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2. Quantum theory: techniques and applications
2.1.Translational motion
2.1.1 Particle in a box
translation
2.1.2 Tunnelling
2.2. Vibrational motion
vibration
2.2.1 The energy levels
2.2.2 The wavefunctions
2.3. Rotational motion
2.3.1 Rotation in 2 dimensions
2.3.2 Rotation in 3 dimensions
2.3.3 Spin
rotation
The energy in a molecule is stored as molecular
vibration, rotation and translation.
2.1 The translational motion
For a free particle (V=0) travelling in one
dimension, the Schrödinger equation has a
k= Aeikx + Be-ikx
2
2
general solution k, where k is a value      E
2 2
2

k
Ek 
characteristic of the energy (eigenvalue) of  2m  x 
2m
the particle Ek.
For a free particle, all the values of k, i.e. all the energies are possible: there is no quantization
2.1.1 Particle in a box
Particle of mass m is confined in an infinite square
well. Between the walls: V=0 and the solution of the
SE is the same as for a free particle.
k= C sinkx + D coskx
NB: with D= (A+B) C= i(A-B)
A. boundary condition (BC): The difference with
the free particle is that the wavefunction of a confined
particle must satisfy certain constraints, called
boundary conditions, at certain locations.
 BC1: k(0)=0  k (0) = C 0 + D 1=0  D=0
 after BC1: k= C sinkx
 BC2: k(L)=0  k (L) = C sinkL =0
 absurd solution: C=0, it gives k(x)=0 and |k(x)|2=0… the particle is not in the box!
 physical solution: kL= n with n=1,2,… (n0 is also absurd)
The wavefunction n(x) of a particle in an infinite square well is now labeled with “n”
instead of k. Because of the boundary conditions, the particle can only have particular
energies En:
nx
 n ( x)  C sin
; n  1,2,...
L
2
 2 n / L 
n2h2
En 

; n  1,2,...
2
2m
8mL
B. Normalization: Let’s find the value of the constant C such that the wavefunction is
normalized.
L

0
L
L
 nx 
 nx 
2 1
2 L
 n ( x) dx  C 2  sin 2 
x
sin  2
1
 dx  C
 C
2
4n
2
 L 
 L 0
0
L
2
2
C  
 L
1
2
C. Properties of the solutions
 2
 L
1
2
 n ( x)    sin
 The solutions are labeled with n, called
“quantum number”. This is an integer that
specifies the energetic state of the system. In order
to fit into the cavity, n(x) must have specific
wavelength characterized by the quantum number.
With an increase of n, n(x) has a shorter wavelength
(more nodes) and a higher average curvature  the
kinetic energy of the particle increases.
nx
;
L
for 0  x  L
n2h2
En 
; n  1,2,...
8mL2
 The probability density to find the particle at a position x in the box is
 n2 ( x) 
2
 nx 
sin 2 

L
L


More n increases, more uniform is 2n(x): we approach the
case of a ball bouncing between two walls, for which there is
no preferred position between the two walls.
 The classical mechanics emerges from the quantum
mechanics as high quantum numbers are reached.
 The zero-point energy: because n>0, the lowest energy is
not zero but E1=h2/(8mL2).
That follows the Uncertainty Principle: if the location of the
particle is not completely indefinite (in the well), hence the
momentum p cannot be precisely zero and E0.
h2
 The energy level separation E= En 1  En  (2n  1)
increases with n.
2
8mL
E decreases with the size L of the cavity  for a molecule in gas phase free to move in a
laboratory-sized vessel, L is huge and E is negligible: the translational energy of a
molecule in gas phase is not quantized and can be described in classical physics.
2.1.2 Tunnelling
If the energy E of the particle is below a finite barrier
of potential V, the wavefunction of the particle is nonzero inside the barrier and outside the barrier.
 there is certain probability to find the particle
outside the barrier, even though according to classical
mechanics the particle has insufficient energy to
escape: this effect is called “tunnelling”.
X=0
X=L
 Let’s try to estimate the transmission probability of the particle through the barrier.
 For x<0: the wavefunction is that of a free particle: (x<0)= Aeikx + Be-ikx with kħ=(2mE)1/2.
 For 0<x<L: the wavefunction must be solution of the SE for a particle in a constant potential
V.
 2 2



