Central Potentials

Chapter 13

Central potentials

c B. Zwiebach

The use of spherical harmonics, which are eigenfunctions of the differential operators

L2 and Lz, allows us to turn the three-dimensional Schrodinger equation for a particlein a central potential into a one-dimensional Schrodinger radial equation for a particle in

an effective potential. We discuss the free particle solutions, which help us solve for the

spectrum of the infinite spherical well. We then turn to the three-dimensional isotropic

harmonic oscillator. We describe the spectrum, at each energy level, in terms of multiplets

of angular momentum. Finally, we review the main facts about the hydrogen atom and

discuss the classical conservation of the Runge-Lenz vector. This vector, whose magnitude

equals the eccentricity of an elliptical orbit, has a quantum analog that is also conserved and

is useful to understand algebraically the spectrum of the hydrogen atom.

13.1 Comments on spherical harmonics

In the coordinate representation the momentum operator for a particle is a differentialoperator. Momentum eigenstates are then viewed as wavefunctions that are eigenfunctionsof this differential operator. The situation is similar for angular momentum operators. Usingspherical coordinates (r, , ) we have seen that the orbital angular momentum operators aredifferential operators on the angular variables (, ). As we will discuss, spherical harmonicsYm(, ) are eigenfunctions of some suitable combinations of angular momentum operators.They can also be viewed as the wavefunctions for the states |,m discussed in the previouschapter.

The differential operator for L2 was derived before ((12.3.11)) and takes the form

L2 = ~2( 1sin

sin

+

1

sin2

2

2

). (13.1.1)

We also showed that the z-component Lz of angular momentum is a rather simple differential

281

282 CHAPTER 13. CENTRAL POTENTIALS

operator:

Lz =~

i

. (13.1.2)

Instead of calculating the differential operators for Lx and Ly it is simpler and in fact moreuseful to obtain the operators for L = Lx iLy. A calculation gives

L = ~ei

(i cot

). (13.1.3)

Recall now how things work for coordinate representations. For a single coordinate x wehad that the operator p can be taken out of the matrix element as a differential operator:

x|p| = ~i

xx| = p x| . (13.1.4)

The last expression is written with a small abuse of notation: the p there refers to thedifferential operator, the only reasonable thing that could be acting on the wavefunctionx|. With some care, we can use the same symbol p to denote the momentum operatordefined by p|p = p|p in the momentum representation and to denote the differentialoperator used in the coordinate representation. It is the same abstract operator after all!

For angular momentum we will let | denote position states on the unit sphere andthe spherical harmonic Ym are defined as the wavefunction for the state |m:

Ym(, ) |,m . (13.1.5)The states |,m are defined by the conditions (12.5.1), where J is now the orbital angularmomentum L. Therefore:

L2|,m = ~2 (+ 1) |,m ,Lz |,m = ~m |,m .

(13.1.6)

Letting the bra | act on these two equations from the left we have|L2|,m = ~2 (+ 1) |,m ,|Lz |,m = ~m |,m .

(13.1.7)

In analogy to (13.1.4), the operators inside the bra-kets come out in the representation asdifferential operators on wavefunctions:

L2|,m = ~2 (+ 1) |,m ,Lz|,m = ~m |,m .

(13.1.8)

These are equivalent to

L2 Ym(, ) = ~2 (+ 1)Ym(, ) ,

Lz Ym(, ) = ~mYm(, ) .(13.1.9)

13.2. THE RADIAL EQUATION 283

On the unit sphere the measure of integration is sin dd so we postulate that thecompleteness relation for the | position states reads

0

d sin

2

0

d || = 1 . (13.1.10)

The integral will be written more briefly asd || = 1 (13.1.11)

whered =

0

d sin

2

0

d = 11

d(cos )

2

0

d =

1

1

d(cos )

2

0

d . (13.1.12)

Our orthogonality relation,m|,m = ,lm,m , (13.1.13)

gives, by including a complete set of position statesd ,m||,m = ,lm,m . (13.1.14)

This gives the familiar orthogonality property of the spherical harmonics:d Y m(, )Ym(, ) = ,lm,m . (13.1.15)

Note that the equationLzYm = ~mYm , (13.1.16)

together with (13.1.2) implies that

Ym(, ) = Pm() eim . (13.1.17)

The dependence of the spherical harmonics is very simple indeed!One can show that for spherical harmonics, which are related to orbital angular momen-

tum, one can only have integer . While in general angular momentum can be half-integral,any attempt to define spherical harmonics for half-integral fails. For j = 1/2 we havespin, and as we have seen, this is intrinsic angular momentum. Our basis states | do nothave spatial wavefunctions associated to them.

