Commutation rules is little bit of entry which has elaborated the basic axioms that can help to solve the commutators.commutators has a physical significance of telling us whether we can observe the two parameters simultaneously or not ,therefore it signifies the simultaneity in this insight ,there is eight rules of commutation in Quantum mechanics which is one of the Axioms of Quantum theory.These rules are some-sort the Lie Brackett algebra and some of them are Jacobi identity. Therefore through this insight you are going to see how these commutation rules can be solved by another alternative way called Jacobi identity.it is not necessarily to use this Jacobi identity to solve these commutation rules,you can use another approaches including commutator-approach to solve these rules.

1. Commutation rules in Quantum
mechanics.
“You cannot prove scientific theories, concepts, rules, laws with
mathematics. You may use mathematics as a tool, a language to
describe a theory or principle but math alone cannot test a theory and
nobody proves anything in science.” By VILMOS-SHEPARD (Former
captain pilot, US Air Force) ---- (Saturday 19, June, 2021)
“Physics is descriptive not prescriptive.” ---BY TRARUH
SYNRED (P h D Particle physics, university of Illinois)
(Saturday 19, June, 2021)
“ … The quantum commutation rules ultimately are postulated, then
tested and verified by experimental and observations. The
commutation rules are then taken to be among the Axioms of
Quantum theory… “----by HARRY ELLIS (PhD, physics, Georgia institute
of technology) (19, Saturday, June, 2021).

2. Contents
1.0 INTRODUCTION ........................................................................................................3
1.1 AXIOMS OF THE COMMUTATIONS;......................................................................3
1.2 Pin point;...........................................................................................................4
2.0 DEBATE ABOUT THE PROVING OF COMMUTATION AXIOMS (time; Saturday, 19,
2021)...............................................................................................................................4
2.1 SCHOLAR, S VIEW ABOUT THE PROOF OF THESE AXIOMS...................................4
2.3 HARRY ELLIS......................................................................................................4
2.4 TRARUH SYNRED...............................................................................................5
2.5 VILMOS SHEPARD .............................................................................................5
2.6 ALI ABDULLA.....................................................................................................5
3.0 JACOBI IDENTITY PROOF...........................................................................................6
4.0 Annexure/Attachments......................................................................................... 10
5.0 Works Cited ........................................................................................................... 14

3. 1.0 INTRODUCTION
After explaining the concept of commutativity in our previous section
of commutativity in Quantum mechanics, we have to see another section of
the Quantum mechanics which is very important in commutativity of
Quantum mechanics. In last paper we saw the way commutativity in
quantum relied in the operators. Therefore we saw the meaning of the
term operator(s) in shortly and clearly, the linearity of operators, the
multiplicity of operators, the commutators and the commutation relations
as well. So today we are going to see another part called the commutation
rules in Quantum mechanics.
As in the previous section we dealt with operators, yet in this part
we are dealing with operators too as well as the commutativity.
Commutation rules are among of the Axioms in Quantum mechanics
which are being postulated then tested and verified with experimental
results. These Axioms are neither proven mathematically nor derived
mathematically instead are ueclided as Axioms and verified by
experimental observations. Therefore we are going to list these axioms
used to ground the commutativity in Quantum mechanics.
1.1 AXIOMS OF THE COMMUTATIONS;
𝑓𝑜𝑟 𝑨 𝑩 = 𝑩𝑨 𝑡ℎ𝑒𝑛;
[𝑨, 𝑩] = 𝑨𝑩 − 𝑩𝑨 = 𝟎 …………………………………………………………………………01
[𝑨, 𝑨] = 𝟎 ……………………………………………………………………………………………02
[𝑨, 𝑩] + [𝑩, 𝑨] = 𝟎 …………………………………………………………………………..03
[𝑨, 𝑩 + 𝑪] = [𝑨, 𝑩] + [𝑨, 𝑪] ……………………………………………………………04
[𝑨 + 𝑩, 𝑪] = [𝑨, 𝑪] + [ 𝑨, 𝑩 ] …………………………………………………………..05

