The Quantum Theory of Atoms and Molecules Particles in boxes and applications Dr Grant Ritchie
The free particle What is the appropriate form of the time independent Schrödinger equation? V = 0, so that TISE is:
Solutions are of the form:
h 2 d 2ψ − = Eψ 2 2m dx
ψ = Aeikx + Be −ikx k 2h 2 E= 2m
Eigenfunctions (A,B are constants) Eigenvalues
There is no restriction on k or allowed energies ⇒ energy is continuous! Probability density example: Suppose that B = 0, then ψ (x) = A eikx . Where is the particle? ψ *ψ = (A eikx)* (A eikx) = (A* e– ikx)(A eikx ) = |A| 2 . This is independent of x, in other words, we cannot predict where we will find the particle. N.B. k = 2π/λ and as k is known exactly then so is p. cf. Δx Δ px ≥ ħ / 2.
Particle in a 1-d box (or infinite square well!) What are the wavefunctions (eigenfunctions) and energies (eigenstates/eigenvalues) of a particle confined in a very deep potential well? +∞
+∞ V(x) V(x) → ∞
Inside the box: V = 0, so that SE is: V(x) → ∞
V(x) = 0 0
h 2 d 2ψ − = Eψ 2 2m dx ψ = Aeikx + Be −ikx = C cos kx + D sin kx
L
Outside the box: V = ∞, so that outside the box ψ = 0, and because ψ is continuous, ψ must also be 0 at the edges of the box, i.e.ψ(0) = 0 and ψ(L) = 0. For ψ(0) = 0 then C = 0, and so ψ (x) = D sin(kx). For ψ(L) = 0 then either D = 0 (SILLY!) or ψ(L) = D sin(kL) = 0. sin(nπ) = 0 where n = 1, 2, 3,… and so k is now restricted as follows:
nπ k= L
Quantisation of energy Boundary conditions lead to wavefunctions (eigenfunctions) of the form ψ(x) = D sin kx with quantised values of k (= nπ/L) and results in quantised values for the energy E:
k 2h 2 n 2 h 2 E= = 2m 8mL2
Eigenvalues:
n = 1, 2, 3,....
To completely specify the eigenfunctions we need to normalise them: L
L
2 * sin x D ψ ψ d = ∫ ∫ kx dx = 1 ⇒ D = 2
0
Eigenfunctions:
0
ψ ( x) =
2 ⎛ nπx ⎞ sin ⎜ ⎟ L ⎝ L ⎠
n = 1,2,....
2 L
.
Particle in box properties Ψ(x)
1. Increasing n indicates increasing K.E. (increasing curvature of the wavefunction). 2. Lowest energy: E1 = h2/8mL2 > 0 ; ZERO POINT ENERGY. If E → 0, then p, δp → 0 and requires δx → ∞ (Heisenberg) 3. Difference between adjacent energy levels: ΔE = En+1 – En = (2n + 1) h2/8mL2 = (2n + 1) E1 4. The more the particle is confined, i.e. L gets smaller, the greater ΔE and the K.E. – hard to compress matter!
0
L
x
5.
If L → ∞ , ΔE → 0, continuous: translational energy is not quantised. Back to classical physics!
Consistent with De Broglie: The longest wavelength is λ = 2 L and the higher modes have wavelengths given by λ = 2L/n (n = 1,2,...). Therefore the De Broglie relationship yields momentum p = ± (nh/2L) = ± h/λ.
Orthogonality Often we are interested in integrating products of wavefunctions. If the integral turns out to be zero then the functions, ψn and ψm are said to be orthogonal: ∞
∫ψ
-∞
L
* n ( x)
2 ⎛ nπx ⎞ ⎛ mπx ⎞ sin ⎜ ψ m ( x ) dx = ⎟ sin ⎜ ⎟ dx = 1 if n = m L0 ⎝ a ⎠ ⎝ a ⎠
∫
= 0 if n ≠ m
Importantly, note that the eigenfunctions for the particle in a box are orthogonal to one another. Indeed they are said to be orthonormal in that each function is also normalised. Examples: (i) Show pictorially that the hydrogen atom 2s and 2pz orbitals are orthogonal; (ii) Shown that the two lowest energy wavefunctions for a particle in a box are orthogonal.
