quantum runge-lenz vector and the hydrogen …...quantum runge-lenz vector and the hydrogen atom,...

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Quantum Runge-Lenz Vector and the Hydrogen Atom,the hidden SO(4) symmetryPascal Szriftgiser1 and Edgardo S. Cheb-Terrab2

(1) Laboratoire PhLAM, UMR CNRS 8523, Université Lille 1, F-59655, France(2) Maplesoft

Let's consider the Hydrogen atom and its Hamiltonian

H =p 2

2 me

r,

where p is the electron momentum, me its mass, a real positive constant, and r the distance of the electron from the proton located at the origin. We assume that the proton's mass is infinite. Classically,

from the potential r

, one can derive a central force F = rr2 that drives the electron's motion.

Introducing the angular momentum

L = r p ,

one can further define the Runge-Lenz vector Z :

Z =1me

L p rr

.

It is well known that Z is a constant of the motion, i.e. ddt

Z t = 0. Switching to Quantum Mechanics,

this condition reads

H, Z = 0.where, for hermiticity purpose, the expression of Z must be symmetrized

Z =1

2 me L p p L

rr

.

Problem: departing from the basic commutation rules between position r , momentum p and angular momentum L in tensor notation, and the expression of the Hamiltonian H , demonstrate the following commutation rules between the quantum Hamiltonian, angular momentum and Runge-Lenz vector Z

H, Ln = 0 and H, Zn = 0,

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(1)(1)

Lm, Zn = i m, n, o

Zo,

Zm, Zn = 2 i me

H m, n, o

Lo.

Remark: Since H commutes with both L and Z, defining

Mn =me

2 H Zn,

the commutation rules demonstrated can be rewritten asLm, Ln = i

m, n, o Lo,

Lm, Mn = i m, n, o

Mo,

Mm, Mn = i m, n, o

Lo.

This set constitutes the Lie algebra of the SO(4) group (closely related to a Poincaré group in special relativity).

I Commutation rules and useful identities

Quantum commutation rules basics and the Hamiltonian of the hydrogen atom

Set macros for M = me

restart;with Physics : with Library :interface imaginaryunit = i :macro M = me :

Set the context: Cartesian coordinates, 3D Euclidean space, lowercase letters representing tensor indices, use automatic simplification (automatically apply simplify/size on everything). Here

V = 1r

will represent the potential of the central force.

Setup coordinates = cartesian, hermitianoperators = X , realobjects = , , me ,

automaticsimplification = true, dimension = 3, metric = Euclidean, spacetimeindices

= lowercaselatin, mathematicalnotation = true, quiet

automaticsimplification = true, coordinatesystems = X , dimension = 3,

hermitianoperators = X , mathematicalnotation = true, metric = 1, 1 = 1, 2, 2

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(2)(2)

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(5)(5)

(1)(1)

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(4)(4)

(3)(3)

= 1, 3, 3 = 1 , realobjects = , , me, x, y, z , spacetimeindices = lowercaselatin

Setting quantum (Hermitian) operators and related commutators:

- Z is Hermitian, but we derive that property further below. - The potential V X of the hydrogen atom is assumed to commute with position, not with momentum - the commutation rule with p is derived further ahead.- The commutator rules for angular momentum are an easy problem, we take them as the departure point.- The last two commutators involving G X are for the differential operators approach only, not really part of the problem.

Setup quantumoperators = Z ,hermitianoperators = V, G, H, L, X, p ,algebrarules = %Commutator X k , X l = 0, %Commutator p k , p l = 0, %Commutator X k , p l = i g_ k, l , %Commutator L j , L k = i LeviCivita j, k, n L n , %Commutator p j , L k = i LeviCivita j, k, n p n , %Commutator X j , L k = i LeviCivita j, k, n X n , %Commutator X k , V X = 0, %Commutator V X , G X = 0, %Commutator X k , G X = 0

algebrarules = V X , G X = 0, Lj, Lk = i j, k, n

Ln, pj, Lk

= i j, k, n

pn, pk, pl = 0, Xj, Lk = i j, k, n

Xn, Xk, G X = 0, Xk,

V X = 0, Xk, pl = i gk, l, Xk, Xl = 0 , hermitianoperators = G, H, L, V,

p, x, y, z , quantumoperators = G, H, L, V, Z, p, x, y, z ,

Define the tensorsDefine p k = px, py, pz , L k = Lx, Ly, Lz , Z k = Zx, Zy, Zz , quiet

a, Lk, a, Xa, Zk, a, ga, b, pk, a, b, c

CompactDisplay V X , G X V X will now be displayed as V G X will now be displayed as G

