Time independent Schr

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3. Schrödingers Wave Equation
-the replacement for Newtons Laws at the quantum
scale
It cannot be derived.. But it was a great guess!
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We have seen how to make a wavepacket that
represents a particle. Now we need some tools
to explore the mechanics of that wavepacket.
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Reminder about wave equations
T
e.g. Waves on a string Tension T, mass per unit
length ?.
y
T
xA
xB
x
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Solutions of wave equation
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Operators and Observables
An operator, , operates on a
wavefunction, , to produce an observable
, as well as returning the wavefunction,
unchanged.
Momentum operator
Total energy operator
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Schrödingers wave equation
We have seen how to make a wave-packet out of
plane waves like
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Schrödingers equation contd..
Momentum operator
Total energy operator
Now the total energy of a particle is just the
sum of the kinetic energy and the potential
energy E p2/2m V
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Things to note re Schrödingers equation

• The equation is linear in R, that is there are no
  terms like R 2 or (MR /Mx)2. 
• The equation is homogeneous, that is there are no
  terms independent of R.
• Taken together these features mean that if R is a
  solution to the equation, then so is cR, where c
  is any complex number.
• This implies that any linear combination of
  solutions is also a solution

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Differences between Schrödinger and classical
wave equations
K.E. P.E. Total Energy
Eqn. derived for total energy Eqn.
derived from force First derivative wrt time
Second derivative wrt time Complex (note i on
RHS) Real equation implies y
must be complex y is a displacement ? real
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Differences between Schrödinger and classical
wave equations
Reminder about complex numbers CAiB with i.i
-1. Complex conjugate CA-iB. Square modulus
C2 CC C2 (A-iB)(AiB) A2 B2.
Modulus C ?A2 B2
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Interpretation of the wavefunction
What does R (x, t) tell us?
c.f. Youngs slit experiment high probability of
detecting particle at bright fringes, low
probability at dark fringes for light expect
probability of detecting a photon to be
proportional to the intensity of E.M. wave.
phase of R(x, t) cannot be important not
observable. Guess that the probability of finding
a particle in the range x to x dx at time t is
proportional to R (x,t)2 dx Define P(x,t)
dx. ? R (x,t)2 dx Need to normalise
probabilities
(one particle!)
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Normalisation of wavefunction
P(x,t) dx. ? R (x,t)2 dx
To get rid of proportional sign we need a
constant of proportionality P(x,t) dx. A R
(x,t)2 dx. Eliminate A by using the
normalisation
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Time independent Schrödinger equation
Time-dependent, Schrödinger equation
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T.I.S.E. contd
The only way a function of x a function of t,
for all x,t is that each function is equal to a
constant. Call the constant E (we will show
later this is the total energy).
This is called the time-independent Schrödinger
equation
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Solving the Schrödinger equation
What do we want? Usually, the allowed energy
levels. First Assume a solution - eg
u(x)Asin(kx) (TISE) Second Substitute
solution into Schrödinger equation.
That gives relationship between E (and V) and
k Third Use boundary conditions (eg u and
du/dx continuous) to solve for allowed
values of k (eg in terms of well size,
a) Fourth Use earlier relationship between E
(and V) and k to obtain allowed
values of E
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Simple example free particle, V(x)0
We can make a wavepacket representing a free
particle by adding together plane waves of the
type
In this case if and
E T
E 2k2 /2m
This confirms that the constant E is the total
energy
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Example of solution of T.I.S.EParticle in a
box infinite potential well
Box in one dimension with walls at a and a
for x ? a,
for x gt a
For x ? a, T.I.S.E becomes
Boundary condition u(x) must vanish for x gt
a, since otherwise we would have an infinite
potential energy!
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Types of solution (I)
A possible solution is u(x)Asin(kx) Check
Insert into T.I.S.E.
Boundary condition u(x) 0 for x gt
a implies sin(kx) 0 for x a. True if
kam? (m integer)
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Types of solution (II)
Another type of solution is u(x)Bcos(kx) Check
Insert into T.I.S.E.
Boundary condition u(x) 0 for x gt a
implies cos(kx) 0 for x a. True if
kam? /2 (m odd integer)
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Summary of solutions
u(x)Asin(kx)
u(x)Bcos(kx)
kam? (m integer) kan? /2 (n odd
integer) Equivalently kan?/2 (n even integer)
n 2, 4, 6, 8,..
n 1, 3, 5, 7,..
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Energy levels for a particle in a box
n 1,2,3,4,5,6,7,8,. ?
Solutions for energy are called energy eigenvalues
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Wavefunctions for particle in a box
n 1, 3, 5, 7,..
u(x)Bcos(n? x/2a)
n 2, 4, 6, 8,..
u(x)Asin(n? x/2a )
Solutions for wavefunctions are called
eigenfunctions
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Normalisation
Values of A, B ?
Normalisation condition
In the present case we only need to integrate
between a and a since u(x) vanishes outside
this range
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Probability density
For first four eigenfunctions for particle in a
box
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Zero point energy
The lowest energy state for a particle in a box is
Why cant the energy be zero?
Remember Heisenberg uncertainty relation
Particle is confined in box, so ?x a.
Since momentum cannot be zero, minimum energy
must be of order
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Consequences of Zero Point Energy - 1
Helium is the only element which remains liquid
at absolute zero (unless under pressure). This is
due to its large vibrational zero point energy.
N.B. Zero point energy ? (1/mass) of
particle. Molar volume of the isotope 3He in the
liquid phase is larger than that of liquid 4He
since the light isotope has a larger zero point
energy.
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Consequences of Zero Point Energy - 2
Conduction electrons in a metal have very
large zero point motion, with typical electron
velocities of order 106 ms-1. When we pass
a current through a metal we are imposing a very
small drift velocity on top of this random motion.
The zero point energy of a gas of fermions
gives rise to degeneracy pressure. Both
electrons and neutrons are fermions. Degeneracy
pressure contributes to (a) the bulk modulus
of a metal - i.e. the resistance of a metal to
being squashed (b) the stability of white dwarf
and neutron stars.
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Consequences of Zero Point Energy 3Degeneracy
pressure in neutron stars and white dwarves
Pressure, p 1/3 nmc2 2/3n x 1/2mc2
i.e. p E But E a-2 Therefore p a-2
i.e. V-2/3
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Spacing of energy levels