Where is the particle most likely to be? We can calculate the most probable position of the particle from knowledge of ψ*ψ. For example, for n = 1 1
1
ψ
ψ*ψ
n=1
0
-1 0.0
0
1 0.2
0.4
0.6
x/a
0.8
1.0
0.0
0.2
0.4
0.6
0.8
1.0
x/a
i.e. particle is most likely to be found at the centre of the box. This results is clearly at odds with classical expectations where each position in the box is equally likely. However as n increases the wavefunction begins to have so many nodes that in the limit as n → ∞ each position is equally probable ⇒ Correspondence Principle.
Further comparisons between classical and quantum results Classically we expect that the probability density is uniform i.e. all positions in box are equally likely. Thus for a box of length L, the probability density P(x) = ψ*ψ = 1/L and so the average values of x and x2 are: L
L
L 1 x = xP( x) dx = x dx = L0 2 0
∫
∫
L
x
2
L2 = x P ( x ) dx = 3 0
∫
2
Compare with the quantum case where n = 1 i.e. ψ ( x) = L
then
L 2 ⎛ πx ⎞ x = x sin 2 ⎜ ⎟ dx = L0 2 ⎝ L⎠
∫
L
x2
L2 L2 2 2 2 ⎛ πx ⎞ x sin ⎜ ⎟ dx = = − 2 L0 L 3 2π ⎝ ⎠
∫
As n → ∞, find that → classical result.
2 ⎛ πx ⎞ sin ⎜ ⎟ L ⎝ L⎠
Momentum of the particle Given the wavefunction ψ = D sin(kx), what is the particle’s momentum?
pˆ xψ =
Use the momentum operator….
h d h ( D sin kx) = k D cos kx ≠ cons.ψ i dx i
ψ is not an eigenfunction px (see later) but we can evaluate the expectation value (see last lecture). L
pˆ x =
∫ψ * pˆ xψ dx ∫ψ * ψ d x
h i
=
D
2
∫ (sin kx)
L
d dx
(sin kx) dx
kh i
D
=
0
L
D 2 ∫ (sin 2 kx) dx
E=
2m
≠
2m
0
L
0
= 0 – this does not mean that the kinetic energy is 0! pˆ x
∫ (sin kx)(cos kx) dx
D 2 ∫ (sin 2 kx) dx
0
pˆ x2
2
2
=0
Eigenfunctions of the momentum operator Consider the following eigenvalue equation:
pˆ xψ = pψ
(where p is a constant)
h dψ = pψ i dx
General solution:
ψ
( )x = Ae ikx = Ae ip h
NB. k can be positive or negative.
Thus Aeikx are eigenfunctions of the momentum operator with eigenvalues p = ±kћ. The particle in a box wavefunction ψ = D sin kx can be expressed as a linear combination of momentum eigenfunctions, i.e. ψ = D sin kx = D′ (eikx + e-ikx). A single measurement of the particle’s momentum must give a definite result of ±kћ. However, since sin kx contains equal amounts of e±ikx, the average value of the momentum = 0.
An application of the particle in a box problem - the UV absorption spectrum of cyanine dyes +
&& − (−CH = CH −) − CH = N (CH ) (CH 3 ) 2 N k 3 2 b +
&& (CH ) (CH 3 ) 2 N = CH − (−CH = CH −) k − N 3 2 Dye I: k = 1 Dye II: k = 2 Dye III: k = 3
(Taken from www.jce.divched.org )
Electronic structure of cyanine dyes Have both σ and π electrons ⇒ σ electrons have largest probabilities in the plane of the molecules while π electrons are most like to be found above and below plane of molecule.
Decouple π electrons from σ framework ⇒ treat π electrons as being delocalised over the length of the molecule between the N atoms. UV and visible light can then be absorbed and energy used to cause transition of π electrons from one energy level to another. NB: Each carbon contributes one electron to the π system while the two N atoms contribute 3 electrons.
Is the particle in a box model justified for this problem?
Length of the box, L = bβ +2δ, where b is the number of bonds.
Resonance condition Absorption of a photon occurs when the energy of the photon (= hν) matches the difference in the energy between the two states involved in the transition (ΔE):
E photon
h2 2 2 n n = hν = ΔE = E f − Ei = − ( f i ) 8mL2
where ni and nf are quantum numbers for the initial and final states respectively. Which values of ni and nf have to be used? Depends on the number of π electrons and the Pauli Exclusion Principle which allows a maximum of 2 electrons per orbital. Electron pairs must have opposite spins.
