Chapter 16 Quantum Particle in Three Dimensions

P. J. Grandinetti

Chem. 4300

P. J. Grandinetti Chapter 16: Quantum Particle in Three Dimensions Momentum and Kinetic Energy Operators in 3D

In 3D the components of the momentum operator are 휕 휕 휕 p̂ = −iℏ ,p ̂ = −iℏ , and p̂ = −iℏ x 휕x y 휕y z 휕z

and the vector operator is ⃗̂ ̂ ⃗ ̂ ⃗ ̂ ⃗ ℏ ⃗ p = pxex + pyey + pzez = −i ∇ Kinetic energy operator in 3D is ( ) 2 ̂ 2 ̂ 2 p̂ py p ℏ2 휕2 휕2 휕2 ℏ2 K̂ = x + + z = − + + = − ∇2 2m 2m 2m 2m 휕x2 휕y2 휕z2 2m

P. J. Grandinetti Chapter 16: Quantum Particle in Three Dimensions Free particle in 3D

Time independent Schrödinger equation for free particle moving in 3D with V(⃗r) = 0 is

ℏ2 ̂ 휓(⃗r) = − ∇2휓(⃗r) = E휓(⃗r) 2m

Eigenstate of ̂ is Ψ(⃗r, t) = 휓(⃗r)e−iEt∕ℏ

Wave function for free particle with momentum p⃗ = ℏk⃗ is

⃗, i(k⃗⋅⃗r−휔t) ip⃗⋅⃗r∕ℏ −iEt∕ℏ Ψ(r t) = Ae = ⏟⏟⏟Ae e 휓(⃗r)

P. J. Grandinetti Chapter 16: Quantum Particle in Three Dimensions Particle in Inﬁnite 3D Well

P. J. Grandinetti Chapter 16: Quantum Particle in Three Dimensions Particle in Inﬁnite 3D Well Particle bound in inﬁnite 3D well ⎧ ≤ ≤ , 0 x Lx ⎪ ≤ ≤ , ⎪ 0 0 y Ly , , ≤ ≤ , V(x y z) = ⎨ 0 z Lz ⎪ ⎪ ⎩ ∞ otherwise.

Separate variables and write 휓(x, y, z) = X(x)Y(y)Z(z)

Plugging into time independent Schrödinger equation

ℏ2 휕2X(x) ℏ2 휕2Y(y) ℏ2 휕2Z(z) − Y(y)Z(z) − X(x)Z(z) − X(x)Y(y) = E휓 2m 휕x2 2m 휕y2 2m 휕z2

P. J. Grandinetti Chapter 16: Quantum Particle in Three Dimensions Particle in Inﬁnite 3D Well

Dividing both sides by 휓(x, y, z) and rearranging gives

1 휕2X(x) 1 휕2Y(y) 1 휕2Z(z) + + + k2 = 0, where k2 = 2mE∕ℏ2 X(x) 휕x2 Y(y) 휕y2 Z(z) 휕z2

Rearranging to 1 휕2X(x) 1 휕2Y(y) 1 휕2Z(z) = − − − k2 = −k2 X(x) 휕x2 Y(y) 휕y2 Z(z) 휕z2 x

2 Introduce separation constant −kx , and obtain uncoupled ODE for X(x)

d2X(x) + k2X(x) = 0 dx2 x

P. J. Grandinetti Chapter 16: Quantum Particle in Three Dimensions Particle in Inﬁnite 3D Well Leaves us with PDE that can be rearranged to

1 휕2Y(y) 1 휕2Z(z) = − − k2 + k2 = −k2 Y(y) 휕y2 Z(z) 휕z2 x y

2 Introduce separation constant −ky to obtain uncoupled ODE for Y(y)

d2Y(y) + k2Y(y) = 0 dy2 y

Leaves us with uncoupled ODE for Z(z)

1 d2Z(z) d2Z(z) − k2 + k2 + k2 = 0 or + k2Z(z) = 0 Z(z) dz2 x y dz2 z

2 2 2 2 where k = kx + ky + kz .

