Quantum Mechanics in Multidimensions

In this chapter we discuss bound state solutions of the Schr¨odinger equation in more than one dimension. In this chapter we will discuss some particularly straightforward examples such as the particle in two and three dimensional boxes and the 2-D harmonic oscillator as preparation for discussing the Schr¨odinger hydrogen atom. The novel feature which occurs in multidimensional quantum problems is called “degeneracy” where different wave functions with different PDF’s can have exactly the same energy. Ultimately the source of degeneracy is symmetry in the potential. With every symmetry, there is a conserved quantity which can be used to “label” the states. For rotationally symmetric potentials, the conserved quantities which serve to label the quantum states are angular momenta.

Electron in a two dimensional box

By an electron in a two dimensional box, we mean that the potential is zero within the walls of the box and infinity outside the box. For convenience we will place the origin at one corner of the box as illustrated below:

y

V =

V = 0 V = b V =

x a

V =

To go from the one dimensional to the two dimensional time independent SE

1 we simply take ∂2/∂x2 → ∂2/∂x2 + ∂2/∂y2 and obtain for a particle of mass µ:

¯h2 ∂2 ∂2 − + ψ(x, y)+V (x, y) ψ(x, y)=Eψ(x, y)(1) 2µ ∂x2 ∂y2

For the case under discussion here, the electron is confined within the box, where we have V (x, y)=0.

We use a separation of variable technique to solve this partial differential equation and try product wave functions of the form ψ(x, y)=X(x) × Y (x).

¯h2 ∂2X ∂2Y − Y + X = EXY (2) 2µ ∂x2 ∂y2

We can divide sides of Eq. (2) by X(x)Y (y) to obtain:

1 ¯h2 ∂2X 1 ¯h2 ∂2Y − + − = E (3) X 2µ ∂x2 Y 2µ ∂y2 1 2

Eq. (3) says that the sum of the first term and second term add to a constant (as opposed to a function). This suggests that each of these terms is equal to a constant or any functional dependence between them cancels when summed to give a constant. However the first term can only be a function of x ; while the second term can only be a function of y. There is therefore no way that the terms can have a non-constant part but sum to a constant. We therefore conclude that E both terms are individual constants. Lets call these constants ( )1 = x,and E E E E ()2 = y,and x + y = . We then have two ordinary differential equations:

¯h2 d2X ¯h2 d2Y − = Ex X,− = Ey Y (4) 2µ dx2 2µ dy2

d2X d2Y = −k2 X, = −k2 Y dx2 x dy2 y

2 h2 ¯ 2 2 where E = Ex + Ey = kx + ky (5) 2µ The solutions of Eq. (5) are sinusoidal products of the form:

ψ x, y α k x β k x × γ k y δ k y ( )=( sin ( x )+ cos ( x ))1 ( sin ( y )+ cos ( y ))2 (6)

Let us consider the wave function boundary conditions. The wave function must y ψ x, y δ vanish on the line =0,whichmeans ( =0)=0or()2 =0meaning =0. The wave function must vanish on the line x =0whichmeansψ(x =0,y)=0 β or ( )1 =0meaning = 0. We thus know we need sine functions for both x and y:

ψ(x, y)=α sin (kx x) × sin (ky y)(7)

We next consider the boundary conditions on the top and the right of the well.

Since ψ(x = a, y)=0,thesin(kx x) piece must have an argument which is an integral multiple of π which implies kx a = nxπ where nx ∈ 1, 2, 3, .... Because

ψ(x, y = b)=0wemusthaveky b = nyπ where ny ∈ 1, 2, 3, ....Wethushave: 2 nx πx ny πy ¯h π2 nx 2 ny 2 ψ(x, y)=α sin ( ) × sin ( )andE(nx,ny)= + a b 2µ a b (8)

Degeneracy

The particle in the two dimensional box has an energy which is controlled by two integer quantum numbers as opposed to the one dimensional case E = (¯hπ)2 n2/(2µa2) where a single, integer quantum number (n) controls the energy level. We get one integer quantum number per dimension. In the case of the particle in a box, nx + 1 is the number of wave function nodes along x and ny +1 is the number of nodes along y. Each set of quantum numbers {nx,ny} results in a distinguishable wave function. If once selects a square box with a = b,often two different sets of quantum numbers {nx,ny} , with two distinguishable wave functions, will have the same (or degenerate) energies. It easy to confirm the below energy level chart for the case of a square, two dimensional box.

3 (1,5) (5,1) 26E (3,4) (4,3) 25E (2,4) (4,2) 20E (3,3) 18E (1,4) (4,1) 17E a (2,3) (3,2) 13E a (1,3) (3,1) 10E 2 (2,2) 8E ( h π ) E = 2m a2 (1,2) (2,1) 5E

(1,1) 2E 0

It looks like the levels are either unique (degeneracy = 1) or pair degenerate (degeneracy = 2), however if we went up further we would see examples of triply degenerate and degeneracy = 4 levels. An example of a quadruple degeneracy is 65(¯hπ)2/(2µa2) since it corresponds to {1, 8} , {8, 1} , {4, 7} ,and{7, 4}.Here is a plot of the ψ∗(x, y)ψ(x, y)forthe{1, 1} ground state.

