3/PH/SB Quantum Theory - Week 1 - Dr. P.A. Mulheran
OUTLINE OF UNIT
 Previous Courses: Schrodinger’s Wave Equation and his solutions for some
problems
 This course: generalise to describe any situation
 Interpretations: assigning reality to the mathematics
 Experimental tests of measurement paradoxes
 Contrasts to classical physics: restrictions on locality and determinism
Essay Assignment (20% of unit's assessment):
You will be set the following essay during a closed-book departmental examination
towards the end of this unit. Only one of the three options will be set, so you should
prepare for all three cases. You will not be allowed to bring any books or notes into
the exam room!
“Discuss the merits and disadvantages of the Copenhagen Interpretation and one of
the following schemes, indicating which of these you prefer and why:
Many Worlds Interpretation; The de Broglie-Bohm Theory; The Transactional
Interpretation."
Your essay will be marked on the following criteria:
Scholarship; Accuracy; Structure and Style; Originality
Recommended Reading:
“Quantum Mechanics” by Alastair I.M. Rae (IOP): This textbook is recommended for
this whole course and the following one in Part IV.
“Quantum Physics: Illusion or Reality?” by Alastair I.M. Rae (Canto): Ideal reading
for this essay assignment.
“In Search of Schrodinger’s Cat” by John Gribbin (Black Swan): An old favourite.
“Schrodinger’s Kittens” by John Gribbin (Phoenix): Wordy but nice.
“Speakable and Unspeakable in Quantum Mechanics” by J.S. Bell (Cambridge):
Some of these research publications are very readable, notably chapters 16 and 20.
Read about Bell’s Inequality from the master!
Other Assessments:
Continuous assessment of workshop assignments carries 20% of the unit marks and
the formal examination next summer carries 60% of the marks.
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3/PH/SB Quantum Theory - Week 1 - Dr. P.A. Mulheran
REVIEW OF WAVE MECHANICS
1.2 Schrodinger’s Equations

Time Dependent Schrodinger Equation:
 2 2

  ( r , t )  V (r ) ( r , t )  i  ( r , t )
2
t
(r,t) is the wave function of the particle; it is not a physical quantity but a
mathematical one, there is nothing vibrating in space!
 = (reduced) mass, V(r) = potential energy of particle at position r.

For a ‘closed’ system with definite energy E   ,
 (r , t )  u (r ) exp(it ) ,
where u(r) is the spatial part of the wave function.

Time Independent Schrodinger Equation:
 2 2
 u(r )  V (r ) u(r )  E . u(r )
2

Interpretation of wave functions
The probability of finding the particle in the volume d around r is
P(r )d   * d
| u(r )|2 d
where the last relation holds for systems with definite energy E.
It has been assumed that the wave functions are normalised, that is all the
probabilities add up to unity:
2
 | u(r )| d  1 .
allspace

Boundary Conditions
Wave function u(r) is continuous (single-valued) so that probability is uniquely
defined.
The first derivatives u/x (etc.) are continuous where the potential V is finite,
otherwise the total energy E diverges which is not physical.
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3/PH/SB Quantum Theory - Week 1 - Dr. P.A. Mulheran
1.3 Particle in a box

The Infinite Square Well: One dimension
The particle is confined to a region of space 0 < x < L by walls of infinite
potential, so that u(0) = u(L) = 0.
The solutions are standing waves in the well:
un ( x)  2 L sin(n x L) , n  1,2,3,...
and the particle’s energy is quantised into discrete levels
En 
 2 n2 2
.
2 L2

The infinite square well in higher dimensions
Since the potential energy can be written as a sum of one-dimensional functions,
the TISE is separable and the wave function can be written as the product of one
dimensional functions. In this case
n x 2
n y 2
n z
un1n2 n3 ( x , y , z)  2 L sin( 1
)
sin( 2
)
sin( 3 )
L
L
1
2
3
L1
L2
L3

Wells with finite walls
The wave function oscillates in space when the total energy E > V(r), the local
potential energy. However when E < V(r) solutions of the TISE require the wave
function to decay or grow exponentially. Clearly if the particle is to remain
bound inside its well, its wave function must only decay into the finite potential
walls. Because the wave function and its first derivative are continuous here,
only certain values of the total energy E produce these type of solutions; the
energy levels are thus also quantised in these type of wells.
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3/PH/SB Quantum Theory - Week 1 - Dr. P.A. Mulheran
1.4 The Simple Harmonic Oscillator

Examples: atoms bound in molecules and solids

The potential energy of the particle (in one dimension) as it is displaced a
distance x from its equilibrium position is
V ( x) 
1 2 1
Kx   2 x 2 ,
2
2
where the (angular) frequency of the oscillator solved using classical mechanics
is


K

, K being the spring constant and  the reduced mass.
Schrodinger’s solution is that the energy levels of the bound states are
1
E n  ( n  ) , n  0,1,2 ,3.....
2
This solution explains the heat capacitance of solids and the vibrations of
molecules.

