Download The Dirac equation. A historical description.

The Dirac equation. A historical description.
(27. august 2012)
The Schrödinger equation may be written in the form:
Hψ = Eψ .
(1)
For a non-relativistic free particle the Hamiltonian is
H = HS =
(~p)2
,
2m
(2)
where
h̄ ~
∂
∇
, E → i h̄∂t ; ∂t ≡
.
(3)
i
∂t
The Schrödinger equation is based on the non-relativistic relation between
energy E and spatial momentum ~p = m ~v :
~p →
(~p)2
1
m (~v )2 =
.
2
2m
The corresponding relativistic relation is
E =
(4)
E 2 = (c ~p)2 + (m c2 )2 ,
(5)
where c is the speed of light. This relation (5) is the starting point for the
Klein-Gordon equation for the field Φ (at first erronously intepreted as the
wave function):
i
h
~ 2 + (m c2 )2 Φ ,
(ih̄ ∂t )2 Φ = (−cih̄ ∇)
which in covariant form with four-momentum
E
∂
ih̄
h̄ ~
pµ = ( , ~p) → i h̄ ∂ µ = i h̄
,
= ( ∂t , ∇)
c
∂ xµ
c
i
(6)
(7)
~ with ∂~ = − ∇):
~
can be written (note that ∂ µ = (∂0 , ∂)
pµ pµ Φ = (m c)2 Φ .
(8)
The adjoint field satisfy the same equation:
pµ pµ Φ† = (m c)2 Φ† .
(9)
If we multiply the equation (8) with Φ† from left and (9) with (-Φ) from right
and sum we obtain the continuitety equation
∂ ρKG
µ
~ · ~jKG = 0 ; j µ ≡ i Φ† ∂ µ Φ − (∂ µ Φ† ) Φ .
∂µ jKG
=
+ ∇
KG
∂t
(10)
The Klein-Gordon describe a spinless particle (s = 0). But because ρKG
might be negative it cannot be intepreted as a probabillity density such as
in non-relativistic quantum mechanics with probabillity density ρS = ψS† ψS .
The Klein-Gordon equation was therefore rejected as basis for a (possible )
relativistic quantum mechanics. However, in quantum field theory it is fine
that ρKG might be both positive and negative because eρKG can be intepreted
as an operator for charge density (e = electric charge).
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Dirac required an equation of Schrödinger form with a Hamiltonian
HD = c α
~ · ~p + βmc2 ,
(11)
where α
~ and β are operators which do not depend on space-time coordinates
x = (ct, ~x) or of ∂ µ . Because HD must be hermitean, α
~ and β must also be
hermitean:
α
~† = α
~
, β† = β .
(12)
In order that (11) should be the equation for a relativistic particle, HD
must satisfy the equation:
(HD )2 = (c ~p)2 + (mc2 )2 .
(13)
Squaring HD in (11), we see that in order to satisfy (13), β and αi (where
i = 1, 2, 3) must satisfy anti-commutation rules, and their square must be
equal to the unity operator 1 in the space which αi and β act :
β2 = 1 , { α
~ , β } = 0 , {αi , αj } = 2 δ ij 1 .
(14)
Operators this kind of algebra can be represented by matrices. The matrices
must be quadratic according to (12).
One might try to represent β and αi with 2 × 2 matrices by choosing
αi = σ i og β = 1 but then β og αi would not anti-commute. (However, for
mass m = 0 one may use αi = σ i because β is irrelevant in this case).
Because the matrices αi and β have a square equal to 1 they must have
eigenvalues equal to ±1. Further, because the four matrices anti-commutes
one may easily show that they have a trace equal to zero:
T r(β) = 0
, T r(αi) = 0 .
(15)
If one for example choose β diagonal there must be equally many +1 as −1
on the diagonal. We have already excluded 2 × 2 matrices. 3 × 3 matriser
with eigenvalues ±1 cannot be traceless, and therefore the next possibillity
is to try 4 × 4 matriser. There are infinitely many sets of β and αi which
satisfy (14). If one finds one set which fits, one may find another one by
using a similarity-transformation. The representation which is mostly used
is the Pauli-representation which has β diagonal and can be written in block
form
!
!
0 ~σ
1 0
β =
, α
~ =
,
(16)
0 −1
~σ 0
where
1 ≡ 12×2 =
1 0
0 1
!
,
0 ≡ 02×2 =
0 0
0 0
!
.
(17)
The Pauli-representation is for instance useful when studying the non-relativivtic
limit for the Dirac equation. Another representation is the Kramer-Weyl representation:
!
!
0 1
~σ 0
β =
, α
~ =
(18)
1 0
0 −~σ
2
which is useful when studying the Lorentz transformations of the Dirac equation. Note, however, that the spin operator
h̄
S = Σk
2
k
σk 0
0 σk
k
; Σ =
!
= γ5 α k .
(19)
is the same in all representations. For the pseudoscalar matrix γ5 we have
γ5 =
i
εµνσρ γ µ γ ν γ σ γ ρ
4!
(γ5 )2 = 1 .
;
(20)
~ describing spin is an axial vector
Note that α
~ is a vector operator, while Σ
operator. For the two representations, Pauli- og Kramer-Weyl- , we have
0 1
1 0
(γ5 )P =
!
;
(γ5 )KW =
1 0
0 −1
!
(21)
Multiplying the Dirac equation (11) from left β and using space-time
derivatives, we find the covariant version of the Dirac equation:
ih̄ γ µ ∂µ ψ = m c ψ
;
γ 0 ≡ β ; ~γ ≡ β α
~
(22)
Taking the hermittean conjugate of this equation we find the adjoint Dirac
equation
; ψ ≡ ψ† β .
(23)
−ih̄ (∂µ ψ)γ µ = m c ψ
In Quantum Field Theory (QFT) one uses units such that h̄ = c = 1.
The motivation is simplification of formulae (which often contains powers of
h̄ and c). Using these units causes no problem because the physical units
may be retained in the end.
It is found that the Dirac equation has solutions with both positive and
negative energy. Thus, forqa free particle (and for h̄ = c = 1), these energies
are E = ±Ep , with Ep = (~p)2 + m2 . The plane wave solutions of the Dirac
equation for positive and negative energy have the form
(−)
, ψpσ
= N vσ (~p)e+ip·x ,
(+)
= N uσ (~p)e−ip·x
ψpσ
(24)
where N is a normalisation factor and p · x = Ep t − ~p · ~x , . (Note that the
spatial momentum is quantised to −~p when the energy is negative). Within
the Pauli representation, the u- and v-spinors are given by:
uσ (~p) = K
χσ
~
σ ·~
p
χ
Ep +m σ
!
, vσ (~p) = K
~
σ ·~
p
χ
Ep +m σ
χσ
!
.
(25)
where K is some other normalization factor, and χσ are Pauli-spinors satisfying χ†σ χσ′ = δσσ′ , where σ, σ ′ are spin quantum numbers.
These plane wave solutions satisfy the ortogonality relations
(ε)
(ψpσ
,
(ε′ )
ψp′ σ′ )V
=
Z
V
(ε)
d3 x ψpσ
(x)
3
†
(ε′ )
ψp′ σ′ (x) = δεε′ δpp′ δσσ′ ,
(26)
provided
N ·K =
s
Ep + m
.
2Ep V
(27)
In this course we will choose
N = q
1
2Ep V
,
K =
q
Ep + m .
(28)
For zero spatial momentum,- the plane wave solutions are proportional to
the simple four component spinors