Thus, for Dye I, k =1 and we have a total of (3+3) = 6 π electrons that will pair in levels n = 1, 2 and 3.Therefore the highest occupied molecular orbital (HOMO) has n = 3, while the lowest unoccupied molecular orbital (LUMO) has n = 4. Hence lowest energy transition involves promotion of a π electron from n = 3 → n = 4. Example: If the length of the box L is 8.5Å, what is the peak absorption wavelength for dye I?
Are all transitions possible? – Selection rules Must always obey Pauli exclusion principle. Transitions are electric dipole transitions – the oscillating electric field component of the radiation interacts with electrical charges, i.e. the positive nuclei and negative electrons that comprise an atom or molecule, and cause the transitions observed in uv-visible absorption and emission spectroscopies. The interaction energy, U, between a system of charged particles and an electric field, E, is given by:
U = − μ .E
The dipole moment is defined for a collection of charges by:
μ = ∑ qi ri i
where ri is the position vector of charged particle i. ( See electrostatics lectures in Michaelmas term).
The transition dipole moment In order to obtain the strength of interaction that causes a transition between two states, the transition dipole moment is used rather than the dipole moment. For a transition between and initial state, ψ i , to a final state ψ f , the transition dipole moment integral is.
μ fi = ∫ψ *f μˆψ i dτ
Just like the probability density is given by ψ*ψ, so the probability for a transition (as measured by the absorption coefficient) is proportional to μfi* μfi.. If μfi = 0 then the interaction energy is zero and no transition occurs – the transition is said to be electric dipole forbidden. Conversely, if μfi is large, then the transition probability and absorption coefficient are large. The intensity of the transition is therefore proportional to.
∫ψ
* k
μˆψ j dτ
2
Transition dipole moment integral for particle in a box Need to consider the transition dipole moment integral for one electron. The dipole moment L operator for an electron in one dimension is –ex and so
μ fi = −e ∫ψ *f ( x) xψ i ( x)dx 0
Now evaluate μfi for various wavefunctions to see which are allowed (μfi ≠ 0) and which are forbidden (μfi = 0). Example: Is n = 1 → n = 2 an allowed transition? ⎛ n f πx ⎞ 2e ⎛ n πx ⎞ ⎟⎟ x sin ⎜ i ⎟dx μ fi = − sin ⎜⎜ L 0 ⎝ L ⎠ ⎝ L ⎠ L
(If you have time!) consider the generalised transition ni → nf :
∫
L ⎛ (n + ni )πx ⎞ ⎞ e ⎛⎜ ⎛ ( n f − ni )πx ⎞ ⎟⎟ − cos⎜⎜ f ⎟⎟ ⎟dx =− x cos⎜⎜ ⎟ L 0 ⎜⎝ ⎝ L L ⎠ ⎝ ⎠⎠
∫
If we then define Δn = nf – ni and ntot = nf + ni then the above integral becomes: Looks bad!!!!!
2 ⎫ e⎛L⎞ ⎧ 1 1 1 1 μ fi = − ⎜ ⎟ ⎨ 2 (cos(Δnπ ) − 1) − 2 (cos(ntot π ) − 1) + sin( Δnπ ) − sin( ntot π )⎬ Δn ntot L ⎝ π ⎠ ⎩ Δn ntot ⎭
However, if Δn is even then ntot is even and overall μfi = 0 – Forbidden!
Δn is odd is allowed If Δn is odd then ntot is also odd and overall μfi ≠ 0 and is given by
⎞ 2eL ⎛ 1 1 ⎞ 8eL ⎛⎜ ni n f ⎟ μ fi = − 2 ⎜⎜ 2 − 2 ⎟⎟ = 2 2 2 2 π ⎝ ntot Δn ⎠ π ⎜⎝ (n f − ni ) ⎟⎠ The general selection rule is Δn is odd. However, we only really see a single peak in the absorption spectrum for each dye because other allowed transitions have very much smaller transition moments. Example: For dye 1, compare the values of the transition dipole moment integrals for the two transitions n = 3 → n = 4 and n = 3 → n = 6.
*Also note that longer molecules have larger absorption coefficients because μfi increases with the length of the molecule (see uv spectra earlier in notes).
