Features

Example of a problem in which V depends on coordinates

Power series solution

Energy is quantized because of the boundary conditions

Model for vibrational motion of a diatomic molecule

To solve the Schrördinder Eq. for molecules, make the Born-Oppenheimer Approximation:

Since the electrons are moving much faster than the nuclei, assume that the motions are separable & that the wavefunction can be written as a product of nuclear & electronic wavefunctions

y = y _elec y _nucl

This leads to two separate Eq. to solve--

one for electronic energy (assuming fixed nuclei)

one for nuclear energy (in the effective potential field of the electrons)

Assume that the nuclear energy can be divided up into contributions from

translation, rotation, vibration

Or, y _nucl = y _{trans y rot y vib}

with E _nucl = E_trans + E_rot + E_vib

E _total = E_nucl + E_elec

We will see later that the vibrational part of the Schrördinder Eq. for a diatomic molecule is similar to the rq. for the harmonic oscillator.

A single particle of mass m is attracted to the origin by a linear (i.e. first power of x) restoring force F_x

F_x = - k x, k = force constant

Since F = m a a = acceleration

then -k x = m a = m (d²x/dt²)

or (d²x/dt²) + (k/m) x = 0

This is a 2nd order linear differential eq. with constant coefficients p = 0, q = k/m

Then s = + iÖ q

and x(t) = c₁e^{itÖ (k/m)} + c₂e^{-itÖ (k/m)}

= (c₁ + c₂) cos (t Ö (k/m)) + i (c₁ - c₂) sin (t Ö (k/m))

= A’ sin (t Ö (k/m) + b)

= A’ sin (t 2 p n + b), Ö (k/m) = 2 p n

Since force by definition is the negative first derivative of the potential energy, one can derive an expression for V:

F_x = -dV/dx = - k x

Integrating both sides gives

V(x) = (k/2) x² + C parabolic function

Choosing C = 0, sets the minimum of V at x=0

Kinetic energy: T = (m/2) v², v = velocity = dx/dt

T = (m/2) (dx/dt)²

If x(t) = A sin (t 2p n + b),

then dx/dt = A 2p n cos (t 2p n + b),

and T = (m/2) A² (2p n )² cos² (t 2p n + b)

= (k/2) A² cos² (t 2p n + b)

Also V = (k/2) x²

= (k/2) A² sin² (t 2p n + b)

Since E = T + V

E = (k/2) A² {sin² (t 2p n + b) + cos² (t 2p n + b)}

= (k/2) A²

H = T + V

= (-h²/2m) d²/dx² + V(x)

= (-h²/2m) d²/dx² + (k/2) x², k = m (2p n )²

= (-h²/2m) d²/dx² + 2p ^{2g 2}mx²

= (-h²/2m) (d²/dx² - a x²), a = 2p n m/ h

H y = E y

d^2y /dx² + ( 2mE/ h² - a ² x²) y = 0

This is a 2nd order homogeneous liner differential eq. with

P(x) = 0 and Q(x) = ( 2mE/ h² - a ² x²)

Since we don’t have constant coefficients, must try a different method of solution- power series

Let y (x) = e^{-a x**2/2} f(x), where F(x) is a power series & x**2 is x squared.

Prove to yourself that

d^2y /dx² = e^{-a x**2/2} (d²f/ dx² - 2a x df/dx - a f + a ² x²f)

Then d^2y /dx² + ( 2mE/ h² - a ² x²) y = 0 gives

d²f/ dx² - 2a x df/dx + (2mE/ h² - a ) f = 0

Assume f can be expanded in a Taylor Series around x=0:

¥

f(x) = S c_n xⁿ

Solve for a recursion relation for the c_n from

d²f/ dx² - 2a x df/dx + (2mE/ h² - a ) f = 0

First write df/dx & d²f/ dx² in terms of power series:

f = c₀ 1 + c₁ x + c₂ x² + c₃ x³ + ...

df/dx = c₁ + c₂ 2x + c₃ 3x² + ... = S c_n nx^n-1 = S c_n nx^n-1

¥

d²f/ dx² = c₂ 2 + c₃ 6x + ... = S c_n n(n-1)x^n-2

Convert to a sum from 0 to ¥ :

Let n = k+2, then when n=2, k=0 and

¥

d²f/ dx² = S c_k+2 (k+1)(k+2)x^k

Since n & k are dummy indices,

¥

d²f/ dx² = S c_n+2 (n+1)(n+2)xⁿ

Plug these series into

d²f/ dx² - 2a x df/dx + (2mE/ h² - a ) f = 0

and collect terms in xⁿ:

