Top page (correct Bohr model )

Strange "spin" is NOT a real thing

*(Fig.1) n × de Broglie wavelength.*

It is known that Bohr model **agrees** with experimental results and Schrodinger equation in hydrogen atom, as shown on this site.

In Bohr model, the circumference of the orbit is just equal to an **integer** number of de Broglie wavelength.

Surprisingly, also in Schrodinger equation, both radial and angular wavefunctions satisfy an integer number of de Broglie wavelength, as shown in their **nodes** ( see this site (p.3). )

*(Fig.2) From "circular" Bohr model to "elliptical" Sommerfeld model.*

In 1916, Arnold Sommerfeld extended Bohr's circular orbit to "**elliptical**" one.

Sommerfeld elliptical orbit also satisfies an integer number of de Broglie wavelength.

This condition is expressed as Sommerfeld **quantization** rule in **both** radial and angular directions, as shown on this site (p.119) and this site (p.12).

*(Fig.3) n × de Broglie wavelength.*

where "r" is radius, and φ is azimuthal angle.

Different from simple circular Bohr's orbit, elliptical orbit contains movement in "**radial**" directions.

We can **separate** each momentum at each point into **angular** (= tangential ) and **radial** components, which are perpendicular to each other.

"Perpendicular" means these two components are **independent** from each other.

So we need to calculate **de Broglie** waves in these two directions independently.

Numbers (= n ) of de Broglie wavelength in each tangential and radial directions are given by dividing one orbital length in each direction by each de Broglie wavelength (= λ ).

*(Fig.4) n × de Broglie wavelength is separated into angular and radial parts.*

When the circumference of **elliptical** orbit is **n** × de Broglie wavelemgth, this integer "n" can be expressed as the **sum** of angular and radial waves. See also this section.

In Fig.4, n is equal to "**6**", which is also equal to **principal** quantum number.

The whole path in **tangential** (= angular ) direction is n_{φ} (= **4** ) × λ_{φ} (= tangnetial de Broglie wavelength ).

And the **radial** path is equal to n_{r} (= 2 ) × λ_{r} (= radial de Broglie wavelength ).

This is Sommerfeld quantization rule in radial and angular directions.

*(Fig.5) Simple circular orbit = Only tangential (= angular ) movement.*

As shown in Fig.5, when an electron is moving in **circular** orbit, it moves **Only** in tangential (= angular ) direction.

So in circular orbit, the concept of **radial** momentum and de Broglie wavelength do **NOT** exist.

So the orbital length becomes n (= n_{φ} ) × **tangential** de Broglie wavelength.

We **don't** need to think about the radial wave quantization, different from elliptical orbit.

*(Fig.6) n (= n _{φ} ) × angular de Broglie wavelength.*

In Fig.6, the circumference of this circular orbit is **5** × de Broglie wavelength.

( So, the **principal** quantum number in energy is "5", too. About the calculation, see this section. )

This is a circular orbit, so **all** these "5" de Broglie waves are in **tangential** (= angular ) direction.

"Radial" momentum and waves are **NOT** included in circular orbit at all.

*(Fig.7) Schrodinger's hydrogen also consists of radial and angular parts.*

Schrodinger equation also depends on an integer number of de Broglie wavelength in radial and angular directions.

See this site and this site. ( "node" is related to radial waves. )

Comparing the energies of Bohr model and Schrodinger equation, they are just **consistent** with each other.

And the principal quantum number "n" is the **sum** of radial and angular ( wave ) numbers in both models.

*(Fig.8) 1s wavefuntion = 1 × wavelength. 2s = 2 × wavelength.*

As shown on this site, Schrodinger wavefunctions of 1s and 2s orbitals become like Fig.8.

Radial wavefunction of χ = "**rR**" obeys de Broglie relation, as shown in Eq.1 and this section.

One orbit means the circumference from a point to the **same** point.

In Fig.8, from r = ∞ (= start ) to r = ∞ (= end ), 1s wavefunction contains 1 × de Broglie wavelength.

And 2s wavefunction contains 2 × de Broglie wavelength.

