So far, we have studied simple models in which a particle is subjected to a force in one dimension (particle in a box, harmonic oscillator) or forces in three dimensions (particle in a 3-dimensional box). We were able to write the Laplacian, Ñ ², in terms of Cartesian coordinates, assuming y to be a product of 1-dimensional wavefunctions. By separation of variables, we were able to separate the Schrördinger Eq. into three 1-dimensional eqs. & to solve them.

In order to discuss the motion of electrons in atoms, we must deal with a force that is spherically symmetric:

V(r) µ 1/r,

where r is the distance from the nucleus. In this case, we can solve the Schrördinger Eq. by working in spherical polar coordinates (r, q , j ), rather than Cartesian coordinates. This allows us to separate the Schrördinger Eq. into three eqs. each depending on one variable--r, q , or j (See Fig. 6.5 for definition of r, q , and j ).

y = f(x) g(y) h(z) or y = R(r) Q (q ) F (f )

From Fig. 6.5:

r = x² + y² + z²

x = r sin q cos f

y = r sin q sin f

z = r cos q

tan q = r/x

cos q = z/( x² + y² + z²)^1/2

Since Ñ ² = ¶ ²/¶ x² + ¶ ²/¶ y² + ¶ ²/¶ z² , by using the above functional relationships, one can transform Ñ ² into

Ñ ² = ¶ ²/¶ r² + (2/r) ¶ /¶ r + 1/(r²h²) L²

where

L² = - h² (¶ ²/¶ q ² + cot q ¶ /¶ q + (1/ sin² q ) (¶ ²/¶ f ²)

L² is the orbital angular momentum operator.

Orbital Angular Momentum is the momentum of a particle due to its complex (non-linear) movement in space. This is in contrast to linear momentum, which is movement in a particular direction.

Consider the classical picture of a particle of mass m at distance r from the origin. Let r (here bold type indicates a vector) be written as

r = i x + j y + k z

where i, j, & k are unit vectors in the x, y, & z-directions, respectively. Then velocity, v, is given by

v = dr/dt = i dx/dt + j dy/dt + k dz/dt

= i v_x + j v_y + k v_z

ans linear momentum, p, is given by

p = m v = i mv_x + j mv_y + k mv_z

= i p_x + j p_y + k p_z

Then L, the angular momentum of a particle, is given by

L = r x p

The definition of a vector cross product is

A x B = A B sin q ,

where A is the magnitude of vector A, etc. One can determine the value of the cross product from a 3x3 determinant:

i j k

½ A_x A_y A_{z ½}

A x B = B_x B_y B_z

A_y A_z

A_x A_y

= i (A_yB_z - A_zB_y) - j (A_xB_z - A_zB_x) + k (A_xB_y - A_xB_y)

So L = r x p = i L_x + j L_y + k L_z

with L_x = y p_z - z p_y

L_y = z p_x - x p_z

L_z = x p_y - y p_x

The torque, t , acting on a particle is

t = r x F = dL/dt

When t = 0, the rate of change of the angular momentum with respect to time is equal to zero, & the angular momentum is constant (conserved).

In Quantum Mechanics there are two kinds of angular momentum:

Orbital Angular Momentum - same meaning as in classical mechanics

Spin Angular Momentum - no classical analog; will be covered in a later chapter

One can obtain the quantum mechanical operators by replacing the classical forms by their quantum mechanical analogs:

x ® x, p_x ® -ih ¶ /¶ x, etc.

So L_x = -ih (y ¶ /¶ z - z ¶ /¶ y)

L_y = -ih (z ¶ /¶ x - x ¶ /¶ z)

L_z = -ih (x ¶ /¶ y - y ¶ /¶ x)

For Ñ ² need L² = L × L

Definition of a dot product:

A × B = (iA_x + jA_y + kA_z) × (iB_x + jB_y + kB_z)

= AB cos q

The unit vectors are perpendicular to each other, so q = 90⁰ and i × j = 0 = i × k, etc. For the dot product of a vector with itself, q = 0⁰, so i × i = 1, etc. Therefore,

A × B = A_xB_x + A_yB_y + A_zB_z

and

A × A = A_x² + A_y² + A_z² = A²

so that

L² = L_x² + L_y² + L_z²

{Note that this is how the expression for the Laplacian is derived, since

Ñ = i ¶ /¶ x + j ¶ /¶ y + k ¶ /¶ z.

Therefore

Ñ ² = Ñ × Ñ = = ¶ ²/¶ x² + ¶ ²/¶ y² + ¶ ²/¶ z²}

Investigate the commutation relationships between the components of the orbital angular momentum:

[L_x, L_y] = ?

