Commutators and the uncertainty principle

doug <sweetser@world.com>

Introduction
Commutators
The uncertainty principle
Implications

Introduction

Commutators and the uncertainty principle are central to quantum mechanics.  Using quaternions in these roles has already been established by others (Horwitz and Biedenharn, Annals of Physics, 157:432, 1984).  The first proof of the uncertainty principle I saw relied solely on the properties of complex numbers, not on physics!  In this notebook I will repeat that analysis, showing how commutators and an uncertainty principle arise from the properties of quaternions (or their subfield the complex numbers).

Commutators

Any quaternion can be written in a polar form.

\(q\  = \ 
      \(\((s, \ V)\)\  = \ 
        \@\(\(q\^*\)\ q\)\ 
          Exp[s\/\@\(\(q\^*\)\ q\)\ V\/\@\(\(V\^*\)\ V\)]\)\)

This is identical to Euler's formula except that the imaginary unit vector i is replaced by the normalized 3-vector.  The two are equivalent if j = k = 0.

To simplify things, use a normalized quaternion, so that q* q = 1.  Collect the normalized 3-vector together with I = V/(V* V)^.5.

The angle s/(q* q)^.5 is a real number.  Any real number can be viewed as the product of two other real numbers.  This seemingly irrelevant observation lends much of the flexibility seen in quantum mechanics :-)  Here is the rewrite of q.

q  =  Exp[a b I] \(where\ \ \ \ \ \(q\^*\)\ q\  = \ 
      \(1\ \ \ \ \ a\ b\  = \ 
        \(s\/\@\(\(q\^*\)\ q\)\ \ \ \ \ \ I\  = \ 
          V\/\@\(\(V\^*\)\ V\)\)\)\)

Define a linear operator capital A that multiplies q by the scalar value small a.

A q  =  a q

Define a linear operator capital B that multiplies q by the scalar value small b.  This operator can be defined in terms of a, another simple observation with big consequences.

\(Let\ \ \ B\  = \ \(-I\)\ d\/da
\)B q  =  -I ((d Exp[a b I])/(da ))  =  b q

Operators A and B are linear.

\(\((A\  + \ B)\) q\  = \ 
      \(A\ q\  + \ B\ q\  = \ \(a\ q\  + \ b\ q\  = \ \((a\  + \ b)\) q\)\)

    \)A (q  +  q') =  A q  +  A q'  =  a q  +  a' q'

Calculate the commutator [A, B], which involves the scalar a and the derivative with respect to a.

\(\([A, \ B]\) q\  = \ 
      \(\((A\ B\  - \ B\ A)\) q\  = \ 
        \(\(-a\)\ I\ \(d\ q\)\/\(da\ \)\  + \ I\ \(d\ a\ q\)\/\(da\ \)
 = \ 
          \(\(-a\)\ I\ \(d\ q\)\/\(da\ \)\  + \ a\ I\ \(d\ q\)\/\(da\ \)\ \ 
               + \ \ I\ q \( d\ a\)\/\(da\ \)
 = \ I\ q\)\)\)\)

The commutator acting on a quaternion is equivalent to multiplying that quaternion by the normalized 3-vector I.

The uncertainty principle

Use these operators to construct things that behave like averages (expectation values) and standard deviations.

The scalar a--generated by the operator A acting on the normalized q--can be calculated using the Euclidean product.

\(\(q\^*\)\ \((A\ q)\)\  = \ 
      \(\(q\^*\)\ a\ q\  = \ \(a\ \(q\^*\)\ q\  = \ a\)\)\)

It is hard to shuffle quaternions or their operators around.  Real scalars (the only sort I ever use :-) commute with any quaternion and are their own conjugates.  Operators that generate such scalars can move around.  Look at ways to express the expectation value of A.

\(\(q\^*\)\ \((A\ q)\)\  = \ 
      \(\(q\^*\)\ a\ q\  = \ 
        \(a\ \(q\^*\)\ q\ \  = \ 
          \(\(a\^*\) \(q\^*\)\ q\  = \ 
            \(\(\((A\ q)\)\^*\)\ q\ \  = \ a\)\)\)\)\)

Define a new operator A' based on A whose expectation value is always zero.

\(\(Let\ \ \ A'\  = \ A\  - \ \(q\^*\)\ \((A\ q)\)
		

        \(q\^*\)\ \((A'\ q)\) = \ 
          \(q\^*\)\ \((A\  - \ \(q\^*\)\ \((A\ q)\))\) q)\) = \ 
      \(a\  - \ a\  = \ 0\)\)

Define the square of the operator in a way designed to link up with the standard deviation.

\(Let\ DA'\^2\  = \ 
      \(\(q\^*\)\ \((A\^\(' 2\)\ q)\) - \ 
          \((\(q\^*\)\ \((\(A\^'\)\ q)\))\)\^2\  = \ 
        \(q\^*\)\ \((A\^\(' 2\)\ q)\)\)\)

An identical set of tools can be defined for B.

