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KleinGordon and Dirac Equations

The Schrodinger equation is:
(h^{2}/2m)∂^{2}ψ/∂x^{2} = ih∂ψ/∂t
Because the time derivative is of first order and
the spacial derivative is of second order, it does
not treat time and space on the same footing. As
a result it is not Lorentz invariant. KG set
out to resolve this by replacing the H = p^{2}/2m
term in the SE with with its relativistic version.
From SR,
H = E = ±√(p^{2}c^{2} + m^{2}c^{4})
Substituting p = ih∂/∂x and E = ih∂/∂t gives:
H = ±√{(ih∂/∂x)^{2}c^{2} + m^{2}c^{4}}
Therefore, the TDSE becomes,
±√{(ih∂/∂x)^{2}c^{2} + m^{2}c^{4}}ψ = ih∂ψ/∂t
However, this is a difficult expression to work
with so Klein and Gordon squared both sides to
get:
{(ih∂/∂x)^{2}c^{2} + m^{2}c^{4})}ψ = (ih∂ψ/∂t)^{2}
Which simplifies to:
h^{2}c^{2}∂^{2}ψ/∂x^{2} + m^{2}c^{4}ψ = h^{2}∂^{2}ψ/∂t^{2}
Rearranging we get,
(1/c^{2})∂^{2}ψ/∂t^{2}  ∂^{2}ψ/∂x + m^{2}c^{2}ψ/h^{2} = 0
(□ + μ^{2})ψ = 0 where μ = mc/h
and □ = (1/c^{2})∂^{2}/∂t^{2}  ∂^{2}/∂x^{2}
The KG equation is used to describe spin 0
bosons in Quantum Field Theory.
The problem with the KleinGordon equation is
that it allows for negative energies since
E = ±√(p^{2} + m^{2}). It also allows for negative
probability densities. In addition, Dirac was
uncomfortable with the ∂^{2}ψ/∂t^{2} term because the
Schrodinger equation is only dependent on the first
derivative with respect to time and not the second.
Instead, he sought an alternative formulation that
was first order in both space and time. Consider,
ψ = exp(i(kx  ωt))
∂ψ/∂x = ikψ
Therefore,
ψ = (1/ik)∂ψ/∂x ... (1)
Now ∂ψ/∂t = iωψ
Substituting (1) into this gives:
∂ψ/∂t = iω(1/ik)∂ψ/∂x
= (ω/k)∂ψ/∂x
Now ω/k = 2πf/2π/λ = fλ = v so:
∂ψ/∂t = v∂ψ/∂x
This is the basic wave equation. We can also
write this as:
iωψ = vikψ
∴ (ω  vk) = 0
This equation can be satisfied in the following
way:
+ω = (+v)(+k) or (v)(k)
ω = (+v)(k) or (v)(+k)
Letting v = c, a plot of ω versus k looks like:
ω (E = hω)

left moving  right moving
\  /
c \  / +c
\  /
\  /
\/
+ k (p = hk)
/\
Right moving /  \ Left moving
Dirac Sea /  \ Dirac Sea
/  \
/  \
For each quantum state possessing a positive energy,
E, there is a corresponding state with energy E.
The problem with this is that a positive energy
electron would be able to lose its energy by
continuously emitting photons, a process that could
continue without limit as the electron descends into
lower and lower energy states. Real electrons clearly
do not behave in this way.
Dirac's solution to this was to hypothesize that
what we think of as the "vacuum" is actually the
state in which all the negative energy states are
filled, and none of the positive energy states.
Therefore, if we want to introduce a single electron
we would have to put it in a positiveenergy state,
as all the negativeenergy states are occupied.
Furthermore, even if the electron loses energy by
emitting photons it would be forbidden from dropping
below zero energy. These negative energy states are
referred to as the Dirac Sea. There is a Dirac Sea
for both right moving and left moving particles as
shown in the plot.
Dirac also pointed out that a situation might exist
in which all the negativeenergy states are occupied
except one. This "hole" in the sea of negativeenergy
electrons would respond to electric fields as though
it were a positivelycharged particle. In later years
this "hole' was determined to be the positron.
Right moving electrons and holes have +ve momentum
and energy. Left moving electrons and holes have ve
momentum and energy?
The idea of left and right handed particles leads to
the idea that there must be 2 types of electrons.
The Massless Case

