1

Runge-Lenz method for the energy level of hydrogen atom Masatsugu Sei Suzuki

Department of Physics, SUNY at Binghamton (Date: January 06, 2015)

Runge-Lentz (or Laplace-Runge-Lentz) vector

In classical mechanics, the Runge–Lenz vector (or simply the RL vector) is a vector used chiefly to describe the shape and orientation of the orbit of one astronomical body around another, such as a planet revolving around a star. For two bodies interacting by Newtonian gravity, the LRL vector is a constant of motion, meaning that it is the same no matter where it is calculated on the orbit; equivalently, the LRL vector is said to be conserved. More generally, the RL vector is conserved in all problems in which two bodies interact by a central force that varies as the inverse square of the distance between them; such problems are called Kepler problems.

The hydrogen atom is a Kepler problem, since it comprises two charged particles interacting by Coulomb's law of electrostatics, another inverse square central force. The RL vector was essential in the first quantum mechanical derivation of the spectrum of the hydrogen atom, before the development of the Schrödinger equation. However, this approach is rarely used today. http://en.wikipedia.org/wiki/Laplace%E2%80%93Runge%E2%80%93Lenz_vector

Wolfgang Pauli in 1926 used the matrix mechanics of Heisenberg to give the first derivation of the energy levels of hydrogen and their degeneracies. Pauli's derivation is based on the Runge-Lenz vector multiplied by the particle mass. [W. Pauli, Z. Physik 36, 336 (1926).] 1. Kepler's law of planetary motion

The Kepler's laws (I, II, and III) describe the motion of planets around the Sun, (I) The orbit of a planet is an ellipse with the Sun at one of the two foci. (II) A line segment joining a planet and the Sun sweeps out equal areas during equal interval

of time. (III) The square of the orbital period of a planet is proportional to the cube of the semi-major

axis of its orbit.

The Sun is at the one focus of the ellipse (the planet orbit). The ellipse orbit is described by

12

2

2

2

b

y

a

x.

where a is the semi-major axis, b is the semi-minor axis, and e is the eccentricity (0<e<1).

2

(i) The eccentricity e

From the definition of ellipsoid, we have

aKFKF 20201

When 0201 KFKF , we have .0201 aKFKF We apply the Pythagorean theorem to the

triangle 01OKF ,

2222 eaba ,

Then we have the expression for the eccentricity

2

2

1a

b

Note that

F1 ae,0 F2 ae,0Oa a

a b a

K0

aeae

r1 r2

p0

3

21 eab .

(ii) The perihelion and aphelion

The focus is at (ae,0) and (-ae,0). For simplicity, we assume that Sun is located at focus (-ae,0). The perihelion (r1) the point nearest the Sun

)1(1 ear ,

The aphelion ( r2) the point farthest the Sun

)1(2 ear

(iii) Area of the ellipsoi

The area of the ellipse orbit is given by

22 1 eaabA .

(iv) The mathematical formula the ellipsoid: Here we show that mathematically, an ellipse can be represented by the formula

cos10

e

pr

,

where p0 is the semi-latus rectum, and e is the essentricity of the ellipse, r is the distance from the

Sun to the planet, and is the angle to the planet's current position K0 from its closest point P (perihelion), as seen from the Sun at F1. We use p0 instead of p since p is typically a linear

momentum in physics. We also use instead of , for convenience.

4

In the triangle F1F2K0 (see the above figure), we have

arr 21 , (1)

from the definition of the ellipse. Using the cosine law, we have

cos44)cos(44 22222221 aerearaerearr . (2)

where is the angle between the vector PF1 and 01KF . From Eqs.(1) and (2), we get

cos44)2( 2222 aerearra

or

cos4444 22222 aerearrara

or

02 )1()cos1( peaer

F1 ae,0 F2 ae,0OP a,0

r r1

K0

5

or

cos10

e

pr

,

with

)1( 20 eap ,

where (r, ) are heliocentric polar coordinates for the planet, p0 is the semi latus rectum, and e is the eccentricity, which is less than one. For = 0 the planet is at the perihelion at minimum distance:

)1(1

)1(

1

20

1 eae

ea

e

pr

. (perihelion)

For 2

,

0pr . (semi latus rectum)

For , the planet is at the aphelion at maximum distance:

)1(1

02 ea

e

pr

. (aphelion)

Note that

00021

21111

pp

e

p

e

rr

.