13.2 The radial equation

It is now time to consider the Schrodinger equation for central potentials. Our treatmentof angular momentum will allow us to simplify this equation until we get a radial equation:a differential equation for the radial part of the wavefunction. This radial equation looks


similar to the Schrodinger equation for a particle moving in a one-dimensional potential.There are some differences, however. The range of motion is r [0,) and the potential isan effective potential that includes a contribution from the angular momentum.

Recall that from (12.3.12) we have

H =p2

2m+ V (r) = ~

2

2m

1

r

2

r2r +

1

2mr2L2 + V (r) (13.2.1)

where here L2 refers to the differential operator realization. The Schrodinger equation willbe solved using the following ansatz for an energy eigenstate of energy E:

Em(x) = fEm(r)Ym(, ) . (13.2.2)

We have the product of a radial function fEm(r) times the spherical harmonic Ym. Plug-ging this into the Schrodinger equation H = E, and recalling that the Yms are eigen-functions of L2 with eigenvalue ~2(+1), the Ym dependence can be cancelled out and weget

~2

2m

1

r

d2

dr2(rfEm) +

~2(+ 1)

2mr2fEm + V (r)fEm = EfEm (13.2.3)

We note that this equation does not depend on the quantum number m (do not confusethis with the mass m!) Therefore the label m is not needed in the radial function and welet fEm fE so that we have the more accurate form of the ansatz:

Em(x) = fE(r)Ym(, ) . (13.2.4)

The differential equation, multiplying by r becomes

~2

2m

d2

dr2(rfE) +

(V (r) +

~2(+ 1)

2mr2

)(rfE) = E (rfE) . (13.2.5)

This suggests writing introducing a modified radial function uE(r) by

fE(r) =uE(r)

r, (13.2.6)

so that we have

Em(x) =uE(r)

rYm(, ) , (13.2.7)

with radial equation

~2

2m

d2uEdr2

+ Veff(r)uE = E uE , (13.2.8)


where the effective potential Veff constructed by adding to the potential V (r) the centrifugalbarrier term proportional to L2:

Veff(r) V (r) +~2(+ 1)

2mr2. (13.2.9)

This is a one-dimensional Schrodinger equation in the variable r, but as opposed to ourusual problems with x (,), the radius r [0,] and we will need some attentionto deal with the properties of the wavefunction at r = 0.

The normalization of our wavefunctions proceeds as follows for the energy eigenstateconsidered above. We require

d3x |Em(x)|2 = 1 . (13.2.10)

This gives r2dr d

|uE(r)|2r2

Y m()Ym() = 1 . (13.2.11)

The angular integral gives one and we are left with 0

dr |uE(r)|2 = 1 , (13.2.12)

a rather natural result for the function uE(r) that plays the role of radial wavefunction.

Behavior of solutions as r 0. We claim thatlimr0

uE(r) = 0 . (13.2.13)

This requirement does not arise from normalization: as you can see in (13.2.12) a finite uEat r = 0 would cause no trouble. Imagine taking a solution uE 0 with = 0 that approachesa constant as r 0:

limr0

uE0(r) = c 6= 0 . (13.2.14)The full solution near the origin would then take the form

(x) crY00 =

c

r, (13.2.15)

since Y00 is simply a constant. The problem with this wavefunction is that it simply doesnot solve the Schrodinger equation! You may remember from electromagnetism that theLaplacian of 1/r is a delta function at the origin so that, as a result,

2(x) = 4c(x) . (13.2.16)Since the Laplacian is part of the Hamiltonian, this delta function must be cancelled bysome other contribution, but there is none, since the potential V (r) does not have delta


functions. In fact, delta function potentials in more than one dimension are very singularand require regulation. This confirms that the radial wavefunction uE0 cannot approach aconstant at the origin. We will see that under general conditions uE(r) must vanish at theorigin.

We can learn about the behavior of the radial solution at the origin under the reasonableassumption that the centrifugal barrier dominates the potential as r 0. In this case themost singular terms of the radial differential equation must cancel each other out, leavingless singular terms that we can ignore in this leading order calculation. So we set:

~2

2m

d2uEdr2

+~2 (+ 1)

2mr2uE = 0 , as r 0 . (13.2.17)

or equivalentlyd2uEdr2

=(+ 1)

r2uE . (13.2.18)

The solutions of this can be taken to be uE = rs with s a constant to be determined. We

then find

s(s 1) = (+ 1) s = + 1, s = , (13.2.19)thus leading to two possible behaviors near r = 0:

uE r+1 , uE 1r

. (13.2.20)

For = 0 the second behavior was shown to be inconsistent with the Schrodinger equationat r = 0 (because of a delta function). For > 0 the second behavior is not consistent withnormalization. Therefore we have established that

uE c r+1 , as r 0 . (13.2.21)

Recall that the full radial dependence of the wavefunction is obtained by dividing by r, sothat

fE c r . (13.2.22)This allows for a constant non-zero wavefunction at the origin only for = 0. Only for = 0 a particle can be at the origin. For 6= 0 the angular momentum barrier preventsthe particle from reaching the origin.