Left and right moving waves:
ψ_{R} = exp(i(kx  ωt)) and ψ_{L} = exp(i(kx + ωt))
The wave equations for each are:
∂ψ_{R}/∂t = c∂ψ_{R}/∂x
and
∂ψ_{L}/∂t = c∂ψ_{R}/∂x
The right and left equations can be written
in matrix form as:
     
 ∂ψ_{R}/∂t  = c 1 0  ∂ψ_{R}/∂x 
 ∂ψ_{L}/∂t   0 1  ∂ψ_{L}/∂x 
     
We can also write the above as:
     
 iωψ_{R}  = cα ikψ_{R}  where α =  1 0 
 +iωψ_{L}   ikψ_{L}   0 1 
     
   
 ω  = cα k 
 +ω   k 
   
From this we can write: ω = +/αck or
ω^{2} = c^{2}α^{2}k^{2} ... (2)
The Massive Case

Now introduce relativistic mass,
E = √(p^{2}c^{2} + m^{2}c^{4})
= √(h^{2}k^{2}c^{2} + m^{2}c^{4})
hω = √(h^{2}k^{2} + m^{2}) setting c = 1
ω = √(k^{2} + m^{2}) setting h = 1
Therefore,
ω^{2} = k^{2} + m^{2} ... (3)
How can (2) with c^{2} = 1 be made to look like
(3)? Try something of the form:
ω = αk + βm where α and β are both matrices.
So,
ω^{2} = α^{2}k^{2} + β^{2}m^{2} + km(αβ + βα)
To get this to match with (3) we need:
α^{2} = 1, β^{2} = 1 and αβ + βα = 0
   
α^{2} =  1 0  1 0  = 1
_{ } 0 1  0 1 
   
 
Try β =  0 1 
 1 0 
 
   
β^{2} =  0 1  0 1  = 1
_{ } 1 0  1 0 
   
       
αβ + βα =  1 0  0 1  +  0 1  1 0 
 0 1  1 0   1 0  0 1 
       
   
=  0 1  +  0 1 
1 0   1 0 
   
   
=  0 1    0 1 
1 0  1 0 
   
= 0
This is just the statement that σ_{3}σ_{1} + σ_{1}σ_{3} = 0
The correspondence between (2) and (3) is:
ω = αk corresponds to:
     
 ∂ψ_{R}/∂t  =  1 0  ∂ψ_{R}/∂x 
 ∂ψ_{L}/∂t   0 1  ∂ψ_{L}/∂x 
     
Then,
ω = αk + βm corresponds to:
         
i ∂ψ_{R}/∂t  = i 1 0  ∂ψ_{R}/∂x  + m 0 1  ψ_{R} 
 ∂ψ_{L}/∂t   0 1  ∂ψ_{L}/∂x   1 0  ψ_{L} 
         
The factor of i is required to make the equation
work, i.e.
i(iωψ_{R}) = i(iαkψ_{R}) + βmψ_{R}
∴ ω = αk + βm
The latter results in the following coupled
equations:
i∂ψ_{R}/∂t = i∂ψ_{R}/∂x + mψ_{L}
i∂ψ_{L}/∂t = +i∂ψ_{L}/∂x + mψ_{R}
What happens when the particle is at rest? Under
this condition αk = 0 and ω = βm
This leads to:
.
iψ_{R} = mψ_{L}
and,
.
iψ_{L} = mψ_{R}
How do we decouple these equations? Consider the
linear combinations.
Adding:
. .
i(ψ_{R} + ψ_{L}) = m(ψ_{R} + ψ_{L})
Which can be written as:
.
iψ_{+} = mψ_{+}
i(iωψ_{+}) = mψ_{+}
∴ ω = m
Subtract bottom from top:
. .
i(ψ_{R}  ψ_{L}) = m(ψ_{R}  ψ_{L})
Which can be written as:
.
iψ_{} = mψ_{}
i(iωψ_{}) = mψ_{}
∴ ω = m
ψ_{+} is linear field operator that describes a
particle with +ve energy at rest and ψ_{} is a
linear field operator that describes a particle
with ve energy at rest. The ve energy particles
go into the Dirac Sea leaving just the +ve energy
particles.
ψ_{+} = (1/√2)(ψ_{R} + ψ_{L}) represents a linear coherent
superposition of quantum states that creates a
single particle with a probability of 1/2 of
being leftmoving or right moving. In other
words, it is at rest. Without the mass term,
ψ_{+} would create a linear superposition of a
particle that is actually moving to the left
and one that is actually moving to the right.
The mass effectively mixes the moving ψ_{R} and ψ_{L}
waves to create a wave that is not moving at
all.
Extension to 3 Dimensions