The ground state solution, n=0, does not have zero energy. The particle can
never be stationary since this requires infinite kinetic energy, and instead the
particle is said to have zero-point motion.

The corresponding wave functions for this system are
un ( x)  Pn ( x).exp( x 2 ) ,
where the Pn(x) is a polynomial of order n (scaled versions of the Hermite
polynomials)
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3/PH/SB Quantum Theory - Week 1 - Dr. P.A. Mulheran
1.5 Hydrogen and Hydrogen-like Atoms

Schrodinger also solved his equation for Hydrogen-like atoms, where a single
electron moves in the electrostatic potential of the nucleus
V (r ) 
 Ze 2
,
4 0 r
where Ze is the charge on the nucleus. From our previous work we know that
only certain energies will provide bound state solutions.

The TISE is written in spherical polar co-ordinates to utilise the symmetry of the
atom:
 1 ˆ2

1 ˆ2
L  V ( r ) u ( r,  ,  )  E u( r,  ,  ) .
 2 Pr 
2
2r


2
Here Pˆr is the operator for the square of the radial momentum (it involves
differentiation with respect to r but not  or ) and L̂2 is the operator for the
square of the total angular momentum.

This equation is separable, so that the solution can be written as a product of
three one-dimensional functions. The equation involving the angular momentum
operator has the spherical harmonics as its solution:
Ylm ( ,  )  Clm P |ml| (cos( )) exp(im ) .
Here Clm is a normalisation constant and Pl m is the associated Legendre
function. l and m are quantum numbers that label the possible solutions;
l  0,1,2,3,...
m  l ,l  1,...l  1, l .

The spherical harmonics are eigenfunctions of the L̂2 operator:
L2 Ylm ( ,  )  l (l  1) 2 Ylm ( ,  )
Thus the square of the total orbital angular momentum of the electron in the
atom is quantised. This results solely from the spherical symmetry of the system,
and so all spherically symmetric systems have quantised angular momentum.

Note the simple  dependence of the spherical harmonics; this is related to the
quantisation of the z-component of angular momentum,
lz  m , m  0,1,2..,  l  m  l .
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3/PH/SB Quantum Theory - Week 1 - Dr. P.A. Mulheran

Writing the wave function as a product of a spherical harmonic and a function of
r, u(r, ,) = R(r).Ylm(,), substituting into the TISE and simplifying we find
  2 d 2
 2 l (l  1) 
 2 . dr 2  V ( r )  2 r 2   ( r )  E .  (r ) ,


where (r) = r.R(r) is an effective radial wave function. This has the form of a
simple one dimensional problem with a particle moving in a potential well
defined by the sum of the Coulomb potential plus a centrifugal term due to the
angular momentum when l > 0.

The solutions to this equation require one extra quantum number n = l+1, l+2,…
The fact that n > l is the only effect the centrifugal potential has on the energy of
the solutions is remarkable, and is only true for the 1/r Coulomb potential. In
other atoms where the nucleus is partially screened by the inner electrons, the
potential seen by the outer electron is not of this simple form and the angular
momentum does affect the allowed energy levels.

The energy levels found by solving this equation with the requirement that the
wave function (r) decays to zero exponentially as r increases are as in Bohr’s
famous formula:
En 
 Z 2 e 4
.
2 ( 4 0 ) 2  2 n 2
The corresponding radial wave functions are of the form
 nl (r )  Fnl (r ) exp( Zr na ) ,
0
where Fnl(r) is a polynomial of order n with (n-l-1) nodes, and a0 is the Bohr
radius:
4 0  2
a0 
.
 e2
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3/PH/SB Quantum Theory - Week 1 - Dr. P.A. Mulheran
WORKSHOP QUESTIONS
Hand your solutions to the following questions to Dr. Mulheran at the start of the first
workshop in week 2. Some of your solutions will be marked as part of the continuous
assessment of this unit which contributes 20% of the overall module grade. Your
solutions must be well presented; untidy work will be penalised.
1.1
(a) Consider the ground state solution of a particle in a one dimensional
infinite square well. Write the time-dependent wave function as a sum of two
travelling waves inside the well; how does this form of the wave function
compare to the classical description of the particle inside the well?
[3 marks]
(b) Now estimate the uncertainty in the particle’s momentum. Is it reasonable
to assume that the position-momentum uncertainty principle is obeyed in this
system?
[2 marks]
1.2
(a) A bead of mass  can slide freely around a circular wire hoop of radius a.
To describe this system in quantum mechanics we must write the TISE in the
appropriate cylindrical co-ordinates and solve it using physical boundary
conditions. The Lapacian operator in cylindrical co-ordinates is given by
1     1 2
2
 

