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KleinGordon and Dirac Equations

It can be shown that the Schrodinger equation 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 equivalent.
From SR,
H = E = ±√(p^{2}c^{2} + m^{2}c^{4})
substitute p for ih∇ and E = ih∂/∂t
H = ±√{(ih∇)^{2}c^{2} + m^{2}c^{4}}
Therefore, the TDSE becomes,
±√{(ih∇)^{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∇)^{2}c^{2} + m^{2}c^{4})}ψ = (ih∂ψ/∂t)^{2}
which simplifies to
h^{2}c^{2}∇^{2}ψ + m^{2}c^{4}ψ = h^{2}∂^{2}ψ/∂t^{2}
Rearranging we get,
(1/c^{2})∂^{2}ψ/∂t^{2}  ∇^{2}ψ + m^{2}c^{2}ψ/h^{2} = 0
(□ + μ^{2})ψ = 0 where μ = mc/h
and □ = (1/c^{2})∂^{2}/∂t^{2}  ∇^{2} ... the d'Alembert operator
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 (E = ±√(p^{2} + m^{2})) and negative probability
densities. Also, Dirac was uncomfortable with the ∂^{2}ψ/∂t^{2} term
because the Schrodinger equation only dependent on the first
derivative with respect to time not the second He sought an
alternative formulation that was first order in both space and
time. Consider,
ψ = exp(i(kx  ωt))
∂ψ/∂t = iωψ
∂ψ/∂x = ikψ
Therefore,
ψ = (1/ik)∂ψ/∂x
Substituting back into ∂ψ/∂t = iωψ gives,
∂ψ/∂t = iω(1/ik)∂ψ/∂x
= (ω/k)∂ψ/∂x
= v∂ψ/∂x since ω = 2πf and k = 2π/λ and v = fλ
This is the basic wave equation. We can also write this (from above)
as:
iω = vik
∴ (ω  vk) = 0
This equation can be satisfied with both positve and negative values of
ω and k.
∴ ω = ±vk
A plot of ω versus k looks like:
ω (E)

left moving  right moving
\  /
v \  / +v
\  /
\  /
\/
+ k (p)
/\
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 is that a
positive energy electron would be able to shed 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.
Left and right moving waves:
ψ_{R} = exp(i(kx  ωt)) and ψ_{L} = exp(i(kx + ωt))
The wave equations for each are:
i∂ψ_{R}/∂t = iv∂ψ_{R}/∂x
and
i∂ψ_{L}/∂t = iv∂ψ_{R}/∂x
where we have multiplied both sides by i.
The right and left equations can be written in matrix form as:
     
i ∂ψ_{R}/∂t  = iv 1 0  ∂ψ_{R}/∂x 
 ∂ψ_{L}/∂t   0 1  ∂ψ_{L}/∂x 
     
Also,
∂ψ_{R}/∂t = iωψ_{R} and ∂ψ_{R}/∂x = +ikψ_{R}
and
∂ψ_{L}/∂t = +iωψ_{L} and ∂ψ_{L}/∂x = +ikψ_{L}
Thus, we can write the above matrix equation as:
     
i iωψ_{R}  = ivα ikψ_{R}  where α =  1 0 
 +iωψ_{L}   ikψ_{L}   0 1 
     
From this we can write: ω = +/αk or
ω^{2} = α^{2}k^{2} ... A.
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} ... B.
How can A be made to look like B? 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 B 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  1 0  +  0 1  1 0 
 0 1  0 1   1 0  0 1 
       
   
=  0 1  +  0 1 
1 0   1 0 
   
   
=  0 1    0 1 
1 0  1 0 
   
= 0
The correspondence between A and B is,
     
ω = αk corresponds to i ∂ψ_{R}/∂t  = iv 1 0  ∂ψ_{R}/∂x 
 ∂ψ_{L}/∂t   0 1  ∂ψ_{L}/∂x 
     
Then,
         
ω = αk + βm corresponds to i ∂ψ_{R}/∂t  = iv 1 0  ∂ψ_{R}/∂x  + m 0 1  ψ_{R} 
 ∂ψ_{L}/∂t   0 1  ∂ψ_{L}/∂x   1 0  ψ_{L} 
         
Which results in the following coupled equations,
i∂ψ_{R}/∂t = iv∂ψ_{R}/∂x + mψ_{L}
i∂ψ_{L}/∂t = +iv∂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 linear combinations.
Adding:
.
iψ_{+} = mψ_{+}
Subtract bottom from top:
.
iψ_{} = mψ_{}
ψ_{R} is linear field operator that describes a particle with +ve energy
at rest and ψ_{L} 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_{+} = ψ_{R} + ψ_{L} creates a particle with a probability of 1/2 of being a
leftmoving or right moving. Thus, ψ_{+} is linear coherent superposition
of quantum states that creates a single particle at rest.
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} ... C.
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 ... D.
In order to match equations C and D we must have,
α_{x}^{2} = α_{y}^{2} = α_{z}^{2} = β^{2} = 1
After adding back constants and making v = c, 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 it is important to realise that the Dirac Gamma
matrices are not fourvectors. They are constant matrices which
remain invariant under a Lorentz transformation.
The γ's can be written as,
   
