Quantum Rigid Rotor

Lecture 11

Quantum rigid rotor

Study Goal of This Lecture

• Classical rigid rotpr

• Quantum rigid rotor: Schr¨odingerequation, spherical coordinate and eigenfunctions

11.1 Classical Rigid Rotor

So far we have restricted ourself to 1-D problem. Now we are ready to go on to treat more complex problems in 3-D and beyond. Before we solve the hydrogen atom problem, we must understood quantum rotation first. Consider a particle rotates around a fixed axis, the radius is fixed and we called this kind of system a rigid rotor. The system should be described by the angular momentum L~ = ~r × ~p, (11.1) and the kinetic energy is

1 L~ 2 Tˆ = mω2 = , where I is moment of inertia. (11.2) 2 2I Again, for many particle system, the rotation can be seperated from the center-of- mass motion and described in reduced coordinate.

1 For example, a rigid diatomic molecule:

Figure 11.1: Transform to the reduced coordinate. with reduced mass µ = m1m2 , so m1+m2 L~ 2 L~ 2 T = = . (11.3) 2I 2µR2 Now we return to consider L~ . It is a vector, deﬁned as:

ˆi ˆj kˆ

~ ˆ L = ~r × ~p = x y z = (ypz − zpy)ˆi + (zpx − xpy)ˆj + (xpy − ypx)k. (11.4)

px py pz Therefore, ~ ˆ ˆ ˆ ~ 2 2 2 2 L = Lxi + Lyj + Lzk and L = Lx + Ly + Lz. (11.5)

11.2 Quantum Rigid Rotor

11.2.1 Angular Momentum Operator

h ∂ i In quantum mechanics, we use the corresponding principle and obtain Note y, ∂y = 0 Lˆ2 Lˆ2 Lˆ2 Hˆ = Tˆ + Vˆ = = = , (11.6) 2µRˆ2 2µRˆ2 2I

Lˆ = Lˆxˆi + Lˆyˆj + Lˆzk,ˆ (11.7) ∂ ∂ Lˆ =y ˆpˆ − zˆpˆ = −i (y − z ), (11.8) x z y ~ ∂z ∂y ∂ ∂ Lˆ = −i (z − x ), (11.9) y ~ ∂x ∂z ∂ ∂ Lˆ = −i (x − y ). (11.10) z ~ ∂y ∂x

2 Figure 11.2: Spherical coordinate.

11.2.2 Spherical Coordinate

For rotation motions, these operators are most conveniently studied in spherical coordinate:

|~r| = r : ﬁxed. (11.11)

x = r sin θ cos φ y = r sin θ sin φ (11.12) y = r cos θ to transform operators from Cartesian Coordinate, we need: z z cos θ = = , r p 2 2 2 x + y + z (11.13) y tan φ = . x

Therefore, we want to write Lˆx, Lˆy, Lˆz in spherical coordinates. We know how to ∂ rewrite x, y, z, but how about the operator ∂x ? → the chain rule! ∂ ∂θ ∂ ∂φ ∂ = + , (11.14) ∂x ∂x y,z ∂θ ∂x y,z ∂φ

3 and ∂ ∂θ ∂ z cos θ = − sin θ = [p ] ∂x ∂x y,z ∂x x2 + y2 + z2 z −1 = 3 × × 2x (x2 + y2 + z2) 2 2 (11.15) 1 = − × x × z r3 1 = − × r sin θ cos φ × r cos θ. r3 ∂θ 1 ∴ = cos θ cos φ. (11.16) ∂x y,z r ∂ We can similarly prove: by ∂x tan φ, that ∂φ 1 sin φ = − × . (11.17) ∂x y,z r sin θ h ∂ ∂ i Lˆ = i sin φ + cot θ cos φ (11.18) ∴ x ~ ∂θ ∂φ Also h ∂ ∂ i Lˆ = −i cos φ − cot θ sin φ , (11.19) y ~ ∂θ ∂φ ∂ Lˆ = −i , (11.20) z ~∂φ and 2 2 2h 1 ∂ ∂ 1 ∂ i Lˆ = −~ sin θ + . (11.21) sin θ ∂θ ∂θ sin2 θ ∂φ2

2 Nota that [Lˆx, Lˆy] , [Lˆy, Lˆz] , [Lˆz, Lˆx] 6= 0 and [Lˆ , Lˆx,y,z] = 0.

11.2.3 Spherical Harmonic

Thus, the Schr¨odingerequation:

Lˆ2 Hψˆ = Eψ, Hˆ = . (11.22) 2I 2 h 1 ∂ ∂ 1 ∂2 i − ~ sin θ + ψ(θ, φ) = Eψ(θ, φ). (11.23) ∴ 2I sin θ ∂θ ∂θ sin2 θ ∂φ2 The solution are well known in mathematics, they are the spherical harmonic. They m ˆ2 ˆ are denoted as Yl (θ, φ) and are simultaneous eigenfunctions of L and Lz Why does it require two quantum number to ˆ2 m 2 m L Yl (θ, φ) = l(l + 1)~ Yl (θ, φ), with l = 0, 1, 2, ··· , (11.24) describe the wavefunction? → Because there are 4 two variables (degree of freedoms). l is angular momentum quantum number.