**Neither** 1s nor 2s has angular parts, so the angular quantum number l is **zero**.

*(Eq.1) Schrodinger (radial ) equation for hydrogen atom.*

In the latter part of this site, there are some examples of this radial amplitude of χ = rR.

The radial wavefunction of 2p orbital is 1 × de Broglie wavelength like 1s orbital.

But 2p also has **angular** wavefunction with quantum number **l = 1**.

The principal quantum number n is the sum of radial and angular wave numbers.

So **2p** has n = **1** ( angular ) + **1** ( radial ) = **2** energy level like 2s orbital.

*(Fig.9) 2p wavefunction: n = 1 ( angular ) + 1 ( radial ) = 2.*

If you check hydrogen wavefunction based on Schrodinger equation in some websites or textbooks, you will find 1s radial wavefunction like Fig.10.

rR_{12} shows **1/2** de Broglie wavelength (= Fig.10 right ), so one orbit ( ∞ → 0 → ∞ ) includes 1 × de Broglie wavelength in the radial direction, as shown in Fig.8 upper.

*(Fig.10) 1s hydrogen wavefunction = 1 × de Broglie wavelength.*

As shown in this site and this site, Sommerfeld quantization conditions are

*(Eq.2) Sommerfeld quantization.*

**L** is **angular** momentum, which is constant.

**φ** is azimuthal **angle** in the plane of the orbit.

**p _{r}** is

h, n

*(Eq.3) Sommerfeld quantization = an integer × de Broglie wavelength.*

Using Eq.2 and de Broglie relation ( λ = h/mv ), you find condition of Eq.2 means an **integer** times de Broglie wavelength (= n_{φ} × λ_{φ} ) is contained in the orbit in **tangential** direction.

*(Fig.11) Number of de Broglie wavelength in tangential direction.*

For example, in Fig.11, one orbit in tangential direction contains 5 × de Broglie wavelength.

Of course, this wavelength (= λ_{φ} ) is based **Only** on **tangential** momentum ( λ_{φ} = h / p_{φ} ).

*(Fig.12) Sommerfeld radial quantization. → n _{r} × de Broglie wavelength !*

Radial quantization condition of Eq.2 also means an integer times de Broglie wavelength is contained in one orbit in **radial** direction.

So Sommerfeld quantization conditions of Eq.2 require quantization in **both** radial and angular directions.

*(Fig.13) Electron's movement in "radial" and "angular" directions.*

In elliptical orbit, the electron is moving both in angular and radial directions, as shown in Fig.13.

Using Pythagorean theorem, we can **divide** total momentum "p" into radial (= p_{r} ) and tangential (= p_{φ} ) components at each point.

These two directions are **perpendicular** to each other at each point.

So, radial (= λ_{r} ) and angular (= λ_{φ} ) de Broglie waves are **independent** from each other.

*(Fig.14) *

As shown in this section, total number (= n ) of de Broglie waves can be expressed as the **sum** of angular (= n_{φ} ) and radial (= n_{r} ) de Broglie waves.

In Fig.14 orbit, the **principal** quantum number **n = 6**, which also expresses total number of de Broglie wavelength.

Angular and radial quantum numbers (= wave numbers in each direction ) are "**4**" and "**2**", respectively.

So, the relation of n = **6 = 4 + 2** holds.

*(Fig.15) Schrodinger's radial wavefunction is a integer de Broglie wavelength = Sommerfeld.*

Fig.15 shows the examples of Schrodinger's wavefunctions including **one** de Broglie wavelength in radial direction of one orbit.

( One orbit means from one point to the same point, ∞ → 0 → ∞ )

For example, R_{32} wavefunction is "**3**" principal number ( n = 3 ) and "**2**" angular (= tangential ) momentum ( l = 2 ).
So the radial wave's number is **3-2= 1**.

These wave's numbers have the **same meaning** as Bohr-Sommerfeld model, which also has an integer de Broglie waves.

*(Fig.16) Schrodinger's radial wavefunction, two or three waves.*

In the upper line of Fig.16, the radial one-round orbits are just **two** de Broglie wavelength.

For example, in R_{31}, the principal quantum number is "**3**" and the angular momentum (= tangential ) is "**1**".

As a result, the radial wave becomes **3-1 = 2**.