Behavior of solutions as r . Again, we can make some definite statements once we as-sume some properties of the potential. Let us consider the case when the potential V (r)vanishes beyond some radius or at least decays fast enough as the radius grows withoutbound

V (r) = 0 , for r > r0 , or limr

rV (r) = 0 . (13.2.23)


A bit annoyingly, the above assumptions are violated for the 1/r potential of the hydrogenatom, so our conclusions below require some modification in that case. Under the aboveassumptions, as r we can ignore the effective potential completely (including thecentrifugal barrier) and the equation becomes

~2

2m

d2uEdr2

= EuE(r) . (13.2.24)

The equation is the familiard2uEdr2

= 2mE~2

uE . (13.2.25)

The resulting r behavior follows immediately

E < 0 , uE exp(

2m|E|~2

r),

E > 0 , uE exp(ikr) , k =2mE

~2.

(13.2.26)

The first behavior, for E < 0 is typical of bound states. For E > 0 we have a continuousspectrum with degenerate solutions (hence the ). Having understood the behavior ofsolutions near r = 0 and for r this allows for qualitative plots of radial solutions.

The discrete spectrum is organized as follows. We have energy eigenstates for all valuesof . In fact for each value of the potential Veff in the radial equation is different. Sothis equation must be solved for = 0, 1, . . .. For each fixed we have a one-dimensionalproblem, so we have no degeneracies in the bound state spectrum. We have a set of allowedvalues of energies that depend on and are numbered using an integer n = 1, 2 . . .. Foreach allowed energy En we have a single radial solution un.

Fixed , Energies: En , Radial function: un , n = 1, 2, . . . (13.2.27)

Of course each solution un for the radial equation represents 2+1 degenerate solutions tothe Schrodinger equation corresponding to the possible values of Lz in the range (~, ~).Note that n has replaced the label E in the radial solution, and the energies have nowbeen labeled. This is illustrated in the diagram of Figure 13.1, where each solution of theradial equation is shown as a short line atop an label on the horizontal axis. This is thespectral diagram for the central-potential Hamiltonian. Each line of a given represents the(2+1) degenerate states obtained with m = , . . . , . Because the bound state spectrumof a one-dimensional potential is non-degenerate, the radial equation cant have anydegeneracies for any fixed . Thus the lines on the diagram are single lines! Of course,other types of degeneracies of the spectrum can exist: some states having different valuesof may have the same energy. In other words, the states may match across columns onthe figure. Finally, note that since the potential becomes more positive as is increased,the lowest energy state occurs for = 0 and the energy E1, of the lowest state for a given increases as we increase .


Figure 13.1: The generic discrete spectrum of a central-potential Hamiltonian, showing the angularmomentum multiplets and their energies.

13.3 The free particle

It may sound funny at first, but it is interesting to find the radial solutions that correspondto a free particle! A free particle moves in V (r) = 0. These radial solutions give a verydifferent description of the familiar energy eigenstates of a free particle. In cartesian coor-dinates we would write energy eigenstates as momentum eigenstates, for all values of themomentum. To label such solutions we could use three labels: the components of the mo-mentum. Alternatively, we can use the energy and the direction defined by the momentum,which uses two labels. Here the solutions will be labeled by the energy and (,m), the usualtwo integers that describe the angular dependence of the solutions (of course, also affectsthe radial dependence). This analysis of the free particle is useful to study potentials, likethe infinite square well, that are zero in some spherical region and nonzero elsewhere. Itis also useful for scattering problems to understand the behavior of the solutions far awayfrom the scattering center.

The radial equation for zero potential V (r) is

~2

2m

d2uEdr2

+~2

2m

(+ 1)

r2uE = EuE , (13.3.1)

which is, equivalently,

d2uEdr2

+(+ 1)

r2uE = k

2uE , k

2mE

~2. (13.3.2)

In this equation there is no quantization of the energy! Indeed we can redefine the radialcoordinate in a way that the energy parameter k does not appear. Letting = kr theequation becomes

d2uEd2

+(+ 1)

2uE = uE . (13.3.3)

13.4. THE INFINITE SPHERICAL WELL 289

The solution of this differential equation with regular behavior at the origin is uE = cj(),where c is an arbitrary constant and j() is a spherical Bessel function. This means thatwe can take

uE = rj(kr) . (13.3.4)

