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Relativistic Quantum Mechanics
Special relativity and quantum mechanics fit together nicely. We have already handled one relativistic particle, the photon. In this section, we learn how to treat other particles when their velocities are no longer negligible compared to the speed of light.
Relativity Review Basically, we want to formulate quantum mechanics in a way that relativistic invariance is manifest. In classical physics, this is accomplished by the use of 4-vectors. To establish notation, we briefly review classical relativity in the language of 4-vectors. The coordinate 3-vector and time are the basic components of the 4-vector xμ, spec- ified by xμ^ = (x^0 x,⃗ ),⃗ x = (x^1 , x^2 , x^3 ), x^0 = ct. (1)
This is saying that we will take the familiar 3-vectorx⃗ and regard its three components as part of the 4-vector. What we often call x is now x^1 , y is x^2 , and z is x^3. Together with x^0 ≡ ct, these make up the basic space-time 4-vector. In special relativity, we need to distinguish upper and lower indices. So along with xμ^ we also have xμ, specified by xμ = (x^0 , −x⃗ )
The scalar product of a 4-vector with itself involves both xμ^ and xμ. We define
x · x = xμxμ^ = (x^0 )^2 −x⃗ ·x,⃗ (2)
where the convention is that repeated upper and lower indices are summed, so xμxμ^ really means ∑^3
μ=
xμxμ
More formally, the relation between xμ and xμ^ can be written using the metric tensor gμν , xμ = gμν xν^ ,
or equivalently, xμ^ = gμν^ xν ,
where again there is a sum on the repeated index ν. By making use of Eq.(2) we find that the components of gμν are given by
gμν :
g 00 = 1
gkk = − 1
gμν = 0, μ ̸= ν
and further that gμν = gμν^. It is easy to remember the rule for raising and lower indices; for a 0 or time index, raised and lowered versions are the same, whereas for a spacial index (1,2,3), raised and lowered versions differ by a minus sign, e.g. x 1 = −x^1.
3-vectors and 4-vectors The discussion above was for the space-time 4-vector xμ. How do we find other 4-vectors? In ordinary three dimensional physics, we have a number of 3-vectors. They are usually denoted by an over-arrow, e.g. V .⃗ A given ⃗V has three components. Index placement is of no concern when dealing with ordinary 3-vectors, so the components of V⃗ could be written in several equivalent ways, e.g. V⃗ = (V 1 , V 2 , V 3 ) or V⃗ = (V 1 , V 2 , V 3 ) are equally valid, so here V 1 = V 1 , etc and for writing convenience,
usually use the lower index form, V⃗ = (V 1 , V 2 , V 3 ). The story changes when we go to 4-vectors and incorporate velocity transformations as well as rotations. Here, since 4-vector scalar product involves some minus signs, it is necessary to distinguish upper and lower components, so if V⃗ is part of a 4-vector, in most cases we write V μ^ = (V 0 ,⃗ V ) = (V 0 , V 1 , V 2 , V 3 ). Now it does matter where the indices are placed, so V 0 = V 0 , but Vk = −V k, k = 1, 2 , 3. The 4-vector xμ^ is the prime example of this. Below is a list of other cases where the generalization from 3-vector to 4-vector works just like it does for xμ.
Energy-momentum
pμ^ = (p^0 p,⃗ ), p^0 =
E
c
Scalar-Vector Potential Aμ^ = (A^0 ,⃗ A), A^0 = ϕ
Charge-Current Density Jμ^ = (J^0 ,⃗ J), J^0 = ρc
Examining this list might suggest that all 3-vectors are merely the spacial components of 4-vectors. This is certainly not true for some important 3-vectors. The electric and magnetic fields as well as any form of angular momentum, orbital, spin, or total are not spacial components of 4-vectors, but in fact become parts of second rank tensors in special relativity. The electromagnetic case is discussed in the next section. There is one last important case where the 3-vector does generalize to a 4-vector, but the index placement is different than the list above. That is the case of the gradient operator, ∇⃗. We may write ∇⃗ in various ways, e.g.