V

  E
2
 2m  x

The general solutions are (0<x<L)= Ceqx + De-qx with qħ=[2m(V-E)]1/2. NB: here, the two
exponentials are real!
 For x>L: V=0, it’s like for a free particle: (x>L)= A’eikx + B’e-ikx with kħ=(2mE)1/2.
 Direction of the transmitted wave is (Left Right):
B’=0 since B’e-ikx is a wave travelling in the (Right  Left) direction.
 The wave function must be continuous at the edges of the barrier (for x=0 and L):
(1) For x=0: (x<0)= (0<x<L)= Aeik0 + Be-ik0 = Ceq0 + De-q0  A+B = C+D
(2) For x=L: (0<x<L)= (x>L)= CeqL + De-qL= A’eikL
 The derivative of the wave function must be also continuous (see 1.4.B, no singular point):
(3) For x=0: [(x<0)/x]0= [(0<x<L)/x]0 = ikA-ikB=qC-qD
(4) For x=L: [(0<x<L)/x]L = [(x>L)/x]L = qCeqL- qDe-qL= ikA’eikL
x
X=0
X=L
2.2 The vibrational motion
Classical mechanics
V
F 
  kx
x
V 
1 2
kx
2
A particle undergoes harmonic motion
if it experiences a restoring force
proportional to its displacement
Quantum mechanics
 2 2
1 2


kx   E

2
2
 2m  x

Eigenvalues:
1

E      ;   0,1,2,...
2

1/ 2
 Energy separation : constant = ħω
k
  
m
 Zero-point energy : E(=0)=½ ħω
 classical limit : for a huge mass m, ω is small and the energy levels form a continuum
A. The form of the wavefunctions
  ( x)  N H ( y )e  y
2
/2
;
1/ 4
 2 

y
and   

 mk 
x
 1 ( x)  N1 2 y e y
  12 ( x) 
4 N12
2
2
/2

x 2 e x
2 N1
x e x
2
/ 2 2
 N0 e x
2
/ 2 2

2
/ 2
N is the normalization constant
 0 ( x)  N 0 e  y
2
/2
  02 ( x)  N 02 e  x
NB: < x >= 0  the oscillator is equally likely to be
found on either side of x=0, like a classical oscillator.
1

 x 2     ( x) x 2   ( x) dx  (  )
2 mk 1/ 2
2
/ 2
B. The virial theorem
In a 1-dimensional problem with a potential V(x)= xn, the expectation values of the
kinetic energy <T> and the potential energy <V> verify the following equality:
2 <T> = n <V> ; with the total energy: <E>= <T> + <V>
 The harmonic oscillator, V=½kx2, is a special case of the virial theorem since n=2
and we have seen that
1/ 2
1
1
k
   (  ) 
2
2
m
1

E    
2

1
1
1
 V  k  x 2  (  ) 
2
2
2
we also know that
 V 
1
E
2
<E>= <T> + <V>
<T> = <V>
C. Quantum behavior of the oscillator
The probability to find an oscillator (in its ground state:
=0) beyond the turning point xtp (the classical limit), is:

P    20 ( x) dx  0.08
xtp
V  Vmax
1
 E  kxtp2
2
Quantum
behavior
Quantum
behavior
Classical
behavior
xtp 0 -xtp
1/ 2

 2E 
xtp   

 k 
0
xtp
Quantum
behavior
xtp
Classical
behavior
 In the harmonic approximation, a diatomic molecule
in the vibration state = 0 has a probability of 8% to be
stretched (and 8% to be compressed) beyond its
classical limit. These tunnelling probabilities are
independent of the force constant and the mass of the
oscillator.
 Classical limit: for huge  (the case of macroscopic
object), P 0
2.3 The rotational motion
2.3.1 Rotation in 2 dimensions
2
2
2
p
J
J
z
z
 Classical mechanics: V  0; T  E 