All in all we have

Free particle: Em(x) = j(kr)Ylm(, ) . (13.3.5)

The spherical Bessel functions have the following behavior

x j(x) x+1

(2+ 1)!!, as x 0 , and x j(x) sin

(x

2

)as x . (13.3.6)

which implies the correct behavior for uE as r 0 and r . Indeed, for r wehave

Free particle : uE sin(kr

2

), as r . (13.3.7)

Whenever the potential is not zero but vanishes beyond some radius, the solutions forr take the form

uE sin(kr

2+ (E)

), as r . (13.3.8)

Here (E) is called the phase shift and by definition vanishes if there is no potential.The form of the solution above is consistent with the anticipated behavior, as this is asuperposition of the two solutions available in (13.2.26) for E > 0. The phase shift containsall the information about a potential V (r) available to any observer probing the potentialfrom far away by scattering particles off of it.

13.4 The infinite spherical well

An infinite spherical well of radius a is a potential that forces the particle to be within thesphere r = a. The potential is zero for r a and it is infinite for r > a.

V (r) =

{0 , if r a , , if r > a . (13.4.1)

For r a the Schrodinger radial equation is the same as the one for the free particle

d2uEd2

+(+ 1)

2uE = uE , = kr , (13.4.2)

where k again encodes the energy E 0. It follows that the solutions are the ones wehad before, with spherical Bessel functions, but this time quantization of the energy arisesbecause the wavefunctions must vanish for r = a.


Let us do the case = 0, where the Bessel function is just a a simple trigonometricfunction. The above equation becomes

d2uE,0d2

= uE,0 uE,0 = A sin +B cos . (13.4.3)

Since the solution must vanish at r = 0 we must choose the sin function:

uE,0(r) = sin kr . (13.4.4)

Since this must vanish for r = a we have that k must take values kn with

kna = n , for n = 1, 2, . . . . (13.4.5)

Those values of kn correspond to energies En,0 where the n indexes the solutions and the 0represents = 0:

En,0 =~2k2n2m

=~2

2ma2(kna)

2 =~2

2ma2n22 . (13.4.6)

Note that ~2

2ma2is the natural energy scale for this problem and therefore it is convenient

to define the unit-free scaled energies En, by dividing En, by the natural energy by

En, 2ma2

~2En, . (13.4.7)

It follows that the energies for = 0 are

En,0 = n22 , un,0 = sin(nr

a

). (13.4.8)

We haveE1,0 9.8696 , E2,0 39.478 , E3,0 88.826 , (13.4.9)

Let us now do = 1. Here the solutions are uE,1 = r j1(kr). This Bessel function is

j1() =sin

2 cos

. (13.4.10)

The zeroes of j1() occur for tan = . Of course, we are not interested in the zero at = 0.You can check that the first three zeroes occur for 4.4934, 7.7252, 10.904. For higher valuesof the vanishing condition is more complicated, but they are easily calculated numerically(or read from old-fashioned tables!).

There is notation in which the nontrivial zeroes (zeroes that are not at zero!) are denotedby zn, where

zn, is the n-th zero of j : j(zn,) = 0 . (13.4.11)

The vanishing condition at r = a quantizes k so that

kn, a = zn, , (13.4.12)

13.4. THE INFINITE SPHERICAL WELL 291

and the energies are quantized as

En, =~2

2ma2(kn,a)

2 En, = z2n, . (13.4.13)

With the following set of zeroes,

z1,1 = 4.4934 , z2,1 = 7.7252 , z3,1 = 10.904 ,

z1,2 = 5.7634 , z2,2 = 9.095 ,

z1,3 = 6.9879 , z2,3 = 10.417 ,

(13.4.14)

we obtain the unit-free energies:

E1,1 = 20.191 , E2,1 = 59.679 , E3,1 = 118.89 ,E1,2 = 33.217 , E2,2 = 82.719 ,E1,3 = 48.83 , E2,3 = 108.51 .

(13.4.15)

The main point to be made is that there are no accidental degeneracies: the energies fordifferent values of never coincide. More explicitly, with 6= we have that En, 6= En,for any choices of n and n. This is illustrated in figure 13.2.

Figure 13.2: The spectrum of the infinite spherical square well. There are no accidental degenera-cies.


13.5 The three-dimensional isotropic oscillator

The potential of the 3D isotropic harmonic oscillator is as follows:

V = 12m2(x2 + y2 + z2) = 1

2m2r2 . (13.5.1)

Isotropy, which means that the frequencies for independent x, y, and z oscillations a are thesame, guarantees that V can be written in terms of r, thus defining a central potential. Aswe will see, the spectrum for this quantum mechanical system has degeneracies, that areexplained by the existence of some hidden symmetry, a symmetry that is not obviousfrom the start. Thus in some ways this quantum 3D oscillator is a lot more symmetric thanthe infinite spherical well.

