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129 Lecture Notes
Relativistic Quantum Mechanics
1 Need for Relativistic Quantum Mechanics
The interaction of matter and radiation field based on the Hamitonian
H =
(~p − ec A~)^2 2 m
Ze^2 r
∫ d~x
8 π
( E~^2 + B~^2 ). (1)
(Coulomb potential is there only if there is another static charged particle.) The Hamiltonian of the radiation field is Lorentz-covariant. In fact, the Lorentz covariance of the Maxwell equations is what led Einstein to propose his special theory of relativity. The problem here is that the matter Hami- tonian which describes the time evolution of the matter wave function is not covariant. A natural question is: can we find a new matter Hamiltonian consistent with relativity? The answer turned out to be yes and no. In the end, a fully consistent formulation was not obtained by modifying the single-particle Schr¨odinger wave equation, but obtained only by going to quantum field theory. We briefly review the failed attempts to promote Schr¨odinger equation to a rel- ativistically covariant one. Despite the failure, it resulted in the prediction that anti-matter exists, which was beautifully confirmed experimentally.
2 Klein–Gordon Equation
The Schr¨odinger equation is based on the non-relativisitc expression of the kinetic energy
E =
~p^2 2 m
By the standard replacement
E → i¯h
∂t
, ~p → −i¯h∇~, (3)
we obtain the Schr¨odinger equation for a free particle
i¯h
∂t
ψ = −
¯h^2 ∆ 2 m
ψ. (4)
A natural attempt is to use the relativistic version of Eq. (2), namely
( E c
) 2 = ~p^2 + m^2 c^2. (5)
Then using the same replacements Eq. (3), we obtain a wave equation
( h¯ c
∂t
) 2 φ = (¯h^2 ∆ − m^2 c^2 )φ. (6)
It is often written as (^) (
2 +
m^2 c^2 ¯h^2
) φ = 0, (7)
where 2 = ∂μ∂μ^ = (^1 c ∂t)^2 − ∆ is called D’Alambertian and is Lorentz- invariant. This equation is called Klein–Gordon equation. You can find plane-wave solutions to the Klein–Gordon equation easily. Taking φ = ei(p~·~x−Et)/¯h, Eq. (6) reduces to Eq. (5). Therefore, as long as energy and momentum follows the Einstein’s relation Eq. (5), the plane wave is a solution to the Klein–Gordon equation. So far so good! The problem arises when you try to rely on the standard probability interpretation of Schr¨odinger wave function. If a wave function ψ satisfies Schr¨odinger equation Eq. (4), the total probability is normalized to unity
∫ d~xψ∗(~x, t)ψ(~x, t) = 1. (8)
Because the probability has to be conserved (unless you are interested in seeing 5 times more particles scattered than what you have put in), this normalization must be independent of time. In other words,
d dt
∫ d~xψ∗(~x, t)ψ(~x, t) = 0. (9)
It is easy to see that Schr¨odinger equation Eq. (4) makes this requirement satisfied automatically thanks to Hermiticity of the Hamiltonian. On ther other hand, the probability defined the same way is not conserved for Klein–Gordon equation. The point is that the Klein–Gordon equation is second order in time derivative, similarly to the Newton’s equation of motion in mechanics. The initial condtions to solve the Newton’s equation of motion are the initial positions and initial velocities. Similarly, you have to give both
At this point, we don’t know what α~ and β are. The Dirac further required that this equation gives Einstein’s dispersion relation E^2 = ~p^2 c^2 + m^2 c^4 like the Klein–Gordon equation. Because the energy E is the eigenvalue of the Hamitonian, we act H again on the Dirac wave function and find
H^2 ψ = [c^2 αiαj^ pipj^ + mc^3 (αiβ + βαi)pi^ + m^2 c^4 β^2 ]ψ. (13)
In order for the r.h.s. to give just ~p^2 c^2 + m^2 c^4 , we need
αiαj^ + αj^ αi^ = 2δij^ , β^2 = 1, αiβ + βαi^ = 0. (14)
These equations can be satisfied if αi, β are matrices! Setting the issue aside why the hell we have to have matrices in the wave equation, let us find solutions to the above equations. There are of course infinite number of solutions related by unitary rotations, but the canonical choice Dirac made was
αi^ =
( 0 σi σi^0
) , β =
( 1 0 0 − 1
)
. (15)
They are four-by-four matrices, and σi^ are the conventional Pauli matrices. You can easily check the relations Eq. (14) using the matrices in Eq. (15). Correspondingly, the wave function ψ must be a four-component column vector. We will come back to the meaning of the multi-component-ness later. But the first point to check is that this equation does allow a conserved probability
i¯h
d dt
∫ d~xψ†ψ =
∫ d~x[ψ†(Hψ) − (Hψ)†ψ] = 0, (16)
simply because of the hermiticity of the Hamiltonian (note that ~α, β matri- ces are hermitean). This way, Dirac found a wave equation which satisfies the relativistic dispersion relation E^2 = ~p^2 c^2 + m^2 c^4 while admitting the probability interpretation of the wave function.