The height of the surface is proportional to the PDF for finding the electron

4 anywhere within the box. The electron tends to populate in the center of the box and has nodes at the wall boundary as expected.

Here are plots of ψ∗(x, y)ψ(x, y) for the degenerate excited states {2, 1} and {1, 2}. We note the presence of an extra node in either the x and y directions. It is clear that the PDF for these two cases are related by a 90o rotation about the z axis and hence it is not surprising that they are degenerate with an energy of 5(¯hπ)2/(2µa2).

Electron in a two dimensional harmonic oscillator

Another fairly simple case to consider is the two dimensional (isotropic) har- monic oscillator with a potential of V (x, y)= 1 µω2 x2 + y2 where µ is the 2 electron mass , and ω = k/µ. The Schr¨odinger equation reads: ¯h2 ∂2ψ ∂2ψ 1 − + + µw2 x2 + y2 ψ(x, y)=Eψ(x, y)(9) 2µ ∂x2 ∂y2 2

Following our treatment of the two dimensional box, we insert a product wave function of the form ψ(x, y)=X(x) Y (y): ¯h2 ∂2X ∂2Y 1 − Y + X + µw2 x2 + y2 XY = EXY (10) 2µ ∂x2 ∂y2 2

5 We can then divide both sides by X(x) Y (y) and group into x dependent and y dependent terms: 1 ¯h2 ∂2X 1 1 ¯h2 ∂2Y 1 − + µw2 x2 + − + µw2 y2 = E (11) X 2µ ∂x2 2 Y 2µ ∂y2 2 1 2

Again Eq. (11) says that the sum of the first term and second term add to a constant, suggesting that each of these terms is equal to a constant. There is no way that the terms can have a non-constant part but sum to a constant since the first term can only depend on x while the second can only depend on y. Calling E E E E E these constants ( )1 = x,and()2 = y,wehave x + y = and the two differential equations:

h2 ∂2X h2 ∂2Y ¯ 1 2 2 ¯ 1 2 2 − + µw x X(x)=Ex X(x) , − + µw y Y (Y )=Ey Y (x) 2µ ∂x2 2 2µ ∂y2 2 (12) The Eq. (12) equations are exactly the same as the Schr¨odinger¨rodinger Equa- tion for the harmonic oscillator which we obtained and solved in the chapter on Bound States in One Dimension. The energies are: 1 1 Ex = nx + ¯hω , Ey = ny + ¯hω where nx,ny ∈ 0, 1, 2,.. (13) 2 2

The solutions are: µω 2 Xn (x)=Nn Hn (ξ)exp(−ξ /2) where ξ = x (14) x x x ¯h

µω 2 Yn (y)=Nn Hn (η)exp(−η /2) where η = y (15) y y y ¯h

Putting this all together we have quantized energies of the form:

E =(nx + ny +1)¯hω (16)

6 and wave functions of the form:

2 2 ψ(ξ,η)=N Hnx (ξ)exp(−ξ /2) × Hny (η)exp(−η /2)

µω where (ξ,η)= (x, y) (17) ¯h

We could write the energy spectrum as E =(n +1)hω ¯ where n ∈ 0, 1, 2.... As shown below the degeneracy of the nth level is just equal to n +1.

(0,4) (1,3) (2,2) (3,1) (4,0) 5 h ω (0,3) (1,2) (2,1) (3,0) 4 h ω (0,2) (1,1) (2,0) 3 h ω (1,0) (0,1) 2 h ω (0,0) h ω

0

As you can see, the spectrum for the isotropic harmonic oscillator is more degenerate than for the two dimensional square box. We can construct the wave functions using the explicit forms for the Hermite polynomials: Table 1: Hermite polynomials

3 Ho =1 H3 =8ξ − 12 ξ 4 2 H1 =2ξ H4 =16ξ − 48 ξ +12 2 5 3 H2 =4ξ − 2 H5 =32ξ − 160 ξ + 120 ξ

The wave functions for these solutions is in Table 1.

This table explicitly shows the energy and wave functions for the ground, first, and second excited state in both cartesian and polar coordinates. One gets from cartesian to polar coordinates via x = ρ cos φ and y = ρ sin φ.

7 Table 2: unnormalized 2-d harmonic oscillator wave functions

(nx,ny) E ψ ψ (0,0) ¯hω exp (−γρ2/2) exp (−γρ2/2) (1,0) 2¯hω x exp (−γρ2/2) ρ cos φ exp (−γρ2/2) (0,1) 2¯hω y exp (−γρ2/2) ρ sin φ exp (−γρ2/2) (1,1) 3¯hω xy exp (−γρ2/2) ρ2 cos φ sin φ exp (−γρ2/2) (2,0) 3¯hω 2 γx2 − 1 exp (−γρ2/2) 2γρ2 cos2 φ − 1 exp (−γρ2/2) (0,2) 3¯hω 2 γy2 − 1 exp (−γρ2/2) 2γρ2 sin2 φ − 1 exp (−γρ2/2)

µω where γ = ¯h

The below figure illustrates the two degenerate states ψ(1, 0) and ψ(0, 1).