The semi-major axis a is the arithmetic mean between r1 and r2,

221 rr

a

.

The semi-minor axis b is the geometric mean between r1 and r2,

6

2

021

1 e

prrb

((Note)) The meaning of semi latus rectum

The chord through a focus parallel to the conic section directrix of a conic section is called the latus rectum, and half this length is called the semilatus rectum (Coxeter 1969). "Semilatus rectum" is a compound of the Latin semi-, meaning half, latus, meaning 'side,' and rectum, meaning 'straight.' 2. Kepler problem in classical mechanics (hydrogen atom)

The classical Hamiltonian for the Kepler problem is

rmH

2

2

1p ,

where m is a reduced mass and is a positive quantity. For the case of the hydrogen-like atom,

we can identify 2Ze . This system is invariant under the rotation. So the angular momentum is conserved. The classical orbit of the particle is elliptical. The Runge-Lentz vector is defined as

LprM mr

1,

or

LpeMA rmm ,

where L (= pr ) is the angular momentum and p is the linear momentum.

7

8

Fig. A (=mM) (denoted by the arrows with purple) on the ellipse orbits. The vector A points in

the direction of the perihelion. The magnitude is constant. The angular momentum L is always perpendicular to the orbit. The perihelion, point of the orbit the nearest to the focal point F). The aphelion, the point of orbit from the focal point.

It is obvious from the definition that

0ML , (M lies in the plane of motion).

rpMp r

,

and

9

2

2

1

1

)(1

)1

(

lm

r

mr

mr

mr

L

Lpr

LprrMr

.

or

2lrmm MrAr , since

2)()( LprLLpr , MA m .

Since the angular momentum (along the z axis) is conserved, we can calculate

11 prlL ,

where 1p is the linear momentum at the perihelion. M is a conserved quantity since

0

])(

[])(

[

])()()[(])(

[

)]()([])(

[

)]([1

])(

[

)(11

22

22

22

2

rrmrrm

rmrrrm

mrrrm

mrrm

dt

d

mrdt

d

dt

d

rr

rrrr

rrr

r

reppperp

erprepprpeperp

erppreperp

FrpLFperp

LprM

where

])(

[11

2rrmrdt

d r perpr

,

10

rrdt

deFp

2

and

FLτ rdt

d. (torque).

((Mathematica)) Proof of dM/dt = 0.

______________________________________________________________________________ We note that

prprpprpLp )()( 2

Then the vector M ( MA m ) always points in the direction of the perihelion from the focal point. At the perihelion,

0 rp

We have

Clear"Global`"; rt_ : xt, yt, zt;

Rt_ : rt.rt ;

pt_ : m Drt, t; Lt_ : Crossrt, pt;

Ft_ :

Rt3rt;

eq1

D rtRt, t 1

mCrossFt, Lt

1

mCrosspt, Crossrt, Ft

FullSimplify

0, 0, 0

11

)(

))((

)(

0

1

2

1

2

12

x

x

r

rr

A

r

lm

r

lm

rpm

m

e

e

e

ee

MA

where xr ee and

1

2

0 r

lmA .

In general case, we have the orbital equation as

20 cos lrmrA Ar ,

where is the angle between r and the perihelion direction. From this equation we get

2

0 )cos( lmAr ,

or

m

l

m

Ar

20 )1cos(

cos1cos1

0

0

2

e

p

m

Am

l

r

,

Then we have p0 is given by

m

lp

2

0 . (semi latus rectum)

and

12

emAA 0 .

Since 1

2

0 r

lmA , we have

)1(1

2

emr

l

3. Derivation of the Runge-Lentz vector

The equation of motion of a particle of mass m in the attractive potential is

rrdt

deF

p2

.

We take the cross product of both sides with the angular momentum L.

rLFp

L 3rdt

d .

Since L is constant in time, the left-hand side can be written as

)( pLp

L dt

d

dt

d.

From the explicit form

dt

dmL

rrpr ,

we get

rdt

dm

dt

d

dt

d

r

mdt

d

r

m

rr

r

rrr

rr

rrrLrrL

])[(

)(

23

333

Then we have

13

0)(

rdt

dm

dt

d rpL ,

or

0)( r

mdt

d rpL .