3.2 Solutions to the Dirac Equation
Let us solve the Dirac equation Eq. (12) together with the matrices Eq. (15). For a plane-wave solution ψ = u(p)ei(~p·~x−Et)/¯h, the equation becomes ( mc^2 c~σ · ~p c~σ · ~p −mc^2
) u(p) = Eu(p). (17)
This matrix equation is fairly easy to solve. The first point to note is that the matrix ~σ ·p~ has eigenvalues ±|~p| because (~σ ·~p)^2 = σiσj^ pipj^ = 12 {σi, σj^ }pipj^ = δij^ pipj^ = ~p^2. Using polar coordinates ~p = |~p|(sin θ cos φ, sin θ sin φ, cos θ), we find
~σ · ~p =
( pz px − ipy px + ipy −pz
) = |~p|
( cos θ sin θe−iφ sin θeiφ^ − cos θ
) , (18)
and their eigenvectors
~σ · ~pχ+(~p) = ~σ · ~p
( cos θ 2 sin θ 2 eiφ
) = +|~p|χ+(~p), (19)
~σ · ~pχ−(~p) = ~σ · ~p
( − sin θ 2 e−iφ cos θ 2
) = −|~p|χ−(~p). (20)
Once ~σ · ~p is replaced by eigenvalues ±|~p|, the rest of the job is to diagonalize the matrix (^) ( mc^2 ±|~p|c ±|~p|c −mc^2
)
. (21)
This is easily done using the fact that E =
√ |~p|^2 c^2 + m^2 c^4. In the end we find two eingenvectors
u+(p) =
√ E+mc^2 √ 2 mc^2 χ+(~p) E−mc^2 2 mc^2 χ+(~p)
(^) , u−(p) =
√ E+mc^2 2 mc^2 χ−(p~) −
√ E−mc^2 2 mc^2 χ−(~p)
(^). (22)
In the non-relativistic limit E → mc^2 , the upper two components remain O(1) while the lower two components vanish. Because of this reason, the up- per two components are called “large components” while the lower two “small components.” This point will play an important role when we systematically expand from the non-relativistic limit. An amazing thing is that there are two solutions with the same momen- tum and energy, and they seem to correspond to two spin states. Then the wave equation describes a particle of spin 1/2! In order to make this point clearer, we look at the conservation of angular momentum. The commutator
[H, Li] = [c~α · p~ + mc^2 β, ijkxj^ pk] = −i¯hcijkαj^ pk^6 = 0 (23)
does not vanish, and hence the orbital angular momentum is not conserved. On the other hand, the matrix
Σ =^ ~
( ~σ 0 0 ~σ
) , (24)
reason to prefer positive energy solutions over negative energy ones as far as the Dirac equation itself is concerned. What is wrong with having negative energy solutions? For example, suppose you have a hydrogen atom in the 1s ground state. Normally, it is the ground state and it is absolutely stable because there is no lower energy state it can decay into. But with the Dirac equation, the story is different. There are infinite number of negative energy solutions. Then the 1s state can emit a photon and drop into one of the negative energy states, and it happens very fast (it is of the same order of magnitude as the 2p to 1s transition and hence happens within 10−^8 sec for a single negative energy state. If you sum over all final negative-energy states, the decay rate is infinite and hence the lifetime is zero)! Such a situation is clearly unacceptable.
3.3 Anti-Matter
Dirac is ingenious not just to invent this equation, but also to solve the problem with the negative energy states. He proposed that all the negative energy states are already filled in the “vacuum.” In his reasoning, the 1s state cannot decay into any of the negetive energy states because they are already occupied, thanks to Pauli’s exclusion principle. It indeed makes the 1s state again absolutely stable. Now the equation is saved again. The “vacuum” with all the negative energy states (an infinite number of them) occupied is called the “Dirac sea.”
p
mc^2
2 mc^2
–2 mc^2
–2 mc – mc mc 2 mc
E
Figure 1: Dirac sea. All negative energy states are filled, while the positive energy states are not.
If you put an electron in one of the positive energy states, that is a normal electron with normal dispersion relation E =
p^2 c^2 + m^2 c^4. One the other hand, you can remove one electron from the Dirac sea. Let us remove an
electron of momentum ~p and energy −
~p^2 c^2 + m^2 c^4. Relative to the Dirac sea, the state has momentum −~p because you have removed the momen- tum ~p. The energy is positive, because you have removed a negative energy −
~p^2 c^2 + m^2 c^4. Therefore, this state has a normal dispersion relation. An important point is that you also have removed the electric charge e < 0 of the electron. Therefore this state has the electric charge −e > 0. What it means is that this is a particle of positive charge√ −e with momentum −~p and energy ~p^2 c^2 + m^2 c^4. This is a new particle, “positron.” It is the anti-particle of the electron. Dirac theory hence predicts the existence of an anti-particle for any spin 1/2 particles.^1
p
mc^2
2 mc^2
–2 mc^2
–2 mc – mc mc 2 mc
E
p
mc^2
2 mc^2
–2 mc^2
–2 mc – mc mc 2 mc
E
Figure 2: An electron in a positive energy state is a normal electron. If you remove an electron from one of the negative energy states, the state has a positive electric charge and is a positron.