∇⃗ = ( ∂ ∂x
∂y
∂z
is fine in ordinary 3-dimensional physics. But for generalization to special relativity, we replace x, y, z by x^1 , x^2 , x^3. In this form, ∇⃗ reads
∇⃗ = ( ∂ ∂x^1
∂x^2
∂x^3
The index placement here suggests that ∇⃗ is part of a 4-vector generalization of three dimensional gradient, but with lower indices. We can see that this is correct as follows. Let us write
∇⃗ = ( ∂ ∂x^1
∂x^2
∂x^3
The other components of the magnetic field follow in a similar way, so we have
F 12 = −B 3 , F 23 = −B 1 , F 31 = −B 2
The electric field is in the space-time components. Writing out F 01 , we have
F 01 = ∂ 0 A 1 − ∂ 1 A 0 = −
c
∂A^1
∂t
∂ϕ ∂x^1
, = E 1 ,
so F 01 = E 1 , F 02 = E 2 , F 03 = E 3.
The tensor Fμν is antisymmetric, Fμν = −Fνμ, and so has 6 components, which comprise the two 3-vectors, E,⃗ and B.⃗ With Fμν in hand, it is easy to check that the following covariant equation contains both Ampere’s and Gauss’ laws,
∂μFμν =
4 π c
Jν. (4)
As an example, consider Eq.(4) for ν = 1. We have
∂^0 F 01 + ∂^2 F 21 + ∂^3 F 31 =
4 π c
J 1.
Using results from above, this is the same as
1 c
∂E 1
∂t
− ∇ 2 B 3 + ∇ 3 B 2 = −
4 π c
J^1 ,
which is the 1-component of Ampere’s law.
Relativistic Plane Waves Treating plane waves is a simple way to deduce relativistic wave equations. Suppose we have a one photon state, |⃗k, λ >. From the previous notes on photons, the matrix element of the vector potential is (using box normalization)
< 0 |A⃗ |⃗kλ >=
√ 4 π¯hc^2 2 ωkV
exp(i⃗k ·x⃗ − iω⃗ktϵ)⃗ (^) ⃗k.
The photoelectric effect and the Compton effect establish the relation between ω,⃗ k and energy-momentum. (Prior to the experimental results, de Broglie also used relativistic reasoning to introduce his famous wavelength.) We have
p⃗ = ¯h⃗k, Ep(⃗ ) = ¯hω⃗k = p|⃗ |c.
Rewriting the photon plane wave space-time dependence in terms of energy-momentum gives
exp(i⃗k ·x⃗ − iω⃗kt) = exp(
i ¯h⃗
p ·x⃗ −
i ¯h
Ep(⃗ )t) = exp(−i
p · x ¯h
The middle term in Eq.(5) is very familiar in non-relativistic quantum mechanics, where we use the formula Ep(⃗ ) =p⃗ p/·⃗ 2 m. The example of the photon implies that this general form is also valid for a relativistic particle; we need only replace the non-relativistic formula for Ep(⃗ ) with its relativistic counterpart. For a massive particle, we have
p · p = (
Ep(⃗ ) c
)^2 −p⃗ ·p⃗ = (m 0 c)^2 ,
which gives the Einstein formula,
(Ep(⃗ ))^2 = (m 0 c^2 )^2 +pc⃗ ·pc.⃗ (6)
So as far as its space-time dependence is concerned, a relativistic particle of definite 4-momentum will be described by
exp(−
i ¯h
p · x), where p^0 =
Ep(⃗ ) c
c
√ ((m 0 c^2 )^2 +p⃗ ·pc⃗ 2.
We already know the factors that accompany the plane wave factor for photons. The corresponding factors for massive particles, including spin 1/2, will be deduced later in this section. We will also learn the meaning of taking the minus sign in the energy equation,
Ep(⃗ ) = −
√ ((m 0 c^2 )^2 +p⃗ ·pc⃗ 2.
Operators Once we have a plane wave, in non-relativistic quantum mechanics, we can bring down the 3-momentum by acting with the momentum operator,
P⃗ = h⃗¯∇ i
In order to use this to define a 4-vector operator, we need to do a little index manipulation. Using the rule for that for the momentum 3-vector, its components have upper indices (see Eq.(3), we can write Eq.(7) as
P k^ =
¯h∂k i
, or Pk = i¯h∂k,
where we lowered the index on P k^ and used the fact that ∇k = ∂k. Having gotten the indices in the same position on both sides of the equation, the generalization to a 4-momentum operator is straightforward,
Pμ ≡ i¯h∂μ,
with components P 0 = i¯h∂ 0 , Pk = ih∂¯ k = i¯h∇k.