And in the lower wavefunction, n = **3** and l = **0**, so the radial wave is **3-0 = 3**.

*(Fig.17) Schrodinger's tangential wavefunction (= angular momentum ) = Sommerfeld model.*

In Fig.17, the angular moementum of Spherical Harmonics in Schrodinger's hydrogen shows the tangential de Broglie wave's number in one orbit.

For example, in e^{2iφ} = cos2φ + isin2φ, one round is **two** de Broglie wavelength. See this section.

Because the rotation by **π** returns its phase to the original one.

As a result, the **total number** of radial and tangential de Broglie waves means the principal quantum number ( n = energy levels ), in both Bohr-Sommerfeld and Schrodinger's models.

*(Fig.18) Negative kinetic energy is unreal.*

**Unrealistically**, the radial region of all orbitals in Schrodinger's hydrogen must be from **0** to **infinity**.

Fig.18 shows 2p orbital of Schrodinger's hydrogen.

In 2p orbitals, radial kinetic energy becomes **negative** in two regions ( r < a1, a2 < r )

Of course, in this **real** world, kinetic energy (= 1/2mv^{2} ) **cannot** be negative, which means Schrodinger's hydrogen is **unreal**.

In the area of r < a1, potential energy V is **lower** than total energy E, so, tunnel effect has **Nothing** to do with this 2p orbital.

Furthermore, it is **impossible** for the **bound**-state electron to go away to **infinity** ( and then, **return** ! ) in all **stable** atoms and molecules.

*(Fig.19) "Elliptical" orbit (angular momentum = 1).*

Here we think about the classial elliptical orbit with angular momentum = 1, like 2p orbital.

The distance between the perihelion ( aphelion ) and nucleus is a1 ( a2 ).

In classical orbit, when the electron has angular momentum, it **cannot** come closer to nucleus than perihelion (= a1 ).

*(Eq.4) *

We suppose the angular momentum (= L ) of this orbit is ħ, which is **constant**.

Using Eq.4, **tangential** kinetic energy (= T_{φ} ) becomes

*(Fig.20) Tangential kinetic energy becomes divergent as r → 0.*

As shown in Fig.20, this tangential kinetic energy diverges to **infinity**, as r is closer to zero.

This T_{φ} is the inverse of **r ^{2}**, which

So, to **cancel** this divergence of tangential kinetic energy, radial kientic energy must be **negative** !

Unfortunately, our Nature does **NOT** depend on this **artificial** rule, so Schrodinger equation is **unreasonable**.

*(Fig.21) Picking up only realistic region from a1 to a2 → "discontinuous" radial wavefuction.*

Some people may think we can pick up only **realistic** region ( a1 < r < a2 ) with **positive** kinetic energy from radial wavefucntion.

But it is **impossible**, because this wavefunction becomes **discontinuous** at a1 and a2 in this case.

One orbit means the electron must **U-turn** at a1 and a2, and returns to the original point.

*(Fig.22) Discontinuous → Schrodinger equation does NOT hold.*

As a result, the **discontinuous** wavefunction of Fig.21 **cannot** hold in Schrodinger equation for hydrogen.

Because "discontinuous" means the gradient (= **derivative** ) of this wavefunction becomes **divergent** to infiniy.

At the points of a1 and a2, potential energy V and total energy E must be **finite**.

So sudden U-turn at these points makes this radial wavefunction **unrelated** to the original equation.

*(Fig.23) Schrodinger's 2p radial wavefunction. Unreal "negative" kinetic energy.*

As shown in Fig.23, the original 2p radial wavefunction is continuous at both ends of r = 0 and ∞

Because the **gradient** and amplitude of the wavefunction becomes **zero** at these points.

Instead, this wavefunction must contain unrealistic region with **negative** kinetic energy.

As I said in Fig.18, this region has **Nothing** to do with tunnel effect.

*(Eq.5) Schrodinger's radial equation for hydrogen.*

As shown on this site and this site, they replace the radial parts by **χ = rR** to satisfy de Broglie relation.

As shown in Eq.5, this wavefunction becomes **cosine** ( or sine ) functions with some wavelength **λ _{r}**.

