The Schrödinger equation

doug <sweetser@world.com>

Introduction
The total wave function
Implications

Introduction

The Schrödinger equation gives the kinetic energy plus the potential (a sum also known as the Hamiltonian H) of the wave function psi, which contains all the dynamical information about a system.  Psi is a scalar function with complex values.

H ψ  =  
      -i ℏ ∂(ψ/∂)t =  
        ((-(ℏ^2))/(2 m)) (∇^2) ψ  +  
          V (0, X) ψ

H ψ  =  
      -i ℏ ∂(ψ/∂)t =  
        ((-(ℏ^2))/(2 m)) (∇^2) ψ  +  
          V (0, X) ψ

For the time-independent case, energy is written at the operator -i hbar d/dt, and kinetic energy as the square of the momentum operator, i hbar Del, over 2m.  Given the potential V(0, X) and suitable boundary conditions, solving this differential equation generates a wave function psi which contains all the properties of the system.

In this notebook, I will propose a quaternion form for the total wave function.  The partial  derivatives of the wave function will be analyzed.  Under the time-independent constraint, the reason for the equivalence of energy and momentum to operators will be justified.  These will be used to formulate the Schrödinger equation.

The total wave function

My hypothesis is that the total wave function Psi is the following combination of quaternion operators and fields.

\(\(\(Ψ = \ 
          q[cos, \(V\& ⇀ \)\/\@\(V.V\)\ sin] 
            \((\((q[ω, \(K\& ⇀ \)].q[t, 
                        \(X\& ⇀ \)]
 + \ 
                    \(\((q[ω, \(K\& ⇀ \)].q[t, \ 
                              \(X\& ⇀ \)])\)\^*\))\)/2)\)

        {cos, \((ω\ t\  - \ K.X)\), 
            V\/\@\(V.V\)\ sin, \((ω\ t\  - \ K.X)\))\)}
      \)\)

\(\(Ψ = \ 
          \((cos, \(V\& ⇀ \)\/\@\(V.V\)\ sin)\) 
            \((\((ω, \(K\& ⇀ \))\) 
                    \((t, \(X\& ⇀ \))\) + \ 
                  \(\((\((ω, \(K\& ⇀ \))\) 
                          \((t, \(X\& ⇀ \))\))\)\^*\))\)/2)\)

       = \((cos, 
          \((ω\ t\  - \ 
              \(K\& ⇀ \).\(X\& ⇀ \))\), 
          \(V\& ⇀ \)\/\@\(V.V\)\ sin, 
          \((ω\ t\  - \ 
              \(K\& ⇀ \).\(X\& ⇀ \))\))\)\)

where w is the scalar angular frequency, K is the wave 3-vector, t is the scalar time, X is the position 3-vector, and V is the phase 3-vector.

Now to analyze that definition.  For any quaternion, (q + q*)/2 gives the scalar part of the quaternion.

\(Simplify[
      \((\((q[ω, K].q[t, X] + 

                Transpose[\((q[ω, K].q[t, \ X])\)])\)/2)\).{1, 0}]\){ω t  - K.X, 0}

\(\((\((ω, \(K\& ⇀ \))\) 
              \((t, \(X\& ⇀ \))\) + \ 
            \(\((\((ω, \(K\& ⇀ \))\) 
                    \((t, \(X\& ⇀ \))\))\)\^*\))\)/2
 = 
      \((ω\ t\  - K.X, 0)\)\)

The trig functions act on this scalar.

The phase 3-vector V/(V.V)^.5 is similar to the imaginary number i, except that two phases do not necessarily commute.

(q[0, (V/(Sqrt[V V]))].q[0, (V/(Sqrt[V V]))]).{1, 0}{-1, 0}

(0, (V/(Sqrt[V V]))) (0, (V/(Sqrt[V V]))) = {-1, 0}

The total wave function as defined here is always normalized.

\(\((q[Cos[ω\ \ t\  - \ K.X], 
            V\/\@\(V\ V\)\ Sin[ω\ t\  - \ K.X]].
	
          q[Cos[ω\ t\  - \ K.X], 
            \(-\(V\/\@\(V\ V\)\)\)\ Sin[ω\ t\  - \ K.X]])\).{1, 0}\){Cos[ω t - K.X]^2 + Sin[ω t - K.X]^2, 0}

\(\((Cos \((ω\ \ t\  - \ K.X)\), 
          V\/\@\(V\ V\)\ Sin \((ω\ t\  - \ K.X)\))\)\^2
 = 
      \((Cos \((ω\ t - K.X)\)\^2 + Sin \((ω\ t - K.X)\)\^2, 0)
        \)\)

Extend Euler's formula--Exp[i theta] = Cos[theta] + i Sin[theta]--to quaternions, with
 i -> V/(V.V)^.5, and theta = w t - K.X.

