schrodinger equation is only one part of the bohr

8/12/2019 Schrodinger Equation is Only One Part of the Bohr

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Schrodinger equation is only one part of the

Bohr-Sommerfeld model.

Top page (correct Bohr model including the two-electron atoms)Strange "spin" is NOT a real thing

Table of contents (12/10/14)

Schrodinger equation is one of Bohr Sommerfeld models. Calculation of elliptical Bohr-Sommerfeld model. Reason why Schrodinger equation is wrong and Bohr-Sommerfeld is correct.

Schrodinger equation is one of Bohr Sommerfeld models.

[ Common conditions of Bohr-Sommerfeld and Schrodinger's hydrogens.]

Bohr model is the same as the quantum mechanics in allenergy levels of the

hydrogen atom.

And the fine structureof Bohr-Sommerfeld model is the same as the Dirac equation

as shown onthis pageandthis page.

The ordinary textbooks often say that this is only a "accidental" coincidence.

But in this page we show that Schrodinger's hydrogen is one of Bohr-Sommerfeld

models from the "mathematical" viewpoint.

Furthermore, we show Schrodinger's hydrogen includes "unreal" states, which means

Schrodinger equation itself is wrong.
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(Fig.A-1) From Bohr model to Bohr-Sommerfeld model.

As shown inthis section,simple Bohr model hydrogen uses the condition that one

round orbit is an integertimes de Broglie wavelength.

In 1916, Arnold Sommerfeld extended Bohr's circular orbit to "elliptic" one, which

adds electron's "radial" motion to the original model.

"Radial" and "tangential" momentums (= de Broglie

waves ).

The radial and tangental momentums are perpendicular to each other at each point of

the orbit, so their de Broglie waves should be considered independently.

Actually Sommerfeld model needed two quantizationconditions that each radial and

tangential orbit is an integer times de Broglie wavelength. ( The sum of them means

the energy level n. )
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(Fig.A-2) 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 Alwaysinclude "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.A-2.

So if this constant C is not zero, it divergesto 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.

Schrodinger's hydrogen "satisfies" Sommerfeld's

quantization.


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(Fig.A-3) Schrodinger's hydrogen satisfies Sommerfeld quantization conditions.

The radial region of Schrodinger's hydrogen is from 0 to infinity, so it looks like

a extremely longelliptic orbits.

If you see some quantum mechanical textbooks, you would find that Schrodinger's

radial eigenfunctions (= rRnl) are just an integer times de Broglie wavelength. ( It iseasily understood seeing the probability graphs of |rR|2. )

( One round means 0 or 0 0. )

"P" state of Schrodinger's hydrogen = "Sommerfeld".


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This is the true form of the mysterious uncertainty principle, which concept

clearly obstructsthe development of science.

Bohr-Sommerfeld quantization condition.

(Eq.1) Sommerfeld quantization condition (A) + de Broglie relation (B).

Eq.1 shows Bohr-Sommerfeld quantization condition (A), and de Broglie relation (B).

In Eq.1A, "p" denotes angular momentum, and "pr" denotes the radial momentum.In Eq.1B, the upper equation is the relation between tangentialde Broglie wavelength

and tangential momentum.

The lower is the relation between radialde Broglie wavelength and radial momentum.

If we combine these relations, we can prove Bohr-Sommerfeld quantization is equal

to an integer times de Broglie wavelength in each radial and tangential direction.

Sommerfeld condition = an integral multiple of de Broglie

wavelength.(Eq.1') Sommerfeld quantization = integer times de Broglie wavelength.

If we use the de Broglie relation ( wavelength = h/mv ) of Eq.1, you can easily find

that Bohr Sommerfeld's elliptical orbit is just an integertimes de Broglie wavelength

in each tangential (= Eq.1' upper ) amd radial (= Eq.1' lower ) directions.


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( In Eq.1', pris radial momentum, and "p" means constant angular momentum( p = r

mv ) ).

To make the ends of wave phases the same in one-round orbit, de Broglie waves must

be just integers in both tangential and radial directions.

As shown inthis section,the total de Broglie wave's number is the sum of radial andtangential waves.

Schrodinger's radial wavefunction satisfies n de Broglie

wavelength !

(Fig.1) Schrodinger's radial wavefunction is a integer de Broglie wavelength =

Sommerfeld's model.

Fig.1 shows the examples of Schrodinger's wavefunctions including onede Broglie

wavelength in radial direction of one orbit.