ˆ m m LzYl (θ, φ) = m~Yl (θ, φ), with m = −l, −l + 1, ··· , 0, ··· , l − 1, l , (11.25) m is magnetic quantum number. So, the energy levels of a rigid rotor is Energy only depends on

2 l. E = ~ l(l + 1) = ~ J(J + 1), where l, J = 0, 1, 2, ··· . (11.26) l 2I 2I

Note that each energy level has 2l + 1-fold degeneracy!(Due to for all m = −l, −l + J is often used in 1, ··· , 0, ··· , l − 1, l Y m states have the same energy.) One should think that for a spectroscopy. l m m So what is Yl (θ, φ) state Yl (θ, φ), the amplitude of angular momentum is function? p |L~ | = l(l + 1)~, (11.27) and the projection on the rotational axis z is

|Lz| = m~. (11.28)

Of course |Lz| must smaller than |L~ |, so m = −l, −l +1, ··· , 0, ··· , l −1, l, alway < pl(l + 1). This is summarized by the figure, for example for l = 2, there are five m p √ states: m = ±2, ±1, 0. Each represented by a vector with length = l(l + 1)~ = b~ and z-projection m~ precessing along z. Note the quantization conditions. All five states has the same energy. The first few spherical harmonics will be given later and they are also listed on Silbey’s Table 9.2.

11.2.4 Uncertainty of Quantum Rigid Rotor

Let’s spend some time on the uncertainty of spherical harmonics: Conceptually, given a quantized ratotaional energy level, one determined l rotational energy:

E = ~ l(l + 1), (11.29) l 2I but the direction is not determined, i.e. there are degenerate states with the same energy, this preserves the uncertainty in the system.

5 Figure 11.3: Graphical summary of spherical harmonics.

11.3 The First Few Spherical Harmonics

The ﬁrst few spherical harmonics:

1 0 1 2 Y0 = ( 4π ) , (11.30) 1 0 3 2 Y1 = ( 4π ) cos θ, (11.31) −1 1 3 2 −iφ Y1 = ( 8π ) sin θe , (11.32) 1 1 3 2 iφ Y1 = ( 8π ) sin θe . (11.33)

Let’s look on some features of those spherical harmonics function:

0 • For Y0 , the rotation energy is ”0”. However, it does not violate the uncertainty principle because this state has an wavefunctions that gives ”equal”, ”uniform” probability density in θ and φ. This is quite diﬀerent to the case of harmonic oscillator.

• We ﬁnd the complex number in wavefunction. Practically, we will do superpo-

6 sition to obtain:

1 1 −1 px = √ (Y + Y ), 2 1 1 1 1 −1 py = √ (Y − Y ), (11.34) 2 1 1 0 pz = Y1 . These are p orbitals in hydrogen atom!

Note that, just like the eigenfunctions of harmonic oscillator, we do not need to know m the explicit forms of Yl (θ, φ), we only need to know that: Z m m0 Yl (θ, φ)Yl0 (θ, φ)dτ = δl,l0 · δm,m0 . (11.35)

11.4 More on Magnetic Quantum Number

The emergence of the magnetic quantum number provides useful insights and is analogous to the giving of l, so we will study it a little more, recall: ∂ Lˆ = −i . (11.36) z ~∂φ To ﬁnd eigenfunctions ⇒ T (φ) ∂ −i T (φ) = b · T (φ), (11.37) ~∂φ

−ibφ ∴ T (φ) = Ae ~ . (11.38) If T (φ) is a single-valued function, we must have T (φ) = T (φ+2π) ⇐ the same after rotating one round. Cylic boundary condi-

Derivation of eigenfunction of Lˆz tion. Since T (φ) = T (φ + 2π),

ibφ ib(φ+2π) ibφ i2πb Ae ~ = Ae ~ = Ae ~ e ~ , (11.39)

ib2π 2πb 2πb e ~ = 1, cos + i sin = 1. (11.40) ~ ~ Therefore 2πb = 2πm, m = 0, ±1, ±2, ··· , b = m~, (11.41) ~ 1 T (φ) = √ e−imφ. (11.42) 2π

7 11.4.1 Physical Meaning

• ”Measured” angular momentum along on axis equals to integer of ~!

• For rotation about an axis, i.e. on a plane, or rotation in two directions

Lˆ2 1 Hˆ = z ⇒ Hˆ T (φ) = (m )2T (φ), m = 0, ±1, ±2, ··· , (11.43) 2D 2I 2D 2I ~

for m 6= 0 state, the energy has two-fold degenerarcy. Why it has two-fold degenerarcy? 2 2 Still, we do not explain why l = 0, 1, 2, ··· and why Lˆ has eigenvalues of l(l + 1)~ . ˆ2 ˆ2 But a physical way to understand this is to recognize that: For hL i and hLzi, ˆ2 ˆ2 ˆ ˆ if hL i = hLzi ,then hLxi = hLyi = 0, everything is determined. Therefore, if ˆ2 2 ˆ2 L = l(l + 1)~ , the maximal hLzi is when m = ±l.