﻿ Dirac's equation solution of hydrogen atom

# Truth of Dirac equation for hydrogen atom.

## Dirac's hydrogen = Bohr-Sommerfeld model.

(Fig.1) Accidental coincidence ?

But as shown on this page, famous fine structure was first gotten by Bohr-Sommerfeld model in 1916.

The fine structure used formally now is the hydrogen solution by Dirac equation.
Surprisingly, these solutions by Dirac equation are just equal to those of Sommerfeld model, about which ordinary people do not know.

In Fig.1, nr and nφ mean radial and tangential Sommerfeld quantization numbers, which express de Broglie wavelength in each direction, as shown on this page.
( In Bohr model, one round orbit is an integer times de Broglie wavelength. )
And Dirac's "n" and "j" mean the energy level and total spin angular momentum ( j = l + s ).

In fact, this Dirac hydrogen includes fatal flaws, so wrong.

## Strange "coincidence" in fine structure.

(Fig.2) Fine structure by Sommerfeld and Dirac.

Fig 2 shows the fine structure which shows the difference inside the energy level of "2".
As shown on this page, Sommerfeld model extended Bohr's circular orbit to elliptic one.
Circular orbit contains only tangential motion, and elliptical orbit includes radial momentum, too.

Considering the relativistic effect (= maximum speed "c" ), these two orbits are slightly split in the energy level (= fine structure ).
On the other hand, Dirac hydrogen contains three kinds of states ( 2s1/2, 2p1/2, 2p3/2 ) in the energy level "2".
But the spectrum result shows ONLY two split lines.

## Examples of Dirac = Sommerfeld in energy levels.

(Fig.3) Dirac = Sommerfeld !?

Fig.3 upper shows the energy level of Sommerfeld elliptic orbit ( nr = 1, nφ = 1 ).
Surprisingly, Dirac's 2s1/2 and 2p1/2 states are completly equal to Sommerfeld's elliptic otbit, though their sources are completely different from each other, as shown in Fig.3 lower.
( Dirac hydrogen includes "spin" effects and Sommerfeld's does not. )

"n = 2" represents the energy level = 2, and "j = 1/2" represents total angular momentum = 1/2.
Dirac's 2s1/2 has no orbital angular momentum, but has spin angular momentum 1/2, so j = 0 + 1/2 = 1/2.
On the other hand, 2p1/2 has orbital angular momentum ( l = 1 ), and spin, so j = 1 - 1/2 = 1/2.
As a result, these are the same also in the energy level. This is too good to be true.

(Fig.4) Dirac = Sommerfeld !?

Fig.4 upper shows Sommerfeld's circular orbit ( nφ = 2, nr = 0 ).
( Circular orbit means it does not include radial motion. )
Fig.4 lower expresses Dirac's 2p3/2 state in which total angular momentum j = 1 + 1/2 = 3/2.

The maximum orbital angular momentum in Dirac's energy level "2" is l = 1 ( p state ), so j = 2/3 is the maximum in n = 2.
Surprisingly, these energy levels completely agree with each other, though Sommerfeld model does not have strange spin.

## Dirac's hydrogen includes many accidental coincidences.

(Fig.4') A series of coincidences !

As shown in Fig.1 and Fig.3, when the two different states ( such as 2s1/2 and 2p1/2 ) have the common total angular momentum j, their energy levels completely agree with each other.
2s1/2 state has zero angular momentum, so only 2p1/2 state has spin-orbital interaction.
And 2s1/2 comes closer to nucleus, which causes heavier electron relativistic mass.
So the electron tends to be close to the nucleus and its energy becomes lower than 2p1/2.

Surprisingly, this relativistic mass difference is just cancelled out by spin-orbital interaction of 2p1/2 !
As a result, the energy level of 2s1/2 is just equal to 2p1/2. ( Can you believe this ?? )

Furthermore, Dirac's hydrogen includes many other accidental coincidences such as 3s1/2 = 3p1/2, 3p3/2 = 3d3/2, 4s1/2=4p1/2 ..... !
Can you believe this states ?
Clearly, they aimed to get the same solutions as Bohr-Sommerfeld fine structure and made these unnatural states.