.
r  
r r  r  r 2  2 z 2
2
Now the bead cannot change its distance r from the origin, and has no zdependence, so only the middle term is nonzero in the TISE:
  2 d 2 u( )
 E. u( ) .
2  a 2 d 2
Solve this equation to find the functional form of the wave functions u().
[2 marks]
(b) Impose the boundary condition that the wave function must be singlevalued, i.e. that u(+2) = u(). Hence derive the energy levels for this
system.
[3 marks]
1.3
A particle of mass  moving in one dimension is attached by a spring (with
spring constant K) to a wall at x=0.
(a) What are the energy levels of this system?
[2 marks]
Hint: this situation is similar to the simple harmonic oscillator except that the
wave function must be zero at the wall.
(b) If the above particle is now free to move in three dimensions with x>0
derive its energy spectrum.
[3 marks]
Hint: use the fact that the TISE is separable into the three co-ordinates x, y
and z, so that the wave function may be written as a product of three one
dimensional waves which themselves are solutions of 1D harmonic
oscillators.
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3/PH/SB Quantum Theory - Week 1 - Dr. P.A. Mulheran
1.4
Show that the wave function

Zr 
 exp(  Zr
 ( r )  C r 1 
)
2a 0
2a 
0 

satisfies the TISE for the radial part of the electron’s wave function in a
Hydrogen-like atom (as given in the lecture notes), where the quantum
numbers are n=2 and l=0.
[5 marks]
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3/PH/SB Quantum Theory - Week 1 - Dr. P.A. Mulheran
WORKSHOP SOLUTIONS
1.1
 (x , t ) 

2
sin( kx ).exp(i t )
L
i
 exp[i (kx  t  ]  exp[i ( kx  t )])
2L
where k=/L and w = E1/. This linear superposition of travelling waves
represents the classical situation where the particle bounces back-and-forth
inside the well without losing energy. The momentum of the particle is  k, so
the uncertainty delta p ~ k = /L. The uncertainty in the position delta x ~
L/2, so that the product ~ /2 > /2 as required by HUP.
1.2
Need functional form that when differentiated twice we get negative constants
times original form. The complex exponential satisfies this (as does the
sinusoidal), so
u( )  C.exp(im  ) .
Now u(+2) = u() means exp(im2)=1, so that m must be an integer. Finally
substituting this solution into the TISE we have
Em 
 2 m2
, m = 0, +/-1, +/-2,... for the allowed energy levels of the
2 a 2
system.
1.3
(a) From the 1D simple harmonic solutions, the first, third, fifth…etc excited
state wave functions are zero at the origin. Thus the x>0 parts are
eigenfunctions of the given problem, so the energy levels are
1

E n1    n1   , n1  1,3,5...
2

(b)


H ( x , y , z ) u ( x , y , z )  h1 ( x )  h2 ( y )  h3 ( z ) u ( x , y , z )  E . u ( x , y , z ) ,
where
 2  2
1 2

h1 ( x ) 

Kx , etc.
2  x 2 2
Putting u = u1(x).u2(y).u3(z) into the TISE and dividing through by u we find
1 
1 
1 
h1 u1 ( x ) 
h2 u 2 ( y ) 
h3 u 3 (z )  E .
u1 (x )
u2 ( y)
u3 (z )
Since each variable can vary independently, each group on the left must itself
be equal to a constant, say E1, E2 and E3 respectively. Then
h1 u1 (x )  E1 . u1 (x ) , etc., which is a one dimensional SH problem whose
solutions we already know. Thus in total
E  E1  E 2  E 3 
.
(n1  n 2  n 3  3 / 2) , n1  1,3,5..., n 2  0,1,2,3..., n 3  0,1,2,3...
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3/PH/SB Quantum Theory - Week 1 - Dr. P.A. Mulheran
1.4
  2 d 2
2l (l  1) 
.

V
(
r
)


  (r )  En .  (r )
2
2 r 2 
 2  dr
with n=2 and l=0.
4 0  2
Substituting in, and using a 0 
, the LHS of the TISE becomes just
 e2
 Z 2 e 4
 which equals the RHS for E2.
8( 4 0 ) 2  2
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