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

For example, R32wavefunction 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 meaningas Bohr-Sommerfeld model, which

also has an integer de Broglie waves. ( This reason is explained later. )

"De Broglie wave" form = Change of variable ( rR = ).
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(Eq.2) de Broglie wave's form in Schrodinger's hydrogen.

The important point is that we need to express Schrodinger equation as de Broglie

wave's form in radial and tangential directions.

(Eq.3)

As shown in Eq.2 and Eq.3, if we use = rRnlinstead of Rnl, this radial wavefunction

comes to express de Broglie waves, as shown Fig.1.

(Eq.4)

We can easily show Eq.2 transformation is right, using Eq.4.

Schrodinger's "2" or "3" de Broglie wavelength.


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(Fig.2) Schrodinger's radial wavefunction, two or three waves.

In the upper line of Fig.2, the radial one-round orbits are just twode Broglie

wavelength.

For example, in R31, the principal quantum number is "3" and the angular momentum

(= tangential ) is "1".As a result, the radial wave becomes 3-1 = 2( which reason is explained later ).

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

Schrodinger's "tangential" wavefunction also satisfies n de

Broglie wavelength !


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(Fig.3) Schrodinger's tangential wavefunction (= angular momentum ) =

Sommerfeld model.

In Fig.3, the angular moementum of Spherical Harmonics in Schrodinger's hydrogen

shows the tangential de Broglie wave's number in one orbit.

For example, in e2i= cos2 + isin2, one round istwode Broglie wavelength.

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

As a result, the total numberof radial and tangential de Broglie waves means the

principal quantum number ( n = energy levels ), in both Bohr-Sommerfeld andSchrodinger's models.

In this page, we prove the common relations between Bohr-Sommferld and

Schrodinger's hydrogens later.

Total de Broglie waves = radial + tangential waves.


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(Fig.4)

In the elliptical orbits, we can divide the particle's motion into radial and tangential

directions at each point.

(Eq.4)

where p means "momentum", and "m" means the particle's mass.

( Only in this section ,"p" means total momentum, which is different from angular

momentum of Eq.1 )

And "dq", "dr" and "rd" means the distance travelled for the short time dt in each

direction.

Of course, the radial and tangential vectors satisfy Pythagorean theorem.

Using de Broglie relation ( wavelength = h/p = h/mv ) and Eq.4, the number of de

Broglie waves included in the infinitesimal section in each direction is

(Eq.5)


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As a result,

(Eq.6)

So the total number of de Broglie wave in one orbit is just the sum of radial and

tangential waves, ( see also Eq.1' )

(Eq.7)

Simple Bohr model ( circular ) hydrogen.

( Bohr model hydrogen. )

In this section, we review ordinary Bohr model.

First, as shown onthis page,Bohr's single electron does NOT fall into nucleus just by

acceleration.

(Eq.8)

In Eq.8, 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 one circular orbit is an integer (= n ) times de Broglie

wavelength.
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Substituting the first equation into the second one,

(Eq.9)

Inserting v of the third equation into the first equation of Eq.8, we have

(Eq.10)

where r0( n = Z = 1 ) means "Bohr radius".

Substituting Eq.10 into Eq.9, the total energy E of Bohr model is(Eq.11)

which is the same as Schrodinger's hydrogen.

From Eq.10, the ratio of the particle's velocity v ( n=Z=1 ) to the light speed c is

(Eq.12)

This is the famous fine structure constant

Calculation of elliptical Bohr-Sommerfeld model.

( Bohr-Sommerfeld model. )


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Here we suppose one electron is orbiting (=rotating or oscillating)

around +Zenucleus.

(Of course, also in the Schrodinger equation of hydrogen, one electron is moving

around +Ze nucleus by the Coulomb force.)

Change the rectanglar coordinates into the polar coordinates as follows, (Annalen derPhysik [4] 51, 1-167, A. Sommerfeld.)

(Eq.13)

When the nucleus is at the origin, the equation of the electron's motion is, ( Coulomb

forcecondition )

(Eq.14)

Here we define as follows,

(Eq.15)

If one electron is moving around one central positive charge, this angular momentum

( = p ) is constant(= law of constant areal velocity).

The coordinate ris a function of , so using Eq.15, we can express the differentiation

with respect to t (=time) as follows;

(Eq.16)

Here we define

(Eq.17)


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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 sameresult of

(Eq.20)

The solution of this in Eq.20 becomes,

(Eq.21)


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where we suppose that the electron is at the perihelion(=closest point), when

is zero, as follows,

(Eq.22)

(Fig.5) "Elliptical" orbit of hydrogen-like atom.

where the nucleus is at the focus (F1), and eccentricity (=) is,

(Eq.23)

Here we prove the equation of Eq.21 ( B=0 ) means an ellipsewith the nucleus at itsfocus.