P. J. Grandinetti Chapter 16: Quantum Particle in Three Dimensions Particle in Inﬁnite 3D Well

d2X(x) d2Y(y) d2Z(z) + k2X(x) = 0, + k2Y(y) = 0, + k2Z(z) = 0 dx2 x dy2 y dz2 z 3D boundary conditions constrain normalized ODEs solutions to √ √ √ 2 , 2 , 2 X(x) = sin kxx Y(y) = sin kyy Z(z) = sin kzz Lx Ly Lz

kx, ky, and kz have discrete values given by n 휋 n 휋 n 휋 x , y , z , , , , , , kx = ky = kz = where nx ny nz = 1 2 3 … Lx Ly Lz

and total energy ( ) ( ) 2 2 2 n2 n n2 ℏ 2 2 2 h x y z E , , = k + k + k = + + nx ny nz x y z 2 2 2 2m 8m Lx Ly Lz

P. J. Grandinetti Chapter 16: Quantum Particle in Three Dimensions Degeneracy and density of states

When Lx = Ly = Lz = L, that is, box is cube, energy expression becomes

2 ( ) h 2 2 2 E , , = n + n + n nx ny nz 8mL2 x y z

When diﬀerent states lead to the same energy we say that those states are degenerate states. Particle in a 3D Box has degenerate (identical energy) states

6h2 E , , = E , , = E , , = 2 1 1 1 2 1 1 1 2 8mL2

Degeneracies play an important role in application of statistical mechanics to quantum mechanics.

P. J. Grandinetti Chapter 16: Quantum Particle in Three Dimensions Degeneracy and density of states Number of discrete states of an inﬁnite well represented as histogram as function of energy for 1D Particle on a line 5 Count 0 0 10 20 30 40 50 60 70 80 90 100

2D Particle in a square

5 Count 0 0 10 20 30 40 50 60 70 80 90 100

3D Particle in a cube

10 Count 5

0 0 10 20 30 40 50 60 70 80 90 100

P. J. Grandinetti Chapter 16: Quantum Particle in Three Dimensions Degeneracy and density of states Number of states in E to E + dE is g(E) dE, where g(E) ≡ density of states. Density of states in 1D in given dE decreases with increasing E

5 Count 0 0 10 20 30 40 50 60 70 80 90 100

Density of states in 2D in given dE stays roughly constant with increasing E

5 Count 0 0 10 20 30 40 50 60 70 80 90 100

Density of states in 3D in given dE increases with increasing E

10 Count 5

0 0 10 20 30 40 50 60 70 80 90 100

P. J. Grandinetti Chapter 16: Quantum Particle in Three Dimensions Derive g(E) for particle in 1D infinite well For particle in 1D infinite well the energy is n2h2 E = one energy state for each n n 8mL2 In 1D case, number of states associated with given energy interval, dE, is dn g (E) = 1D dE Rearrange energy expression √ 8mL2E n = h and calculate ( ) dn 1 8m 1∕2 L g (E) = = √ 1D 2 dE 2 h E Density of states decreases with inverse square root of energy - Consistent with 1D energy histogram plot. P. J. Grandinetti Chapter 16: Quantum Particle in Three Dimensions Derive g(E) for particle in 2D infinite well For particle in 2D infinite well the energy is √ ( ) √ 2 2 mLE h 2 2 h 2 2 2 8 E , = n + n = n where n = n + n = nx ny 8mL2 x y 8mL2 x y h

Deﬁnes circle passing through positive n1 and n2 quadrant. Take 1/4 of circle circumference (2휋r) times dn as number of states that lie in annular region of n to n + dn

Calculate number of states associated with given dE in 2D as ( ) 1 (2휋n)dn (2휋n) dn (휋n) dn 휋 8m g (E) = = = = L2 2D 4 dE 4 dE 2 dE 4 h2

Density of states is independent of energy – Consistent with 2D energy histogram plot. P. J. Grandinetti Chapter 16: Quantum Particle in Three Dimensions Derive g(E) for particle in 3D inﬁnite well

Imagine spherical shell in 3D space of nx, and ny, and nz with radius of √ √ 8mLE n = n2 + n2 + n2 = x y z h and thickness of dn associated with states in interval E + dE. Take 1/8 of surface area of sphere (4휋r2) times dn as number of states that lie in n to n + dn 1 (4휋n2)dn 휋n2 dn g (E) = = 3D 8 dE 2 dE Substituting expressions for n and dn∕dE gives ( ) √ 휋 8m 3∕2 g (E) = L3 E 3D 4 h2 Density of states increase with square root of energy – Consistent with 3D energy histogram plot.