As you know, for the 3D oscillator we can use creation and annihilation operatorsax, a

y, a

z and ax, ay, az associated with 1D oscillators in the x, y, and z directions. The

Hamiltonian then takes the form:

H = ~(N1 + N2 + N3 +

3

2

)= ~

(N + 3

2

). (13.5.2)

where we defined N N1 + N2 + N3.We now want to explain how tensor products are relevant to the 3D oscillator. We have

discussed tensor products before to describe two particles, each associated with a vectorspace and the combined system associated with the tensor product of vector spaces. Buttensor products are also relevant to single particles, if they have degrees of freedom thatlive in different spaces, or more than one set of attributes, each of which described by statesin some vector space. For example, if a spin 1/2 particle can move, the relevant states livein the tensor product of momentum space and the 2-dimensional complex vector space ofspin. States are obtained by superposition of basic states of the form |p (|+ + |)

For the 3D oscillator, the Hamiltonian is the sum of commuting Hamiltonians of 1Doscillators for the x, y, and z directions. Thus the general states are obtained by tensoringthe state spaces Hx,Hy, and Hz of the three independent oscillators. It is a single particleoscillating, but the description of what it is doing entails saying what is doing in each ofthe independent directions. Thus we write

H3D = Hx Hy Hz . (13.5.3)

Instead of this tensor product reflecting the behavior of three different particles, this tensorproduct allows us to describe the behavior of one particle in three different directions. Thevacuum state |0 of the 3D oscillator can be viewed as

|0 |0x |0y |0z . (13.5.4)

The associated wavefunction is

(x, y, z) = x| y| z| |0 = x|0xy|0yz|0z = 0(x)0(y)0(z) , (13.5.5)

13.5. THE THREE-DIMENSIONAL ISOTROPIC OSCILLATOR 293

where 0 is the ground state wavefunction of the 1D oscillator. This is the expected answer.Recalling the form of (non-normalized) basis states for Hx,Hy, and Hz:

basis states for Hx : (ax)nx |0x , nx = 0, 1, . . .basis states for Hy : (ay)ny |0y , ny = 0, 1, . . .basis states for Hz : (az)nz |0z , nz = 0, 1, . . .

(13.5.6)

We then have that the basis states for the 3D state space are

basis states ofH3D : (ax)nx |0x (ay)ny |0y (az)nz |0z , nx, ny, nz {0, 1, . . .} (13.5.7)

This is what we would expect intuitively, we simply pile arbitrary numbers of ax, ay, and a

z

on the vacuum. It is this multiplicative structure that is the signature of tensor products.Having understood the above, for brevity we write such basis states simply as

(ax)nx(ay)

ny(az)nz |0 . (13.5.8)

Each of the states in (13.5.7) has a wavefunction that is the product of x, y, and z-dependentwavefunctions. Once we form superpositions of such states, the total wavefunction can-not any longer be factorized into x, y, and z-dependent wavefunctions. The x, y, and z-dependences become entangled. These are precisely the analogs of entangled states ofthree particles.

We are ready to begin constructing the individual states of the 3D isotropic harmonic os-cillator system. The key property is that the states must organize themselves into represen-tations of angular momentum. Since angular momentum commutes with the Hamiltonian,angular momentum multiplets represent degenerate states.

We already built the ground state, which is a single state with N eigenvalue N = 0. Allother states have higher energies, so this state must be, by itself a representation of angularmomentum. It can only be the singlet = 0. Thus we have

N = 0 , E = 32~ , |0 = 0 . (13.5.9)

The states with N = 1 have E = 52~ and are

ax|0 , ay|0 , az |0 . (13.5.10)

These three states fit precisely into an = 1 multiplet (a triplet). There is, in fact, no otherpossibility. Any higher multiplet has too many states and we only have 3 degenerate ones.Moreover, we cannot have three singlets, this would mean a degeneracy in the bound statespectrum of the radial Schrodinger equation, which is impossible (as discussed at the endof section 13.2). The = 0 ground state and the = 1 triplet at the first excited level areindicated in Figure 13.4.


Let us proceed now with the states at N = 2 or E = 72~. These are, the following six

states:(ax)

2|0 , (ay)2|0 , (az)2|0 , axay|0 , axaz|0 , ayaz|0 . (13.5.11)To help ourselves in trying to find the angular momentum multiplets recall that that thenumber of states # for a given are

#

0 1

1 3

2 5

3 7

4 9

5 11

6 13

7 15

Since we cannot use the triplet twice, the only way to get six states is having five from = 2and one from = 0. Thus

Six N = 2 states : ( = 2) ( = 0) . (13.5.12)Note that here we use the direct sum (not the tensor product!) the six states define a sixdimensional vector space spanned by five vectors in = 2 and one vector in = 0. Had weused a tensor product we would just have 5 vectors.