[L_x, L_y] = L_x L_y - L_y L_x

= - ih (y ¶ /¶ z - z ¶ /¶ y) (-ih) (z ¶ /¶ x - x ¶ /¶ z)

- (-ih) (z ¶ /¶ x - x ¶ /¶ z) (- ih) (y ¶ /¶ z - z ¶ /¶ y)

= - h² {y ¶ /¶ z (z ¶ /¶ x - x ¶ /¶ z) - z ¶ /¶ y (z ¶ /¶ x - x ¶ /¶ z)

- z ¶ /¶ x (y ¶ /¶ z - z ¶ /¶ y) + x ¶ /¶ z (y ¶ /¶ z - z ¶ /¶ y)}

= - h² {y (¶ /¶ x + z ¶ /¶ z ¶ /¶ x - x ¶ ²/¶ z²)

- z (z ¶ /¶ y ¶ /¶ x - x ¶ /¶ y ¶ /¶ z)

- z ( y ¶ /¶ x ¶ /¶ z - z ¶ /¶ x ¶ /¶ y)

+ x ( y ¶ ²/¶ z² - ¶ /¶ y - z ¶ /¶ z ¶ /¶ y)}

= - h² { (-yx + xy) ¶ ²/¶ z² + ( yz ¶ /¶ z ¶ /¶ x - zy ¶ /¶ x ¶ /¶ z)

+ ( -z^{2 ¶} /¶ y ¶ /¶ x + z^{2 ¶} /¶ x ¶ /¶ y)

+ ( zx ¶ /¶ y ¶ /¶ z - xz ¶ /¶ z ¶ /¶ y) + (y ¶ /¶ x - x ¶ /¶ y)}

Since the first four terms are zero,

[L_x, L_y] = (ih)² (y ¶ /¶ x - x ¶ /¶ y)

= (ih) {-ih (x ¶ /¶ y - y ¶ /¶ x)}

= ih L_z

The other expressions can be given by symmetry & cyclic permutation: (x, y, z) ® (y, z, x) ® (z, x, y)

[L_x, L_y] = ih L_z [L_y, L_z] = ih L_x [L_z, L_x] = ih L_y

[L², L_x] = ?

[L², L_x] = [L_x² + L_y² + L_z², L_x]

= [L_x², L_x] + [L_y², L_x] + [L_z², L_x]

But [L_x², L_x] = L_x² L_x - L_x L_x² = L_x L_x L_x - L_x L_x L_x = 0

So [L², L_x] = [L_y², L_x] + [L_z², L_x]

= L_y² L_x - L_x² L_y + L_z² L_x - L_x² L_z

= L_y L_y L_x - L_x L_x L_y + L_z L_z L_x - L_x L_x L_z

Lets look at some related forms which can be used to simplify the above expression:

[L_y , L_x] L_y + L_y [L_y , L_x]

= (L_y L_x - L_x L_y) L_y + L_y (L_y L_x - L_x L_y)

= L_y L_x L_y - L_x L_y L_y + L_y L_y L_x - L_y L_x L_y

The first & fourth terms cancel, giving

[L_y , L_x] L_y + L_y [L_y , L_x] = L_y L_y L_x - L_x L_y L_y

Similarly, [L_z , L_x] L_z + L_z [L_z , L_x] = L_z L_z L_x - L_x L_z L_z

So, [L², L_x] = [L_y , L_x] L_y + L_y [L_y , L_x]

+ [L_z , L_x] L_z + L_z [L_z , L_x]

= - ih L_z L_y - ih L_y L_z + ih L_y L_z + ih L_z L_y = 0

One can also show that

[L², L_y] = 0 = [L², L_z]

If A & B commute, Y can simultaneously be an eigenfunction of both operators. That means that the observables a & b can be measured simultaneously if AY = a Y

& BY = b Y .

Example: position & momentum operators. In problem 3.11 we showed that

[x, p_x] = ih.

That means that position & momentum cannot be measured simultaneously--i.e. can’t know definite values for x & p_x.

Example: position & energy. Since

[x, H] = (ih/m) p_x,

can’t assign definite values to position & energy. A stationary state Y has a definite energy, so it shows a spead of possible values of x.

Example: Derive the Heisenber g Uncertainty Principle--from the product of the standard deviation of property A & the standard deviation of property B.

<A>: average value of A

A_i - <A> : deviation of the i-th measurement from the average value

s _A = D A : standard deviation of A; measure of the spread of A or uncertainty in the values of A.