P. J. Grandinetti Chapter 16: Quantum Particle in Three Dimensions Quantum Theory of Angular Momentum

P. J. Grandinetti Chapter 16: Quantum Particle in Three Dimensions Quantum Theory of Angular Momentum Angular momentum of particle with respect to origin is ( ) ⃗ L̂ = ⃗r̂ × p⃗̂ = ⃗r̂ × −iℏ∇⃗ = −iℏ⃗r̂ × ∇⃗ Recalling procedure for expanding cross product | | | e⃗ e⃗ e⃗ | | x y z | ⃗̂ ⃗̂ | ̂ ̂ ̂ | r × p = | x y z | | ̂ ̂ ̂ | | px py pz | Operators do not commute. Be careful with order when expanding. | | | | | | | ̂ ̂ | | ̂ ̂ | | ̂ ̂ | ⃗̂ ⃗̂ ⃗̂ ⃗ | y z | ⃗ | x z | ⃗ | x y | L = r × p = ex | ̂ ̂ | −ey | ̂ ̂ | +ez | ̂ ̂ | | py pz | | px pz | | px py | ⏟⏞⏞⏟⏞⏞⏟ ⏟⏞⏞⏟⏞⏞⏟ ⏟⏞⏞⏟⏞⏞⏟

Lx Ly Lz and we ﬁnd ̂ ̂ ̂ ̂ ̂ , ̂ ̂ ̂ ̂ ̂ , ̂ ̂ ̂ ̂ ̂ Lx = y pz − z py Ly = z px − x pz Lz = x py − y px

P. J. Grandinetti Chapter 16: Quantum Particle in Three Dimensions Angular Momentum of a Single Particle

̂ ̂ ̂ ̂ ̂ , ̂ ̂ ̂ ̂ ̂ , ̂ ̂ ̂ ̂ ̂ Lx = y pz − z py Ly = z px − x pz Lz = x py − y px Unlike linear momentum operators which all commute:

̂ , ̂ , ̂ , ̂ , ̂ , ̂ [px py] = 0 [py pz] = 0 [pz px] = 0 ̂ ̂ ̂ Not true for Lx, Ly, and Lz.

̂ , ̂ ℏ̂ , ̂ , ̂ ℏ̂ , ̂ , ̂ ℏ̂ . [Lx Ly] = i Lz [Ly Lz] = i Lx and [Lz Lx] = i Ly

Notice cyclic permutation of subscripts, x → y → z → x ⋯. ̂ ̂ ̂ Commutators tell us Lx, Ly, and Lz are incompatible observables. ℏ ΔL ΔL ≥ |⟨L ⟩|. x y 2 z

P. J. Grandinetti Chapter 16: Quantum Particle in Three Dimensions Quantum Theory of Angular Momentum The total angular momentum operator is ̂ 2 ̂ 2 ̂ 2 ̂ 2 L = Lx + Ly + Lz ̂ ̂ ̂ It commutes with all 3 components, Lx, Ly, and Lz

̂ 2, ̂ , ̂ 2, ̂ , ̂ 2, ̂ , ̂ 2, ⃗̂ [L Lx] = 0 [L Ly] = 0 [L Lz] = 0 or [L L] = 0

̂ 2 ̂ ̂ ̂ ̂ ̂ ̂ L commutes with Lx, Ly, and Lz. (But Lx, Ly, and Lz don’t commute with each other.) ̂ 2 ̂ ̂ ̂ L eigenstate cannot simultaneously be eigenstate of Lx, Ly, and Lz. ̂ 2 ̂ 2 ̂ ̂ 2 ̂ ̂ 2 ̂ L eigenstate can only be eigenstate of L and Lx, or L and Ly, or L and Lz.

We cannot simultaneously know all 3 components of angular momentum vector in QM. At best, we can know angular momentum vector length and one vector component. ̂ 2 ̂ Convention is to work with eigenstates of L and Lz