2m 2mr 2 2 I
The angular momentum Jz = pr
The moment of inertia I= mr2

 In quantum mechanics: not all the values of Jz are permitted,
and therefore the rotational energy is quantized. Where does
this quantization come from?
h
hr
J z   pr and p 
 Jz  

No physical meaning

 The wavelength  of the wavefunction () cannot have
any value. When  increases beyond 2, we must have ()=
(+2), such that the wavefunction is single-valued: |()|2 is
then meaningful.
 The wavelength  should fit to the circumference 2r of
the circle. The allowed wavelengths are = 2r/ml ; where ml
is an integer that is the quantum number for rotation.
m hr
mh
hr
J z     l   l   ml 

2r
2
A. Schrödinger equation for rotation in 2D
Go to cylindrical coordinates:
2  2
2 
 2 

H 
2 
2m   x
y 
x= r cos; y= r sin
2  2 1 
1 2 
2 2
 2 

H 
 2
2 
2m   r
r  r r  
2mr 2   2
because r  0
 2 2 
 ( )  E ( )
 Schrödinger equation: 
2
2
I




 The normalized general solutions:  ( ) 
e iml
2 1 / 2
have to fulfill the cyclic boundary condition ()= (+2):
 (  2 ) 
e iml (  2 )
2 1 / 2
ei 2ml  ei 
2 ml

e iml e i 2ml
2 1 / 2
 e i 2ml  ( )
 cos   i sin  
2 ml
  1
2 ml
1
2ml = an even integer  ml = 0, 1, 2, 3, ...
J z2 ml2  2
 The eigenvalues are given by E 

2I
2I
NB: With ml2, the energy does not depend on the sense of rotation
 ( ) 
e iml
2 1 / 2
 For an increasing ml, the real part of the
wavefunction has more nodes
 the wavelength decreases and consequently, the
momentum of the particle that travels round the
ring increases (de Broglie relation): p=h/
 The probability density to find the particle in 
is a constant: |()|2=1/2
 knowing the angular momentum precisely
eliminates the possibility of specifying the
particle’s location: the operator position and
angular momentum do not commute: uncertainty
principle.
Plots of the real part of
the wavefunction ()
B. The angular momentum operator lz
Correspondence
cylindrical coordinates:
Classical mechanics
x= r cos; y= r sin
principles (chap 11)
ux u y uz
 
 
  

l z   x
y
lz  r  p  x
y
z  xpy  yp x
l z   
i  y
x
i   
px p y pz
Quantum mechanics
ux, uy, uz are
because p z  0, z  0
unitary vectors
What are the eigenfunctions and eigenvalues of lz?
Let’s apply lz to the wavefunctions that are solutions
of the Schrödinger equation:
l z ml 
  
 1
  ml  iml
e iml  ml  ml
1/ 2
i   
i 2 
Vector representation of angular momentum: the magnitude
of the angular momentum is represented by the length of the
vector, and the orientation of the motion in space by the
orientation of the vector
 The solutions of the Schrödinger equation, eigenfunctions of the Hamiltonian operator,
are also eigenfunctions of the angular momentum operator lz: H and lz are commutable: the
energy and the angular momentum can be known simultaneously.
 ml() is an eigenfunction of the angular momentum operator lz and corresponds to an
angular momentum of mlħ.
z
2.3.2 Rotation in 3 dimensions
A particle of mass m free to travel (V=0) over a sphere of radius r.
2  2
2
2 
 2 
H 
 2 
2
2m   x
y
z 
spherical coordinates:
r
x= r sincos; y= r sin sin; z= r cos

2  2
2 
1
 2 
H 
 2 2 
2m   r
r r r

r = 0
x
1
2
1  
 

 is the Legendrian
 

sin

  
sin 2    2 sin    
2
 The Schrödinger equation is :
Since I = mr2, we can write:
1 2
2mE


(

,

)