## Fatal defects in Dirac's hydrogen.

(Fig.5) Origin of Dirac equation.

As shown on this page, Dirac equation originates in relativistic energy and momentum relation.
Klein-Gordon equation was first gotten from special relativity, and they are second-order equations.

And they tried to get the first-order equation, and it leads to Dirac equation.
Instead of being first-order, Dirac equation has to include 4 × 4 γ matrices and spinors.

(Fig.6) Spinor includes Degeneracy ?

As I explain later, Dirac's Hamiltonian does not commute with orbital angular momentum operator, so Dirac's hydrogen solution must have more than one angular momentum.
They tried to take advantage of 4 × 1 spinor, and express the degeneracy such as 2s1/2 and 2p1/2.
( For example, upper two component spinor is 2p1/2 and lower is 2s1/2. )
As a result, Dirac's hydrogen can agree with experimental results and Bohr- Sommerfeld's hydrogen.

But this type of wavefunction is peculiar to Dirac's hydrogen atom.
Original Dirac's equation does NOT use this type.
So it is safe to say these special types were introduced to get the same solution as Sommerfeld's model.

## "Fictitious" energy levels in Dirac's hydrogen.

(Fig.7) Side effects ? = "Wrong" states in Dirac's hydrogen.

But this side effects appeared.
The case of 2s1/2 and 2p1/2 is good, but in the case of 2p3/2, it does NOT have its right pair.
2p3/2 is the energy level n =2 and the total angular momentum j = 3/2.

According to the above logic, the pair of 2p3/2 becomes unreal 2d2/3 having orbital angular momentum l = 2, which does NOT exist in quantum mechanical hydrogen.
Because Schrodinger's hydrogen must not include the angular momentum "2" in n = 2 energy.

As I explain later, these contradictory states clearly show Dirac's hydrogen is wrong.
Dirac's hydrogen includes a lot of these wrong states (= 1p1/2, 2d3/2, 3f5/2 .... )

And if Dirac's hydrogen is wrong, we need to change the interpretation of very tiny lamb shift, because Lamb shift completely depends on the assumption that Dirac's hydrogen is right. See also QED Lamb shift.

## Dirac's hydrogen is wrong → Lamb shift is wrong, too.

(Fig.8) Interpretaion of Lamb shift.

Lamb shift of hydrogen atom is very tiny (= 0.0000043 eV ) and one-tenth of fine structure ( almost same as nuclear hyperfine structure ).
Very tiny Lamb shift cannot be observed in the usual spectrum lines.

Though this Lamb shift is very small, they tried to measure this value, believing 2s1/2 state is "metastable" and the collision between excited hydrogen atom and plates is a precise method for Lamb shift.

In this experiment, there is no guarantee that modified anomalous Zeeman effect is always linearly effective, and excited metastable states really mean 2s1/2, as shown on this page. ( These are only assumptions. )
And of course, the collision method is rough and not precise to measure this very tiny value.

Even the latest optic methods cannot confirm these states really express the energy difference between 2s1/2 and 2p1/2. They "just" estimate it.
Considering Lamb shift is almost same as nuclear hyperfine structure, some nuclear or electron's vibrations may influence this very tiny data.

## Hamiltonian (= energy ) of Dirac equation.

(Fig.9) Dirac hydrogen energy.

On this page, we get the Dirac Hamiltonian (= energy ) of free particles
In Dirac's hydrogen atom, we add Coulomb potential to this Hamiltonian (= Fig.9 ).

Of course, γ matrices in Hamiltonian is 4×4 matrices, so these become simultaneous equations mixing upper and lower spinors, and we solve them.

## Division of Dirac equation into upper and lower spinors.

(Fig.10) Simultaneous equations to get energy solution.

I explain about these meanings later.
Hamiltonian consists of kinetic energy, rest mass energy (= mc2 ), and Coulomb potential energy (= V ).
Like usual Schrodinger equation, kinetic energies are divided into radial and tangential directions.
Of course, the solutions can be divided into radial and tangential parts, too.

