\ \\ sr^+=\sqrt[4]{n+\frac{1}{2}}\sqrt[4]{n+1} \\ rs^+=sr^+, Herm.. \\ r^+s=\sqrt[4]{n}\sqrt[4]{n+\frac{1}{2}} \\ s^+r=r^+s, Herm...$

So we have mixed hermitian eigenoperators of state $$|n>$$.

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 operator with no coefficients $\nu^+|n>=|n+1> \\ \nu|n>=|n-1>$

which just alone is not compatible with a QFT as it allows one to descend into negative occupations. But if we make $$(\nu)^{\dagger}=\nu^+$$ then we can explore how the $$a$$ and $$a^+$$ operators behave. $r^+r|n>=\sqrt{n}|n> \\ \nu r^+r|n>=\sqrt{n}|n-1> \equiv a|n> \\ (a)^{\dagger}=a^+=(\nu(r^+r))^{\dagger} \\ a^+|n>=r^+r\nu^+|n>=\sqrt{n+1}|n+1>$

so there is consistency in this. We can take the conjugate of the definition using $$s$$ and $$s^+$$ $(a^+)^{\dagger}=a=(\nu^+ ss^+)^{\dagger} \\ a = ss^+\nu \\ ss^+\nu|n> = \sqrt{n}|n-1>$ again there is consistency here.

Comparing this result to the operators, we can link the sub fields $$r$$ and $$s^+$$ by $s^+\nu\equiv r \\ r\nu^+\equiv s^+ \\$

which, agree if $$[\nu,\nu^+]=0$$, which is true from the lack of coefficient. Taking the conjugate leads to the relationship $r^+ \equiv \nu^+s \\ s \equiv \nu r^+$

If we assume there is an inverse of $$r^+$$ we can eliminate $$\nu$$ and come to the identity $$s^+s \equiv rr^+$$ which was true from definition before. We have also $s\nu^+s=s^+\nu s^+ \\ r\nu^+r=r^+\nu r^+$

From the old definitions $N=a^+a \\ N=r^+s^+sr=r^+rr^+r$

These two actions are equivalent. So say a quantum had a left and a right part, I can take the right part of a quantum out ($$r$$) and then the left part of a quantum out ($$s$$) then put back the left part ($$s^+$$) then put back the right part ($$r^+$$). But according to the rules, I could also take the right part out and put it back, and then repeat that action and still extract the same information.

We can determine some commutators $[\nu,\nu^+]=0 \\ [a,a^+]=1=[sr,r^+s^+] \\ \\ sr^+[r,s^+]+s[r,r^+]s^++r^+[s,s^+]r+[s,r^+]s^+r=1\\ [r,r^+rr^+]=N+\frac{1}{2}-N=\frac{1}{2} \\ [s,s^+ss^+]=\frac{1}{2} \\ [r,r^+]=\sqrt{n+\frac{1}{2}} - \sqrt{n} \\ [s,s^+]=\sqrt{n+1}-\sqrt{n+\frac{1}{2}} \\ [r,s^+]=[s,r^+]=\sqrt[4]{n+\frac{1}{2}}\sqrt[4]{n+1}-\sqrt[4]{n+\frac{1}{2}}\sqrt[4]{n}$

It would be nice if the fields were still independant entities such that $[r_k,r_l]=f(n)\delta_{kl} \\ [s_k,s_l]=f(n)\delta_{kl} \\ [r^+_k,r^+_l]=f(n)\delta_{kl} \\ [s^+_k,s^+_l]=f(n)\delta_{kl}$

where $$f(n)$$ is some relationship between the $$|n>$$ state the commutator is applied to. It would be nicer still if $$f(n)$$ was a constant, however this does not appear to be true.

In some respects, with the correct left and right divisions $\frac{a^+a}{r^+r}=s^+s=rr^+=\frac{aa^+}{ss^+}$ But does that imply $r^+a^+ar=saa^+s^+? \\ \sqrt{n+1}(n+\frac{3}{2})\ne(n-\frac{1}{2})\sqrt{n}$ There are no solutions, it is never true.

Based on some of the above definitions, instead of writing the Hamiltonian as $H=\hbar \omega(a^+a +\frac{1}{2})$

we could simply use the fundamental operators although there are a few degenerate expressions $H=\hbar \omega rr^+rr^+ \\ H=\hbar \omega s^+ss^+s \\$

If we usually have in a dynamical system $X=\sqrt{\frac{\hbar}{2\omega m}}(a^+ +a) \\ P=i\sqrt{\frac{\hbar \omega m}{2}}(a^+ -a) \\ H=\frac{P^2}{2m} + \frac{m\omega^2X^2}{2}$

Using the fact $$rr^+=s^+s$$ Can end up writing $H=\frac{\hbar \omega}{2}(srr^+s^++r^+s^+sr) \\ H=\frac{\hbar \omega}{2}(ss^+ss^++r^+rr^+r)$

For interaction Hamitonians we can write $H=\sum_{k} \hbar \omega a^+_ka_k + \sum_{k,l} <l|V_1|k> a^+_la_k + \sum_{k,l,m,n} <m,n|V_2|k,l> a^+_ma^+_na_ka_l$