*(Eq.6) de Broglie relation holds in radial direction.*

As a result, Schrodinger equation gives **de Broglie** relation ( radial momentum **p _{r} = h/λ_{r}** ).

Here λ

*(Eq.7) Sommerfeld + de Broglie relation = Schrodinger equation.*

Using Eq.6, Sommerfeld quantization condition of Eq.2 becomes like Eq.7.

Eq.7 shows number of ( radial ) de Broglie wavelength in the whole **radial** path becomes "n_{r}".

This means Schrodinger's hydrogen **also** satisfies an integer times de Broglie wavelength like Bohr model !

*(Fig.24) 2p radial wavefunction consists of different wavelengths.*

Like in elliptical orbit, radial momentum (= wavelength ) is **changing** at different points.

So, 2p radial wavefunction consists of **infinite** kinds of de Broglie wavelengths, as shown in Fig.24.

Within each **small** segment (= dr ), we can consider the de Broglie wavelength is a **constant** value.

*(Fig.25) Quantization of radial wavelength.*

As a result, using **different** wavelengths (= λ_{r} ) at different points, Sommferfeld quantization condition in one orbit can be expressed lie Fig.25.

*(Eq.8) Azimuthal Schrodinger equation.*

Eq.8 is azimuthal Schrodinger equation and wavefunction (= Φ ).

In this section, we show this quantized azimuthal wavefunction means an **integer** times ( **tangential** ) de Broglie wavelength.

*(Eq.9) Schrodinger's azimuthal wavefunction = n _{φ} × de Broglie wavelength.*

Applying usual momentum operator ( in **tangential** direction ) to this azimuthal wavefunction (= Φ ), we obtain the tangential momentum (= p_{φ} ), as follows,

*(Eq.10) *

Using de Broglie relation, tangential de Broglie **wavelength** (= λ_{φ} ) becomes

*(Eq.11) *

Here ħ = h/2π ( h is Planck constant ) is used.

Using Eq.11, the number of de Broglie wavelength in one ( tangential ) orbit becomes

*(Eq.12) *

This equation holds true.

As a result, the azimuthal wavefunction of Eq.8 **also** means an integer number of de Broglie wavelength like Sommerfeld model.

*(Fig.25) Bohr-Sommerfeld model. *

Here we suppose one electron is orbiting (= *rotating or oscillating* ) around *+Ze* nucleus.

( Of course, also in the Schrodinger equation of hydrogen, one electron is moving around +Ze nucleus by the **Coulomb** force. )

This section is based on Sommerfeld's original paper in 1916 ( Annalen der Physik [4] 51, 1-167 ).

Change the rectanglar coordinates into the polar coordinates as follows,

*(Eq.13)*

When the nucleus is at the origin and interacts with an electron through **Coulomb** force, the equation of the electron's motion becomes

*(Eq.14)*

Here we define as follows,

*(Eq.15) p = "constant" angular momentum.*

Only in this section, we use "**p**" as

If one electron is moving around one central positive charge, this angular momentum ( = p ) is **constant** (= *law of constant areal velocity*).

The coordinate *r* is a function of φ, so using Eq.15, we can express the differentiation with respect to t (=time) as follows;

*(Eq.16)*

Here we define

*(Eq.17)*

Using Eq.13 and Eq.16, each momentum can be expressed by,

*(Eq.18)*

Using Eq.14, Eq.16, and Eq.18, the equation of motion becomes,

*(Eq.19)*

From Eq.19, we obtain the *same* result of

*(Eq.20)*

The solution of this σ in Eq.20 becomes,

*(Eq.21)*

where we suppose that the electron is at the *perihelion* (=closest point), when φ is *zero*, as follows,

*(Eq.22)*

*(Fig.26) "Elliptical" orbit of hydrogen-like atom.*

where the nucleus is at the focus (F_{1}), and eccentricity (=ε) is,

*(Eq.23)*

Here we prove the equation of Eq.21 ( B=0 ) means an **ellipse** with the nucleus at its focus.