Cylindrical Symmetry

We note that the two dimensional, isotropic harmonic oscillator has has cylin- drical symmetry both in the potential and boundary condition. By a cylin- drically symmetric potential, we mean that in polar coordinates (ρ, φ)where

8 ρ = x2 + y2 and φ =tan−1 (y/x), the potential has no φ dependence but rather is a function of ρ only:

1 1 V = µω2 (x2 + y2)= µω2 ρ2 (20) 2 2

Given that the potential is cylindrically symmetric, it is surprising that the ψ(1, 0) and ψ(0, 1) PDFs depicted above break this symmetry by orientating their two probability antinodes along the x or y axis. Because the ψ(1, 0) and ψ(0, 1) are degenerate, any linear combination of them will also be a stationary state

with energy 2¯hω.Theψ± = ψ(1, 0) ± iψ(0, 1) combinations have cylindrically symmetric PDFs.

−γ(x2+y2)/2 −γ(x2+y2)/2 ψ+ ∝N (x + iy) e ,ψ− ∝ (x − iy) e (21)

Some very simple computation shows that we can write these states in polar coordinates as:

x ± iy = ρ(cos φ ± i sin φ)=e±iφ

Thus

−γρ2/2 ±iφ ψ±(ρ, φ)=N ρe e (22)

where it is clear that ψ∗ψ is a function of ρ only since the azimuthal (φ) depen- dence only affects the phase of of the wavefunction. The PDF for either the ψ± combination is illustrated below:

9 We note that the original ψ(1, 0) and ψ(0, 1) were real, since it is always possible with a real potential to have a real stationary state. These real wave

functions carry no current density. The combinations ψ± , on the other hand, are complex stationary states, with position dependent phases. As such the wave functions carry current densities which (as you can show in the exercises) circulate along the φˆ direction. This situation seems to be a bit contradictory. Since ψ± are stationary states they have a static PDF, and yet there is a probability density current flow.

The way around this apparent contradiction is that the electron described by ψ± is moving in a direction which leaves its PDF constant. Since ψ± wave function has an azimuthally symmetric PDF, the simplest motion would be that the electron is rotating with an angular velocity of Ω about thez ˆ axis. Classically a charged disk spinning about its symmetry axis, has a constant charge density, but still carries an electrical current density. This current density would be in the φˆ direction which is the same direction which applies to our wave function.

Angular momentum

Given that ψ± wave function involves a circulating electron current which

10 classically would correspond to a rotating charge density, it is not surprising that the ψ± wave function describes an electron with angular momentum about the symmetry (ˆz) axis. Classically the angular momentum L about a point O is related to its momentum (p ) and displacement ( r from the point O according to L = r × p . This means (after converting to momentum operators): h ∂ ∂ L xp − yp ¯ x − y z = y x = i ∂y ∂x (23)

One can convert this operator form to polar coordinates by making use of of the chain rule to transform the Cartesian derivatives to polar derivatives.

∂ ∂ρ ∂ ∂φ ∂ ∂ ∂ρ ∂ ∂φ ∂ , ∂x = ∂x ∂ρ + ∂x ∂φ ∂y = ∂y ∂ρ + ∂y ∂φ

ρ = x2 + y2 ,φ=tan−1 (y/x)

∂ρ x ∂φ − 2 φ , sin ∂x = ρ ∂x = y

∂ρ y ∂φ 2 φ , cos ∂y = ρ ∂y = x (24)

In the exercises you will find it straightforward to assemble the pieces in Eq. (24) to obtain the fairly elegant result: h ∂ ∂ h ∂ Lˇ ¯ x − y ¯ z = i ∂y ∂x = i ∂φ (25)

As the below calculation shows, the ψ± wave functions are eigenstates of Lˇz

with eigenvalues of Lz = ±¯h: To recapitulate, the original ψ(0, 1) and ψ(1, 0) were purely real wave functions with a zero current density, which broke the cylindrical symmetry of the potential. Since these two wave functions were degenerate we

could combine them in the form ψ± = ψ(1, 0) ± iψ(0, 1) to form two alternative

11 wave functions whose PDF exhibits cylindrical symmetry. Because these wave functions have position dependent complex phases they have either clockwise and

counter-clockwise current flow. We saw that the ψ± wavefunctions are eigenstates of Lˇz and thus have unique values of Lz = ±¯h. Each of the original ψ(0, 1) and

ψ(1, 0) wave functions can in turn be written as linear combinations of ψ± with equal probability of being in either ψ± state. This means that half the time they will be measured to have Lz =¯h and half the time they will be measured to have

Lz = −¯h.