P. J. Grandinetti Chapter 16: Quantum Particle in Three Dimensions Eigenvalues of angular momentum operators

P. J. Grandinetti Chapter 16: Quantum Particle in Three Dimensions ̂ ̂ Meet the Raising and Lowering operators L+ and L− Start with ̂ 2휓 휆휓 ̂ 휓 휇휓 L = and Lz = 휆 and 휇 are yet-to-be-determined eigenvalues. Introduce related raising and lowering operators ̂ ̂ ̂ ̂ ̂ ̂ ← L+ = Lx + iLy and L− = Lx − iLy What do these operators do? Similar approach taken for harmonic oscillator 휓′ ̂ 휓 ̂ 휓′ ̂ ̂ 휓 ̂ ̂ 휓. If = L+ then Lz = Lz(L+ ) = LzL+ Little math slight of hand gives L̂ 휓′ = L̂ L̂ 휓 −L̂ L̂ 휓 +L̂ L̂ 휓 = (L̂ L̂ −L̂ L̂ ) 휓+L̂ L̂ 휓 = ℏL̂ 휓 + L̂ 휇휓 z z + ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟+ z + z ⏟⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏟z + + z + z + + zero ̂ ,̂ ℏ̂ [Lz L+]= L+ Then L̂ 휓′ = ℏL̂ 휓 + L̂ 휇휓 = (ℏ + 휇) L̂ 휓 ← L̂ increase 휇 (eigenvalue of L̂ ) by ℏ z + + ⏟⏟⏟ + + z eigenalue of 휓′ ̂ 휓 휇휓 ̂ 휓 휓′ ̂ 휓′ 휇 ℏ 휓′ Summary: Lz = , L+ = , and Lz = ( + ) P. J. Grandinetti Chapter 16: Quantum Particle in Three Dimensions ̂ 2 ̂ What are eigenvalues of angular momentum operators L and Lz? ̂ ̂ ℏ Similarly, L− on eigenstate of Lz makes new eigenstate with eigenvalue reduced by . ̂ ̂ L+ and L− are called the raising and lowering operators, respectively. L̂ 2 gives square of total angular momentum, L̂ 2휓 = 휆휓 ̂ ̂ 휓 휇휓 Lz gives z component of angular momentum, Lz = ℏ is has units of angular momentum, ℏ = 1.054571800139113 × 10−34m2⋅kg/s

̂ ̂ L+ cannot create new Lz eigenstate with eigenvalue greater than total angular momentum. No single component of vector cannot exceed total length of vector. 휓 ̂ 퓁ℏ If max is eigenstate of Lz with highest possible eigenvalue, , then ̂ 휓 , ̂ 휓 퓁ℏ휓 ̂ 2휓 휆휓 L+ max = 0 while Lz max = max and L max = max

where 퓁 and 휆 are to be determined.

P. J. Grandinetti Chapter 16: Quantum Particle in Three Dimensions ̂ 2 ̂ What are eigenvalues of angular momentum operators L and Lz? Determine 퓁 and 휆 from

̂ 휓 , ̂ 휓 퓁ℏ휓 ̂ 2휓 휆휓 L+ max = 0 while Lz max = max and L max = max ̂ 2 휓 Start by applying L to max, ( ) ̂ 2휓 ̂ 2 ̂ 2 ̂ 2 휓 휆휓 . L max = Lx + Ly + Lz max = max

Next, use identity ̂ 2 ̂ 2 ̂ ̂ ℏ̂ ̂ ̂ ℏ̂ Lx + Ly = L−L+ + Lz = L+L− − Lz

P. J. Grandinetti Chapter 16: Quantum Particle in Three Dimensions ̂ 2 ̂ 2 ̂ ̂ ℏ̂ Prove Lx + Ly = L−L+ + Lz ( ) ( ) ( ) ( ) L̂ + L̂ L̂ + L̂ L̂ − L̂ L̂ − L̂ L̂ 2 + L̂ 2 = + − + − + + − + − x y 2 2 2i 2i ⏟⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏟ ⏟⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏟ ⏟⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏟ ⏟⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏟ ̂ ̂ ̂ ̂ Lx Lx Ly Ly ( ) ( ) ̂ 2 ̂ ̂ ̂ ̂ ̂ 2 ̂ 2 ̂ ̂ ̂ ̂ ̂ 2 ( ) L + L−L+ + L+L− + L L − L−L+ − L+L− + L 1 L̂ 2 + L̂ 2 = + − + + − = L̂ L̂ + L̂ L̂ . x y 4 −4 2 − + + −

̂ , ̂ ̂ ̂ ̂ ̂ ℏ̂ ̂ ̂ ℏ̂ ̂ ̂ From [L+ L−] = L+L− − L−L+ = 2 Lz get L+L−− = 2 Lz + L−L+ 1 ( ) L̂ 2 + L̂ 2 = L̂ L̂ + 2ℏL̂ + L̂ L̂ = L̂ L̂ + ℏL̂ x y 2 − + z − + − + z

̂ 2 ̂ 2 ̂ ̂ ℏ̂ Similar approach to prove Lx + Ly = L+L− − Lz

P. J. Grandinetti Chapter 16: Quantum Particle in Three Dimensions ̂ 2 ̂ What are eigenvalues of angular momentum operators L and Lz?