Which would correspond to a Feynman diagram with incoming states $$k,l$$ and outgoing states $$n,m$$ respectively. When we expand the normal self interaction terms in the fractional operators $H= \sum_{k} \hbar \omega a^+_ka_k \\ H= \sum_{k} \hbar \omega r^+_ks^+_ks_kr_k \\$

the form of the term is the same! The labels are just the same state but it appears to be an interaction of four fields $$s,r,s^+$$ and $$r^+$$. The fields scatter into the same state they started in.

suppose that $$a=x(yz)$$ and $$a^+=z^+y^+x^+$$ or similar. We could define fractional operators for example $z|n>=\sqrt[6]{n}|n-\frac{1}{3}> \\ y|n>=\sqrt[6]{n+\frac{1}{3}}|n-\frac{1}{3}> \\ x|n>=\sqrt[6]{n+\frac{2}{3}}|n-\frac{1}{3}>$

with this definition $xyz|n>=a|n>$

Then a similar definition could be made $x^+|n>=\sqrt[6]{n+1}|n+\frac{1}{3}> \\ y^+|n>=\sqrt[6]{n+\frac{2}{3}}|n+\frac{1}{3}> \\ z^+|n>=\sqrt[6]{n+\frac{1}{3}}|n+\frac{1}{3}>$

# Fermionic Half Operators

Seek to create the fermionic creation and annihilation operators $b|n>=\sqrt{n}|n-1> \\ b^+|n>=\sqrt{1-n}|n+1>$

which only accept occupancies of 0 and 1. It can be seen that the definition of the annihilation operator is the same, and it will still be such that $$b=sr$$, however for $$b^+=r^+s^+$$ we must define the fermionic half operators $r|n>=\sqrt[4]{n}|n-\frac{1}{2}> \\ s|n>=\sqrt[4]{n+\frac{1}{2}}|n-\frac{1}{2}> \\ s^+|n>=\sqrt[4]{1-n}|n+\frac{1}{2}> \\ r^+|n>=\sqrt[4]{\frac{3}{2}-n}|n+\frac{1}{2}>$

Manipulations with these quantities have interesting results $s^+s|n>=\sqrt[4]{\frac{3}{2}-n}\sqrt[4]{n+\frac{1}{2}}|n> \\ r^+r|n>=\sqrt[4]{2-n}\sqrt[4]{n}|n> \\ ss^+|n>=\sqrt[4]{n+1}\sqrt[4]{1-n}|n> \\ rr^+|n>=\sqrt[4]{n+\frac{1}{2}}\sqrt[4]{\frac{3}{2}-n}|n>$

Then by considering that ideally n = 0 or 1 for a fermionic state $s^+s|1>= \sqrt[4]{\frac{1}{2}}\sqrt[4]{\frac{3}{2}}|1> =A|1>\\ r^+r|1>= \sqrt[4]{1}\sqrt[4]{1}|1> =1|1> \\ ss^+|1>= \sqrt[4]{2}\sqrt[4]{0}|1> =0|1>\\ rr^+|1>= \sqrt[4]{\frac{1}{2}}\sqrt[4]{\frac{3}{2}}|1> =A|1>\\ \\ s^+s|0>= \sqrt[4]{\frac{1}{2}}\sqrt[4]{\frac{3}{2}}|0> =A|0>\\ r^+r|0>= \sqrt[4]{2}\sqrt[4]{0}|0> =0|0>\\ ss^+|0>= \sqrt[4]{1}\sqrt[4]{1}|0> =1|0>\\ rr^+|0>= \sqrt[4]{\frac{1}{2}}\sqrt[4]{\frac{3}{2}}|0> =A|0>\\$

where A is $$\sqrt[4]{\frac{3}{4}}$$ however, there could be fractional occupancy states and these have interesting values again $s^+s|\frac{1}{2}>= \sqrt[4]{1}\sqrt[4]{1}|1> =1|\frac{1}{2}> \\ r^+r|\frac{1}{2}>= \sqrt[4]{\frac{1}{2}}\sqrt[4]{\frac{3}{2}}|1> =A|\frac{1}{2}>\\ ss^+|\frac{1}{2}>= \sqrt[4]{\frac{1}{2}}\sqrt[4]{\frac{3}{2}}|1> =A|\frac{1}{2}>\\ rr^+|\frac{1}{2}>= \sqrt[4]{1}\sqrt[4]{1}|1> =1|\frac{1}{2}> \\ \\ s^+s|\frac{-1}{2}>= \sqrt[4]{2}\sqrt[4]{0}|1> =0|\frac{-}{2}> \\ r^+r|\frac{-1}{2}>= \sqrt[4]{\frac{5}{2}}\sqrt[4]{\frac{-1}{2}}|1> =B|\frac{-1}{2}>\\ ss^+|\frac{-1}{2}>= \sqrt[4]{\frac{1}{2}}\sqrt[4]{\frac{3}{2}}|1> =A|\frac{-1}{2}>\\ rr^+|\frac{-1}{2}>= \sqrt[4]{0}\sqrt[4]{2}|1> =0\left|\frac{-1}{2}\right\rangle \\$