To understand what to write for our scalar particle it is useful to review the concept of so-called “probability current.” Non-relativistic particles are conserved and as such carry a conserved quantum number, and there is a conserved current which goes along with any such conserved quantity. In non-relativistic quantum mechanics, we have
ρ(t,⃗x ) ≡ Ψ∗(t,⃗x )Ψ(t,⃗x ),⃗ J(t,⃗x ) ≡
¯h 2 mi
Ψ∗(t,⃗x )
↔ ∇ Ψ(t,⃗x^ ),^ (11)
where Ψ∗(t,⃗x )
↔ ∇ Ψ(t,⃗x^ )^ ≡^ Ψ∗(t,⃗x^ )∇⃗ Ψ(t,⃗x^ )^ −^ (^ ∇⃗ Ψ∗(t,⃗x^ ))Ψ(t,⃗x^ )
The Schr¨odinger equation guarantees that
∂ρ ∂t
+ ∇ ·⃗ J⃗ = 0, (12)
and the scalar product of Eq.(10) is the same as
< Ψ|Ψ >=
∫ dxρ^3 ⃗ (t,⃗x ). (13)
To find the correct definition of scalar product for our charged scalar field ϕ, we follow a similar route. Particle number is not a good concept in relativistic quantum mechanics, but for our current we may simply use a multiple of the electromagnetic current. Guided by the non-relativistic formula, we define probability current
Jμ = i¯hϕ∗^
↔ ∂μ ϕ, (14)
where ϕ∗^
↔ ∂μ ϕ = ϕ∗∂μϕ − (∂μϕ∗)ϕ.
(The relation of Jμ as defined in Eq.(14) to the actual electromagnetic current will become clear later.) It is easy to check that if ϕ satisfies the Klein-Gordon equation, Eq.(9), that Jμ is conserved, i.e.
∂μJμ = ∂μJμ^ = ∂ 0 J 0 + ∇ ·⃗ J⃗ = 0. (15)
(Note that again the lower index nature of ∇μ is playing a role here; there is a plus sign
before ∇ ·⃗ J⃗ in Eq.(15).) The individual components of Jμ^ are
J^0 = J 0 = i¯hϕ∗^
↔ ∂ 0 ϕ,⃗ J =
¯h i
ϕ∗
↔ ∇ ϕ.^ (16)
What we want in the scalar product is ρ. Noting that
∂ 0 J^0 = ∂t
J^0
c
we have that ρ = J^0 /c. We now can define the quantum mechanical scalar product as
< ϕ|ϕ >=
∫ dxρ^3 ⃗ =
c^2
∫ dxϕ^3 ⃗ ∗i¯h
↔ ∂t ϕ (17)
An important special case of Eq.(17) is using it to correctly normalize a plane wave. Going back to the non-relativistic case, we would write
Ψp⃗ (t,⃗x ) =
(2π¯h)^3 /^2
exp(−
iEpt ¯h
pi⃗ ·x⃗ ¯h
where here of course Ep = p^2 / 2 m. The standard normalization is that
∫ dx^3 ⃗ Ψ∗ p′ Ψp = δ^3 p(⃗ − p⃗′). (19)
For our relativistic scalar particle, let us write
ϕp =
Np (2πh¯)^3 /^2
exp(−
iEpt h ¯
pi⃗ ·x⃗ ¯h
(here Ep =
√ (m 0 c^2 )^2 + (pc)^2 .) We will determine the normalization factor Np by de- manding that
< ϕp′ |ϕp >= i
¯h c^2
∫ dx^3 ⃗ (ϕ∗ p′
↔ ∂t ϕp) = δ^3 p(⃗ − ⃗p′)
Using the formula Eq.(20), we have
< ϕp′ |ϕp >=
(2π¯h)^3 /^2
(Np′ Np)(
Ep′ + Ep c^2
∫ dx^3 ⃗ exp(
i ¯h
p(⃗ − p⃗′) ·x⃗ )) = N (^) p^2
2 Ep c^2
δ^3 p(⃗ − ⃗p′). (21) Demanding that the result be δp^3 ⃗ − p⃗′), we find
Np =
vu ut c^2 2 Ep
so a correctly normalized plane wave for the Klein-Gordon equation is
ϕp(x 0 x,⃗ ) =
(2πh¯)^3 /^2
vu ut c^2 2 Ep
exp(−
iEpt ¯h
pi⃗ ·x⃗ ¯h
for Ep =
√ (m 0 c^2 )^2 + (pc)^2 .) The result in Eq.(22) is more general than the present discussion would imply. Imagine a real process in which a scalar particle of the type we are discussing is incident. The particle may be destroyed, or other particles may be produced. Nevertheless, before the interaction occurs, the particle is in a plane wave state, and Eq.(22) would be the correct factor for such a particle in the initial state. To
change normalization to a delta function in wave vector ⃗k instead of momentump,⃗ we merely replace (2π¯h)^3 /^2 by (2π)^3 /^2 , and to go to box normalization replace (2π)^3 /^2 by 1 /
V.