̂ 휓 , ̂ 휓 퓁ℏ휓 ̂ 2휓 휆휓 L+ max = 0 while Lz max = max and L max = max Use these equations to determine values of 퓁 and 휆. ̂ 2 휓 Start by applying L to max, ( ) ̂ 2휓 ̂ 2 ̂ 2 ̂ 2 휓 휆휓 . L max = Lx + Ly + Lz max = max

Next, use identity ̂ 2 ̂ 2 ̂ ̂ ℏ̂ ̂ ̂ ℏ̂ Lx + Ly = L−L+ + Lz = L+L− − Lz and obtain ( ) ( ) ̂ 2휓 ̂ ̂ ℏ̂ ̂ 2 휓 퓁ℏ2 퓁2ℏ2 휓 퓁 퓁 ℏ2휓 휆휓 , L max = L−L+ + Lz + Lz max = 0 + + max = ( + 1) max = max

learn that 휆 = 퓁(퓁 + 1)ℏ2.

P. J. Grandinetti Chapter 16: Quantum Particle in Three Dimensions ̂ 2 ̂ What are eigenvalues of angular momentum operators L and Lz? Similarly, ̂ 휓 , ̂ 휓 퓁′ℏ휓 ̂ 2휓 휆휓 L− min = 0 while Lz min = min and L min = min As before, ( ) ̂ 2휓 ̂ 2 ̂ 2 ̂ 2 휓 휆휓 . L min = Lx + Ly + Lz min = min giving ( ) ( ) ̂ 2휓 ̂ ̂ ℏ̂ ̂ 2 휓 퓁′ℏ2 퓁′ℏ 2 휓 퓁′ 퓁′ ℏ2휓 휆휓 L min = L+L− − Lz + Lz min = 0 − + ( ) min = ( − 1) min = min

learn that 휆 = 퓁′(퓁′ − 1)ℏ2 = 퓁(퓁 + 1)ℏ2 ← Since L̂ 2휓 = 휆휓 for all 휓 can only be true if 퓁′ = −퓁. L̂ 2휓 = 퓁(퓁 + 1)ℏ2휓, thus, 퓁(퓁 + 1)ℏ2 is eigenvalue of L̂ 2 ̂ 퓁ℏ 휓 퓁ℏ 휓 ℏ Eigenvalues of Lz range from − for min to + for max, and increase in steps of

P. J. Grandinetti Chapter 16: Quantum Particle in Three Dimensions ̂ 2 ̂ What are eigenvalues of angular momentum operators L and Lz? ̂ 2 ̂ Using raising and lowering operators, determined behavior and eigenvalues of L and Lz without explicit expression for 휓.

L̂ 2휓 = 퓁(퓁 + 1)ℏ2휓

and ̂ 휓 ℏ휓 퓁, 퓁 , , 퓁 , 퓁 Lz = m where m = − − + 1 … − 1

If there are N steps between m = −퓁 and m = 퓁 then 퓁 = −퓁 + N gives 퓁 = N∕2. 퓁 much have an integer or half-integer value,

퓁 = 0, 1∕2, 1, 3∕2, … .

̂ Notice√ that maximum eigenvalue of Lz is smaller than total angular momentum, i.e., 퓁ℏ < 퓁 퓁 ℏ ̂ ̂ ( + 1) . This is due to uncertainty in Lx and Ly.

P. J. Grandinetti Chapter 16: Quantum Particle in Three Dimensions Eigenstates of angular momentum operators

P. J. Grandinetti Chapter 16: Quantum Particle in Three Dimensions ̂ 2 ̂ Eigenstates of angular momentum operators L and Lz ̂ 2 ̂ To determine eigenstates of L and Lz, go back to ⃗ L̂ = ⃗r̂ × p⃗̂ = −iℏ ⃗r̂ × ∇⃗ To go further we are better oﬀ working in spherical coordinates, z

x P. J. Grandinetti Chapter 16: Quantum Particle in Three Dimensions ̂ 2 ̂ Eigenstates of angular momentum operators L and Lz In spherical coordinates