• N2 molecule in a box of size 1cm cubed
• Electron in a box of size 1nm cubed
• c.f. Thermal energy at room temperature
  0.025eV

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Quantum dots and nanocrystals
Small semiconductor structures can be built to
confine electrons into boxes of size 10 100nm.
These behave like artificial atoms, with discrete
energy levels. They have interesting potential
for electro-optical devices and computer memory.
Two samples of powdered CdSe, a semi-conductor,
differing only in the particle size. Upper part
larger crystals, lower part smaller crystals.
The colour difference is due to the difference in
spacing of the energy levels of electrons trapped
in the nanocrystals
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Temporary Lecture Changes

• Additional single lectures
• Tuesday 12th Feb, 1300-1400, Shackleton
  (B44)
• LTA (ie room 1041)
• Wednesay 13th Feb, 1000-1100, Music (B2),
• LTH (ie room 2065)
• No lecture on Thursday 14th Feb
• or in week Monday 18th - Friday 22nd Feb.
• Back to normal on Monday 25th Feb.

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Differences between Schrödinger and classical
wave equations- Complex exponential or
sine/cosine form?
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Classical wave equation
Here either form works well as y is real, eg
So
i.e.
, thus
We get exactly the same result with
sines/cosines. (Generally, we can use complex
form and take real part of the answer makes
treatment of phase easier.)
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Schrödinger wave equation
Complex equation Note i on RHS, so time
dependence of y must be complex, MUST use
complex notation