The Hamiltonian in Polar Coordinates

One can solve for the wave functions in cylinderically symmetric systems using the direct separation of variable approach in polar coordinates in essentially the same way as we used separation of variables in cartesion coordinates (Eq. (1) - Eq. (8)). Here we will use a slightly different approach taking advantage of some insights based on classical orbit theory. We begin by writing Schr¨odinger’s equation in polar coordinates. Using the same sort of tedious procedure described by Eq. (24) one can show one can write ∇2 in polar coordinates which allows us 2 Tˇ − ¯h ∇2 to write the kinetic energy operator = 2µ as: ¯h2 ¯h2 ∂2 ∂2 ¯h2 1 ∂ ∂ 1 ∂2 Tˇ = − ∇2 = − + = − ρ + (26) 2µ 2µ ∂x2 ∂y2 2µ ρ ∂ρ ∂ρ ρ2 ∂φ2

The underlined portion can be written in terms of the angular momentum oper- ator:   2 2 ¯h 1 ∂ ∂ L T = − ρ +   2µ ρ ∂ρ ∂ρ 2µρ2

The entire Hamiltonian is then:      2 2 ¯h 1 ∂ ∂ L H = − ρ +   + V (ρ) (27) 2µ ρ ∂ρ ∂ρ 2µρ2 radial centifugal eff pot We have written the Hamiltonian in a form inspired by classical mechanics: For the a classical orbit of a satellite about the earth the classical energy expression

12 is given by: 1 E = µ r˙2 +(rφ˙)2 + V (r) 2

Herer ˙ is the radial velocity and rφ˙ is the tangential velocity. The angular 2 momentum, given by L = µvφ r = µr φ˙, is a constant of motion. This allows us to express φ˙ in terms of L, insert it in Eq. (28) to obtain an expression in r only:   L 1 L2 φ˙ = ,E= µr˙2 + + V (r) (28) µr2 2 2 µr2 centrifugal eff pot

The quantum effective potential expression Eq. (27) is identical to the classi- cal effective potential expression Eq. (28). The centrifugal barrier piece of the effective potential plays the pivotal role in preventing a satellite from crashing into the earth as first realized by Isaac Newton. We illustrate this role for a V = −K/r coulomb potential well (as a classical model of the atom):

V eff

2 -K L V = + 2 eff r 2µ r

dominated by centrifugal potential

r E Perigee Apogee dominated by Coulomb potential

A general principle of both quantum mechanics and ordinary mechanics is sometimes called Noether’s Theorem which says that every symmetry can be associated with a conserved quantity. By symmetry, we generally mean a sym- metry of the classical or QM Hamiltonian. In particular the spacial symmetries of the Hamiltonian such as translation or rotation symmetry are associated with

13 with conservation of linear or angular momentum. If , for example, a potential has translational invariance, the potential is independent of position and has no gradients and therefore no force acts. A particle with a given momentum P will continue traveling with this momentum in a straight line according to Newton.

Similarly a central force law about an origin, will lead to a cylindrically sym- metric or spherically symmetric potential which will be the same independent of rotation. A particle under the interaction of a central force, experiences no torques and therefore has a constant angular momentum. If the Hamiltonian has rotation symmetry, the angular momentum operator will commute with the Hamiltonian ([H, Lˇ] = 0), and it is possible to find wave functions which are si- multaneous eigenfunctions of commuting operators. If [H, Lˇ] = 0 the stationary states can always be arranged as eigenstates of the angular momentum operator which in turn implies that stationary states can be arranged to have a unique (non-varying) value of angular momentum. Hence the classical statement that angular momentum is conserved for the case of particle acted upon by a cen- tral force has an exact quantum mechanical analogy : the stationary states can be chosen to have unique angular momentum. Since this angular momentum is “constant” we can assign each stationary state a conserved angular momen- tum quantum number which helps specify or “label” the state. Inspired by the separation of variable solutions we used for the particle in the two-dimensional box or two-dimensional harmonic oscillator, we assume that for a cylindrically symmetric potential, we are looking for cylindrically symmetric wave functions are of the form ψ = R(ρ)Φ(φ).

An eigenfunction of Lˇz obeys the eigenvalue equation:

h ∂ Lˇ ψ L ψ ¯ ψ L ψ z = z or i ∂φ = z

h ∂ L φ, ¯ L Since z only operates on i ∂φΦ= z Φ (29)

14 where Lz is just a constant eigenvalue. Re-arranging this we have: ∂Φ i = Lz ∂Φ (30) Φ ¯h Integrating both sides we have:

iφLz ln Φ = → Φ=eiLzφ/¯h (31) ¯h Eq. (31) has an important boundary condition, if one goes from φ to φ +2π one is back to the same space point hence: Φ(φ +2π)=Φ(φ) Hence Lzφ Lz(φ +2π) Lzφ Lz exp i =exp i =exp i × exp 2iπ (32) ¯h ¯h ¯h ¯h