The Runge-Lentz vector is obtained as

LpepLeMA rr mmm .

which is a constant of motion. 4. Energy of the system We note that

222 2 LMm

H.

The proof is given by using the Mathematica. ((Mathematica))

The energy can be derived as follows.

Clear"Global`"; r x, y, z; p px, py, pz;

L Crossr, p; R r.r ;

M1

Rr

1

mCrossp, L;

H1 1

2 mp.p

R;

eq1 M1.M1 2

mH1 L.L FullSimplify

2

14

222222222 2 emmHmlm MA , or

ae

l

mEH

2)1(

22

2

2 (<0),

with emA , 22 lL , and EH . 4. Construction of the diagram by Mathematica

Fig. Simplified version of the diagram. is the angle between A and r. The direction of A is

from the focal point to the perihelion (P). We consider a point (K0) of the orbit of the ellipsoid which is expressed by

P

p L

m

A

F

2

O H1

H2

K0

K1

K2K3

K4

K5

P

15

12

2

2

2

b

x

a

x,

where the essentricity is given by

2

2

1a

be .

where a>b. The co-ordinate of the focal point F is )0,( ae . We also consider a point (H2) on the

circle which is expressed by

222 ayx .

We assume that the angle 12OHH is . The co-ordinates of the points K0 and H2

)sin,cos(2 aaOH , )sin,cos(0 baOK .

The slope of the tangential line (K0K4) at the point K0 on the ellipsoid is given by

slope=

tancotsin

cos

a

b

a

b

d

dxd

dy

,

where the angle 104 KKK is ;

)cotarctan( a

b .

The unit vector along the vector 20KK ,

2220

202

)sin()cos(

)sin,cos(

baea

baea

KK

KK

u ,

The unit vector along the vector 20KK ,

16

)cos,(sin30

303

KK

KKu .

We note that

xAmKKKKsKK euuLp 232203330 ,

where

emA , ).0,1(xe

Then we have

)( 3323 ueuu xms .

The co-ordinate of the point is given by

333003 )sin,cos( usbaKKOKOK

((Mathematica))

17

18

19

20

5. Runge-Lenz vector in quantum mechanics

Here we introduce a Runge-Lenz vector which in quantum mechanics is defined by

)(2

12

pLLprM mr

Ze.

This operator is Hermitian since . )()( pLLp Note that r [=(x, y, z)] is the position

vector, L is the orbital angular momentum,

prL .

M commutes with the Hamiltonian H,

0],[ iMH ,

P

p L

m

A

F

2

O H1

H2

K0

K1

K2K3

K4

K5

21

where H is the Coulomb Hamiltonian

r

Ze

mH

22

2

1 p .

The commutation relations between the angular momentum L and linear momentum p are given by

kijkji pipL ],[ ,

which leads to the expression

pLppL i2 .

Then we have

pLprMm

i

mr

Ze )(

12

.

We note that

0 LMML . We also note that

2)( LLpr , rpLrLp i2)( 2 ,

222)( LpLp .

We find that

)(2 22422 LM Hm

eZ .

The 2 term arises from the non-commutativity of quantum operators. These equations can be easily proved using the Mathematica (see below in detail).

We know that the angular momentum operators satisfy the commutation relations,

22

kijkji LiLL ],[ .

We also get

kijkji MiLM ],[ ,

and

kijkji HLm

iMM 2

],[ .

The Li’s generate rotations and define a closed algebra. But Li.s and Mi’s do not form a closed algebra since the last relation involves the Hamiltonian. However, we consider the case of specific bound states. In this case, the vector space is truncated only those that are eigenstates of H, with eigenvalue E<0. In this case we replace H with –E, and the algebra is closed.

We introduce a vector N such that

ME

m 2/1)2

(N .

In this case we have the closed algebra,

kijkji LiLL ],[ ,

kijkji NiLN ],[ ,

and

kijkji LiNN ],[ ,

0],[ 2 LiM .

6. Dynamic Symmetry operation

We consider the symmetry operation generated by the operators L and N, which corresponds

to the rotation in four spatial dimensions. We introduce ),,,( 4321 xxxx and ),,,( 4321 pppp . Note

that x4 and p4 are fictitious and cannot be identified with dynamical variables.