Using symmetry to evaluate integrals An alternative to evaluating integrals is to use symmetry…… Firstly, consider the parity of the particle in a box wavefunctions by shifting the positions of the potential barriers from 0 and L to –L/2 to L/2. * For symmetric potentials V(x) = V(-x) the ψ has a definite parity, i.e. ψ n ( x) = ±ψ n (− x) V ( x) = 0 for − L / 2 ≤ x ≤ L / 2 V ( x) = ∞ otherwise
x → x' = x − L / 2
ψ n ( x) =
2 ⎛ nπx ⎞ cos ⎜ ⎟ = +ψ n (− x) ; odd n − even parity L ⎝ L ⎠
=
2 ⎛ nπx ⎞ sin ⎜ ⎟ = −ψ n (− x) ; even n − odd parity L ⎝ L ⎠
n=1
−
L 2
0
n=2
L 2
−
L 2
0
L 2
Symmetry II For transition n =1 → n = 2 the transition dipole moment integral is: L/2
L/2
e e ⎛ πx ⎞ ⎛ πx ⎞ sin ⎜ ⎟ x cos⎜ f (x )dx μ 21 = − ⎟ dx = − 2 L L L −L / 2 ⎝ L ⎠ ⎠ ⎝ −L / 2
∫
n=1
∫
x
0
L 2
L 2
−
0
L 2
−
f ( x) ≠ 0
→
×
L 2
∫
−L/ 2
n=2
× −
L/2
L 2
0
L 2
−
L 2
0
L 2
Clearly ∫ f (x)dx ≠ 0 and the transition is allowed. In contrast, for the transition n = 1 → n = 3, ∫ f(x) L/2 dx = 0 and transition is forbidden. ∫ f (x) = 0 x
n=1
× −
L 2
0
L 2
−
L 2
−L/2
n=3
→
× 0
L 2
−
L 2
0
L 2
−
L 2
0
L 2
Conclusion: if integrand is odd / antisymmetric / ungerade then ∫ f(x) dx = 0 and transition is forbidden.
Particles in “round” boxes (or on a ring!) z
Hamiltonian for a particle of mass m constrained to move in a circular path of radius r in the xy-plane (V = 0 everywhere) is:
Jz
φ x
y
r
m
∂2 ⎞ h2 ⎛ ∂2 ˆ ⎜ 2 + 2⎟ H =− 2m ⎜⎝ ∂x ∂y ⎟⎠
p
In polar co-ordinates x = r cosφ , y = r sinφ and the Hamiltonian becomes:
But radius of ring is fixed and so derivatives in r are 0 and Hamiltonian simplifies to:
1 ∂2 ⎞ h2 ⎛ ∂2 1 ∂ ˆ ⎜ 2+ ⎟ + 2 H =− 2 ⎟ ⎜ 2m ⎝ ∂r r ∂r r ∂φ ⎠ 2 2 2 2 1 h ∂ h ∂ Hˆ = − =− 2 2 2m r ∂φ 2 I ∂φ 2
where I = mr2 is the moment of inertia of the mass m on the ring of radius r. SE is of the (familiar) form:
2 IE d 2ψ = − ψ = −ml2ψ 2 2 dx h
ψ ml (φ ) = Ae imlφ
with ml = ±
2 IE h2
N.B. ml has nothing to do with mass m, it is the angular momentum quantum number.
Particle on a ring solutions Cyclic boundary conditions:
ψ ml (φ ) = ψ ml (φ + 2π )
ψ has to be single-valued!
Ae imlφ = Ae iml (φ + 2π ) = Ae imlφ e iml 2π i.e. e iml 2π = (−1) 2 ml = 1
Therefore ml can only take the values 0, ±1, ±2, ±3,… Again, boundary conditions lead to wavefunctions (eigenfunctions) with a restricted range of values for ml (eigenfunctions) and results in quantised values for the energy E: |mL| = 2
|mL| = 1
Eigenvalues:
ml2 h 2 E= 2I
ml = 0, ± 1, ± 2, ± 3,....
Normalisation of wavefunctions: 2π
2π
∫ψ *ψ dφ = A * A ∫ e 0
mL = 0
Eigenfunctions :
−imlφ imlφ
e
2π
0
ψ ml (φ ) =
∫
dφ = A * A dφ ⇒ A = 0
1 imlφ e 2π
1 2π
ml = 0, ± 1, ± 2, ± 3,....
Properties of the solutions 1. Energy gaps decrease as moment of inertia increases. ΔE → 0 as I →∞ and so recover classical mechanics. 2. Zero point energy is 0 (when ml = 0). Does this contravene Heisenberg?
