lecture 4 2010
TRANSCRIPT
The Quantum Theory of Atoms and Molecules
Particles in boxes and applications
Dr Grant Ritchie
The free particle
ψψ Edxd
m=− 2
22
2h
ikxikx BeAe −+=ψ
What is the appropriate form of the time independent Schrödinger equation?
V
= 0, so that TISE is:
Solutions are of the form: Eigenfunctions (A,B
are constants)
mkE2
22h= Eigenvalues
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.
There is no restriction on k
or allowed energies ⇒ energy is continuous!
Particle in a 1-d box (or infinite square well!)
kxDkxCBeAe
Edxd
mikxikx sincos
2 2
22
+=+=
=−
−ψ
ψψh
What are the wavefunctions
(eigenfunctions) and energies (eigenstates/eigenvalues) of a particle confined in a very deep potential well?
0 L
V(x)+ ∞
V(x)
→∞
+ ∞
V(x)
= 0
V(x)
→∞
Inside the box: V
= 0, so that SE is:
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: L
nk π=
Quantisation of energy
,....3,2,182 2
2222
=== nmL
hnm
kE h
. d d 00 L
DxkxDxLL 21sin* 22 =⇒== ∫∫ ψψ
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:
To completely specify the eigenfunctions
we need to
normalise
them:
Eigenfunctions:
Eigenvalues:
,....2,1sin2)( =⎟⎠⎞
⎜⎝⎛= n
Lxn
Lx πψ
Particle in box properties
Ψ(x)
0x
L
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!
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
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.
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:
mnif
mnifxa
xma
xnL
xxxL
mn
≠=
==⎟⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛= ∫∫
∞
∞
d d 0-
0
1sinsin2)()(* ππψψ
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
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.
0.0 0.2 0.4 0.6 0.8 1.0
1
0
1
x/a0.0 0.2 0.4 0.6 0.8 1.0
-1
0
1
ψ
x/a
n
= 1ψ*ψ
Further comparisons between classical and quantum results
3)(
21)(
2
0
22
00
LxxPxx
LxxL
xxxPx
L
LL
==
===
∫
∫∫
d
d d
⎟⎠⎞
⎜⎝⎛=
Lx
Lx πψ sin2)(
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:
Compare with the quantum case where n
= 1 i.e.
then
2
22
0
222
0
2
23sin2
2sin2
πLLx
Lπxx
Lx
LxLπxx
Lx
L
L
−=⎟⎠⎞
⎜⎝⎛=
=⎟⎠⎞
⎜⎝⎛=
∫
∫
d
d
As n
→∞, find that <x2> → classical result.
Momentum of the particle
ψψ .cos)sin(ˆ conskxDi
kkxDdxd
ipx ≠==
hh
0)(sin
))(cos(sin
)(sin
)(sin)(sin
*
ˆ*ˆ
0
22
0
2
0
22
0
2
====
∫
∫
∫
∫
∫∫
L
L
ik
L
L
dxd
ix
x
xkxD
xkxkxD
xkxD
xkxkxD
x
xpp
d
d
d
d
d
d hh
ψψ
ψψ
Given the wavefunction
ψ
= D sin(kx),
what is the particle’s momentum?
Use the momentum operator….
ψ
is not an eigenfunction
px
(see later)
but we can evaluate the
expectation value (see last lecture).
<px
>
= 0 –
this does not mean that the kinetic energy is 0!
mp
m
pE xx
2ˆ
2
ˆ 22
≠=
Eigenfunctions
of the momentum operator
ψψψψ
pdxd
i
ppx
=
=h
ˆConsider the following eigenvalue
equation:
(where p is a constant)
General solution: ( ) negative.or positive becan NB. kAeAe ikxxip
== hψ
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 <px > = 0.
