The Dirac Equation : Intro to RQM

Introduction

"When I realized that the equation contained the spin of the electron, and also the magnetic moment - everything needed for the properties of the electron - it was a surprise. A complete surprise." P.A.M. Dirac

This is written for those who know QM and wish to see a peak at RQM, or are just curious about spin. Also, it's a testimony to the genious of Dirac, and the beauty of this Equation. Finally, I use an interesting notation which veterans may be amused by, involving relativistic indices and "matrix composition" (a tool of my own invention). Oh, and I also make a (new to me, anyway) interpretation of the Dirac equation, which is kind of amusing.

See also a novel (?) and pretty derivation of the dirac equation in 2d.

BTW I note that it is no coincidence that Dirac got spin by taking the square root of the hamiltonian. In fact, spinors are the roots of one. You may think of spinors as directions, x,y,z in 3-space. The pi-rotations in 3-space (x,y,z) are nothing but roots of (-1) (the 2pi rotation is minus one (see the Candle Dances), since 4pi is one). This is why Dirac got spin by taking the square root.

First, some conventions for this text-based equation writing. # means "compose". dx(y) means dy/dx. d[x](y) also means dy/dx, and is used when x is compound. Ocassionally, rather than writing squares (exponents), I will write pp for "p squared". T = ct, the x-dimensional version of time. There's some confusion because i is the imaginary, and is also a favorite summation index. This shouldn't be terrible.

The Beginning : Energy and Momentum

We wish to write a quantum mechanics in the form:

H|y> = E|y>

Now, in classical quantum, we had

H = (p^2)/2m

(for the free particle).  In relativistic quantum we must
use the correct relativistic energy equation.

H^2 = (pc)^2 + (mc^2)^2

This provides our equation:

((pc)^2 + (mc^2)^2)^.5 ) |y> = E |y>

Now lets switch to a nicer notation.  First, what is
momentum?  

pj = - ih dj		(j = x,y,z,T)

An operator.  What is E?  E = c * pT.  Energy is momentum
in the time direction.  We want a quantum mechanics which
is linear in E, and therefore linear in dt.  

(why linear in dt ?  it's easy to see that probability
current conservation will not work with a positive-definite
probability unless the equation of motion is only linear in dt)

Let's switch to relativistic summation, now with indices
x,y,z, and m.  So when I write (aibi) it means
(axbx + ayby + azbz + ambm).

So energy is (pt)^2 = (pjpj)  (j=x,y,z,m for review).

We want a theory linear in pt, but without having to deal with

pt = sqrt(pjpj)

So, we seek a new formulation, with the same results.

A Linear Guess

We guess a linear form, i.e. pt linear in pj.  The general
form is:

pt = ajpj

Where the aj are numbers, yet to be determined. (note, these
are quantum numbers, not classical numbers, and do not
necessarily commute (i.e. they are matrices, rank 1 tensors not
rank 0 tensors)).  We now have:

pt - ajpj |y> = 0

Now, since both sides are zero, we can multiply by anything.
We choose (pt + akpk).

(pt + akpk)(pt - ajpj) |y> = 0

(pt)^2 - (akpk)(ajpj) |y> = 0

Now, this is exactly the same relativistic energy equation, if
and only if

(akpk)(ajpj) = pipi

So, we can say two things:

the coefficient of each (pi)^2 term is 1 for all i
the coefficient of each pipj term is 0 for all i != j (not equal)

Now, expanding the sums, we have that the (pi)^2 coefficient is (ai)^2
The (pipj) coefficient is (aiaj + ajai).

Putting this all together, we have:

{ai,aj} = 2 delta(i,j)

where {x,y} is the anticommutator (xy + yx) and delta(i,j) is the
kronicher delta.

Finding the "a" vector

So, we need only solve this for the ai vector.  This is easy if
we know the algebra of "matrix composition".

In that case, we see that this algebra for ai is identical to that
for spin.  These behave just like spin, si, the only problem is that we
have 4 components of ai, but only 3 of si.  This is why we use
the matrix composition, which allows you to generally extend the
properties of any matrices to higher dimensions.

In this context, we just "compose" (# means compose).

a1 = sx # sx;
a2 = sx # sy;
a3 = sx # sz;
a4 = sy # I;

(there are many other posibilities).

It is a property of composition that

{ai,aj} = {sx,sx} # {si,sj} 	(for i,j in 1->3, i not = j)
		= 2 # 0 = 0

{ai,a4} = {sx#si,sy#I} = {sx,sy}#si = 0

{ai,ai} = 2			(i,j in 1-4)

This confirms it.  Now, lets write the answer.

pt = aipi

pt = (sx#si)pi + (sy#I)mc

That's it.  That's the Dirac equation. 

Is "s" really spin?

So, we have this relativistic quantum equation, and
it has the spin right in it.  But, we don't know
that this is "spin" in the physical sense - all we
know is that the Pauli spin matrices are >>useful<<
in this context.

So, we need to show that this really is spin.

How do we do that?  Well, what does spin do.  It adds
to the total angular momentum (which we know is
conserved), and it gives the electron a magnetic
moment.

So, we do two things: look at (d/dt) of the
angular momentum, and look at the behaviour of an
electron in an E&M field using this equation.