The Hamiltonian for the hydrogen atom

H =p l 2

2 MV X

H =pl

2

2 me V

(7)(7)

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(13)(13)

(10)(10)

(6)(6)

(1)(1)

(11)(11)

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(12)(12)

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(9)(9)

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Identities (I): n

V = V3 X

n, V

3 X

l

2= V and V = 0

For a more compact calculus, we use the dimensionless potential V X

V X =1

sqrt X o 2

V =1

Xo2

12

The gradient of V X is d_ n (6)

nV =

1

Xo2

32

Xn

So thatsubs rhs = lhs (6)3 , (7)

nV = V3 Xn

Equivalently, V X can be writtenSumOverRepeatedIndices (6)

V =1

x2 y2 z2

12

from which one can deduce V 3 Xl2 = V , that will often be used afterwards

(9)3 x2 y2 z2

V3 x2 y2 z2 =1

x2 y2 z2

12

subs rhs = lhs (9) , x2 y2 z2 = X l 2 , (10)V3 Xl

2 = V

And finally V = 0SumOverRepeatedIndices dAlembertian (9)

V =

3 1

x2 y2 z2

52

4 x2 4 y2 4 z2

43

1

x2 y2 z2

32

Factor (12)V = 0

(20)(20)

(19)(19)> >

(17)(17)

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> > (15)(15)

(1)(1)

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(21)(21)

(22)(22)

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(16)(16)

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(23)(23)

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(14)(14)

(18)(18)

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Identities (II): the commutation rules between L, p and the potential V XOne has

L q = LeviCivita q, m, n X m p nLq =

m, n, q Xm pn

Commutator (14), V XLq, V =

m, n, q Xm pn, V

These two commutators cannot be computed until providing more information - we set the equations here for use further below with a test function G X

%Commutator = Commutator p q , V Xpq, V = pq, V

%Commutator = Commutator p q , V X 3

pq, V3 = pq, V3

Set now differentialoperators and some commutators with the arbitrary test function G X , to be used in the alternative demonstrations based on using differentialoperators and to derive the commutation rules between L , p and V X

Setup differentialoperators = p k , x, y, zdifferentialoperators = pk, X

Now, apply the differential operators found in the commutators above to G Xlhs = ApplyProductsOfDifferentialOperators@rhs (15) G X

Lq, V G =m, n, q

Xm pn V G V pn G

lhs = ApplyProductsOfDifferentialOperators@rhs (16) G Xpq, V G = pq V G V pq G

lhs = ApplyProductsOfDifferentialOperators@rhs (17) G Xpq, V3 G = pq V3 G V3 pq G

The result of pl G X is not known to the system at this point, we set equation to be used furtherbelow

pl G

ApplyProductsOfDifferentialOperators p l G X = p l G Xpl G = pl G

Define now the momentum operator as an indexed procedure

p proc local Ind op procname ;

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(29)(29)

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(25)(25)

(32)(32)

(1)(1)

(26)(26)

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(30)(30)

(24)(24)

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(27)(27)

(34)(34)

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(28)(28)

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(31)(31)

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(33)(33)

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return i

Physics:-d_ Ind args ;

end:

With this definition, we have(19)

Lq, V G = i m, n, q

Xm n

V G

So thatSubstituteTensor (8), (24) Inverse G X

Lq, V = i m, n, q

Xm V3 Xn

Finally we get the first commutation rule:Simplify (25)

