quantum mechanics - Rigorous Mathematical Proof of the Uncertainty Principle from First Principles

While looking at an intuitive explanation for the Heisenberg Uncertainty Principle (related question below), there was a mention of an axiomatic approach to establishing the uncertainty principle. Could someone please point out a source with detailed steps and explanations from first principles?

Related Question

Can the Heisenberg Uncertainty Principle be explained intuitively?

Some (re)-search will reveal the below proof (and others) which are perhaps not immediately grasped by people unfamiliar with certain terminology / concepts.

Related Proof

Heisenberg Uncertainty Principle scientific proof

In the above answer, It is not clear

1. 
   

   How the product of vectors, $PQ$, is decomposed into real and imaginary parts?

   
   


2. 
   

   How the expected value of $PQ$ squared is the square of the imaginary and real parts separately?

   
   


3. 
   

   How both square things are positive since a complex portion is involved?

   
   


4. 
   
   

   How square things being positive means that the left hand side is bigger than one quarter the square of the commutator?

   
   


5. 
   

   Why the commutator is unchanged by the shifting $[P,Q]=[p,x]=ℏ$ ?

   
   

Please note, I was a decent physics student (perhaps not, but am still deeply interested) who has wandered off into the social sciences for graduate studies. Hence, I am a bit rusty on the notation and terminology. Any pointers to brush up the concepts and fill the gaps in existing explanations would be much appreciated. I understand my questions might seem very trivial or obvious to experts, hence please pardon my ignorance of any basic concepts.

Answer

1) It is a product of operators. And they are not so much decomposed into real and imaginary parts, but rather into self-adjoint and antiself-adjoint parts.

If we take self-adjoint $A,B$ linear operators on some suitable Hilbert space, it is clear that $$ AB=\frac{1}{2}(AB+BA)+\frac{1}{2}(AB-BA), $$ since $$ \frac{1}{2}(AB+BA)+\frac{1}{2}(AB-BA)=\frac{1}{2}AB+\frac{1}{2}BA+\frac{1}{2}AB-\frac{1}{2}BA=2\frac{1}{2}AB=AB. $$ Now, since $A$ and $B$ are self-adjoint, $$ (AB+BA)^\dagger=B^\dagger A^\dagger+A^\dagger B^\dagger=BA+AB=AB+BA, $$ so this "anticommutator", $AB+BA$, is self-adjoint if $A$ and $B$ are.

However, if we look at the commutator, $AB-BA$, $$ (AB-BA)^\dagger=B^\dagger A^\dagger-A^\dagger B^\dagger=BA-AB=-(AB-BA), $$ the commutator of self-adjoint operators $A,B$ is antiself-adoint.

Now, the reason he called them "real" and "imaginary", is because in the space of all linear operators of a unitary vector space, self-adjoint operators are analogous to real numbers within the complex number field, and antiself-adjoint operators are analogous to imaginary numbers.

2, and the rest) First we should note that we take expectation values with respect to quantum states. If our particle is in the state $|\psi\rangle$, then the expectation value of $A$ with respect to the state $|\psi\rangle$ is $\langle A\rangle_\psi=\langle\psi|A|\psi\rangle$, from which we can see that the expectation value is linear.

Now then, $$ \langle AB\rangle=\left\langle\frac{1}{2}(AB+BA)+\frac{1}{2}(AB-BA)\right\rangle=\frac{1}{2}\langle AB+BA\rangle+\frac{1}{2}\langle AB-BA\rangle .$$

We should note that the expectation value of an operator is related to its eigenvalues. The expectation value of a self-adjoint operator is real, because its eigenvalues are real, and the expectation value of an antiself-adjoint operator is imaginary, because the eigenvalues are imaginary. Also, because the commutator $[A,B]=AB-BA$ is antiself-adjoint, there exists a self-adjoint operator $C$, for which $[A,B]=iC$, since $iC$ is then antiself-adjoint ($C$ is self-adjoint, but the $i$ swaps sign).

Note now, that the post you were quoting was wrong in the sense we do not take the square of the expectation value, but the square of the absolute value of the expectation value.

But since $\langle AB-BA\rangle=\langle[A,B]\rangle=\langle iC\rangle=i\langle C\rangle$, and then we take the absolute value square of $\langle AB\rangle$: $$ |\langle AB\rangle|^2=\frac{1}{4}\langle AB+BA\rangle^2+\frac{1}{4}\langle C\rangle^2,$$ but then $$ |\langle AB\rangle|^2\ge \frac{1}{4}\langle C\rangle^2 ,$$ and everything here is less than $(\Delta A)^2(\Delta B)^2$, which means that $$(\Delta A)(\Delta B)\ge\frac{1}{2}\langle C\rangle,$$ but for $x$ and $p$, $C=-i[x,p]=-ii\hbar\mathbb{I}=\hbar\mathbb{I},$ so $\frac{1}{2}\langle C\rangle=\hbar/2$.

Do note, that the proof you linked is slightly wrong, based on my first glance, or it used some implicit algebraic manipulations I find nontrivial, but the general line of thought is the same.