Top page (correct Bohr model including helium )

Special relativity is wrong.

Electron spin is illusion.

QED Lamb shift is wrong.

- Dirac's hydrogen = Bohr-Sommerfeld model.
- Calculation of Dirac equation for hydrogen atom.
- Reason why Dirac equation is wrong.

*(Fig.1) Accidental coincidence ? *

You may often hear that Bohr model couldn't give answers about the fine structure about hydrogen.

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, n_{r} and n_{φ} 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**.

*(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.

*(Fig.3) Dirac = Sommerfeld !?*

Fig.3 upper shows the energy level of Sommerfeld elliptic orbit ( n_{r} = 1, n_{φ} = 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 ( n_{φ} = 2, n_{r} = 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.

*(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.

*(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.

*(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.

*(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.

*(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.

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

I explain about these meanings later.

Hamiltonian consists of kinetic energy, rest mass energy (= mc^{2} ), 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.

*(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.

*(Fig.12) Radial wavefunction. *

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.

*(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 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.

About the basic explanation of the quantum field theory and Dirac equation, see first this page.

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

As shown on this page (Eq.5-1), Dirac equation is

*(Eq.1) Dirac equation*

Original relativistic energy and momentum relation is second-degree. ( p^{2} - E^{2}/c^{2} + (mc)^{2} = 0. )

To agree with this relativistic equation, γ matrices must satisfy

*(Eq.2)*

So, γ^{0} 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

σ_{j} mean 2 × 2 **Pauli** matrices of

*(Eq.6)*

Multiplying Eq.1 by cγ^{0} from left side, and using γ^{0}γ^{0} = I (= identity matrix ), we get the Dirac's Hamiltonian of

*(Eq.7)*

this result is the same as Eq.5-39 of this page.

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.

So this is **self-contradiction**.

*(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 σ_{3} does not commute with α_{1} and α_{2} 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**.

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 J^{2} = 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.

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)*

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.

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)*

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 a_{0}= b_{0} = 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 ω_{1} and ω_{2} as follows,

*(Eq.78)*

Substituting Eq.78 into Eq.76, coefficient of x to the power of **n - 1** ( = x^{n-1} ) becomes,

*(Eq.79)*

In the same way, substituting Eq.78 into Eq.77, the coefficient of x^{n-1} is

*(Eq.80)*

Summing Eq.79 and Eq.80, we obtain

*(Eq.81)*

Substituting c_{n} and c_{n-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, ω_{1} and ω_{2} (= Eq.78) can be expressed using the common c_{n},

*(Eq.86)*

Eq.86 means if we decide the power of c_{n}, the upper and lower spinors become the same as c_{n}.

The power of c_{n} 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 ....

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, (n_{r}=1, n_{φ}=1)

2p3/2 (n=2, j=3/2) -------- circular, (n_{r}=0, n_{φ}=2)

Why does such a *accidental coincidence* occur ?

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)*

where r_{0} is Bohr radius.

Eq.98 means the exponential part of Dirac's hydrogen 1S is equal to R_{10} 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, a_{0} 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. )

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*.

How about 2S1/2 ?

The eigenfunction of 2S1/2 becomes

*(Eq.106)*

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

Schrodinger's 2S1/2 radial part is

*(Eq.107)*

where r_{0} 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.