Let us continue to figure out the pattern. At N = 3 with E = 92~ we actually have 10

states (count them!) It would seem now that there are two options for multiplets

( = 3) ( = 1) or ( = 4) ( = 0) . (13.5.13)We can see that the second option is at the very least problematic. If true, an = 3 multiplet,which has not appeared yet, would not arise at this level. If it appeared eventually, it woulddo so at a higher energy, and we would have the lowest = 3 multiplet with higher energythan the lowest = 4 multiplet, which is not possible. You may think that perhaps = 3multiplets never appear and the inconsistency is avoided, but this is not true. At any ratewe will give below a rigorous argument that shows the first option is the true one. Therefore,

Ten N = 3 states : ( = 3) ( = 1) . (13.5.14)Let us do the next level! At N = 4 we find 15 states. Instead of writing them out let uscount them without listing them. In fact, we can easily do the general case of arbitraryinteger N 1. The states we are looking for are of the form

(ax)nx(ay)

ny(az)nz |0 , with nx + ny + nz = N . (13.5.15)

We need to count how many different solutions there are to nx + ny + nz = N , withnx, ny, nz 0. This is the number of states #(N) at level N . To visualize this think of

13.5. THE THREE-DIMENSIONAL ISOTROPIC OSCILLATOR 295

nx+ny+nz = N as the equation for a plane in three-dimensional space with axes nx, ny, nz.Since no integer can be negative, we are looking for points with integer coordinates in theregion of the plane that lies on the positive octant, as shown in Figure 13.3. Starting at oneof the three corners, say (nx, ny, nz) = (N, 0, 0) we have one point, then moving towardsthe origin we encounter two points, then three, and so on until we find N +1 points on the(ny, nz) plane. Thus, the number of states #(N) for number N is

#(N) = 1 + 2 + . . .+ (N + 1) =(N + 1)(N + 2)

2. (13.5.16)

Figure 13.3: Counting the number of degenerate states with number N in the 3D simple harmonicoscillator.

Back to the N = 4 level, #(4)=15. We rule out a single = 7 multiplet since stateswith = 4, 5, 6 have not appeared yet. By this logic the highest multiplet for N = 4 mustbe the lowest that has not appeared yet, thus = 4, with 9 states. The remaining six mustappear as = 2 plus = 0. Thus, we have

15 N = 4 states : ( = 4) ( = 2) ( = 0) . (13.5.17)Thus we see that jumps by steps of two, starting from the maximal . This is in factthe rule. It is quickly confirmed for the #(5)=21 states with N = 5 would arise from( = 5) ( = 3) ( = 1). All this is shown in Figure 13.4.

Some of the structure of angular momentum multiplets can be seen more explicitly usingthe aL and aR operators introduced for the 2D harmonic oscillator:

aL =12(ax + iay) , aR =

12(ax iay) . (13.5.18)


Figure 13.4: Spectral diagram for angular momentum multiplets in the 3D isotropic harmonicoscillator.

L and R objects commute with each other and we have [aL, aL] = [aR, a

R] = 1. With

number operators NR = aRaR and NL = a

LaL we then have H = ~(NR + NL + Nz +

3

2)

and, more importantly, the z component Lz of angular momentum takes the simple form

Lz = ~(NR NL) . (13.5.19)

Note that az carries no z-component of angular momentum. States are now build actingwith arbitrary numbers of aL, a

R and a

z operators on the vacuum. The N = 1 states are

then presented as

aR|0 , az|0 , aL|0 . (13.5.20)

We see that the first state has Lz = ~, the second Lz = 0 and the third Lz = ~, exactlythe three expected values of the = 1 multiplet identified before. For number N = 2 thestate with highest Lz is (a

R)

2|0 and it has Lz = 2~. This shows that the highest multipletis = 2. For arbitrary positive integer number N , the state with highest Lz is (a

R)

N |0and it has Lz = ~N . This shows we must have an = N multiplet. This is in fact whatwe got before! We can also understand the reason for the jump of two units from the topstate of the multiplet. Consider the above state with maximal Lz/~ equal to N and then

13.6. HYDROGEN ATOM AND RUNGE-LENZ VECTOR 297

the states with one and two units less of Lz/~:

Lz/~ = N : (aR)

N |0 ,

Lz/~ = N 1 : (aR)N1az|0 ,

Lz/~ = N 2 : (aR)N2(az)2|0 , (aR)N1 aL|0 .