Ψ  =  Exp[(V/(Sqrt[V V])) (ω t - K.X)];

Ψ  =  
      Exp ((V/(Sqrt[V V])) (ω t - K.X))

According to my hypothesis, the total wave function is the normalized quaternion exponential with a phase 3-vector V/(V.V)^.5 and an angle determined by the scalar wt - K.X.  The angular frequency w is related to the energy E by Einstein's photon hypothesis, E = h nu.

ω  =  
      2 π ν  =  ((2 π E)/h)  =  (E/ℏ)

ω  =  
      2 π ν  =  ((2 π E)/h)  =  (E/ℏ)

The wave 3-vector maps to the momentum 3-vector P by de Broglie's wavelength equation, P = h/lambda.

K  =  
      ((2 π)/λ)  =  
        ((2 π P)/h)  =  (P/ℏ)

K  =  
      ((2 π)/λ)  =  
        ((2 π P)/h)  =  (P/ℏ)

Substitute these into the total wave function

Ψ  =  Exp[(V/(Sqrt[V V])) (((E t - P.X))/ℏ)];

Ψ  =  
      Exp ((V/(Sqrt[V V])) (((E t - P.X))/ℏ))

This is starting to look like physics, specifically the plane wave solution for a free particle.  Examine the partial time derivative.

\(∂Ψ\/∂t = \ 
      \(V\/\(\(\@\(V\ V\)\) ℏ\)\) 
          \((E\  + \ t\ ∂E\/∂t\  - \ 
              ∂P.X\/∂t)\) Ψ\ 

        \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  + \ 
        \(\((E\ t - P.X)\)\/ℏ\) 
          \(∂\(V\/\@\(V\ V\)\)\/∂t\) Ψ\)

\(∂Ψ\/∂t = \ 
      \(V\/\(\(\@\(V\ V\)\) ℏ\)\) 
          \((E\  + \ t\ ∂E\/∂t\  - \ 
              ∂P.X\/∂t)\) Ψ\ 

        \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  + \ 
        \(\((E\ t - P.X)\)\/ℏ\) 
          \(∂\(V\/\@\(V\ V\)\)\/∂t\) Ψ\)

Multiply both sides by (V/(V.V)^.5)^-1 hbar = -V/(V.V)^.5 hbar.  Why does the inverse of the phase equal its negative?  The inverse of any quaternion is the transpose divided by the norm.  The transpose of a 3-vector is the negative of the 3-vector.  The norm is one.

-((V ℏ)/(Sqrt[V V])) 
        ∂(Ψ/∂)t = 
      (E  +  t ∂(E/∂)t  -  
            ∂P.(X/∂)t) Ψ

-((V ℏ)/(Sqrt[V V])) 
        ∂(Ψ/∂)t = 
      (E  +  t ∂(E/∂)t  -  
            ∂P.(X/∂)t) Ψ

In the time-independent case, E, P and X are not functions of time.

-((V ℏ)/(Sqrt[V V])) ∂(ψ/∂)t =  
      E ψ

-((V ℏ)/(Sqrt[V V])) ∂(ψ/∂)t =  
      E ψ

For the time-independent wave function psi, energy is identical to the operator -I hbar d/dt!  The equivalence of energy and this operator is know as the first quantization.

Repeat this process for the spatial derivative of the total wave function.

\(∇\ Ψ\  = \ 
      \(V\/\(\(\@\(V\ V\)\) ℏ\)\) 
          \((\(-P\)\  - \ X ∇\(.P\)\  + \ ∇\ Et)\) Ψ\ 

        \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  + \ 
        \(\((E\ t - P.X)\)\/ℏ\) ∇x \( V\/\@\(V\ V\)\) 
          Ψ\)

\(∇\ Ψ\  = \ 
      \(V\/\(\(\@\(V\ V\)\) ℏ\)\) 
          \((\(-P\)\  - \ X ∇\(.P\)\  + \ ∇\ Et)\) Ψ\ 

        \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  + \ 
        \(\((E\ t - P.X)\)\/ℏ\) ∇x \( V\/\@\(V\ V\)\) 
          Ψ\)

If Psi is time-independent, then Del Et = 0.  I think X Del.P must also be zero, perhaps by a continuity argument, but I am not sure.  Multiply through by -V/(V.V)^.5 hbar.

