Schrdingers wave equation

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Schrödingers 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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The wave function
Wave properties Particle properties
Momentum p k
Energy E T
We can extract the momentum from the
wavefunction by the operation
Similarly we can extract the energy by
operating with
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Schrödingers wave equation
total energy kinetic energy potential
energy E p2/2m V
In operator form Rearranging we get
Schrödingers equation (one dimensional version)
Note complex equation so needs complex wave
function
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Interpretation of the wavefunction
Probability of finding a particle in range
x to x dx at time t is
P(x,t) dx. ? R (x,t)2 dx Normalise
probabilities
(one particle)
i.e.
Single valued - Y2 is an observable
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Time independent Schrödinger equation
Separation of variables Insert into
Schrödinger equation
Divide through by
constant, E
LHS RHS Only depends on x Only depends
on t
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T.I.S.E. contd
RHS becomes
Integrating LHS becomes
Time-independent Schrödinger equation
Solution Probability density
is independent of time
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Symmetry and Parity
When the potential has inversion symmetry,
ie V (-x) V (x),
the
eigenfunctions always have a definite parity
If wavefunction has
even parity, i.e. a parity of 1 (n odd,
in infinite square well) If
wavefunction has odd
parity, i.e. a parity of -1 (n even)
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Boundary conditions

• The wavefunction u(x) must be continuous at the
  boundary
• Mu(x)/Mx must be continuous at the boundary

II
I
N.B. If there is an infinite step the second BC
does not apply
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Particle in a FINITE potential well
V(x) 0 for x 0, region I  V(x) V0
for x gt 0, region II
EltV0 . Region I (inside), TISE
Solutions and
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Outside the well (region II)
This bracket is positive since EltV0
Solutions or with
To avoid u(x) diverging for x? ? we use
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Inside and outside solutions together
Potential is symmetric under inversion, so
eigenfunctions have a definite parity
We now have to match the wavefunctions at the
boundaries
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Even parity solutions
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Solving equation
where
k0
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Odd parity gives
y cot(ka)
y
k0
Using same notation as before for l this becomes
k
p/a
2p/a
3p/a
Solve graphically by plotting
y -(k02 k2)1/2/k
and
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Eigenfunctions for particle in finite well
k0 2.6 (B/a)
Eigenfunctions for particle in finite well. Note
that these have alternately even and odd
parity. Note amplitude outside well (xgta)
increases in states near top of well, ie higher
energies
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Limit of shallow well
V0 ? 0. This is equivalent to k0 ? 0. Always at
least one even parity solution
y (k02 k2)1/2/k
ka/?
Always at least one bound state for the 1-D
finite well N.B. not true in 3-D!
Graphical solution for k0 0.6? /a
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Harmonic Oscillator
Restoring force is proportional to the
displacement. E.g. simple pendulum, mass on a
spring Mechanical system with small
displacement from equilibrium

Taylor expansion of potential
(Remember force is )

Therefore
Spring constant
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Harmonic Oscillator Classical solution
From Newton F Try solutions of type
or
This is the natural frequency of the oscillator
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Harmonic oscillator in quantum mechanics
Potential is quadratic So TISE We will see
that the energy is measured in multiples of Take
out as a factor Now make a
substitution TISE becomes
with
P32..
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Trial solutions
Try solutions of the type Then H(y) must satisfy
P33..
Some trivial solutions (i) H(y) A (constant)
O.K. if i.e with eigenfunction (ii) H(y)
By O.K. if i.e. with
eigenfunction
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Trial solutions contd.
Solutions have alternately even and odd parity
(expected since V(x) is symmetric), H(y) is a
polynomial in either even or odd powers of
y. Each successive solution has extra term in
polynomial. Energy may be written as with n 0,
1, 2, 3, .... ,
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General solution
H(y) is a polynomial of order n (p goes from 0
to n) Substituting into differential eqn. for
H(y) gives i.e. If polynomial is of order n,
coefficient an2 must vanish. From the recursion
relation an2 0 if That is or with n
0, 1, 2, 3, .... ,
Recursion relation
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Summary of general solution
9 8 7 6 5 4 3 2 1 0 n
Infinite ladder of energy levels with a constant
spacing. Zero point energy of the oscillator is
½ T Eigenfunctions where Hn (y) are the
Hermite polynomials. Examples H0 (y) 1 H1
(y) 2 y H2 (y) 4 y 2 - 2 H3 (y) 8 y 3 -
12 y H4 (y) 16 y 4 - 48 y 2 12 etc
T
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Form of eigenfunctions
First six eigen-functions for the quantum
harmonic oscillator. Note alternate even and odd
parity.
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Form of probability density
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Scattering from potential steps, etc
(i) E gt V0
I
II
with and
From TISE
Boundary conditions
(i) (ii)
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Form of solution
Imaginary part
Real part
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Probability density
Red curve
Blue curve
Partial standing wave due to interference between
incident and reflected wave.
Uniform density for transmitted wave
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Reflection and transmission coefficients
(use k or v as )
Flux of particles u 2 H k
R T 1
cf classical case
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Second case, when E lt V0
(ii) E lt V0
with and
Boundary conditions
(i) (ii)
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Reflection coefficient
For E lt V0 particle is totally reflected
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In this case both wavefunctions are real
Exponential decay
Standing wave
e.g. for E/V00.25
x
x
Wavefunction
Probability density
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Total reflection of ultra-cold neutrons
When neutrons enter matter they experience an
average potential V0 , where
scattering length number density neutron mass
Energy of neutron is
If E V0,
the neutron will be totally reflected from
surface.
i.e.
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Particles incident on a barrier, E lt V0
I
II
III
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Boundary conditions at xa
So we have four simultaneous equations and five
coefficients, A, B, C, D, F. We can therefore
calculate the ratios F/A and B/A allowing us to
work out the transmission and reflection
coefficients.
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Solving equations
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Transmission through barrier
Simplifying (
)
Transmission coefficient
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Quantum tunnelling, I Alpha Decay
repulsive
Many unstable nuclei decay by emitting alpha
particles. Particles in states with Egt0 but
EltV0 may tunnel through the potential barrier.
attractive
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Quantum tunnelling, II Nuclear Fusion
The source of energy in the Sun is nuclear
fusion. Two protons collide
to form a deuteron. Barrier is 1 MeV, but thermal
energies of protons is 1keV ( 107K ).
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Quantum tunnelling, III Cold Electron Emission
Conduction electrons in metal tunnel through
barrier formed by work function and external
electrostatic potential. Apply several hundred
volts to a very fine needle point.
Exponential dependence of current on 1/V
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Scanning tunnelling microscope
Tip is scanned across surface, height adjusted so
that tunnelling current is kept constant.
Exponential dependence of tunnelling current on
distance from surface gives extremely good
resolution 0.01nm, sufficient to reveal
individual atoms.
Surface reconstruction of (111) surface of
silicon. Peaks are silicon atoms
(Nobel Prize, 1986, Binnig and Rohrer)