4. [ 𝑨, 𝑩𝑪 ] = [𝑨, 𝑩]𝑪 + 𝑩[𝑨, 𝑪] ……………………………………………………………06
[𝑨 𝑩 , 𝑪] = [ 𝑨, 𝑪]𝑩 + 𝑨 [𝑩, 𝑪] …………………………………………………………..07
[ 𝑨, [𝑩, 𝑪]] + [ 𝑪 , [ 𝑨 , 𝑩 ]] + [ 𝑩, [ 𝑪, 𝑨 ]] = 𝟎 …………………………………..08
These Axioms cannot be proved because they are being postulated, tested
and then verified by experimental observations;
NOTE; I know how much I do not know anything and nor that I am solipsist
,therefore my qualia is that there must be a crowd that trust nothing about
these but we cannot force the reality to be what a solipsist believe.
1.2 Pin point;
Point; we cannot prove these axioms in physics because are Euclid
2.0 DEBATE ABOUT THE PROVING OF COMMUTATION
AXIOMS (time; Saturday, 19, 2021)
2.1 SCHOLAR, S VIEW ABOUT THE PROOF OF THESE AXIOMS
Question; How can I prove mathematically the commutation rules in
commutation of Quantum mechanics?
2.3 HARRY ELLIS
( P h D, Physics, Georgia Institute of technology (1974)).
“ … The quantum commutation rules ultimately are postulated, then
tested and verified by experimental and observations. The
commutation rules are then taken to be among the Axioms of
Quantum theory… “---- (19, Saturday, June, 2021).

5. 2.4 TRARUH SYNRED
(P h D, Particle physics at university of Illinois at Urbana
Champaign (1974) )
“That’s physics not mathematics, they are defined and
they work and give answers that agree with experiment.
Physics is descriptive not prescriptive.” --- (Saturday 19,
June, 2021)
2.5 VILMOS SHEPARD
(Former Captain Pilot at US Air Force (1981-1987) )
“You cannot prove scientific theories, concepts, rules, laws with
mathematics. You may use mathematics as a tool, a language to
describe a theory or principle but math alone cannot test a theory and
nobody proves anything in science.” ---- (Saturday 19, June, 2021)
2.6 ALI ABDULLA
(Professor of Nuclear physics, University of Baghdad, Iraq 1975)
“Commutation of A and B operators’ means 𝑨𝑩 = 𝑩𝑨
.Therefore;[𝑨, 𝑩] = 𝑨𝑩 − 𝑩𝑨 = 𝟎 “---- (Saturday 19, June, 2021)
You can see the annexure provided below the way a debate goes on with
other scholars about this issue of proving these commutation rules in
quantum mechanics.
Do you still want to prove it?
Ohm! Well yes, it is actually yes but it is not easy. These rules worked
as commutation rules are Lie algebras and Jacobi identities, so if you
want to prove them yet, you need to prove and solve Lie algebras and
Jacobi identities. These two theorems are mathematical theorems which
can be proved through other theorems called Poisson Brackett theorems,

6. Lie Brackett, Leibniz algebra, Group theory and the product rule as the part
of partial differentiation. Also we can use the matrices called symplectic
matrices.
The basis of these mathematics relied on the classical mechanics which
are Langragian operation, Hamiltonian operation and Hermitian as well.
3.0 JACOBI IDENTITY PROOF
[ 𝑨, [𝑩, 𝑪]] + [ 𝑪 , [ 𝑨 , 𝑩 ]] + [ 𝑩, [ 𝑪 , 𝑨 ]] = 𝟎 This is Jacobi identity
As I said before these mathematics are formulated from Poisson Brackett
which is derived from the Langragian mathematics and Hamiltonian
mathematics and is given as;
∑ (
𝜕𝑓
𝜕𝑞𝑖
𝑖
𝜕ℋ
𝜕𝑝𝑖
−
𝜕𝑓
𝜕𝑝𝑖
𝜕ℋ
𝜕𝑞𝑖
) Where ℋ is Hamiltonian operator
This means for two operators’ commutators [𝐁, 𝐂] we find that
∑ (
𝜕𝐵
𝜕𝑞𝑖
𝜕𝐶
𝜕𝑝𝑖
−
𝜕𝐵
𝜕𝑝𝑖
𝜕𝐶
𝜕𝑞𝑖
)
𝑖 …………………………………………………………………09
But look dude! The Jacobi’s identity we are talking about is that with
three operators as shown above so we need to write the Poisson algebra
with three operators as follows;
For [𝐴, [𝐵, 𝐶]]
Multiply by
𝜕𝐴
𝜕𝑞𝑗
𝜕
𝜕𝑝𝑗
throughout equation 09 to obtain;
∑ (
𝜕𝐴
𝜕𝑞𝑗
𝜕
𝜕𝑝𝑗
(
𝜕𝐵
𝜕𝑞𝑖
𝜕𝐶
𝜕𝑝𝑖
−
𝜕𝐵
𝜕𝑝𝑖
𝜕𝐶
𝜕𝑞𝑖
))
𝑖𝑗 …………………………………………………10
This in another way is correctly written as;