Now, from above we had:
ω^{2} = k^{2} + m^{2}
= k_{x}^{2} + k_{y}^{2} + k_{z}^{2} + m^{2} ... (4)
Also, from above we had:
ω = αk + βm
Which becomes:
ω = α_{x}k_{x} + α_{y}k_{y} + α_{z}k_{z} + βm
This leads to,
ω^{2} = α_{x}^{2}k_{x}^{2} + α_{y}^{2}k_{y}^{2} + α_{z}^{2}k_{z}^{2} + β^{2}m^{2} + 12 other
terms that equal 0 ... (5)
In order to match equations (4) and (5) we must
have:
α_{x}^{2} = α_{y}^{2} = α_{z}^{2} = β^{2} = 1
α_{i}α_{j} + α_{j}α_{i} = 2δ_{ij}
α_{i}β + α_{j}β = 0
After adding back c and h we get the complete
Dirac Equation as written by Dirac:
ih∂ψ/∂t = ihcα_{i}∂ψ/∂x + βmc^{2}ψ
Which can be rewritten in terms of the momentum
operator, p, as:
ih∂ψ/∂t = (cα_{i}p + βmc^{2})ψ
Covariant Form

Restate the Dirac equation with h = c = 1.
i∂ψ/∂t + iα_{i}∂ψ/∂x  βmψ = 0
Now define γ^{0} = β and multiply everything by γ^{0}.
Therefore,
iγ^{0}∂ψ/∂t + iγ^{0}α_{i}∂ψ/∂x  (γ^{0})^{2}mψ = 0
Now (γ^{0})^{2} = 1. Therefore,
iγ^{0}∂ψ/∂t + iγ^{0}α_{i}∂ψ/∂x  mψ = 0
Now define γ^{i} = γ^{0}α^{i}. Therefore,
iγ^{0}∂ψ/∂t + iγ^{i}∂ψ/∂x  mψ = 0
This can be written as:
iγ^{μ}∂_{μ}ψ  mψ = 0
Or,
(iγ^{μ}∂_{μ}  m)ψ = 0
This is the covariant form.
Noting that p > i∂/∂x and p_{0} > ∂/∂t we can
write:
(iγ^{μ}p_{μ}  m)ψ = 0
This is the momentum form.
The Gamma Matrices

Dirac realized that the solution to this equation
involved 4 x 4 matrices. These became known as
the GAMMA MATRICES.
 
_{ }  1 0 : 0 0 
_{ }  0 1 : 0 0   
γ^{0} =  .....:......  =  I 0 
_{ }  0 0 : 1 0   0 I 
_{ }  0 0 : 0 1   
 
 
_{ }  0 0 : 0 1 
_{ }  0 0 : 1 0   
γ^{1} =  ......:......  =  0 σ_{x} 
_{ }  0 1 : 0 0   σ_{x} 0 
_{ }  1 0 : 0 0   
 
 
_{ }  0 0 : 0 i 
_{ }  0 0 : i 0   
γ^{2} =  ......:......  =  0 σ_{y} 
_{ }  0 i : 0 0   σ_{y} 0 
_{ }  i 0 : 0 0   
 
 
_{ }  0 0 : 1 0 
_{ }  0 0 : 0 1   
γ^{3} =  ......:......  =  0 σ_{z} 
_{ }  1 0 : 0 0   σ_{z} 0 
_{ }  0 1 : 0 0   
 