Applying P · γ again, we have
P · γP · γΨ = m 0 cP · γΨ(x) = (m 0 c)^2 Ψ(x). (24)
Since at this point, we are describing a free particle with mass m 0 , every component of Ψ must satisfy the Klein-Gordon equation,
P · P Ψ = (m 0 c)^2 Ψ(x). (25)
Comparing Eqs.(24) and (25), we must have
P · γP · γ = P · P,
or in terms of differential operators after canceling factors of i¯h, we have
∂μγμ∂ν γν^ = ∂μ∂ν γμγν^ = ∂ · ∂ = ∂μ∂ν gμν^.
Now ∂μ∂ν = ∂ν ∂μ. Using this and relabeling gives
∂μ∂ν γμγν^ =
∂μ∂ν (γμγν^ + γν^ γμ) = ∂μ∂ν gμν^.
Comparing the two terms in this equation, we finally have
γμγν^ + γν^ γμ^ = 2gμν^ I. (26)
where I is the identity matrix. At this point, it is not obvious how many components Ψ has. It turns out that to find a set of four matrices satisfying Eq.(26) requires that Ψ have at least four components, or that the γμ^ be 4×4 matrices. There are many representations of the γμ, just as we might use other representations than the one introduced by Pauli for 2 × 2 matrices. The representation of the γμ^ which is most useful for seeing how the Dirac equation reduces to the Schr¨odinger equation in the non-relativistic limit is the following:
γ^0 =
( I 0 0 −I
) (27)
γk^ =
( 0 σk −σk^0
) ,
where the γμ^ are represented as blocks of 2 × 2 matrices, and the σk^ are Pauli matrices. Let us now explore the solution of the Dirac equation,
i¯h∂μγμΨ(x) = m 0 cΨ(x) (28)
for a free particle. For a free particle with positive energy in a plane wave state, we may write Ψ(x) as
Ψ(x) = up(⃗ ) exp(−
i ¯h
p · x).
It is important to keep in mind that p^0 is not a free variable ( as isp⃗ ) but is constrained by p^0 c = Ep(⃗ ), where Ep(⃗ ) is given by the Einstein formula, Eq.(6). This is why u is denoted as up(⃗ ). Since the γμ^ are represented in 2×2 blocks, it is natural to represent Ψ, or equivalently, up(⃗ ) in terms of 2-component spinors. The standard notation for this is
up(⃗ ) = N
( ϕp(⃗ ) χp(⃗ )
) ,
where ϕp(⃗ ) and χp(⃗ ) are 2-component spinors or column vectors, exactly like those used to describe spin 1/2 in non-relativistic quantum mechanics, and N is a normalization factor. (N will be determined below. Explicit spin labels will also be added further on.) Now i¯h∂μ brings down pμ, so Eq.(28) becomes
p · γup(⃗ ) = m 0 c up(⃗ ),
or (^) ( γ^0 Ep(⃗ ) −γ⃗ ·p⃗
) up(⃗ ) = m 0 c up(⃗ ).
Using the representation of Eq.(27) for the γ−matrices, we have
Ep(⃗ ) −σ⃗ ·pc⃗⃗
σ ·pc⃗ −Ep(⃗ )
ϕ
χ
= m 0 c 2
ϕ
χ
.
Writing this out as two equations for the spinors ϕ and χ, we have
Ep(⃗ )ϕ −σ⃗ ·pc χ⃗ = m 0 c^2 ϕ,
and −Ep(⃗ )χ +σ⃗ ·pc ϕ⃗ = m 0 c^2 χ.
These two equations contain the same information, and show that ϕ and χ are not independent. Solving for χ, we have
χ =
σ ·pc⃗ m 0 c^2 + Ep(⃗ )
ϕ. (29)
This equation shows that in the non-relativistic limit,
p |⃗ | m 0 c
χ becomes O(v/c) relative to ϕ, where v/c = p|⃗ |/m 0 c. This means that in the extreme non-relativistic limit, we have only a two-component spinor as expected for a spin 1/ particle.