## Tangential operator can distinguish angular momentum.

(Fig.11) Tangential parts.

Upper and lower two compoment spinors have the different angular momentum, as shown in Fig.6.
"Tangential" kinetic operator acts on the tangential parts in the wavefunction.
As I explain later, this operator discriminates orbital angular momentums l = j ± 1/2.

If the upper spinor includes the angular momentum l = j + 1/2, the lower's angular momentum must always be l = j - 1/2.
And they have the common total angular momentum j, so this difference can be expressed using the common "k", like Fig.11.

As a result, Dirac's hydrogen always contains many wrong states such as 1p1/2 (= pair of 1s1/2 ), and 2d3/2 (= pair of 2p3/2 ).
This means Dirac's hydrogen atom is wrong, and Bohr-Sommerfeld fine structure in 1916 is right.

## Radial wavefunction of Dirac's hydrogen.

As shown on this page, the radial wavefunction "rR" of Schrodinger hydrogen consists of an integer times de Broglie wavelength.
Dirac's hydrogen uses the same methods as Schrodinger's hydrogen to get the radial parts.

So the radial wavefunction becomes almost same as Schrodinger's hydrogen.
But the lower spinor includes wrong and fictional states, which are completely different from the usual ones.
The problem is these wrong states are indispensable to get the solution of Dirac's hydrogen.

## Solution of Dirac hydrogne is equal to Sommerfeld's model.

(Fig.13) Energy = radial + angular momentum.

Like usual Schrodinger's hydrogen, radial wavefunction diverges to infinity as it is.
To stop this divergence, a coefficient of the radial polynomal expressions must be zero somewhere.

Using this degree " n' " and j, the energy level " n " is determined.
This energy solution of Dirac's hydrogen is completely equal to Bohr-Sommerfeld hydrogen (= Fig.14 ).
This is too good to be true.

(Fig.14) Bohr-Sommerfeld hydrogen.

To be precise, Hamiltonian of Fig.9 is not relativistic covariant form.
Because Coulomb potential energy transforms complicatedly under Lorentz transformation like Fig.15.

(Fig.15) Dirac's hydrogen = Non relativistic.

This means the true relativistic Dirac's hydrogen ( if it exists ) gives different energy solution from Fig.13.
Lamb shift is very tiny, so if this form of Coulomb potential changes under the proton or electron's slight motion, its value changes.
So the calculation of Lamb shift, which completely depends on Fig.13, is wrong.

## Dirac himself started to doubt noncommutative algebra.

Dirac equation is known to be able to express various phenomena.
Dirac was a genius in the physics and mathematics.
So he could combine the Pauli matrices and the Relativity properly to match the experimental results.
But unfortunately, we can not imagine "concrete" images from Dirac equation.

Surprisingly, later Dirac himself started to doubt noncommutative things such as matrices.
( P.A.M. Dirac (1972) Relativity and quantum mechanics. Fields and Quanta Vol.3, 139-164. )
In 1972, he doubted noncommutativity itself.

And he thought that there would be some profound change someday.
He considered "wave phase", which causes interference as the most important thing.
This insistence is very profound and meaningful, I think.
Probably, his wave phase meant a thing like de Broglie's wave, I think.

## Calculation of Dirac equation for hydrogen atom.

### [ Dirac's hydrogen Hamiltonian . ]

First I warn you that Dirac equation relying on matrices has strong "mathematical" property, so Dirac's hydrogen also does.