Lq, V = 0

Likewise, from (20) we get the second commutation rule:(20) Inverse G X

pq, V = i q

V

SubstituteTensor (8), (27)pq, V = i V3 Xq

From (21) we get the third commutation rule we were looking for:(21) Inverse G X

pq, V3 = i q

V V2 V q

V V V2 q

V

lhs (29) = Simplify SubstituteTensor (8), rhs (29)pq, V3 = 3 i V5 Xq

In addition, we rewrite (23) as an identity to be used further below to remove l

G X . Equation (23), after defining p as a procedure in (24), becomes

(23)i

lG = pl G

isolate (23),l

G X

lG =

i pl G

Add now these new commutation rules to the setup of the problem so that they are taken into account when using Simplify

(26), (28), (30)Lq, V = 0, pq, V = i V3 Xq, pq, V3 = 3 i V5 Xq

Setup (33) ;

(35)(35)

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(39)(39)

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(1)(1)

(37)(37)

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(36)(36)

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(40)(40)

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(34)(34)

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(38)(38)> >

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algebrarules = Lj, Lk = i j, k, n

Ln, Lq, V = 0, Xj, Lk = i j, k, n

Xn, Xk,

Xl = 0, Xk, pl = i gk, l, Xk, G = 0, Xk, V = 0, pj, Lk = i j, k, n

pn,

pk, pl = 0, pq, V3 = 3 i V5 Xq, pq, V = i V3 Xq, V, G = 0 ,

Undo differentialoperators to work using two different approaches, with and withoutdifferentialoperators

Setup differentialoperators = nonedifferentialoperators = none

II H, Ln

= 0

Departing from the Hamiltonian of the hydrogen atom (5) and the definition of angular momentum (14)

(5); (14);

H =pl

2

2 me V

Lq =m, n, q

Xm pn

by taking their commutator we getCommutator (5), (14)

H, Lq =i

m, n, q Xm V3 Xn me pl pn gl, m

me

Simplify (37)H, Lq = 0

III H, Zn

= 0

This one is less easy. Start from the definition of the quantum Runge-Lenz vector

Z k =1

2 MLeviCivita a, b, k L a p b p a L b V X X k

Zk = a, b, k La pb pa Lb

2 me V Xk

Since the system now knows about the commutation rule between linear and angular momentum,%Commutator = Commutator L a , p b

La, pb = i a, b, n

pn

the expression (39) for

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(1)(1)

(42)(42)

(47)(47)

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(41)(41)

(44)(44)

(45)(45)

(49)(49)

(34)(34)

(48)(48)

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(46)(46)

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(43)(43)

Zk can be simplified

Simplify (39)

Zk =i pk

me V Xk

a, b, k pa Lb

me

and the angular momentum removed from the the right-hand side so that Zk gets expressed entirely in terms of pk , X and V

(14)Lq =

m, n, q Xm pn

Simplify SubstituteTensor (14), (41)

Zk =i pk V Xk me Xm pk pm Xk pn

2

me

Taking the commutator between the Hamiltonian (5) and this expression for Zk we have the starting point towards showing that H, Zk = 0

Simplify Commutator (5), (43)

H, Zk =1

2 me V3 Xk V5 Xa

2 Xk 2 i pk V 2 i V Xa Xk pa V2

2 i Xb2 pk V

3 2 i Xa Xk pa V3

For the term with V5 we use the derived identity (11)(11)

V3 Xl2 = V

V X 2 (11) X kV5 Xl

2 Xk = V3 Xk

SubstituteTensor (46), (44)

H, Zk = V3 Xk i pk V i Xb

2 pk V3 i V Xa Xk pa V2 i Xa Xk pa V3

me

Simplify (47)

H, Zk = V5 Xa

2 Xk V3 Xk i pk V i Xb2 pk V

3

me

Another term with V5 appearedSubstituteTensor (46), (48)

H, Zk = 2 V3 Xk i pk V i Xb

2 pk V3

me

Make Xb and V be contiguous to further apply (11)

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(53)(53)

(57)(57)

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(56)(56)

(1)(1)

(50)(50)

(51)(51)

(52)(52)

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(34)(34)

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(55)(55)

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SortProducts (49), p k , X b

H, Zk = 2 V3 Xk i pk V i pk Xb

2 2 i gb, k Xb V3

me

Simplify (50)

H, Zk =i pk V pk V

3 Xb2

me

p k (45)pk V

3 Xl2 = pk V

SubstituteTensor (52), (51)H, Zk = 0

And this is the result we wanted to prove. In the next section there is an alternative derivation that couldbe seen as more abstract or more direct, depending on the point of view.