 ( , )
2
2
r

2   
with

2 IE
2
y
 We consider that (,) can be separated in 2 independent functions: ( , )  ( )( )
 the Hamiltonian can be separated in 2 parts  the SE is divided into 2 equations
1 d 2  sin  d 
d 
2



sin




sin
 + ml2 -ml2
2


 d
 d 
d 
At the moment, ml2 is just
introduced as an arbitrary constant
sin  d 
d 
 sin 
   sin 2   ml2
 d 
d 
1 d 2
 ml2
2
 d
d 2
2


m

l
2
d
Same as for the rotation in 2-D
with
 ml ( ) 
e iml
2 1 / 2
 l ,m
The solutions  should also fulfill the cyclic
boundary condition: ()=(+2); because of
that another quantum number “l” appears and
is linked to ml. Plm(cos ) is a polynomial called
the associated Legendre functions. Nlm is the
normalization constant.
( )  N lm Pl m (cos  )
N lm
iml
m
 ( )( ) 
e
P
(cos  )  Yl ,m
l
1/ 2
(2 )
l = 0, 1, 2, 3,…
|m|l
 The normalized functions lm(,)=Ylm(,) are called spherical Harmonics
The figure represents the amplitude of the
spherical harmonics at different points on the
spherical surface.
Note that the number of nodal lines (where
lm(,)=0) increases as the value of l
increases: a higher angular momentum
implies higher kinetic energy.
2
 From the solution of the SE, the energy is restricted to: E  l (l  1) ; l  0,1,2,...
2I
 The energy is quantized and is independent of ml. Because there are (2l+1) different
wavefunctions (one for each value of ml) that correspond to the same energy, the energetic
level characterized by “l” is called “(2l+1)-fold degenerate”.
Spherical harmonics
ml = 0: a path around the vertical z-axis of the
sphere does not cut through any nodes. For
those functions, the kinetic energy arises from
the motion parallel to the equator because the
curvature is the greatest in that direction.
http://www.sci.gu.edu.au/research/laserP/livejava/spher_harm.html
http://mathworld.wolfram.com/SphericalHarmonic.html
Vector representation of the angular momentum
 The comparison between the classical energy E=J2/2I and
the previous expression for E, shows that the angular
momentum J is quantized and has the values ( length of the
vector):
J={l(l+1)}1/2 ħ ;
l= 0, 1, 2,...
 As for the rotation in 2-D, the z-component Jz is also
quantized, but with the quantum number ml ( orientation of
the vector J):
Jz= ml ħ ;
ml= l, l-1, …, -l
 For a particle having a certain energy (e.g. characterized by
l=2), the plane of rotation can only take a discrete range of
orientations (characterized by one of the 2l+1 values ml)
 The orientation of a rotating body is quantized
Cone representation of the angular momentum
While J2 and Jz commute, Jz and Jx (or Jy) do not commute
 Jz and Jx (or Jy) cannot be measured accurately and simultaneously
 If Jz is known precisely, Jx and Jy are completely unknown: representation with
a cone is more realistic than a simple vector. It means that once the orientation of
the rotation plane is known, Jx and Jy can take any value.
2.3.2 Spin of a particle
The spin “s” of a particle is an angular momentum characterizing the
rotation (the spinning) of the particle around its own axis.
 The wavefunction of the particle has to satisfy specific boundary conditions for this
motion (not the same as for the 3D-rotation). It follows that this spin angular momentum is
characterized by two quantum numbers:
 s (in place of l) > 0 and   the magnitude of the spin angular momentum:
{s(s+1)}1/2ħ
 ms  |s|  the projection of the spin angular momentum on the z-axis: msħ
NB: In this course the spin is introduced as such. But in the Relativistic Quantum Field Theory, the
spin appears naturally from the mathematics.
 Electrons: s = ½  the magnitude of the spin angular
momentum is 0.8666 ħ. The spins may lie in 2s+1= 2 different
orientations (see figure). The orientation for ms= -½, called  and
noted ; the orientation for ms= -½ is called  and noted .
 Photons: s = 1  the angular momentum is 21/2 ħ