The only way that Eq. (32) can be satisfied is if Lz/¯h is an integer which we will call m. Hence we in general we write

imφ Φ=e where Lz = m ¯h, m=0, ±1, ±2, ±3,... (33)

Hence in general , the eigenstates of a cylindrically symmetric Hamiltonian can be written as:

ψ = R(ρ)exp(imφ) (34)

Lets reconsider previous example :

−γ x2+y2 /2 −γρ2/2 ±iφ ψ± = N (x ± iy) e ( ) = N ρe e

This is of the expected form given by Eq. (34) with m = ±1andR(ρ)= 2 N ρe−γρ /2. As you will shown in the exercise these wave functions satisfy the Hamiltonian of Eq. (27). 2 γρ2 ¯h 1 ∂ ∂ m2 1 γρ2 H ρ exp − = − ρ + + µω2ρ2 ρ exp − 2 2µ ρ ∂ρ ∂ρ 2µρ2 2 2 γρ2 =2¯hω ρ exp − 2 (35) as long as |m| =1andγ = µω/¯h.

15 Three Dimensions

Particle in a 3-d box

The reasoning for a particle in a two dimensional box can be easily extended to three dimensions.

n πx n πy n πz ψ x, y N x × y × z ( )= sin ( a ) sin ( b ) sin ( c ) (36)

h h2π2 n 2 n 2 n 2 ¯ 2 2 2 ¯ x y z E(nx,ny,nz)= kx + ky + kz = + + (37) 2µ 2µ a b c where nx,y,z ∈ 1, 2, 3, .... The above wave functions are valid if the lower left corner of the box is at the origin. Three quantum numbers are required in 3-d and the system has a great deal of degeneracy for the case of a cube (a = b = c).

There will be even more degeneracy for the case of a spherically symmetric potential. The 3 D harmonic oscillator will have a potential of the form: V = (1/2) µω2(x2 + y2 + z2). The solution described by Eq. (9) - Eq. (17) can be totally recycled with the addition of the Z(z) terms. The 3 D energy is described by: 3 E = nx + ny + nz + ¯hω (38) 2 where nx,ny,nz ∈ 0, 1, 2.... In scaled variables the solution in Cartesian coordi- nates is:

2 2 2 ψ(ξ,η)=N Hnx (ξ)exp(−ξ /2) × Hny (η)exp(−η /2) × Hnz (ζ)exp(−ζ /2)

µω where (ξ,η,ζ)= (x, y, z) (39) ¯h

This table explicitly shows the energy and wave functions for the ground and first excited state in both Cartesian and spherical coordinates in unscaled coordinates.

16 Table 3: 3-d harmonic oscillator solutions

(nx,ny,nz) E ψ ψ (0,0,0) 3¯hω/2 exp (−γr2/2) exp (−γr2/2) (1,0,0) 5¯hω/2 x exp (−γr2/2) r sin θ cos φ exp (−γr2/2) (0,1,0) 5¯hω/2 y exp (−γr2/2) r sin θ sin φ exp (−γr2/2) (0,0,1) 5¯hω/2 z exp (−γr2/2) r cos θ exp (−γr2/2) (1,1,0) 7¯hω/2 xy exp (−γr2/2) r2 cos φ sin φ sin2 θ exp (−γr2/2) (1,0,1) 7¯hω/2 xz exp (−γr2/2) r2 cos φ sin θ cos θ exp (−γr2/2) (0,1,1) 7¯hω/2 yz exp (−γr2/2) r2 sin φ sin θ cos θ exp (−γr2/2) (2,0,0) 7¯hω/2 (0,2,0) 7¯hω/2 (0,2,2) 7¯hω/2

µω where γ = ¯h

This table is constructed using the following relationships between spherical and Cartesian coordinates.

z Spherical coordinates r θ φ

r = x222 + y + z

x = r sin θ cos φ r y = r sin θ sin φ θ θ r cos z = r cos θ

y φ

r sin θ

x

17 As we shall shortly, these cartesian wave functions can be rearranged to satisfy the Hamiltonian for a spherically symmetric potential. Spherically Symmetric Hamiltonian

Except for the form of the radial kinetic energy, the Eq. (40) Hamiltonian is very similar to the Eq. (27) Hamiltonian: −¯h2 1 ∂ ∂ Lˇ2 Hˇ = r2 + + V (r) (40) µ r2 ∂r ∂r µr2 2 radial 2 eff pot

Another important difference is Lˇ2 involves all three component of L in Eq. (40) 2 2 but involves Lˇz alone in Eq. (27). In particular Lˇ operates on both θ and φ , where as Lˇz operates on φ alone.