(13.5.21)

While there is only one state with one unit less of Lz/~ there are two states with two unitsless. One linear combination of these two states must belong to the = N multiplet, butthe other linear combination must be the top state of an = N 2 multiplet! This is thereason for the jump of two units.

For arbitrary N we can see why #(N) can be reproduced by multiplets skipping bytwo

N odd : #(N) = 1 + 2 =1

+3 + 4 =3

+5 + 6 =5

+7 + 8 =7

+ . . .+N + (N + 1) =N

,

N even : #(N) = 1=0

+2 + 3 =2

+4 + 5 =4

+6 + 7 =6

+ . . .+N + (N + 1) =N

.(13.5.22)

The accidental degeneracy of the spectrum is explained if we identify an operator thatcommutes with the Hamiltonian (a symmetry) and connects the various multiplets thatappear for a fixed number N . One such operator is

K aRaL . (13.5.23)

You can check that it commutes with the Hamiltonian and, with a bit more work, thatacting on the top state of the = N 2 multiplet it gives the top state of the = Nmultiplet.

13.6 Hydrogen atom and Runge-Lenz vector

The hydrogen atom Hamiltonian is

H =p2

2m e

2

r. (13.6.1)

The natural length scale here is the Bohr radius a0, which is the unique length that canbe built using the constants ~,m, and e2 that appear in this Hamiltonian. We determinea0 by setting p ~/a0 and equating magnitudes of kinetic and potential terms, ignoringnumerical factors:

~2

ma20

=e2

a0 a0 = ~

2

me2 0.529A . (13.6.2)


Note that if the charge of the electron e2 is decreased, the attraction force decreases and,correctly, the Bohr radius increases. The Bohr radius is the length scale of the hydrogenatom. A natural energy scale E0 is

E0 =e2

a0=

e4m

~2=

( e2~c

)2mc2 = 2(mc2) , (13.6.3)

where we see the appearance of the fine-structure constant that, in cgs units, takes theform

e2

~c 1

137. (13.6.4)

We thus see that the energy scale of the hydrogen atom is about 2 1/18770 timesthe rest energy mc2 0.511MeV of the electron. This gives about E0 = 27.2eV. In factE0/2 = 13.6eV is the exact bound state energy of the electron in the ground state ofthe hydrogen atom.

An elegant way to approach the calculation of the ground state energy and ground statewavefunction is to factorize the Hamiltonian. One can show that H can be rewritten as

H = +1

2m

3k=1

(pk + i

xkr

)(pk i xk

r

), (13.6.5)

for suitable constants and that you can calculate. The ground state |0 is then thestate for which (

pk i xkr

)|0 = 0 . (13.6.6)

The spectrum of the hydrogen atom is described in Figure 13.5. The energy levels areE, where we used = 1, 2, . . ., instead of n to label the various solutions for a given .This is because the label n is reserved for what is called the principal quantum number.The degeneracy of the system is such that multiplets with equal n + have the sameenergy, as you can see in the figure. Thus, for example, E2,0 = E1,1, which is to say thatthe first excited solution for = 0 has the same energy as the lowest energy solution for = 1. It is also important to note that for any fixed value of n the allowed values of are

= 0, 1, . . . , n 1 . (13.6.7)

Finally, the energies are given by

E = e2

2a0

1

( + )2, n + . (13.6.8)

The large amount of degeneracy in this spectrum asks for an explanation. The hydrogenHamiltonian has in fact some hidden symmetry. It has to do with a conserved quantumRunge-Lenz vector operator. In the following we discuss the classical Runge-Lenz vector andits conservation. In the problems you will learn about the quantum Runge-Lenz operator.


Figure 13.5: Spectrum of angular momentum multiplets for the hydrogen atom. Here E with = 1, 2, . . ., denotes the energy of the -th solution for any fixed . States with equal values ofn + are degenerate. For any fixed n, the values of run from zero to n 1. Correction: then = 0 in the figure should be n = 1.

In the following chapter all this knowledge will be used to give a fully algebraic derivationof the hydrogen atom spectrum.

Consider the energy functional for a particle moving in a central potential

E =p2

2m+ V (r) . (13.6.9)

The force on the particle is given by

F = V = V (r)rr, (13.6.10)

where primes denote derivatives with respect to the argument. Newtons equation is

dp

dt= V (r)r

r, (13.6.11)

and it is simple to show (do it!) that in this central potential the angular momentum isconserved

dL

dt= 0 . (13.6.12)


We now calculate (all classically) the time derivative of p L:d

dt(p L) = dp

dt L = V

(r)

rr (r p)

= mV(r)

rr (r r)

= mV(r)

r

[r(r r) r r2] .