2332231 pxpxLLL x ,

23

3113312 pxpxLLL y ,

1221123 pxpxLLL z ,

1441141 pxpxLN ,

2442242 pxpxLN ,

3443343 pxpxLN .

This algebra is isomorphic (equivalent) to that of rotations in four dimensions [or the O(4) group]. 7. Energy level of the hydrogen-like atom

We define two sets of new generators as

)(2

1iii NLI , )(

2

1iii NLK .

and

iii KIL .

Then we get

kijkji IiII ],[ ,

kijkji KiKK ],[ ,

0],[ ji KI .

Furthermore we have

0],[],[ HKHI ii .

24

So that these operators are also conserved. In this basis the algebra becomes equivalent to that of two decoupled algebras of angular momenta. The eigenvalues of the operators I2 and K2 will have the eigenvalues,

I2; )1(2 ii , K2; )1(2 kk .

where i and k are either integer or half integers. We note that

)(2

1 2222 NLKI C ,

0.' 22 NLKIC . The second relation implies that

)1()1( kkii ,

or

i = k. Correspondingly, the allowed values of C are

C: )1(2 2 kk ,

)(22 224222 LNM Em

eZm

E,

or

)(2

2242

2 LNE

emZ,

or

)1(224

)(2

1 2242

22 kkE

emZ

NL ,

or

25

22242

)12(2

]1)1(4[24

kkkE

emZ ,

22

42

22

42

2)12(2 n

emZ

k

emZE

. (energy level of hydrogen atom).

where

12 kn . Since KIL with i= k,

l = i + k, i + k - 1,…….., |l - k|, or

l = 2k, 2k - 1,…….., 0. or

l = n - 1, n – 2,…….., 0. ______________________________________________________________________________ ((Mathematica))

26

Clear"Global`";

ux 1, 0, 0; uy 0, 1, 0; uz 0, 0, 1; r x, y, z;

R x2 y2 z2 ;

L : — Crossr, Grad, x, y, z & Simplify;

P : — Grad, x, y, z &; Lx : ux.L &; Ly : uy.L &;

Lz : uz.L &; Px : ux.P &; Py : uy.P &;

Pz : uz.P &; PSQ : PxPx PyPy PzPz &;

LSQ : LxLx LyLy LzLz &;

Mx :1

mPyLz PzLy —

mPx Z1 e12

Rx &;

My :1

mPzLx PxLz —

mPy Z1 e12

Ry &;

Mz :1

mPxLy PyLx —

mPz Z1 e12

Rz &;

M1 ux Mx uy My uz Mz &;ProofMx Lx + My Ly + Mz Lz = 0

MxLxx, y, z MyLyx, y, z MzLzx, y, z FullSimplify

0

ProofLx Mx + Ly My + Lz Mz = 0

LxMxx, y, z LyMyx, y, z LzMzx, y, z FullSimplify

0

Proof

[Mi ,L2]∫0

27

MxLSQx, y, z LSQMxx, y, z Simplify

1m

2 —2 e12 m x Z1 x, y, z

x2 y2 z2

e12 m x z Z1 0,0,1x, y, zx2 y2 z2

x —2 0,0,2x, y, z e12 m x y Z1 0,1,0x, y, zx2 y2 z2

x —2 0,2,0x, y, z e12 m y2 Z1 1,0,0x, y, zx2 y2 z2

e12 m z2 Z1 1,0,0x, y, zx2 y2 z2

—2 1,0,0x, y, z

3 z —2 1,0,1x, y, z x2 —2 1,0,2x, y, z y2 —2 1,0,2x, y, z 3 y —2 1,1,0x, y, z 2 y z —2 1,1,1x, y, z x2 —2 1,2,0x, y, z z2 —2 1,2,0x, y, z 2 x —2 2,0,0x, y, z 2 x z —2 2,0,1x, y, z 2 x y —2 2,1,0x, y, z

y2 —2 3,0,0x, y, z z2 —2 3,0,0x, y, z

Proof

x P μ Lx + y P μ Ly +z P μ Lz = L2

28

x PyLzx, y, z PzLyx, y, z y PzLxx, y, z PxLzx, y, z z PxLyx, y, z PyLxx, y, z LxLxx, y, z LyLyx, y, z LzLzx, y, z FullSimplify