Now from Eq.(31), we can write Ψ†^ = ΨΓ¯ −^1. Using this and Eq.(33) in Eq.(32), we have as the equation satisfied by Ψ¯,
−i¯h∂μ Ψ¯γμ^ = Ψ¯m 0 c. (34)
Returning to the Dirac current, we have finally have the current conservation equation,
i¯h∂μJμ^ = i¯h∂μ Ψ¯γμΨ = Ψ(¯ m 0 c − m 0 c)Ψ = 0.
Probability and Scalar Product The current we have just defined allows defini- tion of a probability density for a Dirac spinor. As in the non-relativistic case, the 0th component of the current can be thought of as a probability density. We have
J^0 = Ψ¯γ^0 Ψ,
but since Γ = γ^0 , we have J^0 = Ψ†Ψ.
This resembles the result in non-relativistic quantum mechanics for spin 1/2, but here we use the full 4-component Dirac spinor. The quantity Ψ†Ψ is clearly non-negative and allows a probability interpretation. If we have two Dirac spinor states, Ψ 1 and Ψ 2 , we may define a scalar product of these two states by
∫ d⃗x Ψ† 2 x, t(⃗ )Ψ 1 x, t(⃗ ).
Note that the scalar product or probability amplitude is defined just as in the non- relativistic case at a given time.
Spin Indices and Normalization of Dirac Spinors Let us first deal with spin indices. Since ϕ is a 2-component spin 1/2 spinor, it would seem natural to give it an index s, referring to “up” or “down” along some arbitrarily chosen z-axis. While this can be done, it is much more useful to choose the axis of spin quantization to be along the 3-momentum of the particle. This is a “helicity” basis, completely analogous to circular polarization states for a photon. We will use λ to denote the helicity basis index, where λ = ± 1 /2 are the allowed values. The 2-spinor ϕλ then satisfies by definition,
pˆ ·σϕ⃗ (^) λ = 2λϕλ,
and we will choose ϕ† λϕλ = 1. Eq.(29) determining χ now becomes very simple. We have
χλ =
( p|⃗ |c m 0 c^2 + Ep(⃗ )
) 2 λϕλ
Using the helicity basis, both ϕ and χ have definite spin components alongp.⃗
Finally, let us normalize the plane wave spinor, which we now denote as uλp(⃗ ). Various choices are possible, but the most common choice of norm is
u¯λp(⃗ )uλ′^ p(⃗ ) = δλλ′^.
Writing this out, we have
|N |^2
( ϕ† λ χ† λ
) ( I 0
0 −I
) ( ϕλ χλ
) = 1.
Using Eq.(29) this gives
|N |^2 ϕ† λϕλ
( 1 −
pc ·pc⃗ (m 0 c^2 + Ep(⃗ ))^2
)
. = 1
Since ϕλ is normalized to unity, this leaves
|N |^2
( 2 m 0 c^2 m 0 c^2 + Ep(⃗ )
) = 1,
where we used the Einstein formula Eq.(6). Solving for N , we have
uλp( ⃗) =
√ m 0 c^2 + Ep(⃗ ) 2 m 0 c^2
( ϕλ χλ
) .
For a free particle, we will normalize states as usual to δ functions in 3-momentum. So if we have two solutions Ψpλ⃗ x, t(⃗ ) and Ψp⃗′λ′ x, t(⃗ ), we require
< Ψ (^) p⃗′λ′ |Ψpλ⃗ >=
∫ d⃗x Ψ†p⃗′λ′ x, t(⃗ )Ψpλ⃗ x, t(⃗ ) = δ^3 p(⃗ − ⃗p′)δλλ′ (35)
The spinor Ψpλ⃗ (x) must contain a factor of uλp(⃗ ), but since we have normalized the uλp(⃗ ) in a way which takes no account of the overall normalization of our plane wave states, we must allow for another normalization constant, N .˜ We may write
Ψpλ⃗ (x) = N u˜ λp(⃗ ) exp(−
i ¯h
p · x).
Using this form for both Ψpλ⃗ x, t(⃗ ) and Ψ⃗p′λ′x, t (⃗ ), in Eq.(35) we arrive at
(2π¯h)^3 δ^3 p(⃗ − p⃗′)| N˜ |^2 u† λ′ p(⃗ )uλp(⃗ ) = δ^3 p(⃗ − p⃗′)δλλ′^ (36)
Now, using formulas established above it is easy to show that
u† λ′ p( ⃗)uλp(⃗ ) =
( Ep(⃗ ) m 0 c^2
) δλλ′
The Dirac equation for a particle in an external 4-potential is then ( i¯h∂μ −
q c
Aμ
) γμΨ = m 0 cΨ
We will explore the consequences of this in the next section.