γ^{i} =  0 σ^{i}  where i = x, y, z and γ^{0} =  I 0 
_{ } σ^{i} 0  _{ }  0 I 
   
Now,
   
γ^{i}.p =  0 σ.p  and γ^{0}m =  mI 0 
_{ }  σ.p 0 _{ }  0 mI 
   
Thus,
 
(γ^{i}.p + βm) =  mI σ.p 
_{ }  σ.p mI 
 
Rembering that E = i∂/∂t. we can write the eigenvalue equation as:
    _{ } 
E φ  =  mI σ.p  φ 
 χ   σ.p mI  χ 
    _{ } 
Where φ and χ are the 2 component left and right handed spinors.
These are just 2 coupled equations:
Eφ = mφ + σ.pχ ∴ φ = (σ.p/(E  m))χ ≡ (σ.p/(E + m))χ
Eχ = σ.pφ  mχ ∴ χ = (σ.p/(E + m))φ
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} 
 
Since the γ's; are 4 x 4 matrices the solutions to the Dirac
equation has to be a 4component quantity. This quantity is
referred to as the DIRAC SPINOR. The eigenvalue equation yields
4 eigenspinors:
   
_{ }  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) 
v_{1} = N (p_{x}  ip_{y})/(E + m)  v_{2} = N p_{z}/(E + m) _{ } 
_{ }  1 _{ }  _{ }  0 _{ } 
_{ }  0 _{ }  _{ }  1 _{ } 
   
Where E = +/√(p^{2}c^{2} + m^{2}c^{4}) and 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. Two spin states for E = +√(p^{2} + m^{2}) and 2 states for
E = √(p^{2} + m^{2}).
For a FREE fermion the wavefunction is the product of a plane
wave and a spinor, u(p)/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) = v_{1}(p)exp(ipx)exp(iEt)
spin down antiparticle: ψ = u_{4}(p)exp(ipx)exp(iE) = v_{1}(p)exp(ipx)exp(iEt)
Note: Some notations use ψ = u(p)exp(ip_{μ}x^{μ}) where p_{μ} = (E, p_{x}, p_{y}, p_{z})
and x^{μ} = (t, x, y, z) so that ip_{μ}x^{μ} = i(Et  x.p)
Note: exp(iωt) ≡ exp(iEt/h) ≡ exp(im^{2}ct)
Note: For stationary particles p = 0 and the eigenvalue equation
becomes:
    _{ } 
E φ  =  mI 0  φ 
 χ   0 mI  χ 
    _{ } 
This is equivalent to:
Eψ = γ^{0}mψ
Or, since (γ^{0})^{2} = 1,
Eγ^{0}ψ = mψ
Now,
ψ = u(p)exp(iEt)
Therefore,
Eγ^{0}u(p)  mu(p) = 0
Or,
 
 1 0 0 0 
E 0 1 0 0 u(p) = mu(p)
 0 0 1 0 
 0 0 0 1 
 
The eigenspinors become:
_{ }  _{ }  _{ }   _{ }  
_{ }  1  _{ }  0 _{ }  0 _{ }  0 
u_{1} =  0  u_{2} =  1  v_{1} =  0  v_{2} =  0 
_{ }  0 _{ }  0 _{ }  1 _{ }  0 
_{ }  0 _{ }  0 _{ }  0 _{ }  1 
_{ }  _{ }  _{ }  _{ }  
with eigenvalues m, m, m, m
This describes 4 different spin states. Two spin states for E = +m
and 2 states for E = m.
The KleinGordon Equation Again

Consider:
(iγ^{Κ}p_{Κ}  m)(iγ^{λ}p_{λ} + m)ψ = 0
This equals:
(γ^{Κ}γ^{λ}p_{Κ}p_{λ}  m^{2})ψ = 0
γ^{Κ}γ^{λ} 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^{Κ},γ^{λ}} = 2η^{Κλ}I = 0 if Κ ≠ λ
Knowing this we are left with:
(γ^{Κ}γ^{λ}p_{Κ}p^{λ}  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^{Κλ}p_{Κ}p_{λ}  m^{2}]ψ = 0 where η^{Κλ} = (+, , , )
Or,
[η^{Κλ}p_{Κ}p_{λ} + m^{2}]ψ = 0
Or,
[p_{μ}p^{μ}  m^{2}]ψ = 0
Or,
[p_{μ}p^{μ} + m^{2}]ψ = 0
This is the KleinGordon equation.