(Eq.1) Dirac equation

Original relativistic energy and momentum relation is second-degree. ( p2 - E2/c2 + (mc)2 = 0. )
To agree with this relativistic equation, γ matrices must satisfy
(Eq.2)

So, γ0 matrix, which satisfies Eq.2, becomes
(Eq.3)

where I means 2 × 2 identity matrix.
So
(Eq.4)

And 1-3 components of γ matrices are
(Eq.5)

where j = 1, 2, 3
σj mean 2 × 2 Pauli matrices of
(Eq.6)

Multiplying Eq.1 by cγ0 from left side, and using γ0γ0 = I (= identity matrix ), we get the Dirac's Hamiltonian of
(Eq.7)

From γ matrices of Eq.3 and Eq.5, we get
(Eq.8)

From Eq.7 and Eq.8, Dirac's Hamiltonian operator ( which means the total energy ) including the Coulomb potential energy V is
(Eq.9)

Here p means the electron's momentum operator as follows,
(Eq.10)

So, using Eq.3, Eq.5 and Eq.8, Hamiltonian of Eq.9 becomes
(Eq.11)

Here we divide the wavefunction ψ into two 2 × 1 matrices.
(Eq.12)

As I explain later, the upper and lower spinors must have different angular momentums.

Substituting Eq.12 into Eq.11, we get
(Eq.13)

The upper part of Eq.13 is
(Eq.14)

And the lower part of Eq.13 is
(Eq.15)

And potential V of Eq.13 means Coulomb potential as follows,
(Eq.16)

First we omit 4πε, and add this later.

And we solve the simultaneous equations of Eq.14 and Eq.15 and get the hydrogen's energy E.
Of course, the energy "E" of Eq.14 and Eq.15 is common value.
And total angular momentum j is also common value.
But orbital angular momentum "l" is different in the upper and lower spinors.

As a result, Dirac's hydrogen wavefunction contains many contradictory states such as
(Eq.17)

1p1/2 is the energy level is n = 1, and total angular momentum is j = 1/2, which is just equal to 1s1/2 state.
But 1p1/2 state has unrealistic l = 1 angular momentum.
Because 1s1/2 has orbital angular momentum of l = j - 1/2 = 0, and 1p1/2 has l = j + 1/2 = 1.
Of course, the quantum mechanical hydrogen atom does NOT have orbital angular momentum l = 1 in the energy level of n = 1.

## Total angular momentum J ( not L ) commutes with Dirac Hemiltonian.

(Fig.16) Total angular momentum J and Hamiltonian.

In this section, we demonstrate total angular momentum J commutes with Dirac's Hamiltonian, which means J and Hamiltonian can have the same eigenfunctions.
Because each operator acting on the common eigenfunction can give each eigenvalue, which can commute, as shown Fig.16.

(Fig.17) Orbital angular momentum L and Hamiltonian.

On the other hand, orbital angular momentum L does NOT commute with Hamiltonian.
It means L and Hamiltonian can NOT have their common eigenfunction.

As a result, for example, in the states of the energy level n = 2 and j = 3/2, different angular momentums l = j ± 1/2 must be included.
Of course, 2d3/2 state (= pair of 2p3/2 ) having l = 2, does NOT exist in quantum mechanical hydrogen.
These facts clearly proves Dirac's hydrogen is wrong.

From Eq.9 and Eq.11, Hamiltonian is
(Eq.18)

Orbital angular momentum (L) can be expressed as
(Eq.19)

First, we think about 3 (= z ) component of this L,
(Eq.20)

Using Eq.20, we show this angular momentum operator (L) commutes with the Coulomb potential operator (V) as follows,
(Eq.21)

where we use the relations such as
(Eq.22)

where the momentum derivative acts on before and after V.

This angular momentum L doesn't commute with the momentum "p" in Dirac's Hamiltonian.
(Eq.23)

This is de Broglie relation.

So the commutation between orbital angular momentum L and Hamiltonian becomes
(Eq.24)

As a result, L and Hamiltonian doesn't commute.

Next we think about spin operator S.
(Eq.25)

where σ is 4 × 4 Pauli matrix.

This Pauli matrices of spin does NOT commute with α matrices (= Eq.8 ) of Hamiltonian, as follows,
(Eq.26)

Here we think about 3 component of spin, like orbital angular momentum of Eq.20.
This σ3 does not commute with α1 and α2 of Hamiltonian, as follows,
(Eq.27)

where Pauli matrix's relations of Eq.6 are used.