We work in the "Dirac Picture" or "Heisenberg Picture"
in which the wave function does not change in time,
but the opreators do.  The important quantity,
{y|x|y} does not change in time, regardless of whether
it is x or y which changes.  In fact, the unitary
transforms used for "Picture" invariance is the whole
reason a linear-Hamiltonian theory is necessary.  It
was Dirac's discovery of the "Picture" transform
between the Heisenberg and Shrodinger and Dirac formalisms
which made him believe so strongly in a linear dt
theory, which made him find the RQM equation.  Anyway,
that's our picture.  In that picture, we have:

i h dT( x ) = [x,pt]

for all x.  So to get dt( L ) the angular momentum, we use this.

i h dT( L ) = [L,pt]

[L,pt] = [L,(sx#si)pi + (sy#I)mc]
		= [L,(sx#si)pi]
		= (sx#si) [L,pi]
		= (sx#si) [r x p,pi]
		= ih (sx#si) x pi

Well, that's cute, but not zero.  (by si x pi , we mean the vector s cross
the vector p, = si pj Eijk (Eijk is the completely antisymmetric tensor) )

[L,pt] = h (sx # si) pj Eijk

Now, imagine we know the answer, that spin is an angular momentum.  In
fact, we know it has magnitude (h/2).  In that case the total angular
momentum is:

J = L + (h/2) s

This should be conserved. dt(J) = 0.  So, we need to know dt(s).  Well,

ih dT( s ) = [s,pt]

[si,pt] = [si,(sx#si)pj]

		= pj (sx#) [si,sj]
		= pj (sx#) sk (2i Eijk)

(h/2) [si,pt] = ih (sx# sk pk Eijk)

The conclusion is : dt(J) is indeed zero.  The spin and angular
momentum cancel.

Ok, so we got that part.  Now let's check the magnetic moment.
We assume you know how to write E&M in quantum : it's actually
just the same way we did it in Classical Hamiltonian E&M.  

We change pi to (pi + Ai) ; where Ai is the 4-vector for the
field. (we absorb the charge e into the field A)

pt + At = Sx#Si ( pi + Ai) + am m

square it.

(pt + At)^2 = (SxSx)#( (Si ( pi + Ai))^2 ) + (am m)^2 + {Sx,Sy}#Si pi m

now it is a property of a spin-like vector that

(Si Yi)^2 = YiYi + i Si Yj Yk Eijk

so:

(pt + At)^2 - (am m)^2 = (I)#( (Si (pi + Ai))^2 )


(pt + At)^2 - (am m)^2 = I#( (pi + Ai)(pi + Ai) + i Si (pj + Aj) (pk + Ak) Eijk )

We can drop the I# , since that never does anything.

(pt + At)^2 - (am m)^2 - (p + A)^2 = i Si (pj + Aj) (pk + Ak) Eijk

Now the stuff on the left is the classical E&M , i.e. quantum mechanics,
non-relativistic, no spin.  (pt + At)^2 - (am m)^2 - (p + A)^2 = 0

(pj + Aj) (pk + Ak) Eijk = pj pk Eijk + Aj Ak Eijk + (pjAk + Ajpk)Eijk

						= (pj Ak + Aj pk) Eijk

						= -ih e Bi

where B is the magnetic field.

e B = V X A + A X V 	, using V for "del".

(pt + At)^2 - (am m)^2 - (p + A)^2 = i Si ( -ih e Bi )

								= eh Si Bi = eh S*B

So, we have the "classical" E&M with an extra term : ehS*B

This extra term is identical to a magnetic moment of:

u = (eh/2m) S

Amazingly enough, this is the heuristic value for the spin which
you are told in connections with the Stern-Gerlach experiment.
It even has the correct X2 "g-factor" which you are told comes
from relativity.  Well, it does.

Conclusions and Summary

We started with E^2 = (pc)^2 + (mc^2)^2 = (pipi)c^2 , the relativistic
energy relation.

We turned this into a linear-hamiltonian Quantum Theory H|y> = E|y>

H = ptc				pt = aipi

ai = sx # si;		i = 1-3
a0 = sy # I;

This "a" vector came only from the requirement that our E^2 have the
correct form, the relativistic energy given above.

Then we showed that the s in this "a" really is a spin (and not just
something using the spin matrices) by showing that the total angular
momentum is :

J = L + (h/2)S

and that there is a magnetic moment

u = (eh/2m) S

And these are exactly the properties of the spin.  Thus, we have
found the spin - it is a necessary requirement of the relativistic
form of the quantum theory.

Actually, we have found even more.  If you look at "a", there are
really TWO spins in it: we have a 4-d composition of a pair of 2-d
spin matrices : we can have an eigenvector of two spins
simultaneously.  "s" |y> = "s" |a # b> = a b |y>

We have spin eigenstates a and b.  One is the "real" spin (this
is the second of the two, the Si in ai i=1,2,3).  The other is
an isospin.  A "spin" in the Hilbert space of particle-type.  This
one switches the particle between an electron and a positron.
(or parity, or chirality, if you prefer; these things are all tied up)
This is done by making the positron energy negative, and you have
a nice sea of filled negative-energy positron states, and then the
hole in this sea moves around.  This is easy to explain in the
context of excitations of many-body systems, in which we can talk
about creating "excitations" in a vacuum, or "holes" in a full sea.
Anyhoo, I won't go into details about this bit here.

A modern interpretation

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