The Lˇ2 isbasedonallthreecomponents:

2 2 2 2 Lˇ ≡ Lˇx + Ly + Lz where : h ∂ ∂ h ∂ ∂ Lˇ ¯ y − z , Lˇ ¯ z − x x = i ∂z ∂y y = i ∂x ∂z h ∂ ∂ Lˇ ¯ x − y z = i ∂y ∂x (41)

These three Cartesian components are based on the classical expression L = r×p . By using the chain rule technique one can show: 1 ∂2 1 ∂ ∂ Lˇ2 = −¯h2 + sin θ (42) sin2 θ ∂φ2 sin θ ∂θ ∂θ

Hence Lˇ2 operates on both φ and θ. The eigenfunctions that satisfy Eq. (40) canbewrittenas:

m ψ(r, θφ)=N R(r) × Y (θ, φ) (43)

m The Y (θ, φ) functions are called spherical harmonics and they are simulta- neous eigenfunctions of Lˇ2 and Lz. If two operators have simultaneous eigenfunc- tions the two operators must commute. It is easy to verify that [Lˇz, Lˇ2]=0by

18 direct computation. In fact [Lˇx, Lˇ2]=[Lˇy, Lˇ2] = 0 as well. On the other hand, none of the three components of L mutually commute–that is :

[Lˇx, Lˇy]=i¯hLˇz , [Lˇy, Lˇz]=i¯hLˇx , [Lˇz, Lˇx]=i¯hLˇy (44)

By the reasoning of the chapter on Quantum Measurements this means that it is possible to find simultaneous eigenfunctions of either Lˇ2 and Lˇz ,orLˇ2 and

Lˇy ,orLˇ2 and Lˇx. But it is not possible to find simultaneous eigenfunctions of the three operators Lˇ2 and Lˇz and Lˇy.

Before using the spherical harmonics we should summarize their properties. Because of lack of time, we will not be able to prove these important proper- ties but unfortunately will just have time to inventory them. There is a rather beautiful and elegant way of proving these properties using angular momentum ladders. These proofs appear in my Physics 386 Notes. Properties of Spherical Harmonics

The eigenvalues for the spherical harmonics are defined by the  and m quan- tum numbers:

2 m 2 m m m Lˇ Y (θ, φ)=¯h ( +1)Y (θ, φ) , Lˇz Y (θ, φ)=¯hmY (θ, φ) (45)

Both  and m are integer quantum numbers with  ∈ 0, 1, 2 ... and m ∈ 0, ±1, ±2,... but − ≤ m ≤ . The condition that − ≤ m ≤  enforces the condition the square of the z component of L should be smaller than the squared 2 2 2 2 of the total length of L or Lz < Lx +Ly +Lz. Our quantum number condition insures this since m2 <( +1).

From our experience with Lˇz we already know that the azimuthal part of the wavefunction is of the form exp (imφ) which means that the spherical wave function must be of the form:

m m imφ Y (θ, φ)=P (θ) e (46)

m The P (θ) functions are called associated Legendre polynomials.

19 Here is a table of some of the low lying spherical harmonics: Table 4 : Spherical Harmonics

Y 0 Y 0 Y ±1 0 1 1 1 3 θ ∓ 3 e±iφ θ 4π 4π cos 8π sin Y 0 Y ±1 Y ±2 2 2 2 5 2 θ − ∓ 15 e±iφ θ θ 15 e±2iφ 2 θ 16π 3cos 1 8π sin cos 32π sin

2 m 2 m Lˇ Y = ( +1)¯h Y m m Lˇz Y = m¯hY

Quantum Numbers

We are now in a position to flesh out Eq. (43). Our wave functions are of m the form ψ(r, θφ)=N R(r) × Y (θ, φ). It we insert this form into Eq. (40), we m can essentially factor out the Y (θ, φ) piece leaving a differential equation in R m alone. The Y (θ, φ) factor only affects the centrifugal barrier term:

−h2 ∂ ∂ Lˇ2 m ¯ 1 2 m HRˇ (r) Y = r + + V (r) R(r) Y 2µ r2 ∂r ∂r 2µr2

2 −¯h2 1 ∂ ∂ ¯hˇ  ( +1) = r2 + + V (r) R(r) Y m = ER(r) Y m 2µ r2 ∂r ∂r 2µr2  

2 −¯h2 1 ∂ ∂ ¯hˇ  ( +1) Hence r2 + + V (r) R(r)=ER(r) (47) 2µ r2 ∂r ∂r 2µr2

We thus see that both the E of the state and the form of R will depend in general on the  quantum number. In three dimensional problems there are three quantum numbers. The radial wave function and energy will thus depend on a third quantum number n as well. As illustrated in Eq. (47), because of the

20 symmetry of the potential, the radial wave function cannot depend on m which describes the orientation of L .