The total angular momentum operator (J) is the sum of the angular momentum (L) and spin (S) as follows,
(Eq.28)

From Eq.24 and Eq.27, this J commutes with Dirac Hamiltonian.
(Eq.29)

This means both Dirac Hamiltonian and J can have the common eigenfunction.
But Dirac Hamiltonian and L can not.

As a result, Dirac's wavefunction must always include pair state which has the same " j " and energy, but different L.
So for example, the pair of 2p3/2 becomes the fictional 2d3/2.

## How we distinguish different orbital angular momentums ?

From Eq.28, the square of the total angular momentum (J) becomes
(Eq.30)

Here we use the relation of Pauli matrices ( σσ = I ).
Eq.30 means
(Eq.31)

where J2 = j(j+1).

So when l = j + 1/2,
(Eq.32)

where we define k = j + 1/2.

And when l = j - 1/2, ( substituting it into Eq.31 )
(Eq.33)

As you see in Eq.32 and Eq.33, the sign of k becomes opposite, when the angular momentum L is different.
( If k is negative, Eq.32 and Eq.33 are exchanged. )

As I explain later, to get Dirac's hydrogen's solution, the upper and lower spinors must have different L with respect to the common J.
Here, the same "k" is used in the simultaneous equation in Eq.32 and Eq.33.
The same "k" means the same "j" and different "L".
So this Dirac's hydrogen can NOT avoid many fictional states such as 1P1/2, 2D3/2 ....
Because Dirac equation always has to combine the upper and lower 2 component spinors to get the answer.

## Eigenfunctions of Dirac equation for hydrogen.

As I said in Eq.29 and Fig.16, the total angular momentum (J) and Dirac Hamiltonian can have the common eigenfunction, so we can decide one j value in the Dirac spinor.
(As the angular momentum (L) doesn't commute with Hamiltonian, we can not fix a single L value in the eigenfunction.)
(Eq.34)

As shown in Eq.34, it must contain more than one different orbital angular momentum values (= l ), because it doesn't commute with Hamiltonian.
So the angular momentums (= l ) under common j need to be l = j ± 1/2.

For example, when the angular momentum of the upper part is l = j + 1/2, the lower part becomes l = j - 1/2 as follows,
(Eq.35)

And when the upper part is l = j - 1/2, the eigenfunction becomes
(Eq.36)

The upper and lower parts of 4-spinors consist of 2-component spinor with spherical harmonics functions (= Y ) as follows,
(Eq.37)

where l = j + 1/2.

Actually these spherical harmonics are not used in calculation, so you may consider Eq.37 and Eq.38 as rules.
And when l = j -1/2,
(Eq.38)

## Parity switching operator.

Space parity operator causes the inversion of the space coordinates.
And the polar coordinates change under the space inversion as follows,
(Eq.39)

The spherical harmonics function (Y) is known to change under the space inversion as follows,
(Eq.40)

So when the angular momentum (= l ) of Y changes by ± 1, it shows different property under the space inversion.

Some examples of Eq.40 are,
(Eq.41)

A unit vector can be expressed using the polar coordinates as follows,
(Eq.42)

And the inner product of Pauli matrices (Eq.6) and the unit vector becomes
(Eq.43)

The operator Eq.43 satisfies
(Eq.44)

And under the space inversion of Eq.39, the Eq.43 changes as follows,
(Eq.45)

Eq.45 means that when the operator Eq.43 is added, the property under the space inversion changes as +1 → -1 (or -1 → +1)
So when Eq.43 is added to the spherical harmonics functions (Y), the angular momentum (= l ) in it changes by ± 1 as follows,
(The radial part (r) doesn't change under the unit vector.)

(Eq.46)

The operator of Eq.43 appears in Dirac's Hamiltonian, which is explained in the next section.
And the result of Eq.46 is a little difficult to imagine, but if we don't use the relation of Eq.46, we can not solve the Dirac equation for hydrogen atom.