1
0
0
0









, 
0
1
0
0









, 
0
0
1
0









, 
0
0
0
1





,
(29)
corresponding to two energy signs and two spin states for each energy sign.
The Dirac equation gives the correct answer for “relativistic hydrogen”, if
the Coulomb potential V = −e2 /|x| was added to the Dirac-Hamiltonian in
(11). But still it was considered problematic with the negative energy states
which had to be there because of mathematical reasons.
In order to cure the “negative energy problem”, one tried to use the
Dirac “sea -theory” where vacuum should correspond to a state where all
the negative energy states were filled . Abscence of a negativ energy state (a
“hole in the sea”) should then be intepreted as an anti-particle. But by doing
so, one is leaving the intepretation of the Dirac equation as a description of
one particle. Even if one can find correct answers by using the “sea-theory”,
the basic problem can only be solved within QFT. A consistent relativistic
quantum mechanics does not exist, because the negative energy solutions
cannot be ignored and because the number of particles are not conserved in
relativistic theory, only the energy ( cfr. E = mc2 ).
Multiplying the equation (22) with ψ from left and (23) with (-ψ) from
right and sum, we find the continuity equation
∂ ρD
~ · ~jD = 0 ; j µ ≡ ψ † γ µ ψ
+∇
D
∂t
0
jD
≡ ρD = ψ † ψ
; j~D = ψ † α
~ψ
µ
∂µ jD
=
(30)
The Dirac equation describes a particle with spin s = 1/2. It was thought
that ρD should not be negativ, and this was good news for a possible relativistic quantum mechanical description. But in QFT this is not a point.
There the charge density eρD could also be negative because the operators
ψ og ψ contain anti-commuting components.
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