7. ∑ ([
𝜕𝐴
𝜕𝑞𝑗
𝜕
𝜕𝑝𝑗
lg(10𝜋) (
1
𝜋
𝜕𝐵
𝜕𝑞𝑖
𝜕𝐶
𝜕𝑝𝑖
−
𝜕𝐵
𝜋2𝜕𝑝𝑖
lg(10𝜋)
𝜕𝐶
𝜕𝑞𝑖
)] − [
𝜕𝐴
𝜕𝑝𝑗
𝜕
𝜕𝑞𝑗
eγn
(
1
𝜋
𝜕𝐵
𝜕𝑞𝑖
𝜕𝐶
𝜕𝑝𝑖
−
𝑖𝑗
𝜕𝐵
𝜋2𝜕𝑝𝑖
eγn 𝜕𝐶
𝜕𝑞𝑖
)]) ………………………………………………………………………………………..11
Now we have got a complex conjugate equation 11 which can lead as to our
expected solution of Jacobi identity.
Where by the value of 𝑛 is always zero. 𝑛 = 0
Then enter in the inner brackets all the parameters which are out of them and
take a differentiation.
∑ ([
𝜕𝐴
𝜕𝑞𝑗
𝜕
𝜕𝑝𝑗
lg(10𝜋) (
1
𝜋
𝜕𝐵
𝜕𝑞𝑖
𝜕𝐶
𝜕𝑝𝑖
−
𝜕𝐵
𝜋2𝜕𝑝𝑖
lg(10𝜋)
𝜕𝐶
𝜕𝑞𝑖
)] −
𝑖𝑗
[
𝜕𝐴
𝜕𝑝𝑗
𝜕
𝜕𝑞𝑗
eγn
lg(10𝜋) (
1
𝜋
𝜕𝐵
𝜕𝑞𝑖
𝜕𝐶
𝜕𝑝𝑖
− lg(10𝜋)
𝜕𝐵
𝜋2𝜕𝑝𝑖
eγn 𝜕𝐶
𝜕𝑞𝑖
)]) This becomes;
∑ ([
𝜕𝐴
𝜕𝑞𝑗
(
lg(10𝜋)
𝜋
[
𝜕𝐵
𝜕𝑞𝑖
𝜕2𝐶
𝜕𝑝𝑗𝜕𝑝𝑖
+
lg(10𝜋)
𝜋
𝜕2𝐵
𝜕𝑞𝑗𝜕𝑝𝑖
𝜕𝐶
𝜕𝑝𝑖
−
𝜕2𝐵
𝜕𝑝𝑗𝜕𝑝𝑖
𝜕𝐶
𝜕𝑝𝑖
+
𝜕𝐵
𝜕𝑞𝑖
𝜕2𝐶
𝜕𝑞𝑖𝜕𝑝𝑖
])] −
𝑖𝑗
[
𝜕𝐴
𝜕𝑝𝑗
(lg(10𝜋) .
eγn
𝜋
[
𝜕𝐵
𝜕𝑞𝑖
𝜕2𝐶
𝜕𝑝𝑖𝜕𝑞𝑗
+
𝜕2𝐵
𝜕𝑞𝑖𝜕𝑞𝑗
𝜕𝐶
𝜕𝑝𝑖
] − e2γn
. lg(10𝜋) [
𝜕2𝐵
𝜋𝜕𝑞𝑗𝜕𝑝𝑖
𝜕𝐶
𝜕𝑞𝑖
+
𝜕𝐵
𝜕𝑝𝑖
𝜕2𝐶
𝜕𝑞𝑖𝜕𝑞𝑗
])]) ………………………………………………………………………………..12
Note;
lg(10𝜋)
𝜋
= 𝜀 , e2γn
. lg(10𝜋) =∈𝑛
, lg(10𝜋) .
eγn
𝜋
= 𝜌𝑛
Then equation 12 reduces to;
∑ ([
𝜕𝐴
𝜕𝑞𝑗
(𝜀 [
𝜕𝐵
𝜕𝑞𝑖
𝜕2𝐶
𝜕𝑝𝑗𝜕𝑝𝑖
+ 𝜀
𝜕2𝐵
𝜕𝑞𝑗𝜕𝑝𝑖
𝜕𝐶
𝜕𝑝𝑖
−
𝜕2𝐵
𝜕𝑝𝑗𝜕𝑝𝑖
𝜕𝐶
𝜕𝑝𝑖
+
𝜕𝐵
𝜕𝑞𝑖
𝜕2𝐶
𝜕𝑞𝑖𝜕𝑝𝑖
])] −
𝑖𝑗
[
𝜕𝐴
𝜕𝑝𝑗
(𝜌𝑛
[
𝜕𝐵
𝜕𝑞𝑖
𝜕2𝐶
𝜕𝑝𝑖𝜕𝑞𝑗
+
𝜕2𝐵
𝜕𝑞𝑖𝜕𝑞𝑗
𝜕𝐶
𝜕𝑝𝑖
] −∈𝑛
[
𝜕2𝐵
𝜋𝜕𝑞𝑗𝜕𝑝𝑖
𝜕𝐶
𝜕𝑞𝑖
+
𝜕𝐵
𝜕𝑝𝑖
𝜕2𝐶
𝜕𝑞𝑖𝜕𝑞𝑗
])]) To;