## Dividing Dirac's Hamiltonian into radial and tangential parts.

As shown in Eq.11 - Eq.16, Dirac Hamiltonian of Eq.9 can be expressed as
(Eq.47)

The upper part of Eq.47 is
(Eq.48)

And the lower part of Eq.47 is
(Eq.49)

Using the relations of Pauli matrices (= Eq.6 ),
(Eq.6)

We obtain
(Eq.50)

Using Eq.44 and Eq.50, the kinetic term ( see Eq.48 ) of Hamiltonian becomes
(Eq.51)

The operator of Eq.43 appears here.

From Eq.32, we define
(Eq.52)

From Eq.33, we define
(Eq.53)

Using Eq.35, Eq.51 and Eq.53, the equation of Eq.48 can be expressed as
(Eq.54)

Using Eq.46,
(Eq.46)

The sperical harmonics parts (Φ) of Eq.54 becomes the same and can be eliminated as follows,
(Eq.55)

The existence of Eq.43 means if there is only one angular momentum (l) in Dirac spinor, we cannot eliminate Φ.
In the same way, Eq.49 can be expressed as
(Eq.56)

where Eq.52 is used.

And using Eq.46, Φ can be eliminated as follows,
(Eq.57)

## Calculation of radial parts of eigenfunctions.

We replace the radial functions of f(r) and g(r) by
(Eq.58)

Substituting Eq.58 into Eq.55, we have
(Eq.59)

Coulomb potential V of Eq.16 is used here.
In the same way, substituting Eq.58 into Eq.57, we have
(Eq.60)

We expand u(r) and v(r) as follows,
(Eq.61)

Substituting Eq.61 into Eq.59 and Eq.60, and seeing the power of γ-1, we have
(Eq.62)

To avoid the solution of a0= b0 = 0 in Eq.62, the following relation must be satisfied.
(Eq.63)

To avoid the divergence at the origin (see Eq.61), the γ must be plus as follows,
(Eq.64)

To converge at r → ∞, we suppose u(r) and v(r) satisfiy
(Eq.65)

And at r → ∞, Eq.59 and Eq.60 become
(Eq.66)

Substituting Eq.65 into Eq.66, we have
(Eq.67)

From Eq.67, λ (= plus) becomes
(Eq.68)

Considering Eq.61 and Eq.65, we can express u(r) and v(r) as follows,
(Eq.69)

As you may notice, the replacement of Eq.69 is very similar to Schrodinger's hydrogen.
So also in Dirac's hydrogen, unrealistically, radial wavefunctions are always from zero to infinity.

If we add 4πε (see Eq.16) to γ of Eq.64, it can be expressed by the fine structure constant (=α)
(Eq.70)

where α is
(Eq.71)

Substituting Eq.69 into Eq.59, we can get the following relations ( using Eq.68 and the green part of Eq.67 ).
(Eq.72)

and Substituting Eq.69 into Eq.60, we have
(Eq.73)

Here we define (using Eq.64 and Eq.68)
(Eq.74)

and replace r by x as follows,
(Eq.75)

Using Eq.74 and Eq.75, Eq.73 become
(Eq.76)

And Eq.72 becomes
(Eq.77)

Here we expand ω1 and ω2 as follows,
(Eq.78)

Substituting Eq.78 into Eq.76, coefficient of x to the power of n - 1 ( = xn-1 ) becomes,
(Eq.79)

In the same way, substituting Eq.78 into Eq.77, the coefficient of xn-1 is
(Eq.80)

Summing Eq.79 and Eq.80, we obtain
(Eq.81)

Substituting cn and cn-1 of Eq.81 into a and b of Eq.79, and using Eq.74, we get the relation of
(Eq.82)

When we use the replacements of
(Eq.83)

Eq.82 beomes
(Eq.84)

From the relation of Eq.84, we can define the function F(x) as follows,
(Eq.85)

And using Eq.81, ω1 and ω2 (= Eq.78) can be expressed using the common cn,
(Eq.86)

Eq.86 means if we decide the power of cn, the upper and lower spinors become the same as cn.
The power of cn is related to the energy level (= radial part ) of the hydrogen, as I explain later.
( This method is similar to Schrodinger's hydrogen. )

As a result, unrealistic 1p1/2 state and 1s1/2 has the same energy level, but different orbital angular momentum.
So Dirac's hydrogen cannot avoid many wrong states such as 1p1/2, 2d3/2 ....

