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Last modified: January 27, 2022 ✓
Klein-Gordon and Dirac Equations
--------------------------------
The Schrodinger equation is:
-(h2/2m)∂2ψ/∂x2 = 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. K-G set
out to resolve this by replacing the H = p2/2m
term in the SE with with its relativistic version.
From SR,
H = E = ±√(p2c2 + m2c4)
Substituting p = -ih∂/∂x and E = -ih∂/∂t gives:
H = ±√{(-ih∂/∂x)2c2 + m2c4}
Therefore, the TDSE becomes,
±√{(-ih∂/∂x)2c2 + m2c4}ψ = -ih∂ψ/∂t
However, this is a difficult expression to work
with so Klein and Gordon squared both sides to
get:
{(-ih∂/∂x)2c2 + m2c4)}ψ = (-ih∂ψ/∂t)2
Which simplifies to:
-h2c2∂2ψ/∂x2 + m2c4ψ = -h2∂2ψ/∂t2
Rearranging we get,
(1/c2)∂2ψ/∂t2 - ∂2ψ/∂x + m2c2ψ/h2 = 0
(□ + μ2)ψ = 0 where μ = mc/h
and □ = (1/c2)∂2/∂t2 - ∂2/∂x2
The K-G equation is used to describe spin 0
bosons in Quantum Field Theory.
The problem with the Klein-Gordon equation is
that it allows for negative energies since
E = ±√(p2 + m2). It also allows for negative
probability densities. In addition, Dirac was
uncomfortable with the ∂2ψ/∂t2 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 positive-energy state,
as all the negative-energy 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 negative-energy states are occupied
except one. This "hole" in the sea of negative-energy
electrons would respond to electric fields as though
it were a positively-charged 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:
ψR = exp(i(kx - ωt)) and ψL = exp(i(kx + ωt))
The wave equations for each are:
∂ψR/∂t = -c∂ψR/∂x
and
∂ψL/∂t = c∂ψR/∂x
The right and left equations can be written
in matrix form as:
- - - - - -
| ∂ψR/∂t | = -c| 1 0 || ∂ψR/∂x |
| ∂ψL/∂t | | 0 -1 || ∂ψL/∂x |
- - - - - -
We can also write the above as:
- - - - - -
| -iωψR | = -cα| ikψR | where α = | 1 0 |
| +iωψL | | ikψL | | 0 -1 |
- - - - - -
- - - -
| -ω | = -cα| k |
| +ω | | k |
- - - -
From this we can write: ω = +/-αck or
ω2 = c2α2k2 ... (2)
The Massive Case
----------------
Now introduce relativistic mass,
E = √(p2c2 + m2c4)
= √(h2k2c2 + m2c4)
hω = √(h2k2 + m2) setting c = 1
ω = √(k2 + m2) setting h = 1
Therefore,
ω2 = k2 + m2 ... (3)
How can (2) with c2 = 1 be made to look like
(3)? Try something of the form:
ω = αk + βm where α and β are both matrices.
So,
ω2 = α2k2 + β2m2 + km(αβ + βα)
To get this to match with (3) we need:
α2 = 1, β2 = 1 and αβ + βα = 0
- - - -
α2 = | 1 0 || 1 0 | = 1
| 0 -1 || 0 -1 |
- - - -
- -
Try β = | 0 1 |
| 1 0 |
- -
- - - -
β2 = | 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 σ3σ1 + σ1σ3 = 0
The correspondence between (2) and (3) is:
ω = αk corresponds to:
- - - - - -
| ∂ψR/∂t | = -| 1 0 || ∂ψR/∂x |
| ∂ψL/∂t | | 0 -1 || ∂ψL/∂x |
- - - - - -
Then,
ω = αk + βm corresponds to:
- - - - - - - - - -
i| ∂ψR/∂t | = -i| 1 0 || ∂ψR/∂x | + m| 0 1 || ψR |
| ∂ψL/∂t | | 0 -1 || ∂ψL/∂x | | 1 0 || ψL |
- - - - - - - - - -
The factor of i is required to make the equation
work, i.e.
i(-iωψR) = -i(iαkψR) + βmψR
∴ ω = αk + βm
The latter results in the following coupled
equations:
i∂ψR/∂t = -i∂ψR/∂x + mψL
i∂ψL/∂t = +i∂ψL/∂x + mψR
What happens when the particle is at rest? Under
this condition αk = 0 and ω = βm
This leads to:
.
iψR = mψL
and,
.
iψL = mψR
How do we decouple these equations? Consider the
linear combinations.