0

Proof

P μ Lx x + P μ Ly y + P μ Lz z = L2+ 2 i Ñ (Px x+Py y +Pz z)

PyLzx x, y, z PzLyx x, y, z PzLxy x, y, z PxLzy x, y, z PxLyz x, y, z PyLxz x, y, z LxLxx, y, z LyLyx, y, z LzLzx, y, z 2 — Pxx x, y, z 2 — Pyy x, y, z 2 — Pzz x, y, z FullSimplify

0

Proof Px P μ Lx + Py P μ Ly + Pz P μ Lz =0

P. (P x L)=0

29

PxPyLzx, y, z PxPzLyx, y, z PyPzLxx, y, z PyPxLzx, y, z PzPxLyx, y, z PzPyLxx, y, z

FullSimplify

0

Proof

P μ Lx Px + P μ Ly Py + P μ Lz Pz = 2 i Ñ (Px2 + Py2 + Pz2)

(P x L). P. = 2 i Ñ P2

PyLzPxx, y, z PzLyPxx, y, z PzLxPyx, y, z PxLzPyx, y, z PxLyPzx, y, z PyLxPzx, y, z 2 — PxPxx, y, z PyPyx, y, z

PzPzx, y, z FullSimplify

0

30

Proof

P L2 P2 L2

K1 PyLzPy Lzx, y, z PyLzPz Lyx, y, z Pz LyPy Lzx, y, z PzLyPzLyx, y, z PzLxPzLxx, y, z PzLxPx Lzx, y, z Px LzPzLxx, y, z PxLzPx Lzx, y, z PxLyPxLyx, y, z PxLyPy Lxx, y, z Py LxPxLyx, y, z PyLxPy Lxx, y, z

FullSimplify;

K2 PSQLSQx, y, z Simplify;

K1 K2 Simplify

0

Definition of Hamiltonian

H = 1

2 m (Px2 + Py2 + Pz2) - Z2 e2

r

31

H1 :1

2 mPSQ Z1 e12

R &;

Proof [H, Mx]=0, [H, My]=0, [H, My]=0,

H1Mxx, y, z MxH1x, y, z FullSimplify

0

H1Myx, y, z MyH1x, y, z FullSimplify

0

Proof Mx2 + My2 + Mz2 - Z2 e4 = 2

m H (L2 + Ñ2)

MSQ1 MxMxx, y, z MyMyx, y, z

MzMzx, y, z Z12 e14 x, y, z Simplify;

eq1 2

m H1LSQx, y, z —2 H1x, y, z;

MSQ1 eq1 FullSimplify

0

32

Proof[Lx, Ly] = i Ñ Lz,

[Mx, Ly] = i Ñ Mz,[My, Lz] = i Ñ Mx,

LxLyx, y, z LyLxx, y, z — Lzx, y, z Simplify

0

MxLyx, y, z LyMxx, y, z — Mzx, y, z Simplify

0

MyLzx, y, z LzMyx, y, z — Mxx, y, z Simplify

0

Proof

[Mx, My] =- i 2Ñm

H Lz,

[My, Mz] =- i 2Ñm

H Lx,

MxMyx, y, z MyMxx, y, z

2 —

mH1Lzx, y, z Simplify

0

MyMzx, y, z MzMyx, y, z

2 —

mH1Lxx, y, z Simplify

0

33

_________________________________________________________________________ REFERENCES L.I. Schiff, Quantum Mechanics (McGraw-Hill Book Company, Inc, New York, 1955). E. Merzbacher, Quantum Mechanics, third edition (John Wiley & Sons, New York, 1998). Ramamurti Shankar, Principles of Quantum Mechanics, second edition (Springer, New York,

1994). A. Das, Lectures on Quantum Mechanics, 2nd edition (Hindustan Book Agency, 2012). S. Weinberg, Lectures on Quantum Mechanics (Cambridge University Press, 2013). J.J. Sakurai and J. Napolitano, Modern Quantum Mechanics, second edition (Addison-Wesley,

New York, 2011). J.M.Finn, Classical Mechanics Infinity Science Press LLC (Hingham, Ma 2008). E.D. Commins, Quantum Mechanics: An Experimentalist’s Approach (Cambridge, 2014).