-((V ℏ)/(Sqrt[V V])) ∇ ψ  =  
      -P ψ

-((V ℏ)/(Sqrt[V V])) ∇ ψ  =  
      -P ψ

Momentum is equivalent to the operator I hbar Del.  This operator can be squared to yield:

- (ℏ^2) (∇^2)  =  -(P^2)

- (ℏ^2) (∇^2)  =  -(P^2)

Define the kinetic energy in terms of this momentum.

KE  = 
      (( m (v^2))/2)  =  
        ((((m v))^2)/(2 m))  =  
          ((P^2)/(2 m))  =  
            -(( (ℏ^2) (∇^2))/(2 m))

KE  = 
      (( m (v^2))/2)  =  
        ((((m v))^2)/(2 m))  =  
          ((P^2)/(2 m))  =  
            -(( (ℏ^2) (∇^2))/(2 m))

The Hamiltonian of a system is the sum of the kinetic energy and the potential energy.  Write out the Hamiltonian of the time-independent wave function psi.

H ψ  =  
      -(V/(Sqrt[V.V])) ℏ ∂(ψ/∂)t =  
        ((-(ℏ^2))/(2 m)) (∇^2) ψ  +  
          V (0, X) ψ

H ψ  =  
      -(V/(Sqrt[V.V])) ℏ ∂(ψ/∂)t =  
        ((-(ℏ^2))/(2 m)) (∇^2) ψ  +  
          V (0, X) ψ

This is the Schrödinger equation written as a quaternion.  One change involves the replacement of the imaginary number i by the non-commuting phase 3-vector.  Another important difference is that Psi is now a quaternion, not a scalar with complex values.

By choosing different potentials V(0, X), it should be possible to solve the standard collection of problems that confront the Schrödinger equation using quaternions (but I haven't done that work yet :-)

Implications

We ended up very close to the Schrödinger equation, a difference involving the commutativity of i.  A subtle difference with subtle consequences?  That will have to await further work.

The lesson of first quantization--that energy and momentum are operators--holds only under the assumption of time-independence.  When E, P, and X are functions of time, other terms in the partial time derivative of the total wave function survive, and make the math much more difficult to interpret.

What is most interesting about this line of work is where I start, often one step back from more typical derivations.  At the center of the total wave function is the scalar, wt - K.X.  This scalar gets packaged into a polar form to ease analysis.  There are a number of equivalent ways to write this scalar.

\(\(Et\  - \ P.X\)\/ℏ\  = \ \ 
      \(ω\ t\  - \ K.X\ 
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 
         = \ \(2\ π \((νt\  - \ \(1\/λ\).X)\) = \ 
          2\ π \((t\/t\_ν\  - \ \(1\/X\_λ\).X)\)\)\)\)

\(\(Et\  - \ P.X\)\/ℏ\  = \ \ 
      \(ω\ t\  - \ K.X\ 
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 
         = \ \(2\ π \((νt\  - \ \(1\/λ\).X)\) = \ 
          2\ π \((t\/t\_ν\  - \ \(1\/X\_λ\).X)\)\)\)\)

There is no difference among these four ways of writing this scalar, which is at the heart of the wave function so is the focus of the Schrödinger equation.  I find the last way most suggestive:

if ((t/(Subscript[t, ν]))  -  (1/(Subscript[X, λ])).X) is an integer 
      "n",  then\(Ψ\  = \ 
      \(q[Cos[2\ π\ n], \ \(V\/\@\(V\ V\)\) Sin[2\ π\ n]]\ 
 = \ 
        q[1, 0]\)
\)if ((t/(Subscript[t, ν]))  -  (1/(Subscript[X, λ])).X) is a half integer
       "n/2",  then\(Ψ = 
      \(q[Cos[π\ n], \(V\/\@\(V\ V\)\) Sin[π\ n]]
 = 
        q[1, 0]\ or\ q[0, \(±\(V\/\@\(V\ V\)\)\)]\)\)

if ((t/(Subscript[t, ν]))  -  (1/(Subscript[X, λ])).X) is an integer 
      "n",  then\(Ψ\  = \ 
      \(q[Cos[2\ π\ n], \ \(V\/\@\(V\ V\)\) Sin[2\ π\ n]]\ 
 = \ 
        q[1, 0]\)
\)if ((t/(Subscript[t, ν]))  -  (1/(Subscript[X, λ])).X) is a half integer
       "n/2",  then\(Ψ = 
      \(q[Cos[π\ n], \(V\/\@\(V\ V\)\) Sin[π\ n]]
 = 
        q[1, 0]\ or\ q[0, \(±\(V\/\@\(V\ V\)\)\)]\)\)

This may be related to the great boson/fermion dichotomy, then again, maybe not :-)