Transcript Quantum Mechanics

Quantum Mechanics
1
Birthday
of
Quantum Physics
on
14th December, 1900
On this date German physicist Max
Planck first presented his new
quantum concepts.
Planck introduces a new
fundamental constant
Max Karl Ernst Ludwig Planck
1858-1947
h
to explain black-body radiation
E  h
Classical concept
Two distinct categories :
1. Material body (particle)
Newton’s laws of motion
Position and velocity (momentum) are
precisely measurable
Spread over the space, amplitude gives
2. Electromagnetic field (wave)
energy/intensity, frequency is nothing
Maxwell’s equation
but time periodicity of oscillator
Additionally
Laws of thermodynamics
E = kT
Fundamental constants :
1. velocity of light c
2. Avogadro Number N
3. Boltzman constant k
4. Unit of charge e
E = mc2
Velocity << c : non-relativistic
Velocity comparable to c : relativistic
Classical concepts never allow to think that
1. Wave may also behave like particle.
(Planck’s hypothesis)
2. Particle may behave like wave.
(de Broglie hypothesis)
3. Position and momentum of a particle cannot
be measured accurately simultaneously.
(Heisenberg uncertainty principle)
4. Energy of wave is related with frequency
and quantised.
n
E  h
These new concepts are basically quantum concepts
BLACKBODY RADIATION
The mother of Quantum Physics
An ideal black body is an empty cavity whose walls
are maintained at a given temperature T (at
thermodynamic equilibrium)
BLACKBODY RADIATION
The mother of Quantum Physics
According to classical
electrodynamics such a cavity has
infinite number of normal modes.
There is no upper limit to the
frequency of these normal modes
Blackbody Radiation
Classical thermodynamics predicts that at an
equilibrium temperature T, the average amount of
energy in a given normal mode is fixed = kT
Energy density
in the window
υ and υ +d υ
ρ(υ)
3
J/m -Hz
k=:Boltzmann’s const.
= 1.38 x 10
-23
T=1800 K
T=1400 K
T=1000 K
frequency υ
J/K
As there is no upper limit to the frequency of these
normal modes, there would be infinite number of
normal modes, and thus infinite amount of energy
residing in the EM radiation within the cavity at a
finite temperature.
Ultraviolet catastrophe
Classical theory of Rayleigh-Jeans law
The amount of energy per unit volume per unit frequency interval (Spectral
density) should keep on increasing with frequency as 2. (Obeyed well at low
frequencies)
T=1800 K
ρ(υ)
T=1800 K
T=1400 K
T=1000 K
frequency υ
ρ(υ)
T=1800 K
υ dependence
2
frequency υ
Rayleigh-Jeans law
ρ(υ)
Planck’s law
frequency υ
EM radiation in black body is quantum field.
The average energy in a normal mode is of the order of kT.
The probability of emission of high energy photons with
high frequency gets smaller for a given temperature when
h >> kT. There is not enough energy available for
emitting these photons.
Planck’s Radiation Law
The density of radiant energy in the cavity per unit
wavelength interval, at the wavelength , and at the
temperature T
E ( , T ) 
8 hc