D A = < (A - <A>)²>^1/2

= < A² - 2 A <A> + <A>²>^1/2

= (< A²> - 2 <A> <A> + <A>²)^1/2

= (< A²> - <A>²)^1/2

One can show that

(D A) (D B) > (1/2) ½ ò Y ^* [A,B] Y dt ½

If [A,B] = 0, then can have both A = 0 & B = 0, which means both observables can be known precisely.

For (D x) (D p_x) > (1/2) ½ ò Y ^* (ih) Y dt ½

> (1/2) h ½ i½ ½ ò Y ^* Y dt ½

For a normalized wavefunction, ½ ò Y ^* Y dt ½ = 1.

½ i½ = ( -i i )^1/2 = (1) ^1/2 = 1

So (D x) (D p_x) > (1/2) h.

Operators that communte have observables that can be measured simultaneously. So the operators have simultaneous eigenfunctions.

To return to Angular Momentum--

Since L² & L_z commute, we want to find the simultaneous eigenfunctions. Since L² commutes with each of its components (L_x, L_y, L_z) we can assign definite values to pair L² with each of the components

L², L_x L², L_y L², L_z

But since the components don’t commute with each other, we can’t specify all the pairs--only 1. Arbitrarily choose (L², L_z).

Note that L² means the square of the magnitude of the vector L.

One can convert from Cartesian to Spherical Polar coordinates & derive expressions for L_x, L_y, & L_z that depend only on r, q , & f :

L_x = ih (sin f ¶ /¶ q + cos q cos f ¶ /¶ f )

L_y = - ih (cos f ¶ /¶ q - cot q sin f ¶ /¶ f )

L_z = - ih ¶ /¶ f

L² = L_x² + L_y² + L_z²

= - h² (¶ ²/¶ q ² + cot q ¶ /¶ q + (1/sin ^2q ) ¶ ²/¶ f ²)

Read through the derivation of the simultaneous eigenfunctions of L² and L_z in Chapter 5. It involves techniques that we have used--separtation of variables, recursion formulas, etc. The result--the simultaneous eigenfunctions of L² and L_z are the Spherical Harmonics, Y_l^m(q , f ).

L² Y_l^m(q , f ) = l (l + 1) h² Y_l^m(q , f ), l = 0, 1, 2,...

l : quantum number for total angular momentum

L_z Y_l^m(q , f ) = m h Y_l^m(q , f ), m = -l, -l+1,...l-1, l

m : quantum number for angular momentum in the z-direction

The ranges on the quantum numbers result from forcing finite behavior at infinity on the wavefunction, i.e. the wavefunction must be well-behaved in all regions of space

Y_l^m(q , f ) = [(2l+1)/(4p )]^1/2 [(l-½ m½ )!/( l+½ m½ )!]^1/2

x P_l½ ^m½ (cos q ) e^imf

= (1/2p )^1/2 S_l,m(q ) e^imf

Y_l^m are the Spherical Harmonics

P_l½ ^m½ are the Associated Legendre Functions

S_l,m(q ) = [(2l+1)/2] ^1/2 [(l-½ m½ )!/( l+½ m½ )!]^1/2 P_l½ ^m½ (cos q )

Values for S_l,m(q ) are given in Table 5.1:

l = 0 S_0,0(q ) = Ö 2/2

l = 1 S_1,0(q ) = Ö 6/2 cos q

S_1,+1(q ) = Ö 3/2 sin q = S_1,-1(q )

l = 2 S_2,0(q ) = Ö 10/4 (3 cos² q - 1)

S_2,+1(q ) = Ö 15/2 sin q cos q = S_2,-1(q )

S_2,+2(q ) = Ö 15/4 sin² q = S_2,-2(q )

We will use these functions as the angular part of the wavefunction for the hydrogen atom & the rigid rotor.

Since L_x and L_y cannot be specified, we can only say that the vector L can lie anywhere on the surfacr of a cone defined by the z-axis. See Fig. 5.6

The orientations of L with respect to the z-axis are determined by m. See Fig. 5.7

½ L^2½ = L× L = l(l+1) h²

½ L½ = [l(l+1)]^1/2 h

= length of L

m h = projection of L

onto z-axis

For each eigenvalue of L², there are (2l+1) eigenfunctions of L² with the same value of l, but different values of m. Therefore, the degeneracy is (2l+1).

The Spherical Harmonic functions are important in the central force problem--in which a particle moves under a force which is due to a potential energy function that is spherically symmetric, i.e. one that depends only on the distance of the particle from the origin. Then the wavefunction can be separtated as a product

y = R(r) Y_l^m(q , f )

Spherical Harmonics

give the angular dependence of y for the H atom

describe the energy levels of the diatomic rigid rotor, a model for rotational motion in diatomic molecules