Adding:
. .
i(ψR + ψL) = m(ψR + ψL)
Which can be written as:
.
iψ+ = mψ+
i(-iωψ+) = mψ+
∴ ω = m
Subtract bottom from top:
. .
i(ψR - ψL) = -m(ψR - ψL)
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)(ψR + ψL) represents a linear coherent
superposition of quantum states that creates a
single particle with a probability of 1/2 of
being left-moving 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 ψR and ψL
waves to create a wave that is not moving at
all.
Extension to 3 Dimensions
-------------------------
Now, from above we had:
ω2 = k2 + m2
= kx2 + ky2 + kz2 + m2 ... (4)
Also, from above we had:
ω = αk + βm
Which becomes:
ω = αxkx + αyky + αzkz + βm
This leads to,
ω2 = αx2kx2 + αy2ky2 + αz2kz2 + β2m2 + 12 other
terms that equal 0 ... (5)
In order to match equations (4) and (5) we must
have:
αx2 = αy2 = αz2 = β2 = 1
αiαj + αjαi = 2δij
αiβ + αjβ = 0
After adding back c and h we get the complete
Dirac Equation as written by Dirac:
ih∂ψ/∂t = -ihcαi∂ψ/∂x + βmc2ψ
Which can be rewritten in terms of the momentum
operator, p, as:
ih∂ψ/∂t = (cαip + βmc2)ψ
Covariant Form
--------------
Restate the Dirac equation with h = c = 1.
i∂ψ/∂t + iαi∂ψ/∂x - βmψ = 0
Now define γ0 = β and multiply everything by γ0.
Therefore,
iγ0∂ψ/∂t + iγ0αi∂ψ/∂x - (γ0)2mψ = 0
Now (γ0)2 = 1. Therefore,
iγ0∂ψ/∂t + iγ0αi∂ψ/∂x - mψ = 0
Now define γi = γ0αi. Therefore,
iγ0∂ψ/∂t + iγi∂ψ/∂x - mψ = 0
This can be written as:
iγμ∂μψ - mψ = 0
Or,
(iγμ∂μ - m)ψ = 0
This is the covariant form.
Noting that p -> i∂/∂x and p0 -> ∂/∂t we can
write:
(iγμpμ - 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 | - -
γ0 = | .....:...... | = | I 0 |
| 0 0 : -1 0 | | 0 -I |
| 0 0 : 0 -1 | - -
- -
- -
| 0 0 : 0 1 |
| 0 0 : 1 0 | - -
γ1 = | ......:...... | = | 0 σx |
| 0 -1 : 0 0 | | -σx 0 |
| -1 0 : 0 0 | - -
- -
- -
| 0 0 : 0 -i |
| 0 0 : i 0 | - -
γ2 = | ......:...... | = | 0 σy |
| 0 i : 0 0 | | -σy 0 |
| -i 0 : 0 0 | - -
- -
- -
| 0 0 : 1 0 |
| 0 0 : 0 -1 | - -
γ3 = | ......:...... | = | 0 σz |
| -1 0 : 0 0 | | -σz 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, (γo)2 = 1, (γ1)2 = (γ2)2 =
(γ3)2 = -1 form the basis of a CLIFFORD ALGEBRA.
However, Dirac was unaware of this previous work.