If you inject enough energy, you may excite an electron in the negative energy state to a positive energy state. Then you have created a pair of an electron and a positron. However, the excited electron will eventually decay back down to the negative energy state by releasing energy (typically two of three photons). This is a “pair annihilation” process. Pair creation and pair annihilation are common phonemena in high-energy physics. But there is a catch with the “Dirac sea.” We wanted to find a single- particle wave function which is consistent with both relativity and probability interpretation. The Dirac equation indees seems to be consistent both with relativity and probability interpretation. But the correct implementation calls for a multi-body state (actually, an infinite-body state)! We can’t just talk about a single particle wave function ψ(~x) for a single electron, but
(^1) Dirac himself, being afraid of predicting a non-existing particle, initially claimed that this positively charged hole must be the proton. But other people, notably Robert Op- penheimer, pointed out that the hole must have the same mass as the electron.
3.5 Coupling to the Radiation Field
The interaction between the Dirac field and the Electromagnetic Field follows the same rule in the Schr¨odinger theory ~p → ~p− ec A~, or equivalently, −ih¯∇ →~
−i¯h∇ −~ ec A~. Its Lorentz-covariant generalization also determines the time- derivative: i¯h (^) ∂t∂ → i¯h^1 c∂t∂ − ec φ. (The relative sign difference is due to the fact
that Aμ = (φ, − A~) transforms the same way as the derivative ∂μ = (^1 c∂t∂ , ∇~).) Therefore, the Dirac equation is ( i¯h
∂t
− eφ − c~α · (−i¯h∇ −~
e c
A~) − mc^2 β
) ψ. (29)
For stationary states, we are interested in solving the equation
[ c~α ·
( −i¯h∇ −~
e c
A~
)
] ψ = Eψ. (30)
The way we will discuss it is by a sytematic expansion in ~v = ~p/m. It is basically a non-relativisic approximation keeping only a few first orders in the expansion. Let us write Eq. (30) explicitly in the matrix form, and further write E = mc^2 + E′^ so that E′^ is the energy of the electron on top of the rest energy. We obtain
( eφ c~σ · (−i¯h∇ −~ ec A~) c~σ · (−i¯h∇ −~ ec A~) − 2 mc^2 + eφ
) ψ = E′ψ. (31)
The solution lives mostly in the large components, i.e.. the upper two com- ponents in ψ. The equation is diagonal in the absence of ~σ · (−i¯h∇ −~ ec A~), and we can regard it as a perturbation and expand systematically in powers of it. To simplify notation, we will write ~p = −i¯h∇~, even though it must be understood that we are not talking about the “momentum operator” ~p acting on the Hilbert space, but rather a differential operator acting on the field ψ. Let us write four components in terms of two two-component vectors,
ψ =
( χ η
) , (32)
where the large component χ is a two-component vector describing a spin two particle (spin up and down states). η is the small component which
vanishes in the non-relativistic limit. Writing out Eq. (31) in terms of χ and η, we obtain
eφχ + c~σ · (~p −
e c
A~)η = E′χ (33)
c~σ · (~p −
e c
A~)χ + (− 2 mc^2 + eφ)η = E′η. (34)
Using Eq. (34) we find
η =
E′^ + 2mc^2 − eφ
c~σ · (~p −
e c
A~)χ. (35)
Substituting it into Eq. (34), we obtain
eφχ + c~σ · (~p −
e c
A~) 1
E′^ + 2mc^2 − eφ
c~σ · (p~ −
e c
A~)χ = E′χ. (36)
In the non-relativistic limit, E′, eφ mc^2 , and hence we drop them in the denominator. Within this approximation (called Pauli approximation), we find
eφχ +
[~σ · (~p − ec A~)]^2 2 m
χ = E′χ. (37)
The last step is to rewrite the numerator in a simpler form. Noting σiσj^ = δij^ + iijkσk,
[~σ · (~p −
e c
A~)]^2 = (δij^ + iijkσk)(pi^ − e c
Ai)(pj^ −
e c
Aj^ )
= (~p −
e c
A~)^2 + i 2
ijkσk[pi^ −
e c
Ai, pj^ −
e c
Aj^ ]
= (~p −
e c
A~)^2 + ie 2 c
ijkσki¯h(∇iAj^ − ∇j Ai)
= (~p −
e c
A~)^2 − e¯h c
~σ · B.~ (38)
Then Eq. (37) becomes
(~p − ec A~)^2 2 m
χ − 2
e¯h 2 mc
~s · B~ + eφχ = E′χ. (39)
In other words, it is the standard non-relativistic Schr¨odinger equation except that the g-factor is fixed. The Dirac theory predicts g = 2! This is a great success of this theory.