8. ∑ ([
𝜕𝐴
𝜕𝑞𝑗
(𝜀 [
𝜕𝐵
𝜕𝑞𝑖
𝜕2𝐶
𝜕𝑝𝑗𝑝𝑖
+ 𝜀
𝜕2𝐵
𝜕𝑞𝑗𝑝𝑖
𝜕𝐶
𝜕𝑝𝑖
−
𝜕2𝐵
𝜕𝑝𝑗𝑝𝑖
𝜕𝐶
𝜕𝑝𝑖
+
𝜕𝐵
𝜕𝑞𝑖
𝜕2𝐶
𝜕𝑞𝑖𝑝𝑖
])] −
𝑖𝑗
[
𝜕𝐴
𝜕𝑝𝑗
(𝜌𝑛
[
𝜕𝐵
𝜕𝑞𝑖
𝜕2𝐶
𝜕𝑝𝑖𝑞𝑗
+
𝜕2𝐵
𝜕𝑞𝑖𝑞𝑗
𝜕𝐶
𝜕𝑝𝑖
] −∈𝑛
[
𝜕2𝐵
𝜋𝜕𝑞𝑗𝑝𝑖
𝜕𝐶
𝜕𝑞𝑖
+
𝜕𝐵
𝜕𝑝𝑖
𝜕2𝐶
𝜕𝑞𝑖𝑞𝑗
])])
………………………………………………………………………………………………………………….13
So what? We need to simplify the equation 13 because it has been more
complex otherwise we are not too smart to figure it and therefore it is
becoming as below;
For easily understanding, we transform these symbols into the usual
Euclidean coordinates as;
𝑞 = 𝑥 , 𝑝 = 𝑦
∑ ([
𝜕𝐴
𝜕𝑞𝑗
(𝜀 [
𝜕𝐵
𝜕𝑞𝑖
𝜕2𝐶
𝜕𝑝𝑗𝑝𝑖
+ 𝜀
𝜕2𝐵
𝜕𝑞𝑗𝑝𝑖
𝜕𝐶
𝜕𝑝𝑖
−
𝜕2𝐵
𝜕𝑝𝑗𝑝𝑖
𝜕𝐶
𝜕𝑝𝑖
+
𝜕𝐵
𝜕𝑞𝑖
𝜕2𝐶
𝜕𝑞𝑖𝑝𝑖
])] −
𝑖𝑗
[
𝜕𝐴
𝜕𝑝𝑗
(𝜌𝑛
[
𝜕𝐵
𝜕𝑞𝑖
𝜕2𝐶
𝜕𝑝𝑖𝑞𝑗
+
𝜕2𝐵
𝜕𝑞𝑖𝑞𝑗
𝜕𝐶
𝜕𝑝𝑖
] −∈𝑛
[
𝜕2𝐵
𝜋𝜕𝑞𝑗𝑝𝑖
𝜕𝐶
𝜕𝑞𝑖
+
𝜕𝐵
𝜕𝑝𝑖
𝜕2𝐶
𝜕𝑞𝑖𝑞𝑗
])])
Then we are going to observe the better and simple final transformation
results as;
𝐴𝑥(𝜀[ 𝐵𝑥𝐶𝑦𝑦 + 𝜀𝐵𝑥𝑦𝐶𝑦 − 𝐵𝑦𝑦𝐶𝑦 − 𝐵𝑥𝐶𝑥𝑦 ])– 𝐴𝑦 (𝜌𝑛
[𝐵𝑥𝐶𝑥𝑦 + 𝐵𝑥𝑥𝐶𝑥] −
∈𝑛
[
𝐵𝑥𝑦
𝜋
𝐶𝑥 − 𝐵𝑦𝐶𝑥𝑥] )
Assumption;
∈𝑛
𝜋
≅ 1, ∈𝑛
= 1, 𝜌𝑛
= 1, 𝜀 = 1
Hence a simplified equation is;
𝐴𝑥([ 𝐵𝑥𝐶𝑦𝑦 + 𝐵𝑥𝑦𝐶𝑦 − 𝐵𝑦𝑦𝐶𝑦 − 𝐵𝑥𝐶𝑥𝑦 ])– 𝐴𝑦([𝐵𝑥𝐶𝑥𝑦 + 𝐵𝑥𝑥𝐶𝑥] −
[𝐵𝑥𝑦 𝐶𝑥 + 𝐵𝑦𝐶𝑥𝑥] )
[𝐴𝑥[(𝐵𝑥𝐶𝑦𝑦) + (𝐵𝑥𝑦𝐶𝑦) − (𝐵𝑦𝑦𝐶𝑦) − (𝐵𝑥𝐶𝑥𝑦)] − 𝐴𝑦[ (𝐵𝑥𝐶𝑥𝑦) +
(𝐵𝑥𝑥𝐶𝑥) −(𝐵𝑥𝑦 𝐶𝑥 + 𝐵𝑦𝐶𝑥𝑥)]