The Clifford algebra has the defining relationship:
{γμ,γν} = γμγν + γνγμ = 2ημνI
It is important to realise that the Dirac Gamma
matrices are not four-vectors. 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 4-component
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:
- - - -
γi = | 0 σi | where i = x, y, z and γ0 = | I 0 |
| -σi 0 | | 0 -I |
- - - -
Likewise we can decompose ψ into 2 two-component
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 (iγμ∂μ - m)ψ = 0
where c and h have been set to 1 we get:
- -
iγ0∂/∂t = | E 0 |
| 0 -E |
- -
- -
iγi∂/∂xi = | 0 σ.p |
| -σ.p 0 |
- -
Where,
- -
σ.p = σxpx + σypy + σzpz = | pz (px - ipy)|
|(px + ipy) -pz |
- -
Returning to the covariant form:
(iγ0∂/∂t + iγi∂/∂x - m)ψ = 0
∴ iγ0∂ψ/∂t = -iγi∂ψ/∂xi + mψ
∴ γ0E = -γiσ.p + mψ
- - - - - - - -
γ0E| 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:
- - - -
uφ = | 1 |φ uχ = | 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 4-component Dirac spinors, ui:
- - - -
| 1 | | 0 |
u1 = N| 0 | u2 = N| 1 |
| pz/(E + m) | | (px - ipy)/(E + m) |
| (px + ipy)/(E + m) | | -pz/(E + m) |
- - - -
- - - -
| -pz/(-E + m) | | -(px - ipy)/(-E + m) |
u3 = N| -(px + ipy)/(-E + m) | u4 = N| pz/(-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 = +√(p2c2 + m2c4) and 2 states for
E = -√(p2c2 + m2c4) .
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 = u1(p)exp(-ipx)exp(-iEt)
spin down particle: ψ = u2(p)exp(-ipx)exp(-iEt)
spin up antiparticle: ψ = u3(-p)exp(-ipx)exp(-iEt)
spin down antiparticle: ψ = u4(-p)exp(-ipx)exp(-iE)
Note: exp(-iωt) ≡ exp(-iEt/h) ≡ exp(-imc2t/h)
For a particle at rest:
Eγ0ψ = mψ
- - - - - -
| E 0 0 0 || u1 | | u1 |
| 0 E 0 0 || u2 | = m| u2 |
| 0 0 -E 0 || u3 | | u3 |
| 0 0 0 -E || u4 | | u4 |
- - - - - -
Using the characteristic polynomial gives:
- - - -
| E - m 0 0 0 || u1 |
| 0 E - m 0 0 || u2 | = 0
| 0 0 -E - m 0 || u3 |
| 0 0 0 -E - m || u4 |
- - - -
It is a property that the eigenspinors of a diagonal
matrix are:
- - - - - - - -
| 1 | | 0 | | 0 | | 0 |
u1 = | 0 | u2 = | 1 | v1 = | 0 | v2 = | 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, ψi = uiexp(-imc2t/h) which represents
the time evolution of ψi.
The Klein-Gordon Equation Revisited
-----------------------------------
Consider:
(iγμpμ - m)(iγνpν + m)ψ = 0
This equals:
(-γμγνpμpν - m2)ψ = 0
γμγν is rather complicated and consists of
terms involving (γ0)2p02 + (γ1)2p12 + ...
+ (γ0γ1 + γ1γ0) + ... - m2.
Now, from the Clifford algebra,
{γμ,γν} = 2ημνI = 0 if μ ≠ ν
Knowing this we are left with:
(-γμγνpμpν - m2)ψ = [(γ0)2p02 + (γ1)2p12 + (γ2)2p22
+ (γ3)2p32 - m2]ψ = 0
Now, (γo)2 = 1, (γ1)2 = (γ2)2 = (γ3)2 = -1
So we are left with:
[p02 - p12 - p22 - p32 - m2]ψ = 0
Or,
[ημνpμpν - m2]ψ = 0 where ημν = (+, -, -, -)
Or,
[-ημνpμpν + m2]ψ = 0
Or,
[pμpμ - m2]ψ = 0
Or,
[-pμpμ + m2]ψ = 0
Note:
pμ = (E/c,p1,p2,p3)
pμ = (-E/c,p1,p2,p3)
pμpμ = -E2/c2 + |p|2
= (-m2c4 + p2c2)/c2
= -m2c2 + p2
This is the Klein-Gordon equation.