APPENDIX Kepler’s law For convenience, we use the following diagram for the ellipse orbit.

A.1 The angular momentum

In general case

FO

r

P x,y

34

rv

rv

rdt

d

rrra

rra

r

r

)(1

2 2

2

Since the gravitational force is directed toward the origin (so called central field),

0

22

marr

MmGmar

where

mMG . In other words,

rv

rdt

d

rrra

0)(1

2 2

or

mrvr

vmrmrl 22 = constant.

The angular momentum is defined as

zzzrrr Lmrvvvrmm eeeeevrprL )()()( ,

or

rmvLz ,

0)()( FrvrprL

mdt

d

dt

d

dt

d,

35

since F is a central force (r//F), zL is a constant of motion.

2mrmrvl .

The velocity v is given by

mr

l

mr

lrrv 2

.

A.2 Physical meaning

What is the physical meaning of the constant angular momentum? We now consider the dA/dt, where dA is the partial area of the ellipse.

constm

lr

dt

dr

dt

dA

drdA

22

1

2

12

1

22

2

since constmrl 2 . The period T is given by

22 1222

eal

mab

l

mdA

l

mdtT

since dAl

mdt

2 .

A.3. The effective potential

The total energy is a sum of the kinetic energy and the potential energy

36

rvvm

rmvE r

)(

2

1

2

1 222

or

rmr

lrm

rrm

lrmE

2

22

22

22

22

1)(

2

1 , (1)

The energy is dependent only on r (actually one dimensional problem).

2

2

2mr

l

rUeff

. (effective potential)

The effective potential energy Ueff has a local minimum

2

2min

2l

mUeff

,

at

20

2

min

p

m

lr

.

Since E = constant, we have an equation of motion

0)( 3

2

2

3

2

2

rmr

l

rrm

rmr

lr

rrrm

dt

dE

03

2

2

mr

l

rrm

. (equivalent 1D problem)

______________________________________________________________________ Plot of the effective potential as a function of r

37

Fig. The effective potential vs r with l = 0.1 – 0.5 A.4 Perihelion and aphelion

When 0r for the perihelion (nearest from the Sun) and the aphelion (farthest from Sun) r1 and r2 are the roots of Eq.(1).

02

22

mE

lr

Er

.

There are the relations between r1 and r2.

Em

l

mE

lrr

aEE

rr

22

2

22

21

21

where the total energy is given by

aE

2

. (E<0, bound state)

and

0.2 0.4 0.6 0.8 1.0

-5

-4

-3

-2

-1

0

1

38

)1(

)1(

2

1

ear

ear

From this we have

am

la

ma

l

Ema

l

a

rre

2

2

2

2

2

2212 2

221

or p0 is the semi latus rectum,

m

leap

22

0 )1(

A.5 Kepler’s Third Law

m

ma

ama

l

mea

l

mT

2/32222 22

12

,

or

32322

2 44 ma

m

amT ,

or

2222

3

444 sunsun GM

m

GmM

mT

a ,

or

32

2 4a

GMT

sun

or

32 )]([)]([ AUayearT .

39

A.6 Derivation of the Kepler's First Law We start with

tconslmr

rmrrm

tan2

22

Here we have

dmrldt 2 .

Note that r depends only on .

)()(22

2

d

d

mr

l

d

d

mr

l

dt

d

dt

d

d

d

mr

l

dt

d

dt

d

or

232

2

22)(

mrrm

l

d

dr

mr

l

d

d

mr

l

.

We define u as r

u1

,

d

du

rd

d

d

dr

r )

1(

12

.

Then we have

232

2

22

2

)(mrrm

l

d

du

d

d

rm

l

,

or

22

2

l

mu

d

ud

.

40

The solution of this equation is given by

)cos1(1

12

Cl

m

ru ,

where A is constant, )1(1 ear for = 0, and )1(2 ear for =

)1(1

121

Cl

m

r

, )1(

112

2

Cl

m

r

,

or

eea

m

l

eC

)1( 2

2

1 .

Then we have

cos10

e

pr

,

with

m

leap

22

0 )1( .