TheentirewavefunctionwillthereforeinvolvearadialwavefunctionR(r) multiplying the spherical harmonic.

m Ψ=Rn(r) Y (θ, φ) (48)

In general the energy of a spherically symmetric state will depend on just n and  quantum numbers as does the radial wave function. Wave functions with different m quantum numbers will necessarily be degenerate. Normalization

√ / π Y 0 The constant factors such as 1 4 for 0 are connected with the normal- ization condition. For the case of spherical harmonics, the usual one dimensional normalization condition:

+∞ ∗ 1= dx ψ (x)ψ(x) −∞

is replaced by an integral over the solid angle of a sphere. Recall that the solid angle is the area “swept out” by moving through dθ and dφ on a unit sphere. It is most easily visualized by considering the area of an infinitesimal rectangle which is swept out first by dθ followed by sweeping out dφ as illustrated below:

21 z

y

sin θ dθ sin θ r=1 dθ sin θ dφ dφ r=1 x x sin θ z

dφ dθ d Ω d Ω = sin θ dφ dθ dθ r=1 sin θ dφ

2π π m 2 dφ (sin θdθ)|Y (θ, φ)| = 1 (49) 0 0 where we use the fact that an element of spherical area (dΩ) is given by dΩ= θdφdθ Y 0 sin as illustrated above. Lets check this for 0 which is just a constant. The area of a sphere is Ω = 4π and hence to ensure the normalization given Eq. √ |Y 0 θ, φ |2 π Y 0 θ, φ / π (49) we need 0 ( ) =4 or 0 ( )=1 4 . As discussed in the chapter on Quantum Measurements, two eigenfunc- tions with two different eigenvalues are orthogonal. This very useful result means that the spherical harmonics obey:

2π π m
∗ m dφ (sin θdθ)(Y
) Y = δ
 δm
m (50) 0 0

The symbol δij , called the Kronecker Delta, is defined below: 1ifi = j δij = (51) 0ifi = j

Hence the orthogonality condition described by Eq. (50) means that area integral

22 m
∗ m   over (Y
) Y vanishes unless both  =  and m = m. Building Spherically Symmetric Harmonic Oscillator Wave Functions

Exploiting our experience with the two dimensional harmonic oscillator we can obtain φ symmetric combinations of the three degenerate first excited state wave functions as follows: ±iφ −γr2/2 ψ± =(1, 0, 0) ± i(0, 1, 0) = N sin θe re

−γr2/2 ψ3 =(0, 0, 1) = N (cos θ) re (52)

Comparing the wave functions in Eq. (52) to our spherical harmonic table (Table 4) we see:

−γr2 −γr2 ±1 0 ψ± = N r exp( ) Y (θ, φ) ,ψ= N r exp( ) Y (θ, φ) (53) 2 1 0 2 1 Hence these three wave functions all have  =1andhavem values of +1, −1, 0 and are degenerate with an energy of 5¯h/2 Although there isn’t any need to do so, one could check that these wave functions are eigenfunctions Lˇz and Lˇ2 by direct computation. For example 2 2 2 h ∂ 2 Lˇ ρe−γρ /2 Y 0 Lˇ ρe−γρ /2 e±iφ ρe−γρ /2 ¯ e±iφ ±h ρe−γρ /2 e±iφ z 1 = z = i ∂φ = ¯ (54) and 1 ∂2 1 ∂ ∂ Lˇ2 Y ± = Lˇ2 sin θe±iφ = −¯h2 + sin θ sin θe±iθ 1 sin2 θ ∂φ2 sin θ ∂θ ∂θ e±iθ e±iφ ∂ = −¯h2 − + (sin θ cos θ) sin θ sin θ ∂θ e±iθ e±iφ = −¯h2 − + cos2 θ − sin2 θ =2¯h2 sin θe±iφ (55) sin θ sin θ The below vector picture gives a graphical summary of the angular momentum state of ψ± and ψ0.

23 z

1 h

L 2 h

0 2 h

2 h

-1 h

The angular momentum lies on a cone with Lz = ±¯h or a disk with Lz =0. √ √ 2 The magnitude of L2 = 2¯h = 2¯h.TheLx and Ly values are uncertain since ψ± are not eigenstates of these operators. Extracting the angular momentum contents of wave functions

As described in Quantum Measurements, the orthonormality condition (Eq. (50) ) allows one to expand any arbitrary angular wave function f(θ, φ)as a sum of spherical harmonics.

m= m f(θ, φ)= am Y (θ, φ)  m=−

2π π m ∗ where am = dφ (sin θdθ)(Y ) f(θ, φ) (56) 0 0

The number am is the amplitude for finding the electron with an angular wave function f(θ, φ) in an angular momentum state (, m)thatiswithL2 = (+1)h ¯2

and Lz = m¯h.

24 When evaluating solid angle integrals, I find the following trick to be most useful. d θ cos − θ → θdθ −d θ dθ = sin sin = cos (57)

This means that one can replace the sin θdθpart of the solid angle integral by an integral over −d cos θ. We also have to be careful to convert the limits of integration.

π cos π 1 sin θdθ→− d cos θ = d cos θ (58) 0 cos 0 −1

We thus obtain the very useful formula

π 2π 1 2π dθ sin θ dφ ≡ d cos θ dφ (59) 0 0 −1 0

Example

Assume that an angular wave function for an electron in a spherically sym- metric potential is of the form

x2 ψ ∝ = N sin2 θ cos2 φ r2

Find the probability the electron has zero angular momentum  =0.