Alternative approach using differential operators

As done in the previous section when deriving the commutators between linear and angular momentum, on the one hand, and the central potential V on the other hand, the idea here is again to use differential operators taking advantage of the ability to compute with them as operands of a product, that get applied only when it appears convenient for us


So take the starting point (44)(44)

H, Zk =1

2 me V3 Xk V5 Xa


2 i Xb2 pk V

3 2 i Xa Xk pa V3

and to show that the left-hand side is equal to 0, multiply by a generic function G X followed by transforming the products involving pk by the application of this differential operator

(44) G X

H, Zk G =1

2 me V3 Xk V5 Xa


2 i Xb2 pk V

3 2 i Xa Xk pa V3 G

ApplyProductsOfDifferentialOperators (56)

H, Zk G =1me

V k

G k

V G V Xa Xk aV V

(58)(58)

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(64)(64)

(57)(57)

(1)(1)

(50)(50)

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(60)(60)

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(62)(62)

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(34)(34)

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(61)(61)

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(63)(63)

(59)(59)

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V a

V G V2 a

G Xb2

kV V2 V

kV V V2

kV G

V3 k

G Xa Xk aV V2 V

aV V V2

aV G V3

aG

V3 Xk G

2

V5 Xa2 Xk G

2

Simplify (57)

H, Zk G =1me

2 V k

Gk

V G Xb2

kV G V2 V Xb

2 k

V G V

V2 Xb2

kV G V3 Xb

2 k

G Xa Xk aV G V2

G V3 Xk

2

G V5 Xa2 Xk

2

Use now the derived identities for the gradients of V and G and then remove the generic function G from the equation by multiplying by the inverse of G

(8); (32);

nV = V3 Xn

lG =

i pl G

Simplify SubstituteTensor (8), (32), (58) Inverse G X

H, Zk = i V pk i V3 Xa

2 pk

3 V3 Xk

2

3 V5 Xa2 Xk

2me

To show that the right-hand side is actually 0, recalling (11)(11)

V3 Xl2 = V

rhs = lhs (11) p kV pk = V3 Xl

2 pk

This and using (46) againSimplify SubstituteTensor (62), (46) , (60)

H, Zk = 0

Reset differentialoperators in order to proceed to the next section working without differential operators

Setup differentialoperators = nonedifferentialoperators = none

(69)(69)

(66)(66)

2. 2.

3. 3.

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(57)(57)

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(1)(1)

(50)(50)

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(65)(65)

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(34)(34)

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(68)(68)

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1. 1.

(67)(67)

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(70)(70)

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IV Lm

, Zn

= i m, n, k

Zk

Strategy:

Express Lm and Zk in terms of Xa and pb from previous sectionsConstruct the left-hand and right-hand sides of Lq, Zk = i

k, q, u Zu , the formula we want

to proveSimplify the result

Step 1 got accomplished in the previous sections(14)

Lq =m, n, q

Xm pn

(43)

Zk =i pk V Xk me Xm pk pm Xk pn

2

me

Step 2.

The left-hand side of the identity to be proved is the left-hand side of their commutatorCommutator (14), (43)

Lq, Zk =1me

m, n, q i Xm gk, n V V3 Xn Xk me gk, m pn i Xm pk pa ga, n

i Xm pb2 gk, n 2 i Xk pb pn gb, m i Xa ga, m pk gk, m pa pn

The right-hand side of Lq, Zk = i k, q, u

Zu to be proved is given by

i LeviCivita q, k, u SubstituteTensorIndices k = u, (43)

i k, q, u

Zu =i

k, q, u i pu V Xu me Xu pn

2 Xm pu pm

me

Step 3.