(13.6.13)

We now note that

r r = 12

d

dt(r r) = 1

2

d

dtr2 = rr . (13.6.14)

Using this

d

dt(p L) = mV

(r)

r

[r rr r r2] = mV (r)r2 [ r

r r r

r2

]= mV (r)r2

d

dt

(rr

).

(13.6.15)

Because of the factor V (r)r2, the right-hand side fails to be a total time derivative. Butif we focus on potentials for which this factor is a constant we will get a conservation law.So, assume

V (r) r2 = , (13.6.16)

for some constant . Then

d

dt(p L) = m d

dt

(rr

) d

dt

(p Lm r

r

)= 0 . (13.6.17)

We got a conservation law: that complicated vector inside the parenthesis is constant intime! Back to (13.6.16) we have

dV

dr=

r2 V (r) =

r+ c0 . (13.6.18)

This is the most general potential for which we get a conservation law. For c0 = 0 and = e2 we have the hydrogen atom potential

V (r) = e2

r, (13.6.19)

so we haved

dt

(p Lme2 r

r

)= 0 . (13.6.20)

Factoring a constant we obtain the unit-free conserved Runge-Lenz vector R:

R 1me2

p L rr,

dR

dt= 0 . (13.6.21)


Figure 13.6: The Runge-Lenz vector vanishes for a circular orbit.

The conservation of the Runge-Lenz vector is a property of inverse squared central forces.The second term in R is simply minus the unit radial vector.

To gain intuition on the Runge-Lenz vector, we first examine its value for a circularorbit, as shown in figure 13.6. The vector L is out of the page and p L points radiallyoutward. The vector R is thus a competition between the outward radial first term and theinner radial second term. If these two terms would not cancel, the result would be a radialvector (outwards or inwards) but in any case, not conserved, as it rotates with the particle.Happily, the two terms cancel. Indeed for a circular orbit

mv2

r=

e2

r2 mv

2r

e2= 1 (mv)(mvr)

me2= 1 pL

me2= 1 , (13.6.22)

which is the statement that in a circular orbit the first term in R is a unit vector. Sinceit points outward it cancels with the second term. The Runge-Lenz vector indeed vanishesfor a circular orbit.

We now argue that for an elliptical orbit the Runge-Lenz vector is not zero. Considerfigure 13.7, where we have a particle going around the ellipse in the counterclockwise di-rection. At the aphelion (point furthest away from the focal center), denoted as point A,the first term in R point outwards and the second term point inwards. Thus, if R does notvanish it must be a vector along the line joining the focus and the aphelion, a horizontalvector on the figure. Now consider point B right above the focal center of the orbit. Atthis point p is no longer perpendicular to the radial vector and therefore pL is no longerradial. As you can see, it points slightly to the left. It follows that R points to the left sideof the figure. R is a vector along the major axis of the ellipse and points in the directionfrom the aphelion to the focus.

Since R vanishes for circular orbits, the length R of R must measure the deviation ofthe orbit from circular. In fact, the magnitude R of the Runge-Lenz vector is precisely the


Figure 13.7: In an elliptic orbit the Runge-Lenz vector is a vector along the major axis of the ellipseand points in the direction from the aphelion to the focus.

eccentricity of the orbit! To see this we form the dot product of R with the radial vector r:

r R = 1me2

r (p L) r . (13.6.23)

Referring to the figure, let be the angle defined with = 0 the direction to the perihelionand increasing in the clockwise direction. The angle between r and R is then and weget

rR cos =1

me2L (r p) r = 1

me2L2 r . (13.6.24)

Collecting terms proportional to r:

r(1 +R cos ) =L2

me2 1

r=

me2

L2(1 +R cos ) , (13.6.25)

This is one of the standard presentations of an elliptical orbit and R appears at the placeone conventionally has the eccentricity e, thus e = R. If R = 0 the orbit is circular becauser does not depend on . You can verify that the identification of R with e is correct usingthe independent definition of eccentricity as the ratio

e =rmax rminrmax + rmin

. (13.6.26)

Here rmin is the value of r for = 0 and rmax is the value of r for = . A very shortcalculation shows that the above right-hand side is actually equal to R, as claimed.


This whole analysis has been classical. Quantum mechanically we need to change somethings a bit. The definition of R only has to be changed to guarantee that R is a hermitianoperator. As you will verify, the hermitization gives

R 12me2

(p L L p) rr. (13.6.27)

The quantum mechanical conservation of R is the statement that it commutes with thehydrogen Hamiltonian

[R ,H ] = 0 . (13.6.28)

You will verify this; it is the analog of our classical calculation that showed that the time-derivative of R is zero. Moreover, the length-squared of the vector is also of interest. Youwill show that

R2 = 1 +2

me4H(L2 + ~2) . (13.6.29)

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Central Potentials