Using the theorem of cosines in Fig.5, and from the definition of the ellipse,


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(Eq.24)

From Eq.24, we obtain

(Eq.25)

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

(Eq.26)

As a result, becomes,

(Eq.27)


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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)


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where the angular momentum p is constant.

So p becomes an integer times

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 zeroangular momentum is possible. )

The radial quantization condition can be rewritten as

(Eq.35)

Using Eq.27,

(Eq.36)


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And from Eq.28,

(Eq.37)

From Eq.35-37, we have

(Eq.38)


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Doing the integration by parts in Eq.38,

(Eq.39)

Here we use the following known formula (complex integral),

(Eq.40)

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

[ Proof of 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)


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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)


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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.


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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 in this page, the radial quantization number (= nr) means the number of de

Broglie waves included in the radial orbits.

And the nmeans quantized angular momentum (= de Broglie tangential wave'snumber).


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(Eq.50)

Going back to Fig.1 and Fig.2 of Schrodinger's wavefunction,

(Fig.6) Schorodinger's hydrogen = Bohr Sommerfeld model.

Fig.6 shows the energy levels of Schrodinger's wavefunctions just obeyBohr-

Sommerfeld quantization rules.The most important difference is that Schrodinger's solutions are alwaysfrom zero to

infinity, which is unreal.

If Schrodinger's hydrogen is just circular like Bohr's 1s orbit, its radial wavefunction

becomes constant.

In this case, when we normalize it, that wavefunction becomes divergent, when that

constant is not zero.


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So Schrodinger's wavefunction does NOT include circular orbits, instead, "unreal"

zero angular momentum appears.

(Eq.51) Schrodinger's momentum operator = h / (= de Broglie wavelength ).

As you know, the wavelength of "cos r" is just 2.

Schrodinger's (momentum) operator uses the de Broglie wavelength included in the

eigenfunctions such as "cos r" in the kinetic energy ( Eq.51 ).

( Of course, this is de Broglie relation, momentum p = h / )

Even when the potential energy V varies at each position, we can consider the

potential energy V as constant in the infinitesimal region.

So at each infinitesimal region, we can say that Schrodinger's equation uses de

Broglie's wavelength for its kinetic energy.And de Broglie wavelength at each point varies continuously and smoothly.

(Eq.52)

Eq.52 shows when the wavefunction returns to its original state by one rotation (=

ei), its (tangential) wavelength is just 2r. ( r = radius ).

So the tangential direction also satisfies an integer times de Broglie wavelength of

Bohr-Sommerfeld quantization rule.


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Schrodinger's and Sommerfeld's hydrogens are linked by de

Broglie relation.

(C-1) Common condition 1.

Two equations of C-1 are just equal to each other, if we use de Broglie relation ( p = h

/ ).Basically, to determine the energy level of Eq.50, threecommon conditions ( C-1, C-

2, C-3 ) are needed.

Most important condition is C-1, in which almost allinformations of hydrogen's

configuration are included.

For example, constant angular momentum (of Schrodinger's and Sommerfeld's ) is

based on C-1.

( C-2 and C-3 ) Integer times de Broglie wavelength.

We used Sommerfeld quantization conditions of Eq.32 to get the energy solution.

As shown in C-2 and C-3, if we use de Broglie relation, they are just equal to an

integer times de Broglie wavelength in the radial and tangential directions.


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In Schrodinger's hydrogen, these informations of de Broglie wavelength appear

as eigenfunctions themselves( see Eq.51 and Eq.52 ).

( These wavelengths are used by the specific momentum operator. )

C-2 and C-3 conditions onlydemand an integer wavelength of one round orbit.

They don't demand other things such as radial regions.Schrodinger's hydrogen is from zero to infinity, but Sommerfeld model is not.

Both these two models satisfy C-2 and C-3 conditions, because C-2 and C-3 can

change the intervals of integration just by the change of variables. ( Seethis page.)

(Fig.7) Common radial quantization condition.

Fig.7 cases show one "radial" orbit is just onede Broglie wavelength.( Sommerfeld hydrogen in Fig.7 upper can be expressed as a wavefunction form using

C-1 - C-3. )

This means half of one orbit is 0.5de Broglie wavelength in Schrodinger's and

Sommerfeld's models.
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(Eq.53)

If we divide Sommerfeld's (or Schrodinger's ) orbits into infinitesimal sections, and

magnify the integral region in proportion to the elongated wavelength, the

quantization condition of Eq.53 does NOT change. ( A/A = B/B = 1 in Fig.7. )

( To be precise, we do the change of variables in E.53. )

As a result, if Schrodinger's eigenfunction and equation are satisfied, it means Bohr-

Sommerfeld's three conditions needed for energy values are satisfied, too.