S c_n+2 (n+1)(n+2)xⁿ - 2a x S c_n nx^n-1

+ (2mE/ h² - a ) S c_n xⁿ = 0

S { c_n+2 (n+1)(n+2) - 2a c_n n + (2mE/ h² - a ) c_n} xⁿ = 0

Setting {...} = 0 leads to a recursion relationship for the c_n’s:

c_n+2 = {[a + 2a n - 2mE/ h²]/(n+1)(n+2)} c_n

Define c₀ & c₁, then calculate all the other c_n’s

If c₀ = 0 & c₁ = 1, have a power series in odd powers of x

If c₀ = 1 & c₁ = 0, have a power series in even powers of x

The most general solution is a combination of the series

Rewrite S c_n xⁿ as S c_2n+1 x²ⁿ⁺¹

S c_n xⁿ as S c_2n x²ⁿ

Then y = A e^{-a x**2/2} S c_2n+1 x²ⁿ⁺¹ + B e^{-a x**2/2} S c_2n x²ⁿ

Determine A & B from the boundary conditions: i.e. the wavefunction must be well-behaved as x ® ¥ . (As in the particle in a box case, this leads to a quantum condition on the energy.)

The factor e^{-a x**2/2} ® 0 as x ® ¥

But how does the series that multiplies this factor behave as x ® ¥ ? The series may overwhelm the damping effect of the exponential factor as x ® ¥ . Consider the ratio of coefficients of x²ⁿ⁺² and x²ⁿ:

For even n, substitute 2j for n in the recursion formula

c_2j+2/c_2j = {a - 4a j - 2mE/h²}/(2j + 1)(2j + 2)

The series will be dominated by large x as x ® ¥ . So the terms in large j will be important because x is raised to powers of j. For large j,

2j + 1 » 2j, 2j + 2 » 2j, 4a j > 2mE/h²

and c_2j+2/c_2j ® 4a j/(2j)(2j) = a /j

For odd n, substitute 2j + 1 for n in the recursion formula

c_2j+3/c_2j+1 = {a + 2a (2j + 1) - 2mE/h²}/(2j+2)(2j+3)

For large j, c_2j+2/c_2j ® 4a j/(2j)(2j) = a /j

Can we find a power series expansion of some function that has the same ratio of successive coefficients? Yes--

ea ^x**2 = 1 + a x² + ... + a ^j x^2j/j! + a ^j+1 x^2j+1/(j+1)! +...

Coefficient of x^2j+1/ Coefficient of x^2j = a /(j+1) ® a /j

for large j

So each series in y must go as ea ^x**2 as j gets large.

But ea ^{x**2 ®} ¥ for large x

So, as x ® ¥ , y ® A e^{-a x**2/2} ea ^x**2 + B e^{-a x**2/2} ea ^x**2

= (A + B) ea ^x**2/2 ® ¥

If the two series could be truncated after a certain number of terms, the multiplication by e^{-a x**2/2} would force y to 0 as

x ® ¥ . If the highest term in the truncated series is x^p, then by

l’Hospital’s Rule

lim x^p e^{-a x**2/2} = 0

If we could truncate the series at some finite value of n, say v, then y would be finite at ¥ and would be quadratically integrable. Since for both even & odd n,

c_n+2 = {[a + 2a n - 2mE/ h²]/(n+1)(n+2)} c_n,

then c_v+2, c_v+4, etc. = 0 if

This truncation condition leads to a quantum condition on the energy:

E = (h²/2m) a (1 + 2v), a = 2p n m/ h

Or E = h p n (1 +2v) = h g (1/2 + v), v = 0, 1, 2,...

So by requiring that y be finite as x ® ¥ , E becomes quantized.

The general recursion relation can be written by substituting

E = (h²/2m) a (1 + 2v)

into

or c_n+2 = [2a (N-v)/(n+1)(n+2)] c_n

So, which series is truncated by quantizing the energy depends on whether v is even or odd. The only way to have the general solution be well-behaved is to set the coefficient of the other series = 0.

So y = e^{-a x**2/2} S c_2j+1 x^2j+1 or e^{-a x**2/2} S c_2j x^2j

Note that the energy levels of the harmonic oscillator are equally spaced:

E = h n (1/2 + v)

D E = E_v+1 - E_v = h n (v + 3/2 - v - 1/2) = h n

For v = 0, E = h g /2 ground state energy

E is non-zero because E = 0 would violate the Uncertainty Principle (both position & momentum would be known exactly).

If E = 0, then T = 0 & V = 0.

T = 0 means D p_x = 0 (i.e. the momentum is exactly 0)

V = 0 means the particle is at exactly 0

v = 0 The power series is even; c₂ = c₄ = ... = 0

y ₀ = c₀ e^{-a x**2/2} E₀ = (1/2) hn

Find c₀ by normalizing y ₀: ò _{-¥ ¥} c₀² e^{-a x**2} dx = 1

e^{-a x**2} is an even function of x since replacing x by -x gives the same result.