Probability density: ψ* ψ = (1/2π) i.e. independent of position on ring. Position on ring is completely uncertain and so Heisenberg allows us to know precisely the angular momentum ⇒ zero point energy can be zero! 3 Two wavefunctions with different quantum numbers can have the same energy. For example wavefunctions with ml = 1 and −1 have the same energy, ћ2/2I. This is known as degeneracy. Example: Let us treat the π-electrons in benzene as particles of mass m moving around the circumference of a flat disc of radius r. How well does this simple model reproduce the actual behaviour of the π-electrons in benzene?
Particle in a 3-d box • In 3D the particle momentum is a vector with 3 components, px, py and pz. The kinetic energy operator is therefore 2 2 2 2 ⎛ ⎞ ∂ ∂ ∂ h 1 2 2 2 ⎜ 2 + 2 + 2⎟ Tˆ = pˆ x + pˆ y + pˆ z = − 2m 2m ⎜⎝ ∂x ∂y ∂z ⎟⎠
(
)
• Therefore the SE equation becomes h2 ⎛ ∂2 ∂2 ∂2 ⎞ ⎜ 2 + 2 + 2 ⎟ψ ( x, y , z ) + Vˆ ( x, y, z )ψ ( x, y, z ) = Eψ ( x, y , z ) − 2m ⎜⎝ ∂x ∂y ∂z ⎟⎠
And is often abbreviated to
h2 2 − ∇ ψ + Vˆ ( x, y, z )ψ = Eψ 2m Kinetic energy
Potential energy
∇2 is the Laplacian operator, also known as “del squared”.
Total energy
Separation of variables Once again let’s assume that the potential energy inside the the box is 0. Now we assume that the wavefunction is separable:
ψ ( x, y, z ) = X ( x)Y ( y ) Z ( z )
dY 2 dZ 2 ⎞ h 2 ⎛ dX 2 ⎜ YZ 2 + XZ 2 + XY 2 ⎟ = EXYZ − 2m ⎜⎝ dz ⎟⎠ dx dy
and so SE becomes:
Now divide by XYZ:
h 2 ⎛ 1 dX 2 1 dY 2 1 dZ 2 ⎞ ⎜ ⎟=E − + + 2m ⎜⎝ X dx 2 Y dy 2 Z dz 2 ⎟⎠
First term only depends on x, and must be constant because the RHS does not contain x. (Similarly for the other terms). Now we have three 1d equations to solve…. h 2 1 dX 2 − = Ex 2m X dx 2 h 2 1 dY 2 − = Ey 2m Y dy 2
h 2 dX 2 ⇒ − = Ex X 2m dx 2 h 2 dY 2 ⇒ − = E yY 2m dy 2
h 2 1 dZ 2 − = Ez 2m Z dz 2
h 2 dZ 2 ⇒ − = Ez Z 2m dz 2
where
Ex + E y + Ez = E
The interpretation of Ex is the component of kinetic energy from motion in x direction etc..
Eigenfunctions, -values and degeneracy in a 3-d box Each of the 3 separated equations is a 1d equation for a particle in a box, whose solutions are known. Inserting appropriate boundary conditions (e.g. ψ = 0 when x = 0 and Lx) yields: ⎛ n xπx ⎞ ⎛ n yπy ⎞ ⎛ n zπz ⎞ ⎟ sin ⎜ ⎟⎟ sin ⎜ ⎜ L ⎟⎟ ⎜ ⎟ L L ⎝ x ⎠ ⎝ y ⎠ ⎝ z ⎠
ψ ( x, y, z ) = N sin ⎜⎜
2 h 2 ⎛⎜ n x2 n y n z2 ⎞⎟ E (nx , n y , nz ) = + + 8m ⎜⎝ L2x L2y L2z ⎟⎠
n x , n y , n z = 1,2,....
Eigenfunctions
n x , n y , n z = 1, 2, 3,....
Eigenvalues
For the case of a cubic box, Lx = Ly = Lz = L and so the eigenvalues are:
nx 1 2
ny 1 1
nz 1 1
1 1
2 1
1 2
(
h2 2 2 2 n + n + n E (n x , n y , nz ) = x y z 8mL2 1 level
3 levels
)
n x , n y , n z = 1, 2, 3,....
Degeneracy – different wavefunctions which the same energy.
Reflects symmetry of the box.
