ANSWER:

Explanation:

2. Orbital angular momentum

Consider a particle of mass m, momentum ~p and position vector ~r (with respect to a

fixed origin, ~r = 0). In classical mechanics, the particle’s orbital angular momentum is given

by a vector L~ , defined by

L~ = ~r × ~p. (1)

This vector points in a direction that is perpendicular to the plane CONTAINING ~r and ~p,

and has a magnitude L = rp sin α, where α is the angle between ~r and ~p. In Cartesian

coordinates, the components of L~ are

Lx = ypz − zpy;

Ly = zpx − xpz;

Lz = xpy − ypx.

(2)

The corresponding QM operators representing Lx, Ly and Lz are obtained by replacing

x, y, z and px, py and pz with the corresponding QM operators, giving

Lx = −i~

y

∂

∂z − z

∂

∂y

;

Ly = −i~

z

∂

∂x − x

∂

∂z

;

Lz = −i~

x

∂

∂y − y

∂

∂x

.

(3)

In a more compact form, this can be written as a vector operator,

L~ = −i~(~r × ∇~ ). (4)

It is easy to verify that L~ is Hermitian.

Using the commutation relations derived for ~x and ~p, the commutation relations between

the different components of L~ are readily derived. For example:

[Lx, Ly] = [(ypz − zpy),(zpx − xpz)] = [ypz, zpx] + [zpy, xpz] − [ypz, xpz] − [zpy, zpx] (5

– 3 –

Since y and px commute with each other and with z and pz, the first term reads

[ypz, zpx] = ypzzpx − zpxypz = ypx[pz, z] = −i~ypx (6)

Similarly, the second commutator gives

[zpy, xpz] = zpyxpz − xpzzpy = xpy[z, pz] = i~xpy (7)

The third and forth commutators vanish; we thus find that

[Lx, Ly] = i~(xpy − ypx) = i~Lz. (8)

In a similar way, it is straightforward to show that

[Ly, Lz] = i~Lx (9)

and

[Lz, Lx] = i~Ly (10)

The three equations are equivalent to the vectorial commutation relation:

L~ × L~ = i~L. ~ (11)

Note that this can only be TRUE for operators; since, for REGULAR vectors, clearly L~ × L~ = 0.

The fact that the operators representing the different components of the angular momentum do not commute, implies that it is impossible to obtain definite values for all component

of the angular momentum when measured simultaneously. This means that if the system

is in eigenstate of one component of the angular momentum, it will in general not be an

eigenstate of either of the other two components.

We define the operator representing the square of the magnitude of the orbital angular

momentum by

L~ 2 = L

2

x + L

2

y + L

2

z

. (12)

It is easy to show that L~ 2 does commute with each of the three components: Lx, Ly or Lz.

For example (using [L

2

x

, Lx] = 0):

[L~ 2

, Lx] = [L

2

y + L

2

z

, Lx] = [L

2

y

, Lx] + [L

2

z

, Lx]

= Ly[Ly, Lx] + [Ly, Lx]Ly + Lz[Lz, Lx] + [Lz, Lx]Lz

= −i~(LyLz + LzLy) + i~(LzLy + LyLz) = 0.

(13)

Similarly,

[L~ 2

, Ly] = [L~ 2

, Lz] = 0, (14)

– 4 –

which can be summarized as

[L~ 2

, L~ ] = 0. (15)

Physically, this means that one can find simultaneous eigenfunctions of L~ 2 and one of the

components of L~ , implying that both the magnitude of the angular momentum and one of

its components can be precisely determined. Once these are known, they fully specify the

angular momentum.

In order to obtain the eigenvalues of L~ 2 and one of the components of L~ (typically, Lz),

it is convenient to express the angular momentum operators in spherical polar coordinates:

r, θ, φ, rather than the Cartesian coordinates x, y, z. The spherical coordinates are related

to the Cartesian ones via

x = r sin θ cos φ;

y = r sin θ sin φ;

z = r cos θ.

(16)

After some algebra, one gets:

Lx = −i~

− sin φ

∂

∂θ − cot θ cos φ

∂

∂φ

Ly = −i~

cos φ

∂

∂θ − cot θ sin φ

∂

∂φ

Lz = −i~

∂

∂φ;

L~ 2 = −~

2

h

1

sin θ

∂

∂θ

sin θ

∂

∂θ

+

1

sin2 θ

∂

2

∂φ2

i

.

(17)

We thus find that the operators Lx, Ly, Lz and L~ 2 depend on θ and φ only, that is they

are independent on the radial coordinate ~r. All these operators therefore commute with any

function of r,

[Lx, f(r)] = [Ly, f(r)] = [Lz, f(r)] = [L

2

, f(r)] = 0. (18)

Also, obviously, if a wavefunction depends only on r (but not on θ, φ) it can be simultaneously

an eigenfunction of Lx, Ly, Lz and L

2

. In all cases, the corresponding eigenvalue will be

0. (This is the only exception to the rule that that eigenvalues of one component (e.g., Lx)

cannot be simultaneously eigenfunctions of the two other components of L).