ò _{-¥ ¥} even(x) dx = 2 ò _0¥ even(x) dx

ò _0¥ e^{-a x**2} dx = (1/2) Ö (p /a ) (See Appendix for integrals)

So 2 c₀² (1/2) Ö (p /a ) = 1 ® c₀ = (a /p )^1/4

See Fig. 4.3 for the plot of y _v, v = 0, 1, 2, 3

y ₀ is an even function, y ₀ (x) = y ₀ (-x). From the plot

it can be seen that y ₀ = 0 only at x = +¥ so there are no nodes in y ₀

v=1 The power series is; c₃ = c₅ =...= 0

y ₁ = c₁x e^{-a x**2/2} E₀ = (3/2) hn

Find c₁ by normalizing y ₁: ò _{-¥ ¥} c₁² x²e^{-a x**2} dx = 1

= 2 ò _0¥ c₁² x²e^{-a x**2} dx = 2 c₁² (1/4) (p /a ³)^(1/2)

c₁= (4a ³/p )^1/4

y ₁ is odd: y ₁(-x) = - y ₁(x); but ½ y _1½ ² is even; one node-- at x = 0

v=2 The power series is; c₄ = c₆ =...= 0

y ₂ = (c₀ + c₂x²) e^{-a x**2}

Need a recursion relation for c₀ & c₂ for v = 2. Set n = 0 to ontain

c₂ = [2a (0 - 2)/(0 + 1) (0 + 2)] c₀

y 2 = c₀ (1 - 2a x²) e^{-a x**2}

Find c₀ by normalizing y ₂: ò _{-¥ ¥} c₂² (1-2a x)e^{-a x**2} dx = 1

c₀ = (a /(4p )^1/4

Note: the number of nodes in the wavefunction,interior to the boundary points at +¥ , is equal to v.

Note: By squaring the wavefunction it can be seen that there is a non-zero probability of finding the particle anywhere except at the nodes.

Tunneling: As in the particle in a box case, the pariicle can be found outside the classically-allowed region for which E>V. See Fig. 4.4 (i.e. the classically allowed region is that which is inside the parabola defined by V = (1/2) kx². This indicates quantum mechanical behavior.

To measure that a particle is in the classically forbidden region, one must measure its position. But the act of measurement changes the position, by imparting energy to the system. So an accurate measurement of the position leads to a large uncertainly in the momentum (as well as the kinetic energy)

How is the wavefunction for the harmonic oscillator related to that of a diatomic molecule?

Consider the line spectrum of H₂. At low resolution one sees a series of thick lines (or bands) that correspond to transitions between electronic energy states. As with the Balmer, Paschen, etc. series for H, the lines have a certain frequency which corresponds to the energy difference between different electronic energy states. At higher resolution, the bands are seen to be made up of several individual lines. The lines are due to the fact that the nuclei aren’t infinitely heavy (in fact, nuclear motion is quantized like that of the electron).

Spectral transitions take place between particular vibrational & rotational levels of an excited electronic state & vibrational & rotational levels of the ground state. So a single electronic transition (which shows up as a band at low resolution) can have many different types of transsitions between vibrational & rotational levels. These transitions show up as fine structure at high resolution. The simplest spectra to interpret are vibrational-rotational transitions within the same electronic energy level.

Sketch out how to solve for the molecular wavefunction:

Use the Born-Oppenheimer Approximation. Assume that the nuclei are fixed at distance R. This assumes that

y = y _elect y _nucl

and results in separate eqs. for the electronic & nuclear motions. Solve the Schrödinger Eq. for the electronic energy, E(R). For the vibrational part of the Schrödinger Eq.,

-(h²/2m ) d^2c (R)/dR² + E(R) c (R) = E_vib c (R)

where

m = reduced mass of diatomic molecule = m₁m₂/(m₁ + m₁)

R = internuclear separation

E(R) = electronic energy, calculated at various R to give curve

E_vib = vibrational energy

c = vibrational wavefunction

By approximating the form of E(R), one gets an eq. for the vibrational wavefunction similar to that for the harmonic oscillator. Expand E(R) in a Taylor Series around R = R_e = equilibrium internuclear separation:

Since the zero of energy falls at R = R _e, E(R_e) = 0. Also

Then E(r ) = (1/2) k r ² + ...

The first term in the expansion is a parabola with the same minimum & curvature at the minimum as the exact curve E(R). If this form is used in the Schrödinger Eq. for vibrational motion of the nuclei, the resulting eq. is that of the harmonic oscillator (changing variables from R to r :

-(h²/2m ) d^2c (r )/dR² + (1/2) k r ² c (r ) = E_vib c (r )

Therefore, the harmonic oscillator is a good approximation for a diatomic molecule when R » R_e. So

E_vib = E_v = (V + 1/2) hu , v = 0, 1, 2,...

u = (1/2p ) (k/m )^1/2

are determined from the time-dependent Schrödinger Eq. (This will be covered in a much later chapter.)

Only if the dipole moment e (r ) varies with internuclear separation r , as in a heteronuclear diatomic like HCl, is the transition moment non-zero &

D v = +1

For homonuclear diatomics like H₂, m is independent of r , so no vibrational transition is allowed.

Derive the Selection Rules:

One can calculate the expectation value of e using harmonic oscillator wavefunctions:

<e > = ò y _v^{* e} (r ) y _v’ dt

If D v = +1, then <e > is not equal to zero and the transition is allowed. Calculate the wavenumber of the transition:

If