In the notebook on bracket notation, the Schwarz inequality for quaternions was shown.

\(\(\(A\^\('*\)\ \(B\^'\)\  + \ \(B\^\('*\)\) \(A\^'\)\)\/2\  
        ≤ \ |\(A\^'\) | \  | \(B\^'\) | \ \ \)\)

The Schwarz inequality applies to quaternions, not quaternion operators.  If the operators A' and B' are surrounded on both sides by q and q*, then they will behave like scalars.

The left-hand side can be rearranged to form a commutator.

\(\(\(q\^*\) \((A\^\('*\)\ \(B\^'\)\  + \ \(B\^\('*\)\) \(A\^'\))\) q\ 

       = \ \(\(q\^*\) A\^\('*\)\ \(B\^'\)\ q\  + \ 
          \(q\^*\) \(B\^\('*\)\) \(A\^'\)\ q
 = \ 
        \(\(q\^*\)\ a\^\('*\)\ \(B\^'\)\ q\  + \ \ 
            \(q\^*\)\ \(\((\(-I\))\)\^*\) d\/da\ \(A\^'\)\ q
 = \ 
          \(\(q\^*\)\ \(a\^'\)\ \(B\^'\)\ q\  - \ \ 
              \(q\^*\)\ \((\(-I\))\) d\/da\ \(A\^'\)\ q
 = \ 
            \(\(q\^*\) \((\(A\^'\)\ \(B\^'\)\  - \ \(B\^'\)\ \(A\^'\))\) q\ 

               = \ \(q\^*\)[\(A\^'\), \ \(B\^'\)] q\)\)\)\)\ \)\)

The right-hand side can be rearranged to form the square of the operator.

\(\(q\^*\) | \(A\^'\) | \  | \(B\^'\) | q\ 
 = \ 
      \(\(q\^*\) A\^\(*'\)\ \(A\^'\) \(B\^\(*'\)\) \(B\^'\) q
 = \ 
        \(\(q\^*\) A\^\(' 2\)\ \(B\^\(' 2\)\) q
 = \ DA'\^2\ DB'\^2\)\)\)

Plug these back into the Schwarz inequality, stripping the primes which appear on both sides along the way.

\(\(\(q\^*\)[A, \ B] q\ \)\/2\  ≤ \ DA\^2\ DB\^2\)

This quaternion exercise can be mapped to physics

({{Quaternions, Standard physics}, {q, |ψ >   bra}, {\(q\^*\),  < ψ |   ket}, {A, A  operator}, {I, i}, {[A,  B], [A,  B]  commutator}, {\(\(q\^*\)\ q\),  < ψ | ψ >   norm}, {\(\(q\^*\)\ A\ q\),  < ψ | A ψ >   expectation of A}, {\(\(q\^*\)\ \((A\ q)\)\  = \ \(\((A\ q)\)\^*\)\ A\), A is Hermitian}, {(DA^2), \(δA\^2 = 

            \( < ψ\) | A\^2\ ψ > \(-\( < ψ\)\) | 
              \(A\ ψ\( > \^2\)\)\)}})

To get to the position-momentum uncertainty equation, make these specific maps

({{A, X}, {B, P  =  i ℏ (d/dx)}, {[A, B] =  I, [X,  P]  =  i ℏ}, {\(\ \ 
          \(\(q\^*\)[A, \ B] q\ \)\/2\  = \ 
	
            I\/2\  ≤ \ DA\^2\ DB\^2\), \(\ \ \ 
          \(\( < ψ\) | \(\([X, \ P]\) ψ > \)\ \)\/2\  = 
	\ 
            \(i\ ℏ\)\/2\  ≤ \ δX\^2\ δP\^2\)}})

The product of the squares of the standard deviation for position and momentum in the x-direction has a lower bound equal to half the expectation value of the commutator of those operators.  The proof is in the structure of quaternions.

Implications

There are many interpretations of the uncertainty principle.  I come away with two strange observations.  First, the uncertainty principle is about quaternions of the form q = Exp[a b I].  With this insight, one can see by inspection that a plane wave Exp[((Et - P.X)/hbar I], or wave packets that are superpositions of plane waves, will have four uncertainty relations, one for the scalar Et and another three for the three-part scalar P.X.  This perspective should be easy to generalize.

Second, the uncertainty principle and gravity are related to the same mathematical properties.  This proof of the uncertainty relation involved the Schwarz inequality.  It is fairly straightforward to convert that inequality to the triangle inequality.  Finding geodesics with quaternions involves the triangle inequality.  If a complete theory of gravity can be built from these geodesics (it hasn't yet been done :-) then the inequalities may open connections where none appeared before.