The σ matrices are the PAULI SPIN matrices.
The algebraic structure represented by the γ
matrices had been created some 50 years earlier
than Dirac by the English mathematician W. K.
Clifford. In fact, (γ^{o})^{2} = 1, (γ^{1})^{2} = (γ^{2})^{2} =
(γ^{3})^{2} = 1 form the basis of a CLIFFORD ALGEBRA.
However, Dirac was unaware of this previous work.
The Clifford algebra has the defining relationship:
{γ^{μ},γ^{ν}} = γ^{μ}γ^{ν} + γ^{ν}γ^{μ} = 2η^{μν}I
It is important to realise that the Dirac Gamma
matrices are not fourvectors. They are constant
matrices which remain invariant under a Lorentz
transformation.
Since the γ's; are 4 x 4 matrices the solutions
to the Dirac equation have to be 4component
entities. These entities are referred to as
the DIRAC SPINORS. Computation of these spinors
is easier if we note that the γ matrices can be
written in terms of 2 component I and σ matrices
as:
   
γ^{i} =  0 σ^{i}  where i = x, y, z and γ^{0} =  I 0 
_{ } σ^{i} 0  _{ }  0 I 
   
Likewise we can decompose ψ into 2 twocomponent
Lorentz spinors φ and χ as:
 
ψ =  φ 
 χ 
 
This decomposition reflects the fact that the
Dirac spinors form a reducible representation
consisting of irreducible representations, φ
and χ. These are the representations used in
the quantum mechanics of spin.
Now, using the covariant form (iγ^{μ}∂_{μ}  m)ψ = 0
where c and h have been set to 1 we get:
 
iγ^{0}∂/∂t =  E 0 
_{ }  0 E 
 
 
iγ^{i}∂/∂x^{i} =  0 σ.p 
_{ }  σ.p 0 
 
Where,
 
σ.p = σ_{x}p_{x} + σ_{y}p_{y} + σ_{z}p_{z} =  p_{z} (p_{x}  ip_{y})
_{ } (p_{x} + ip_{y}) p_{z} 
 
Returning to the covariant form:
(iγ^{0}∂/∂t + iγ^{i}∂/∂x  m)ψ = 0
∴ iγ^{0}∂ψ/∂t = iγ^{i}∂ψ/∂x^{i} + mψ
∴ γ^{0}E = γ^{i}σ.p + mψ
       
γ^{0}E φ  =  0 σ.p  φ  + m φ 
^{ }  χ   σ.p 0  χ   χ 
       
This gives:
Eφ = σ.pχ + mφ
Eχ = σ.pφ + mχ
∴ φ = (σ.p/(E  m))χ
≡ (σ.p/(E + m))χ
and,
χ = (σ.p/(E  m))φ
≡ (σ.p/(E + m))φ
We can also solve this in terms of the c
characteristic quation, (E  mI)ψ = 0:
     
 E σ.p  φ  = m φ 
 σ.p E  χ   χ 
     
   
 E  m σ.p  φ  = 0
 σ.p E  m  χ 
   
(E  m)φ + σ.pχ = 0
σ.pφ + (E  m) = 0
Which is the same as before. We can write these
2 equations as:
1. 1φ + σ.p/(E  m)χ = 0
2. σ.pφ/(E  m) + 1χ = 0
Combining gives:
   
u_{φ} =  1 φ u_{χ} =  σ.p/(E  m) χ
_{ }  σ.p/(E  m) _{ }  1 
   
If we let:
       
φ =  1  :  0  and χ =  1  :  0 
 0   1   0   1 
       
We get the 4component Dirac spinors, u_{i}:
   
_{ }  1 _{ }  _{ }  0 _{ } 
u_{1} = N 0 _{ }  u_{2} = N 1 _{ } 
_{ }  p_{z}/(E + m)_{ }  _{ }  (p_{x}  ip_{y})/(E + m) 
_{ }  (p_{x} + ip_{y})/(E + m)  _{ }  p_{z}/(E + m) _{ } 
   
   
_{ }  p_{z}/(E + m) _{ }  _{ }  (p_{x}  ip_{y})/(E + m) 
u_{3} = N (p_{x} + ip_{y})/(E + m)  u_{4} = N p_{z}/(E + m) _{ } 
_{ }  1 _{ }  _{ }  0 _{ } 
_{ }  0 _{ }  _{ }  1 _{ } 
   