Take one minus the other one and the right-hand side must be equal to 0(67) (68)

Lq, Zk i k, q, u

Zu =1me

m, n, q i Xm gk, n V V3 Xn Xk me gk, m pn

i Xm pk pa ga, n i Xm pb2 gk, n 2 i Xk pb pn gb, m i Xa ga, m pk gk, m pa pn

i k, q, u

i pu V Xu me Xu pn2 Xm pu pm

me

Simplify (69)Lq, Zk i

k, q, u Zu = 0

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(34)(34)

(71)(71)

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(57)(57)

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(1)(1)

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(50)(50)

(72)(72)

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V Zm

, Zn

= 2 i m

e

H m, n, o

Lo

Here again the starting point is (43), the definition of the quantum Runge-Lenz vectorSubstituteTensorIndices k = q, (43)

Zq =i pq V Xq me Xm pq pm Xq pn

2

me

In this section the strategy is the same as in the previous section: construct the left-hand and right-hand

sides of Zm, Zn =2 i me

H m, n, o

Lo , then take one minus the other one, and show that the

right-hand side is equal to 0.

Start with the left-hand side Zm, Zn

Commutator (43), (71)

Zk, Zq =1

me2 2 i gb, m Xq pb pk pm me Xm i pk gm, q V V3 Xm Xq i

gk, q V V3 Xk Xq pm 2 i gb, k Xq pb pn2 i Xm ga, k pq pa pm

ga, m pk pq pa2 V3 Xk Xq me

2 V3 Xq Xk me i Xa ga, m pq

gm, q pa pk pm 2 i ga, n Xk pn pq pa 2 i gn, q Xk pn pb2 i Xq 2 pb V3 Xb

V3 3 i 2 pb Xb 3 i V5 Xb 2 V3 pb Xb Xk me

2 i V Xq pb me gb, k2 pn

2 gk, q2 pk pm gm, q

2 pq pa ga, k2 pb

2 gk, q

2 i Xk V pn gn, q

2 pn V3 Xn V3 3 i 2 pn Xn 3 i V5 Xn 2 V3 pn Xn Xq

2

i Xa pq V3 Xa V3 Xq pa Xk i V Xa ga, k pq gk, q pa me i Xa ga, k pq

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(77)(77)

(75)(75)

(57)(57)

(76)(76)

(1)(1)

(50)(50)

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(34)(34)

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(73)(73)

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(74)(74)

(72)(72)

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gk, q pa pn2 i Xm gk, q pb

2 pm gm, q pk pb2

Now the right-hand side, constructed from the definition of the angular momentum(14)

Lq =m, n, q

Xm pn

Multiply by 2 i

k, q, u H

me

2 iM

H LeviCivita q, k, u SubstituteTensorIndices q = u, (14)

2 i k, q, u

H Lu

me=

2 i k, q, u

m, n, u

H Xm pn

me

Replace the Hamiltonian H by its expression quadratic in the momentum(5)

H =pl

2

2 me V

lhs (74) = SubstituteTensor (5), rhs (74)

2 i k, q, u

H Lu

me=

2 i k, q, u

m, n, u

pl

2

2 me V Xm pn

me

Now set up the problem, taking (72) minus (76)(72) (76)

Zk, Zq

2 i k, q, u

H Lu

me=

1me

2 i Xa ga, m pq gm, q pa pk pm

i Xm ga, k pq pa pm ga, m pk pq pa 2 i ga, n Xk pn pq pa

2 i k, q, u

m, n, u

pl

2

2 me V Xm pn me 2 i gn, q Xk pn pb

2 i Xa ga, k pq

gk, q pa pn2 i Xm gk, q pb

2 pm gm, q pk pb2 V3 Xk Xq me

V3 Xq Xk me 2 i gb, m Xq pb pk pm 2 i V Xq pb me gb, k i Xa pq V3 Xa

V3 Xq pa Xk me 2 i gb, k Xq pb pn2 i V Xa ga, k pq gk, q pa me

(77)(77)

(78)(78)

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(57)(57)

(80)(80)

(1)(1)

(50)(50)

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(34)(34)

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(72)(72)

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(79)(79)

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gm, q pk pm ga, k pq pa gk, q pb2 gk, q pn