## Dirac's energy solution of hydrogen.

When F(x) is an infinite series, F(x) diverges exponentially at r → ∞ as follows,
(Eq.87)

To make F(x) a finite series, μ of Eq.84 must satisfy
(Eq.88)

where F(x) becomes the n' th degrees with resprct to x.

From Eq.83,
(Eq.89)

Substituting Eq.68 and Eq.74 into Eq.89,
(Eq.90)

Here we return 4πε.

Using the fine structure constant α of Eq.71, Eq.90 becomes
(Eq.91)

When we difine the new integer n as
(Eq.92)

The energy E of Eq.91 becomes ( using Eq.70 and |k| = j + 1/2 )
(Eq.93)

where n means the principal quantum number.

And as shown on this page. the energy values of Eq.93 are completely consistent with those of Bohr Sommerfeld model of Eq.94.
(Eq.94)

This means that the enegy levels of them are just equal to each other, as follows,

2p1/2 (n=2, j=1/2) -------- elliptic, (nr=1, nφ=1)
2p3/2 (n=2, j=3/2) -------- circular, (nr=0, nφ=2)

Why does such a accidental coincidence occur ?

## Reason why Dirac equation is wrong.

### [ Dirac's hydrogen solution contains many wrong and fictional states ! ]

In this section, we calculate the wavefunction of the ground state of hydrogen atom.
As kyou know, the ground state is only 1S1/2 state, which has zero angular momentum.
But Dirac's hydrogen must include unrealistic 1P1/2 state in the lower spinor.
This fact shows Dirac's hydrogen atom is wrong.

(Fig.18) Ground state of hydrogen atom ?

Here we return to the eigenfunctions of Eq.35 or Eq.36, and consider about 1S1/2 state ( =ground state ).
1S1/2 state has the total angular momentum of j = 1/2 and orbital angular momentum of l = 0.
And according to Eq.35 and Eq.36, we must consider another strange state of 1P1/2 ! (See Eq.95.)
This 1P1/2 doesn't exist according to Schrodinger equation and Pauli's spin theory.
Because 1P1/2 has the principal quantum number of n=1 and the angular momentum of l=1
(In Bohr model, this state exists instead of l=0 states.)

(Eq.95)

This strange phenomenon is caused by the 4-component spinors of Dirac equation.
Dirac equation succeeded in expressing the relativistic effect, and is of the first degree with resprct to the energy E and momentum p.
But instead, Dirac equation must use 4 × 4 matirices and 4-component spinor.
If only the half (= 2 ) of the 4 component spinor exists ( which means φ or χ of Eq.35 and 36 becomes zero ), we can not combine the mass energy term and the kinetic energy term.
Dirac equation becomes broken.

And as I said, Dirac Hamiltonian and the angular momentum operator (L) doesn't commute, so we can not fix the L value as a single one.
(So more than one different L values must be included in Dirac's solution.)
As a result, the wrong state of 1P1/2 is indispensable to get the energy value of 1S1/2 !
( l = j ± 1/2. the lower l is 1S1/2, and the upper l is 1P1/2. )

Here we try to get the 1S1/2 eigenfunction ( Z = 1, hydrogen ).
When the principal quantum number n is 1 (and j=1/2 ), n' becomes zero according to Eq.92.
(Eq.96)

From Eq.70 and Eq.93, the total energy E and γ are
(Eq.97)

where |k| = j + 1/2 = 1.

Substituting Eq.97 into Eq.68,
(Eq.98)

Eq.98 means the exponential part of Dirac's hydrogen 1S is equal to R10 of Schrodinger's hydrogen.
Substituting Eq.97 and Eq.98 into Eq.74, A becomes
(Eq.99)

where 4πε is returned.