Where N is a normalization factor.
Mathematically, these spinor objects describe quantum
mechanical spin states. Spinors are different from
vectors because they behave differently under rotations.
The above describes 4 different spin states. There are
2 spin states for E = +√(p^{2}c^{2} + m^{2}c^{4}) and 2 states for
E = √(p^{2}c^{2} + m^{2}c^{4}) .
For a FREE fermion the wavefunction is the product
of a plane wave and a spinor, u(p) or v(p). Thus:
spin up particle: ψ = u_{1}(p)exp(ipx)exp(iEt)
spin down particle: ψ = u_{2}(p)exp(ipx)exp(iEt)
spin up antiparticle: ψ = u_{3}(p)exp(ipx)exp(iEt)
spin down antiparticle: ψ = u_{4}(p)exp(ipx)exp(iE)
Note: exp(iωt) ≡ exp(iEt/h) ≡ exp(imc^{2}t/h)
For a particle at rest:
Eγ^{0}ψ = mψ
     
 E 0 0 0  u_{1}   u_{1} 
 0 E 0 0  u_{2}  = m u_{2} 
 0 0 E 0  u_{3}   u_{3} 
 0 0 0 E  u_{4}   u_{4} 
     
Using the characteristic polynomial gives:
   
 E  m 0 0 0  u_{1} 
 0 E  m 0 0  u_{2}  = 0
 0 0 E  m 0  u_{3} 
 0 0 0 E  m  u_{4} 
   
It is a property that the eigenspinors of a diagonal
matrix are:
_{ }  _{ }  _{ }   _{ }  
_{ }  1  _{ }  0 _{ }  0 _{ }  0 
u_{1} =  0  u_{2} =  1  v_{1} =  0  v_{2} =  0 
_{ }  0 _{ }  0 _{ }  1 _{ }  0 
_{ }  0 _{ }  0 _{ }  0 _{ }  1 
_{ }  _{ }  _{ }  _{ }  
and the eigenvalues are m, m, m, m.
This is same result as before when p = 0;
In this case, ψ_{i} = u_{i}exp(imc^{2}t/h) which represents
the time evolution of ψ_{i}.
The KleinGordon Equation Revisited

Consider:
(iγ^{μ}p_{μ}  m)(iγ^{ν}p_{ν} + m)ψ = 0
This equals:
(γ^{μ}γ^{ν}p_{μ}p_{ν}  m^{2})ψ = 0
γ^{μ}γ^{ν} is rather complicated and consists of
terms involving (γ^{0})^{2}p_{0}^{2} + (γ^{1})^{2}p_{1}^{2} + ...
+ (γ^{0}γ^{1} + γ^{1}γ^{0}) + ...  m^{2}.
Now, from the Clifford algebra,
{γ^{μ},γ^{ν}} = 2η^{μν}I = 0 if μ ≠ ν
Knowing this we are left with:
(γ^{μ}γ^{ν}p_{μ}p^{ν}  m^{2})ψ = [(γ^{0})^{2}p_{0}^{2} + (γ^{1})^{2}p_{1}^{2} + (γ^{2})^{2}p_{2}^{2}
+ (γ^{3})^{2}p_{3}^{2}  m^{2}]ψ = 0
Now, (γ^{o})^{2} = 1, (γ^{1})^{2} = (γ^{2})^{2} = (γ^{3})^{2} = 1
So we are left with:
[p_{0}^{2}  p_{1}^{2}  p_{2}^{2}  p_{3}^{2}  m^{2}]ψ = 0
Or,
[η^{μν}p_{μ}p_{ν}  m^{2}]ψ = 0 where η^{μν} = (+, , , )
Or,
[η^{μν}p_{μ}p_{ν} + m^{2}]ψ = 0
Or,
[p_{μ}p^{μ}  m^{2}]ψ = 0
Or,
[p_{μ}p^{μ} + m^{2}]ψ = 0
Note:
p^{μ} = (E/c,p^{1},p^{2},p^{3})
p_{μ} = (E/c,p_{1},p_{2},p_{3})
p_{μ}p^{μ} = E^{2}/c^{2} + p^{2}
= (m^{2}c^{4} + p^{2}c^{2})/c^{2}
= m^{2}c^{2} + p^{2}
This is the KleinGordon equation.