25 Lets begin by normalizing the wave function.

2π +1 N −2 = dφ d cos θ sin4 θ cos4 φ 0 −1

2π +1 2 3π 16 5 = dφ cos4 φ × d cos θ 1 − cos2 θ = × →N = 4 15 4π 0 −1

The φ integral can be done analytically through two applications of the well known substitution: 1+cos2φ cos2 φ = 2

which leave vanishing integrals over full cycles of trig functions. The integral over cos θ is just an integral over a polynomial in cos θ

a 0 We next evaluate 0 using Eq. (59):

2π +1 ∗ 1 5 a 0 = dφ d cos θ √ sin2 θ cos2 φ 0 4π 4π 0 −1

2π +1 1 5 =(√ )∗ dφ cos2 φ d cos θ 1 − cos2 θ 4π 4π 0 −1

√ 2π +1 √ 5 5 4 5 = dφ cos2 φ × d cos θ 1 − cos2 θ = × π × = 4π 4π 3 9 0 −1 where we used the facts that

2π 2π dφ cos2 φ = = π 2 0

26 +1 1 +1 2 4 d cos θ 1 − cos2 θ = cos θ − cos3 θ =2− = 3 −1 3 3 −1

Hence the probability of finding this electron with zero angular momentum is |a 0|2 / 0 =5 9. Important Points

1. We applied the separation of variables technique to the time independent Schr¨odinger Equation in Cartesian coordinates for both the particle in an infinite box and the isotropic harmonic oscillator for the case of two and three dimensions. The three dimensional expressions are:

Particle in a box

n πx n πy n πz ψ x, y N x × y × z ( )= sin ( a ) sin ( b ) sin ( c )

h h2π2 n 2 n 2 n 2 ¯ 2 2 2 ¯ x y z E(nx,ny,nz)= kx + ky + kz = + + 2µ 2µ a b c where nx,y,z ∈ 1, 2, 3, .... The above wave functions are valid if the lower left corner of the box is at the origin.

Isotropic Harmonic Oscillator

2 2 2 ψ(ξ,η)=N Hnx (ξ)exp(−ξ /2) × Hny (η)exp(−η /2) × Hnz (ζ)exp(−ζ /2)

µω where (ξ,η,ζ)= (x, y, z)andnx,y,z ∈ 0, 1, 2,.. ¯h

To go from three to two dimensions one simply drops the nz terms.

27 2. It is often possible to get degeneracies in the above expressions where several

sets of {nx,ny,nz} quantum numbers give exactly the same energy. This degeneracy is connected with the symmetry of the system. The maximum degeneracy in the infinite box occurs for a cube.

3. When the two dimensional harmonic oscillator is specified by Cartesian

quantum numbers {nx,ny}, the PDF is generally not azimuthally sym- metric although such wave functions are real. One can build azimuthally

symmetric, complex wave functions of the form Rn,m(rho)exp(imφ)which involve position dependent phases. These position dependent phases imply circulating probability currents which give the electron a non-zero angular

momentum (Lz).

4. It is possible to choose stationary states which are simultaneous eigenstates of angular momentum since the angular momentum operator commutes with the Hamiltonian in the case of a cylindrically symmetric or spheri- cally symmetric potential. In two dimensions such eigenstates of angular momentum have a φ dependence of the form ψ ∝ exp (imφ)wherem is an integer.

5. In three dimensions, the stationary states spherically symmetric potentials

can simultaneous eigenfunctions of both Lˇ2 and Lˇz. Because [Lx,Lz] =0

and [Ly,Lz] = 0 they cannot also be eigenfunctions of Lˇx or Lˇy.Theoper-

ators Lˇ2 and Lˇz only operate on angular part of the wave function. Their m 2 eigenfunctions are called spherical harmonics, Y (θ, φ)withLˇ eigenval- 2 ues ( +1)¯h and Lˇz eigenvalues of m¯h.The and m quantum numbers are integers,  ∈ 0, 1, 2,.., m ∈ 0, ±1, ±2... but − ≤ m ≤ +.

6. The spherical harmonics involve powers of sin θ or cos θ which increase with increasing  times a φ dependence of the form exp (imφ). In order to insure a single valued phase at the poles, the spherical harmonics include at least one factor of sin θ whenever m = 0 which causes them to vanish at the poles.

28 7. The spherical harmonics are orthonormal in the sense:

2π +1 m
∗ m dφ d cos θ (Y
) Y = δ
 δm
m 0 −1

which allows one to expand any wavefunction in terms of spherical harmon- ics via: m= m f(θ, φ)= am Y (θ, φ)  m=−

2π +1 m ∗ where am = dφ d cos θ (Y ) f(θ, φ) 0 −1

The amplitudes am determine the probability of finding an electron in an (, m) angular momentum state.

29