Using the theorem of cosines in Fig.26, and from the definition of the ellipse,

*(Eq.24)*

From Eq.24, we obtain

*(Eq.25)*

From Eq.21, Eq.22, and Eq.25, we have

*(Eq.26)*

As a result, σ becomes,

*(Eq.27)*

Using Eq.16 and Eq.27,

*(Eq.28)*

So the kinetic energy (T) becomes,

*(Eq.29)*

From Eq.27, the potential energy (V) is,

*(Eq.30)*

So the total energy (W) is,

*(Eq.31)*

In the Bohr-Sommerfeld quantization condition, the following relations are used,

*(Eq.32)*

where the angular momentum p is constant.

So p becomes an integer times ħ

By the way, what do Eq.32 mean ?

If we use the de Broglie relation in each direction ( tangential and radial ),

*(Eq.33)*

Eq.32 means

*(Eq.34)*

So they express quantization of de Broglie wavelength in each direction.

( In Schrodinger equation, the **zero** angular momentum is possible. )

The radial quantization condition can be rewritten as

*(Eq.35)*

Using Eq.27,

*(Eq.36)*

And from Eq.28,

*(Eq.37)*

From Eq.35-37, we have

*(Eq.38)*

Doing the integration by parts in Eq.38,

*(Eq.39)*

Here we use the following known formula (= complex integral, see this site (p.5) ),

*(Eq.40)*

--------------------------------------------------------------

According to Euler's formula, cosine can be expressed using the complex number z,

*(Eq.41)*

So,

*(Eq.42)*

Using Eq.41 and Eq.42, the left side of Eq.40 is

*(Eq.43)*

where

*(Eq.44)*

So only the latter number of Eq.44 is used as "pole" in Cauchy's residue theorem.

In the residue theorem, only the coefficient of 1/(z-a) is left, like

*(Eq.45)*

and

*(Eq.46)*

From Eq.43 to Eq.46, the result is

*(Eq.47)*

we can prove Eq.40.

---------------------------------------------------------------

Using the formula of Eq.40, Eq.38 (Eq.39) becomes

*(Eq.48)*

where the quantization of Eq.32 is used.

From Eq.48, we have

*(Eq.48')*

Substituting Eq.48' into Eq.31, and using Eq.32, the total energy W becomes

*(Eq.49)*

This result is completely equal to Schrodinger's hydrogen.

Here we confirm Bohr-Sommerfeld solution of Eq.49 is valid also in the Schrodinger's hydrogen.

As shown on this page, the radial quantization number (= n_{r} ) means the number of de Broglie waves included in the radial orbits.

And the n_{φ} means quantized angular momentum (= de Broglie tangential wave's number).

*(Eq.50)*

Some orbitals of Schrodinger's wavefunction are

*(Fig.27) Schorodinger's hydrogen = Bohr Sommerfeld model.*

Fig.27 shows the energy levels of Schrodinger's wavefunctions just **obey** Bohr-Sommerfeld quantization rules.

The most important difference is that Schrodinger's solutions are **always** from zero to infinity, which is **unreal**.

*(Fig.28) Why Schrodinger's hydrogen must NOT have circular orbits ?*

Schrodinger's hydrogen eigenfunctions consist of "radial" and angular momentum (= "tangential" ) parts.

As you know, Schrodinger's radial parts **Always** include "radial" momentum, which means there are **NO** circular orbitals in Schrodinger's hydrogen.

Because if the circular orbitals are included in Schrodinger's hydrogen, the radial eigenfunction needs to be **constant** (= C ), as shown in Fig.28.

So if this constant C is not zero, it **diverges** to infinity, when we normalize it.

( Schrodinger's radial region is from 0 to infinity. )

This is the main reason why Schrodinger's hydrogen includes many **unrealistic** "S" states which have **no angular momentum**.

*(Fig.29) Angular momemtum L = 0 in S wavefunction → destructive interference of de Broglie wave ! *

As shown on this page, Schrodinger's wavefunctions also satisfy an integer number of de Broglie wavelength.

So in **s** orbital **without** angular momentum ( L = 0 ), **opposite** wave phases in one orbit **cancel** each other due to **destructive** interference in Schrodinger wavefunction.

As a result, stable wavefunction is **impossible** in "s" orbitals.

*(Fig.30)*

And electron in s orbital always crash ( or penetrate ) into nucleus ?

Also in multi electron atoms such as sodium and potassium, 3s or 4s electrons always penetrate the inner electrons and nucleus ?