⃗ 휃 휙 ̂ 휃 휙 ⃗ 휃 ⃗ , er = sin cos ex + sin sin ey + cos ez ⃗ 휃 휙 ̂ 휃 휙 ⃗ 휃 ⃗ , e휃 = cos cos ex + cos sin ey − sin ez ⃗ 휙 ̂ 휙 ⃗ , e휙 = − sin ex + cos ey

or the inverse

⃗ 휃 휙 ̂ 휃 휙 ⃗ 휙 ⃗ , ex = sin cos er + cos cos e휃 − sin e휙 ⃗ 휃 휙 ̂ 휃 휙 ⃗ 휙 ⃗ , ey = sin sin er + cos sin e휃 + cos e휙 ⃗ 휃 ̂ 휃 ⃗ , ez = cos er − sin e휃

and can express ∇⃗ as ⃗ 휕 1 휕 1 휕 ∇ = e⃗ + e⃗휃 + e⃗휙 . r 휕r r 휕휃 r sin 휃 휕휙

P. J. Grandinetti Chapter 16: Quantum Particle in Three Dimensions ̂ 2 ̂ Eigenstates of angular momentum operators L and Lz After some quantum algebra, we obtain ( ) 휕 휕 ̂ ℏ 휙 휙 휃 Lx = −i − sin 휕휃 − cos cot 휕휙 ( ) 휕 휕 ̂ ℏ 휙 휙 휃 Ly = −i − cos 휕휃 − sin cot 휕휙

휕 ̂ ℏ Lz = −i 휕휙

Similarly, one can show that [ ( ) ] 1 휕 휕 1 휕2 L̂ 2 = L̂ 2 + L̂ 2 + L̂ 2 = −ℏ2 sin 휃 + x y z sin 휃 휕휃 휕휃 sin2 휃 휕휙2

P. J. Grandinetti Chapter 16: Quantum Particle in Three Dimensions ̂ 2 ̂ Eigenstates of angular momentum operators L and Lz ̂ 2 ̂ To determine eigenfunction of both L and Lz, start with ℏ 휕휓(휃, 휙) L̂ 휓(휃, 휙) = = mℏ 휓(휃, 휙) z i 휕휙 Use separation of variables: 휓(휃, 휙) = Θ(휃)Φ(휙) Substituting into the partial diﬀerential equation, ℏ 휕 ℏ 휕Φ(휙) Θ(휃)Φ(휙) = Θ(휃) = mℏ Θ(휃)Φ(휙) i 휕휙 i 휕휙 and simplifying gives dΦ(휙) dΦ(휙) = imΦ(휙) which rearranges to = imd휙 d휙 Φ(휙) and integrates to Φ(휙) = Aeim휙 Since we require wave functions to be single valued we must have Φ(휙) = Φ(휙 + 2휋) or Aeim휙 = Aeim(휙+2휋) which leads to the constraint eim2휋 = 1 requiring m = 0, ±1, ±2, … P. J. Grandinetti Chapter 16: Quantum Particle in Three Dimensions ̂ 2 ̂ Eigenstates of angular momentum operators L and Lz Next we consider L̂ 2휓(휃, 휙) = 퓁(퓁 + 1)ℏ2 휓(휃, 휙)

Substituting the expression for L̂ 2 gives [ ( ) ] 1 휕 휕 1 휕2 L̂ 2휓(휃, 휙) = −ℏ2 sin 휃 + 휓(휃, 휙) = ℏ2퓁(퓁 + 1)휓(휃, 휙). sin 휃 휕휃 휕휃 sin2 휃 휕휙2 Substituting 휓(휃, 휙) = Θ(휃)Φ(휙) into this PDE and dividing both sides by 휓(휃, 휙) ( ) sin 휃 휕 휕Θ(휃) 1 휕2Φ(휙) sin 휃 + 퓁(퓁 + 1) sin2 휃 = − = m2, Θ(휃) 휕휃 휕휃 Φ(휙) 휕휙2

Identify m2 as separation constant for this PDE. No need to go further. Recognize this PDE from Chapter 8 (Classical Waves) as having spherical harmonic wave solutions √ 퓁 퓁 m (2 + 1)( − m)! m im휙 Y퓁, (휃, 휙) = (−1) P (cos 휃)e m 4휋(퓁 + m)! 퓁

P. J. Grandinetti Chapter 16: Quantum Particle in Three Dimensions