This is the main reason why they give the same energy levels.

One round radial orbit ( r1 r1) is(Eq.54)

Reason why Schrodinger equation is wrong and Bohr-

Sommerfeld is correct.

In Schrodinger's hydrogen, the "radial" kinetic energy (= Tr) becomes minusin someregions ( 1/2 mv2< 0) !

Because Schrodinger's wavefunctions are always spreading from zero to infinity.

This is the reason why the Schrodinger equation is wrong!

As shown in this page, Schrodinger's and Bohr-Sommerfeld's hydrogens obey the

common rules, so these serious defects of Schrodinger's hydrogen strongly

indicates Bohr-Sommerfeld model is right.

(This strange phenomenon is different from the "quantum tunnelling".)

Also in the quantum mechanics, the particle we can find must have the "positive"

kinetic energy ( 1/2 mv2> 0 ).This is quite natural. So the Schrodinger equation includes "self-contradiction".

(The Bohr model, which includes the forces of de Broglie's waves, can explain the

phenomenon of "quantum tunnelling", too.)

For example, in the case of 2S Bohr's orbit, the angular momentum is 1 ( See

Fig.8 ).


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(Fig.8) Bohr's 2S "elliptical" orbit (angular momentum = 1).

where the shortest radius of this ellipse is a1, and the longest radius is a2.

The angular momentum (L) can be expressed as

(Eq.55)

So the "tangential" kinetic energy (T) becomes

(Eq.56)

As shown in Eq.56, as the radius (r) beomes smaller, "tangential" kineticenergy increasesat the inverse square of the radius.

The sum of the "tangential" kinetic energy and the Coulomb potential energy is

(Eq.57)

As shown in Eq.57, the change of the "tangential" kinetic energy (1/r2)

becomes biggerthan that of the Coulomb energy (1/r), as the radius becomes smaller.So in the region of ( r < a1 ), the total energy (E) becomes higherthan the original

value, if the angular momentum is constant.

For example, when the radius r is close to zero, the energy of Eq.57 is close to +, as

follows,


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(Eq.58)

So in the "real" Bohr-Sommerfeld model, the electron's orbital radius (r) can not beshorterthan some "minimum" value ( a1), when the angular momentum is not zero

(Fig.8).

(When the angular momentum is zero, the electrons is oscillating, and the r can be

"theoretically" close to zero.)

But as I said above, the probability density of the Schrodinger equation mustbe from

zero to infinity.

Strange to say, in the region of ( r < a1 ), the "radial" kinetic energy becomes minus

( 1/2 mv2< 0! ) in the Schrodinger equation.

Also in the Schrodinger equation, Kepler's law of the constant areal velocity is

satisfied, which means the angular momentum (L) is constant(L = 0, 1, 2, ..) in each

orbital.

So the "tangential" kinetic energy can not be minus. Instead, "radial" kinetic energy

beomes minus in some region (Fig.9).

(Fig.9) Schrodinger's 2P "radial" wave function (angular momentum = 1).


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As shown in Fig.9, the 2P "radial" wave function ( = rR21) contains the regions of

the minuskinetic energy ( r < a1, a2 < r ).

When the r is bigger than a2, the Coulomb potential energy becomes higherthan

some maximum value.

In this region, to keep the total energy (E < 0) constant, the "radial" kinetic energy

must be minus !

Here we try actually calculating the "radial" kinetic energy in the 2P wave function.

The wave function ( = r R21) of 2P is

(Eq.59)

Substituting Eq.59 into the "radial" kinetic energy term of Schrodinger equation,

(Eq.60)

where Trmeans the "radial" kinetic energy.

When r is close to zeroor infinity, the "radial" kinetic energies (Tr) become minus,

as follows,

(Eq.61)

Onlythe "radial" kinetic energy can be minus, though both "radial" and "tangential"

kinetic energies have the same propertyof the kinetic energy !

It is very strange.


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In Eq.2, the tangential kinetic energy terms are

(Eq.62)

When is fixed at 90 degreesin Eq.62, this wave function is just equal to classical

orbits.

(Eq.63)

When is fixed at 90 degrees, the "strange" number "2" term of Eq.63 vanishes. ( cos

90o= 0)

(Fig.10)


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schrodinger equation is only one part of the bohr

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