2 i Xm pk gm, q V

V3 Xm Xq gk, q V V3 Xk Xq pm 2 i Xk V pn gn, q

2 pn V3 Xn V3 3 i 2 pn Xn 3 i V5 Xn 2 V3 pn Xn Xq

2

i Xq 2 pb V3 Xb V3 3 i 2 pb Xb 3 i V5 Xb 2 V3 pb Xb Xk me

So the starting point to prove that Zk, Zq =2 i

k, q, u H Lu

me is

Simplify (77)2 i

k, q, u H Lu Zk, Zq me

me=

1me

2 V5 Xj2 Xk Xq me i Xb

2 Xq pk V3 me

i Xb2 Xk pq V3 me i Xq pk V me i Xk pq V me i Xj Xk Xq pj V

3 me

i V Xg Xk Xq pg V2 me i gk, q Xc pc pd2 i gk, q Xb pa

2 pb 2 i Xq pd2 pk

2 i Xq pa2 pk

were the proof is achieved showing that the right-hand side of this equation is indeed equal to 0.

Start checking the repeated indices, as we would do by handCheck (78), all

The products in the given expression check ok.The repeated indices per term are: ... , ... , ... , the free indices are: ...

u , k, q = a, b, c, d, g, j , k, q

Check in which terms - that involve V - are these repeated indices appearingfor term in select has, map op, indets (78), ` ` , V do term = Check term, repeated, quiet od

i Xk pq V me =

i Xb2 Xq pk V

3 me = b

i V Xg Xk Xq pg V2 me = g

i Xq pk V me =

i Xb2 Xk pq V3 me = b

i Xj Xk Xq pj V3 me = j

(77)(77)

(84)(84)

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(57)(57)

(80)(80)

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(1)(1)

(50)(50)

(81)(81)

(83)(83)

(87)(87)

(85)(85)

(82)(82)

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(34)(34)

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(86)(86)

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(72)(72)

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V5 Xj2 Xk Xq me = j

We see that there are terms with Xj2 and terms with Xb

2 , so j = b results in a simplification

SubstituteTensorIndices j = b , (78)2 i


me=

1me

2 V5 Xb2 Xk Xq me i Xb

2 Xq pk V3 me

i Xb2 Xk pq V3 me i Xq pk V me i Xk pq V me i Xb Xk Xq pb V3 me

i V Xg Xk Xq pg V2 me i gk, q Xc pc pd2 i gk, q Xb pa

2 pb 2 i Xq pd2 pk

2 i Xq pa2 pk

Simplify (81)2 i


me

=i Xb

2 Xq pk V3 Xb

2 Xk pq V3 Xq pk V Xk pq V

me

There are two terms with V 3 that could be removed using (11)

V3 Xl2 = V

but for that to be possible, in (82) we need to have V to the left of Xb2 . So sort the products indicating

the desired ordering of operands , then simplifySortProducts (82), pk, pq, Xk, Xq, Xb

2 i k, q, u

H Lu Zk, Zq me

me=

1me

i pk Xq Xb2 i gk, q Xb

2

2 i gb, k Xq Xb V3 pq Xk Xb2 i gk, q Xb

2 2 i gb, q Xk Xb V3 pk Xq

i gk, q V pq Xk i gk, q V

Simplify (84)2 i


me

=i pk V Xq pk V

3 Xb2 Xq pq V3 Xb

2 Xk pq V Xk

me

SubstituteTensor (83), (85)2 i


me=

i pk V Xq V pk Xq V pq Xk pq V Xk

me

Simplify (86)

(92)(92)

(77)(77)

> >

(57)(57)

(80)(80)

(1)(1)

(50)(50)

> >

(87)(87)

> >

> >

(90)(90)

(34)(34)

> >

(91)(91)

> >

> >

(89)(89)

(88)(88)

(72)(72)

> >

> >

> >

2 i k, q, u

H Lu Zk, Zq me

me= 0

which is the identity we wanted to prove. In the next section the same result is obtained using differential operators

Alternative approach using differential operatorsThe main idea: make pk be a differential operator, then make (78)

(78)2 i


me=

1me

2 V5 Xj2 Xk Xq me

i Xb2 Xq pk V

3 me i Xb2 Xk pq V3 me i Xq pk V me i Xk pq V me

i Xj Xk Xq pj V3 me i V Xg Xk Xq pg V2 me i gk, q Xc pc pd

2

i gk, q Xb pa2 pb 2 i Xq pd

2 pk 2 i Xq pa2 pk

by a generic function G(X), apply the products of differential operators, then use the tensorial simplification equations