As you feel, this unrealistic **s** state is **impossible** in this real world.

On this page, we have proved rigorously that Schrodinger's energies are equal to Bohr Sommerfeld's model.

*(Ap.1) Bohr model hydrogen. -- Circular orbit. *

In this section, we review ordinary Bohr model ( see also this site (p.9) ).

First, as shown on this page, Bohr's single electron does NOT fall into nucleus just by acceleration.

Because vacuum electromagnetic energy around a **single** electron is **NOT** energy.

*(Ap.2) Bohr model conditions.*

In Ap.2, the first equation means that the centrifugal force is equal to **Coulomb** force.

The second equation is the sum of Coulomb potential energy and kinetic energy.

The third equation means the circumference of one circular orbit is an integer (= n ) number of **de Broglie** wavelength.

Condition "1" is included in the condition "2". ( So two conditions "2" and "3" is needed for Bohr's orbit. )

Substituting equation "1" into equation "2",

*(Ap.3)*

Inserting v of equation "3" into equation "1", we have

*(Ap.4)*

Here, "r_{0}" ( n = Z = 1 ) means "**Bohr radius**".

"Bohr radius" and "Bohr magneton" are **indispensable** for quantum mechanics, which is the reason why Bohr model is **taught** in school even now.

Substituting Ap.4 into Ap.3, the total energy E of Bohr model is

*(Ap.5) Bohr model = Schrodinger equation.*

This energy E is the **same** as Schrodinger's hydrogen.

From Ap.4, the ratio of the particle's velocity v ( n=Z=1 ) to the light speed c is

*(Ap.6) Fine structure constant α*

This is the famous fine structure constant α

So there is a **strong** tie between fine structure constant and Bohr's orbit.

*(Ap.7) Radial (= p _{r} ), tangential (= p_{φ} ) *

In elliptical orbit, we can divide momentum (= p ) into **radial** (= p_{r} )
and **angular** (= p_{φ} ) components.

According to the Pythagorean theorem, we have

*(Ap.8) *

Dividing momentum (= p ) by mass (= m ), we get velocity "**v**".

We suppose the electron travels the distance "**dq**" during the infinitesimal time interval "**dt**".

*(Ap.9) *

Using Phythagorean theorem, the moving distance "dq" can be separated into radial and angular (= tangential ) directions.

*(Ap.10) *

In Ap.10, "r" is the radius, and φ is the angle.

Like Ap.9, we have

*(Ap.11) *

Using de Broglie wavelength (= λ = h/p ), **number** of de Broglie waves contained in small region "dq" becomes

( Using Ap.9 )

*(Ap.12) *

In the same way, using de Broglie wavelength in radial (= λ_{r} ) and tangential (= λ_{φ} ) directions, number of waves contained in each segment becomes ( using Ap.11 )

*(Ap.13) *

From Ap.12 and Ap.13, we get the relation of

*(Ap.14) *

As a result, we have

*(Ap.15) Total number of de Broglie waves n = n _{r} + n_{φ} *

Doing the line integral along the orbit in both sides of Ap.14, we get the result of Ap.15.

"**n**" means the **total** numebr of de Broglie wavelength, and is equal to **principal** quantum number.

"**n _{r}**" denotes the number of radial de Broglie wavelength along

"

The relation of **n = n _{r} + n_{φ}** holds.

*(Ap.16) Actual relation between angular and radial wavelengths. *

We can explain the reason why de Broglie waves can be separated into angular and radial directins.

The matter wave of Ap.16, has de Broglie wavelength of **λ**, so the total momnetum p = h/λ.

As you see Ap.16, angular and radial de Broglie wavelengths are **longer** than the total wavelength.

( λ_{φ}, λ_{r} > λ ).

*(Ap.17) Angular and radial components of momentum "p"*

Angular (= **p _{φ}** ) and radial (=

From Ap.16, you find λ can be expressed as

*(Ap.18) *

Substituting Ap.18 into the total momentum p of Ap.17, we find de Broglie relation satisfies Pythagorean theorem, as follows,

*(Ap.19) *

As a result, de Broglie relation of p = h/λ **agrees** with classical mechanics, too.