5
1

e
hc
 kT
1
Planck’s Formula
Rayleigh-Jeans law
Planck’s postulate
Any physical entity with one degree of freedom
and whose ``co-ordinate” is oscillating
sinusoidally with frequency  can possess only
total energies E as integral multiple of h.
h = Planck’s constant
4
Classical
2
E=0
Waves behaving as particles
Experiments
1. Photoelectric effect (1902)
2. Compton effect (1922)
3. Pair Production
particle wave duality
de Broglie postulate
Particles behaving as waves
Experiments
Electron diffraction
Davisson –Germer (USA)
and Thompson (UK) (1927)
Electron microscope
Photoelectric effect
Photoelectric effect
h  k .e  
h  k .e  
Lenard 1902: Studied energy of the
photoelectrons with intensity of light.
He could increase the intensity
thousand fold.
1. Noticed a well defined minimum
voltage Vstop to stop the current in
the circuit. Vstop was independent of
the intensity of light.
Current vs Voltage
Different intensities: Ib > Ia
Cut-off voltage
2. Increasing the intensity of light
would increase the current
3. He performed the experiment with
various coloured lights and found the
maximum energy of the electrons did
depend on the frequency of light.
Qualitatively he obtained more the
frequency more the energy.
Objections with wave theory
1. Kinetic energy (K) of the
photo-electron should increase
with intensity of the beam.
But Kmax was found to be
independent of the intensity
of the falling light.
2. Effect should occur for any
frequency of light provided
only that light is intense enough
to eject the electron.
But a cut-off frequency
was
observed below which photoelectrons
were not ejected (no matter how
intense was beam).
3. Energy in the classical theory
is uniformly distributed over the
wave front. If light is feeble,
there should be a time lag between
the light striking the plate and
ejection of photoelectrons.
-9
Ejection is instant, t < 10 sec
Einstein equation
Light is wave : Interference, Diffraction, polarisation
Light is a stream of photons/wave packets (particles)
So wave behaves like particle
Compton effect
Collision between photon and electron
Compton Effect
in 1920
Arthur Holly Compton
(1892-1962)
Partial transfer of photon energy
m = m(v) is the relativistic mass
Conservation of momentum
along initial photon direction
Conservation of momentum along
perpendicular to initial photon direction
Square and add the above two expressions
Energy conservation
Compton shift
~
Compton wavelength
= 2.4 pm
Compton
Experiment
Pair production
(Energy into matter)
Nucleus
The rest mass of e- or e+ is 0.51 MeV.
Hence pair production requires a photon
energy of at least 1.02 MeV.
Corresponding max. photon wavelength is
1.2 pm (Gamma rays)
The linear momentum is conserved with
the help of the nucleus
In empty space momentum and energy
cannot be simultaneously conserved
θ
θ
Pair annihilation is inverse of Pair production
Compton scattering
Confirmation of discrete
energy levels in atom
Franck and Hertz experiment (1914)
(James Franck and Gustav Hertz)
Vo
Peaks at 4.9 V and its multiples
300
200
I in
mA
100
5
10
15
Accelerating Voltage in Volts
Continuum
E=0
2nd excited
state
1st excited
state
6.7 eV
Ground
state
4.9 eV
E= -10.4eV
Particles behaving as waves
Experiments
Electron diffraction
Davisson –Germer (USA)
and Thompson (UK) (1927)
Diffraction?
Apply Bragg’s law
From X-ray diffraction
de Broglie wavelength of electron
h