(8); (32); (11);

nV = V3 Xn

lG =

i pl G

V3 Xl2 = V

The goal is again to show that the right-hand side of (78) is equal to 0, so set pk to be a differential operator, multiply (78) by a generic function G X followed by applying pk where it corresponds


(78) G X2 i

k, q, u H Lu Zk, Zq me G

me=

1me

2 V5 Xj2 Xk Xq me

i Xb2 Xq pk V

3 me i Xb2 Xk pq V3 me i Xq pk V me i Xk pq V me

i Xj Xk Xq pj V3 me i V Xg Xk Xq pg V2 me i gk, q Xc pc pd

2

i gk, q Xb pa2 pb 2 i Xq pd

2 pk 2 i Xq pa2 pk G

Apply now pk

ApplyProductsOfDifferentialOperators (91)

(92)(92)

> >

(77)(77)

(93)(93)

(94)(94)

(57)(57)

(80)(80)

> >

(1)(1)

(50)(50)

> >

(87)(87)

> >

(96)(96)

(34)(34)

> >

> >

> >

(95)(95)

(72)(72)

> >

2 i k, q, u

H Lu Zk, Zq me G

me=

1me

2 gk, q 3 Xb b

G

gk, q 3 Xc cG me Xk q

V G V q

G me Xq k

V G

V k

G me V Xg Xk Xq g

V V V g

V G V2 g

G

me Xb2 Xk q

V V2 V q

V V V2 q

V G V3 q

G

me Xb2 Xq

kV V2 V

kV V V2

kV G V3

kG

me Xj Xk Xq j

V V2 V j

V V V2 j

V G V3 j

G

V5 Xj2 Xk Xq me G

We want to show that the right-hand side is equal to 0; start simplifying with respect to algebra rules and using Einstein's sum rule for repeated indices

Simplify (92)2 i


me=

1me

2 Xk Xl Xq l

V G V2

V3 Xa2 Xk q

G V3 Xa2 Xq

kG V2 Xa

2 Xk qV G V2 Xa

2 Xq k

V G

V Xa2 Xk q

V G V V Xa2 Xq

kV G V Xa

2 Xk qV G V2

Xa2 Xq

kV G V2 G V5 Xa

2 Xk Xq V Xk qG V Xq

kG Xk q

V G

Xq k

V G

Next, using (8), the gradient of V can be removed(8)

nV = V3 Xn



me=

1me

2 Xk Xl Xq V3 Xl G V2

V3 Xa2 Xk q

G V3 Xa2 Xq

kG V2 Xa

2 Xk V3 Xq G

V2 Xa2 Xq V3 Xk G V Xa

2 Xk V3 Xq G V V Xa

2 Xq V3 Xk G V

Xa2 Xk V

3 Xq G V2 Xa2 Xq V3 Xk G V2 G V5 Xa

2 Xk Xq V Xk qG

V Xq k

G Xk V3 Xq G Xq V3 Xk G

Simplify (95)2 i


me

(92)(92)

(97)(97)

> >

(77)(77)

(98)(98)

> >

(57)(57)

(80)(80)

(1)(1)

(50)(50)

(99)(99)

(100)(100)

(87)(87)

> >

(96)(96)

> >

(34)(34)

> >

> >

> >

> >

(72)(72)

> >

=2 V3 Xa

2 Xk qG V3 Xa

2 Xq k

G V Xk qG V Xq

kG

me

From (32) the gradient of G can be removed(32)

lG =

i pl G

Simplify SubstituteTensor (32), (96)2 i


me

=i V3 Xa

2 Xk pq G V3 Xa2 Xq pk G V Xk pq G V Xq pk G

me

And there are still two terms of the form X2 V3 that can be removed using (11)(11)

V3 Xl2 = V



me= 0

And since G is just an arbitrary test function, the above is already the result (87) we wanted to obtain.

quantum runge-lenz vector and the hydrogen …...quantum runge-lenz vector and the hydrogen atom,...

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