Hamiltonian for Dirac Equation Let us suppose there is an external vector potential present, as well as an external scalar potential. The latter does not occur in atomic physics, but does occur in nuclear and high energy physics. NOTE For the most part in this section, we will work in natural units with ¯h = c = 1, and we will denote m 0 by m. The Dirac equation now reads
(i∂μ − qAμ) γμΨ = (m + VS )Ψ (40)
Following Dirac’s original treatment, we want to transform this into the form
i∂tΨ = HΨ.
To do so, we write out the various terms in Eq.(40), (note that with c = 1, ∂ 0 = ∂t) ( i∂tγ^0 + (i∂k − qAk)γk)
) Ψ = (m + VS )Ψ.
Now multiply this equation on the left by γ^0 , and move all terms except the term in ∂t to the right hand side. This gives
i∂tΨ = −(i∂k − qAk)γ^0 γkΨ + qA 0 Ψ + (m + VS )γ^0 Ψ
The right hand side is the Hamiltonian acting Ψ. We may re-arrange this as follows. Define
αk^ ≡ γ^0 γk^ =
( 0 σk σk^0
) .
Returning to ordinary 3-vector notation, we have
i∂tΨ =
(
i
− q⃗A) ·α⃗ + qA 0 + (m + VS )γ^0
Ψ.^ (41)
The quantity in { } is the Hamiltonian for the Dirac equation. We may break H up as usual, H = H 0 + V,
where
H 0 = (
i
) ·α⃗ + mγ^0 ,
and V = −q⃗A ·α⃗ + qA 0 + VS γ^0
Scattering may be treated using V and the U (∞, −∞) operator. To first order in V , the matrix element for scattering for a Dirac particle with initial momentum and helicityp, λ⃗ to a final p⃗′, λ′, would be
< f |U (∞, −∞)|i >= −i
∫ (^) ∞
−∞
dt < f |eiH^0 tV e−iH^0 t|i >,
The exponentials in H 0 just give initial and final particle energies, so
< f |eiH^0 tV e−iH^0 t|i >=
∫ d⃗x
√ m E′V
( u† λ′ (p⃗′) V x(⃗ )uλp(⃗ )
) √^ m EV
e−i(p−p
′)·x ,
where we took the case of a static V, although that is not necessary. We again emphasize that p^0 = Ep(⃗ ), and likewise for (p′)^0. Aside from the additional complexity from Dirac spinors rather than 2-component spinors, this formula has an overall similarity to the case of first order scattering of a spin 1/2 particle in non relativistic quantum mechan- ics. Returning to the matrix element of U (∞, −∞), the combination of space and time integrations is an integral d^4 x over all space-time.
Magnetic Moment One of the great successes of the Dirac equation is that it gives a prediction for the magnetic moment of the electron. The value the Dirac equation gives is then corrected by quantum electrodynamic effects, which generate a series of corrections in powers of the fine structure constant. This section treats the prediction of the Dirac equation itself. Returning to Eq.(41), assuming an energy eigenstate and breaking Ψ into 2-component spinors, we have
which is the familiar term in the non relativistic Hamiltonian for coupling a charge to an external magnetic field. The magnetic moment coupling is contained in the term analogous to iϵjklCj^ Dkσl. Writing out the σ^3 term from Eq.(43) gives
i 2 m
( (P 1 −
q c
A^1 )(P 2 −
q c
A^2 ) − (P 2 −
q c
A^2 )(P 1 −
q c
A^1 )
) σ^3
q¯h 2 mc
(∂ 1 A^2 − ∂ 2 A^1 )σ^3 = −
q¯h 2 mc
B^3 σ^3
There are analogous terms involving other components ofσ⃗ and corresponding compo- nents of B.⃗ For an electron, q = −e, so we have for the magnetic moment of the electron
μ⃗ (^) elect = −
e¯h 2 mc⃗
σ.
The factor multiplyingσ⃗ is the Bohr magneton, so if we write
μ⃗ (^) elect = −(
e¯h 2 mc
)gelect S,⃗ where S⃗ =
σ,⃗
we have gelect = 2. The conclusion is that the Dirac equation gives the g factor of the electron, which otherwise has to be fed in from experiment.