9. Therefore;
[ 𝑨, [𝑩, 𝑪]] = 𝐴𝑥 ((𝐵𝑥𝐶𝑦𝑦) + (𝐵𝑥𝑦𝐶𝑦) − (𝐵𝑦𝑦𝐶𝑦) − (𝐵𝑥𝐶𝑥𝑦)) −
𝐴𝑦 ((𝐵𝑥𝐶𝑥𝑦) + (𝐵𝑥𝑥𝐶𝑥) − (𝐵𝑥𝑦 𝐶𝑥) − (𝐵𝑦𝐶𝑥𝑥))
…………………………………………………………………………………………………………………14
[ 𝑪 , [ 𝑨 , 𝑩 ]] = 𝐶𝑥 ((𝐴𝑥𝐵𝑦𝑦) + (𝐴𝑥𝑦𝐵𝑦) − (𝐴𝑦𝑦𝐵𝑦) − (𝐴𝑥𝐵𝑥𝑦)) −
(𝐶𝑦 ((𝐴𝑥𝐵𝑥𝑦) + (𝐴𝑥𝑥𝐵𝑥) − (𝐴𝑥𝑦 𝐵𝑥) − (𝐴𝑦𝐵𝑥𝑥)))
…………………………………………………………………………………………………………………..15
[ 𝑩, [ 𝑪 , 𝑨 ]] = 𝐵𝑥 ((𝐶𝑥𝐴𝑦𝑦) + (𝐶𝑥𝑦𝐴𝑦) − (𝐶𝑦𝑦𝐴𝑦) − (𝐶𝑥𝐴𝑥𝑦)) −
𝐵𝑦 ((𝐶𝑥𝐴𝑥𝑦) + (𝐶𝑥𝑥𝐴𝑥) − (𝐶𝑥𝑦 𝐴𝑥) − (𝐶𝑦𝐴𝑥𝑥))
…………………………………………………………………………………………………………………….16
If you substitute the equation 14,1 and 16 into the Jacob identity you can see
how this issue becomes resolved.
I told you this is not easy, to prove the commutation rule in Quantum
mechanics. It is some sort a damn confusing .But you wanted it yourself. This
is how you can prove.
The rest rules of commutation are solved in the same way, and some of them
are solved by Lie Brackett algebra which is not being solved in this little bit of
entry.

10. 4.0 Annexure/Attachments

14. (L.D.LANDAU, 1974) (Sherrill, 2006) (Wikimedia foundation. inc, 2021)
(gofvonx, 2013) (A.ABDULLA) (Merezbacker).
5.0 Works Cited
A.ABDULLA, A. Quantum Mechanics. USA.
gofvonx, J. a. (2013, November 09). MATHEMATICS. Retrieved Jun 20, 2021, from
Google: https://math.stackexchange.com/questions/557358/showing-jacobi-
identity-for-poisson-bracket
L.D.LANDAU, E. (1974). Quantum Mechanics:A short course of Theoretical physics.
Merezbacker. Quantum Mechanics.
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