From Eq.86, Eq.97 and Eq.99, a0 becomes
(Eq.100)

The fine structure constant α of Eq.71 is used in the last line of Eq.100.
Eq.100 means, in Eq.95 case (= upper part is 1P1/2 ), eigenfunction of Dirac's hydrogen is zero.
( When k = 1, lower spinor is also zero, substituting k = 1 into Eq.102'. )

(Eq.101)

When the sign of k becomes opposite ( k = 1 → -1 ), orbital angular momentums are exchanged, as shown in Eq.101.
So when k = -1, the upper part is 1S1/2 and lower part is 1P1/2.
In this case, the eigenfunction of 1S1/2 is not zero,
(Eq.102)

where k is minus.

In the same way, when k = -1, the lower part of 1P1/2 is NOT zero,
(Eq.102')

where Eq.74, Eq.81, Eq.97, and Eq.98 are used.

(Eq.103) Ground state of Dirac's hydrogen ?

According to advanced quantum mechanics by J.J. Sakurai, the upper part of Dirac's hydrogen is related to Schrodinger's hydrogen.
( About the lower part, I could NOT find what it really means. )
But of course, unreal 1P1/2 is also indispensable, because Dirac equation mixes them.
If 1P1/2 state does not exist in Eq.103, the simultaneous equations are broken.

As I explained above, the angular momentum operator actually acts on 1P1/2 eigenfunction, and get its eigenvalue.
As a result, Dirac equation for hydrogen atom contains many fictional states like this.
(1s1/2 and 1p1/2 ??, 2p3/2 and 2d3/2 ??, 3d5/2 and 3f5/2 ?? ... )

This means Dirac hydrogen includes self-contradiction, and the interpretation of Lamb shift needs to be changed.
Lamb shift is much smaller than very small fine structure level, and cannot be observed directly in the spectrum.
So we have no ways to judge whether it really expresses the difference between 2S1/2 and 2P1/2. ( See this page. )

### [ Can we get the correct 2S1/2 and 2P1/2 eigenfunctions ? ]

OK. so next we consider about the n=2 and j=1/2 states.
Fortunately, in this case, both states 2S1/2 ( L= j-1/2 = 0 ) and 2P1/2 ( L = j+1/2 = 1 ) really exist.
Then can we get the eigenfunctions of these states ?

(Eq.104)

In the case of Eq.104, the eigenfunction of 2P1/2 becomes
(Eq.105)

The result of Eq.105 resembles that of Schrodinger 2P1/2.
(And the part of the exponential function also resembles Schrodinger's n=2 solution.)
The relativistic effect of the hydrogen atom is very small.
So the similar results of the eigenfunctions in Dirac and Schrodinger equations is reasonable.

The eigenfunction of 2S1/2 becomes
(Eq.106)

The result of Eq.106 is completely different from Schrodinger's 2S1/2 solution. !
(Eq.107)

where r0 means the Bohr radius.

As I said, the relativistic effect of the hydrogen atom is very small.
So we can not accept this strange result of 2S1/2.

(Eq.108)

How about the case of Eq.108 ?
(2S1/2 is the upper part of the spinor. Of course, the angular momentum of 2S1/2 is zero.)
In the case of Eq.108, the eigenfunction of 2S1/2 becomes
(Eq.109)

Surprisingly, Eq.109 resembles Schrodinger's 2S1/2 of Eq.107 !

But the 2P1/2 in the case of Eq.108 is
(Eq.110)

this is completely different from Schrodinger's 2P1/2, though the relativistic effect of hydrogen is very small.

As you notice, Dirac equation can not give the correct eigenstates of 2P1/2 and 2S1/2 at the same time.
And it contains the wrong two other states, too.
Unfortunately, Dirac equation can not distinguish these wrong states from the correct ones.

Probably, they tried to match Dirac's hydrogen with Sommerfeld's fine structure using various "mathematical" tricks somehow.

2011/4/17 updated. Feel free to link to this site.