*(Ap.20) Total, angular, radial directions = an integer × de Boglie wavelength.*

In hydrogen-like atom, the orbit becomes elliptical or circular.

So, at two points ( apherion and perihelion ), an electron moves in **angular** direction.

Of course, at the same point ( φ = 0 and 2π ), the wave **phase** must agree with each other.

As shown in Ap.20, when the circumference of the **whole** elliptical orbits is equal to an integer number of de Broglie wavelength, the path in the **angular** direction **also** satisfies an **integer** × ( **tangential** ) de Broglie wavelength.

As I said in Ap.15, the number of "radial" waves is given by n_{r} = n - n_{φ}.

So in this case, "**radial**" direction **also** satisfies an integer (= n_{r} ) × de Broglie wavelength.

*(Ap.21) Schrodinger equation of hydrogen.*

In this section, we solve the Schrodinger equation of hydrogen atom, like on this site and this site.

**ψ** is total **wavefunction** of hydrogen.

Total energy E is the sum of **kinetic** energy and **Coulomb** potential energy (= V ) both in Schrodinger and Bohr's hydrogens.

*(Ap.22) *

Schrodinger equation also uses de Broglie relation ( λ = h/p ), so momentum p is replaced by operater, as shown in Ap.22.

Using spherical coodinate,

*(Ap.23) *

Ap.22 becomes

*(Ap.24) *

They use the form of wavefunction that assumes the separation into **radial** (= R ) and **angular** (= Y ) portions, as follows,

*(Ap.25) *

Dividing both sides of Ap.24 by

*(Ap.26) *

we have

*(Ap.27) *

The left side of Ap.27 is dependent **only** on radial variable "r".

And the right side is **angularly**-dependent (= θ and φ ) equation.

So we can consider both sides as some **constant** (= -l(l+1)ħ^{2} )

As a result, the left side of Ap.27 becomes

*(Ap.28) *

The right side of Ap.27 becomes

*(Ap.29) *

We replace Y by a product of single-variable functions. ( Y → θ and φ )

*(Ap.30) *

Inserting Ap.30 into Ap.29, and dividing it by Φ,

*(Ap.31) *

As a result, the azimuthal wavefunction Φ can be expressed as

*(Ap.32) *

m_{l} must be an integer, otherwise the value of the azimuthal wave function would be **different** for φ = 0 and φ = 2π. , as shown in Ap.33.

So in this case, the equation of Ap.24 is broken.

*(Ap.33) *

As shown in this section, Ap.32 means Schrodinger's hydrogen also **obeys** an integer number of **de Broglie** wavelength in angular direction. ← **Sommerfeld** quantisation.

*(Ap.34) *

For angular equation of Ap.31 to have finite solution, angular constant l must satisfy Ap.34.

*(Ap.28) Radial equation of hydrogen.*

*(Ap.35) *

To make calculation easier, they substitute **χ** for rR in radial equation, as shown in Ap.35.

See also this site and this site.

Using this χ in Ap.28, we have

*(Ap.36) *

where

*(Ap.37) *

As a result, the radial part of Ap.28 becomes

*(Ap.38) *

Ap.38 uses momentum operator based on **de Broglie** relation in radial direction.

So, this wavefunction χ shows de Broglie waves in radial direction.

Substituting ρ for βr,

*(Ap.39) *

they express χ as

*(Ap.40) *

As you see the form of χ in Ap.40, this **radial** wavefunction always becomes **zero** at both ends ( χ = 0, r =0 and ∞ ).

So, at this point, this χ becomes an **integer** times de Broglie wavelength.

L(ρ) is a polynomial in ρ,

*(Ap.41) *

For this polynomial to have **finite** terms, the enrgy E must be **quantized** as follows,

*(Ap.42) *

The principal quantum number n is the **sum** of angular and radial quantum numbers.

*(Ap.43) *

This energy solution just **agrees** with that of Bohr-Sommerfeld model.

In both of Schrodinger and Bohr's hydrogens, the sum of radial and angular ( wave ) numbers are important for energy levels (= n ).

*(Ap.44) The same mechanism is used in Schrodinger and Bohr-Sommerfeld models.*

2014/6/11 updated. Feel free to link to this site.