An application of the particle in a box problem -
the UV absorption spectrum of cyanine dyes
(Taken from www.jce.divched.org
)
3 2 3 2
3 2 3 2
( ) ( ) ( )
( ) ( ) ( )
k
k
CH N CH CH CH N CH
CH N CH CH CH N CH
+
+
− − = − − =
= − − = − −
&&
b
&&
Dye I: k
= 1
Dye II: k
= 2
Dye III: k
= 3
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):
( )2
2 22
8photon f i f ihE h E E E n nmL
ν= = Δ = − = −
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
EU .μ−=
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:
The dipole moment is defined for a collection of charges by: ∑=i
iirqμ
where
ri
is the position vector of charged particle i. ( See electrostatics lectures in Michaelmas
term).
The transition dipole moment
∫= τψμψμ diffi ˆ*
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.
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 thereforeproportional to.
2* ˆ∫ τψμψ djk
Transition dipole moment integral for particle in a box
∫−=L
iffi xxxxe0
* )()( dψψμ
∫
∫
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛ +−⎟⎟
⎠
⎞⎜⎜⎝
⎛ −−=
⎟⎠⎞
⎜⎝⎛
⎟⎟⎠
⎞⎜⎜⎝
⎛−=
Lifif
Lif
fi
xL
xnnL
xnnx
Le
xL
xnxL
xnLe
0
0
)(cos
)(cos
sinsin2
d
d
ππ
ππμ
Need to consider the transition dipole moment integral for one electron. The dipole moment operator for an electron in one dimension is –ex
and so
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?
⎭⎬⎫
⎩⎨⎧
−ΔΔ
+−−−ΔΔ
⎟⎠⎞
⎜⎝⎛−= )sin(1)sin(1)1)(cos(1)1)(cos(1
22
2
πππππ
μ tottot
tottot
fi nn
nn
nn
nn
LLe
(If you have time!)
consider the generalised transition ni
→ nf
:
If we then define Δn
= nf
–
ni
and ntot
= nf
+ ni
then the above integral becomes:
Looks bad!!!!!
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
⎟⎟⎠
⎞⎜⎜⎝
⎛
−=⎟⎟
⎠
⎞⎜⎜⎝
⎛
Δ−−= 2222222 )(
8112
if
fi
totfi nn
nneLnn
eLππ
μ
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.
*Also note that longer molecules have larger absorption coefficients because μfi
increases with the length of the molecule (see uv
spectra earlier in notes).
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.
Using symmetry to evaluate integrals
)()( xx nn −±= ψψ
parity oddeven ; )( sin2
parityeven odd ; )( cos2)(
−−−=⎟⎠⎞
⎜⎝⎛=
−−+=⎟⎠⎞
⎜⎝⎛=
nxL
xnL
nxL
xnL
x
n
nn
ψπ
ψπψ
otherwise )(2/2/for 0)(
∞=≤≤−=
xVLxLxV
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.
02L
−2L
n = 1 n = 2
2L
2L
− 0
* For symmetric potentials V(x)
= V(-x) the ψ
has a definite parity, i.e.
2/' Lxxx −=→
Symmetry II
For transition n
=1 → n
= 2 the transition dipole moment integral is:
( )
dd ∫∫−−
−=⎟⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛−=
2/
2/
2/
2/21 2
cossinL
L
L
L
xxfLex
Lxx
Lx
Le ππμ
Clearly ∫
f (x)dx
≠
0 and the transition is allowed. In contrast, for the transition n
= 1 → n
= 3, ∫
f(x) dx
= 0 and transition is forbidden.
× ×0
2L
−2L
n = 1
2L
2L
− 0
x n = 2
2L
2L
− 0
→2L
2L
− 0
/ 2
/ 2
( ) 0L
L
f x−
≠∫
×0
2L
−2L
n = 1
×2L
2L
− 0
x
2L
2L
− 0
n = 3
→
2L
2L
− 0
/2
/2
( ) 0L
L
f x−
=∫
Conclusion: if integrand is odd / antisymmetric
/ ungerade
then ∫
f(x) dx
= 0 and transition is forbidden.
Particles in “round”
boxes (or on a ring!)
⎟⎟⎠
⎞⎜⎜⎝
⎛
∂∂
+∂∂
+∂∂
−= 2
2
22
22 112
ˆφrrrrm
H h
⎟⎟⎠
⎞⎜⎜⎝
⎛
∂∂
+∂∂
−= 2
2
2
22
2ˆ
yxmH h
2
22
2
2
2
2
21
2ˆ
φφ ∂∂
−=∂∂
−=Irm
H hh
x
y
z
r
mp
Jz
φ
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:
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:
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
222
2
2)(
2
h
h
IEmAe
mIEdxd
lim
m
l
ll
±==
−=−=
withφφψ
ψψψ N.B. ml
has nothing to do with mass m, it is the angular momentum quantum number.
Particle on a ring solutions
,....3,2,1,02
22
±±±== ll mI
mE h
dd d 000 πφφφψψ
ππφφ
π
21***
222
=⇒== ∫∫∫ − AAAeeAA ll imim
,....3,2,1,021)( ±±±== l
imm m e
πl
l
φφψ
Eigenvalues:
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
| = 1
|mL
| = 2
mL
= 0
Normalisation of wavefunctions:
Eigenfunctions :
Cyclic boundary conditions:
ψ has to be single-valued!
1)1(..
)2()(
22
2)2(
=−=
==
+=+
ll
llll
ll
mim
imimimim
mm
eei
eAeAeAeπ
πφπφφ
πφψφψ
Therefore ml
can only take the values 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
( ) ⎟⎟⎠
⎞⎜⎜⎝
⎛
∂∂
+∂∂
+∂∂
−=++= 2
2
2
2
2
22222
2ˆˆˆ
21ˆ
zyxmppp
mT zyx
h
ψψψ EzyxVm
=+∇− ),,(ˆ2
22h
•
In 3D the particle momentum is a vector with 3 components, px
, py
and pz
. The kinetic energy operator is therefore
• Therefore the SE equation becomes
And is often abbreviated to
),,(),,(),,(ˆ),,(2 2
2
2
2
2
22
zyxEzyxzyxVzyxzyxm
ψψψ =+⎟⎟⎠
⎞⎜⎜⎝
⎛
∂∂
+∂∂
+∂∂
−h
Kinetic energy
Potential energy
Total energy
∇2
is the Laplacian
operator, also known as “del squared”.
Separation of variables
)()()(),,( zZyYxXzyx =ψ
EXYZdzdZXY
dydYXZ
dxdXYZ
m=⎟⎟
⎠
⎞⎜⎜⎝
⎛++− 2
2
2
2
2
22
2h
EdzdZ
ZdydY
YdxdX
Xm=⎟⎟
⎠
⎞⎜⎜⎝
⎛++− 2
2
2
2
2
22 1112h
Once again let’s assume that the potential energy inside the the box is 0.
Now we assume that the wavefunction
is separable:
and so SE becomes:
Now divide by XYZ:
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….
ZEdzdZ
mE
dzdZ
Zm
YEdydY
mE
dydY
Ym
XEdxdX
mE
dxdX
Xm
zz
yy
xx
=−⇒=−
=−⇒=−
=−⇒=−
2
22
2
22
2
22
2
22
2
22
2
22
21
2
21
2
21
2
hh
hh
hh
where EEEE zyx =++
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
,....2,1,,sinsinsin),,( =⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛= zyx
z
z
y
y
x
x nnnL
znL
ynL
xnNzyx πππψ
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:
,....3,2,1,,8
),,( 2
2
2
2
2
22
=⎟⎟⎠
⎞⎜⎜⎝
⎛++= zyx
z
z
y
y
x
xzyx nnn
Ln
Ln
Ln
mhnnnE
Eigenfunctions
Eigenvalues
( ) ,....3,2,1,,8
),,( 2222
2
=++= zyxzyxzyx nnnnnnmLhnnnE
For the case of a cubic box, Lx
= Ly
= Lz
= L
and so the eigenvalues
are:
nx ny nz
1 1 12 1 11 2 11 1 2
1 level
3 levels
Degeneracy –
different
wavefunctions
which the same
energy.
Reflects symmetry of the box.