p
p  2mE
KE=54 eV is non relativistic
h

p
p  2mE
Bohr’s Complementarity Principle
“In a situation where the wave aspect of a system is
revealed, its particle aspect is concealed; and,
in a situation where the particle aspect is revealed, its
wave aspect is concealed.
Revealing both simultaneously is impossible; the wave
and particle aspects are complementary.”
W. Heisenberg
Uncertainty relations &
Matrix mechanics (1932)
E. Schrödinger
Wave mechanics
(1933)
P.A. M. Dirac
Relativistic theory
of electron (1933)
W. Pauli
Exclusion principle
(1945)
Heisenberg’s Uncertainty Relations
Minimum Uncertainty (Gaussian wave packets)
Single-slit diffraction: x in slit-width, changes px
This principle is of great importance in understanding many phenomena
Electrons confined in atoms
J
eV
Hydrogen atom ionisation potential : 13.6 eV
Calculate the energy of e- if it were to be in the nucleus: x ~ 10-15 m
This principle is of great importance in understanding many phenomena
particles confined in nucleus
Size of the nucleus
This principle is of great importance in understanding many phenomena
Harmonic oscillator: Ground state
Zero point energy
This principle is of great importance in understanding many phenomena
Hydrogen atom: Ground state
Bohr radius
Minimum energy
Classical physics is deterministic
Quantum physics/mechanics is probabilistic
Consider single-slit diffraction (photon or electron):
- With a single particle the diffraction pattern is absent.
- When the experiment is repeated with large number of
particles the diffraction pattern is observed.
Q: So, where does the particle go after passing through the slit?
Ans: It has a large probability to go where there is a diffraction
maxima.
For a classical system made up of particles:
a) Position and momentum (or equivalently velocity) are specific at any
particular time.
b) If position and momentum is known at an initial time, then if you
know all the forces acting on that particle you can write down
equations which tell you exactly what its position and momentum
will be at any future time.
c) Note that instead of specifying the forces you can specify the
potential energy function, which is equivalent: force is the gradient
of potential.
In quantum mechanics the situation is a little more complicated
The systems that you study are still made of particles, and the basic procedure is in
some ways similar:
a)You measure the state of a particle at some initial time, you specify the forces
acting on that particle (or equivalently, the potential energy function describing those
forces), and quantum mechanics gives you a set of equations for predicting the
results of measurements taken at any later time.
b) There are two key differences between these two theories.
 the state of a particle in quantum mechanics is not just given by its position
and momentum but by something called a "wavefunction."
 knowing the state of a particle (i.e., its wavefunction) does not enable you to
predict the results of measurements with certainty, but rather gives you a set
of probabilities for the possible outcomes of any measurement.
The wave-function is a complex function which is defined everywhere in space.
No matter what state your particle is in, its wave-function has some complex
value at every point in the universe.
Basic postulates of Quantum mechanics
Probabilities in QM are determined
from wave functions (not observable)
Complex Wavefunction
2
Observable
Well behaved
 must be continuous and single-valued everywhere
/x must be continuous and single-valued everywhere
 must go to zero as x  ± 
Probability of finding the particle between
x
2
1
2
=P(1, 2)
x
Superposition
If 1 and 2 are two wavefunctions that satisfy the
equation, then 1+2 is also a solution, where a and b are
constants.
A typical quantum mechanics problem would thus run as follows:
i) You start with a particle in an unknown state subject to a known set of forces, such
as an electron in an electromagnetic field.
ii) You perform a measurement on that particle that tells you its state, i.e., its wavefunction.
iii) You let it evolve for a certain amount of time and then take another measurement.
iv) Quantum mechanics can tell you what result to expect from this second
measurement.
Position probabilities with a discrete wavefunction
• How do you use  to predict the results of a measurement of the
particle's position?
• A simple case
Suppose a particle can exist in only three points: x=1, x=2, and x=3.
Our particle must be on exactly one of those points. It cannot be
anywhere else, including in between them.
So  is just three complex numbers, which might look something like
this.
x = 1:  = 1 + i
x = 2:  = 2 - 2i
x = 3:  = 2 + 2i
• ||2 gives the probability of finding the particle at a particular
position.
• The probability of finding our particle at position x=1 is |1+i|2
which is 2.
The probability of x=2 is 8.
So, for any given measurement, we are four times as likely to find
the particle at position 2 as at position 1.
• Now, if we sum up the probabilities of all three locations, we get
the odds that the particle will be in any one of those three states.
Since we've already said that the particle must be at one of those
states, we should get a probability of 1.
• But at the moment, we find a total probability of 18!
• This wavefunction is unnormalized:
• It can tell us the relative probabilities of two positions, but not the
absolute probabilities of any of them.
• To normalize it, we multiply all the values by a constant, to make
the total probability equal to 1.
• In this case, we have to multiply every value of ||2 by 1/18, which
means we multiply every value of  by the square root of 1/18.
This will keep all the relative probabilities the same, but ensure
that the total probability of finding the particle somewhere is 1.
• So now, at our three places, we have the
following.
x=1, probability=2/18
x=2, probability=8/18
x=3, probability=8/18
• That is a properly normalized description of the
particle's position. You can't say for sure what any
given measurement will find, but you know
exactly what the odds are.
Therefore the wave-functions at the three places can be
written as:
probability=2/18
probability=8/18
probability=8/18
• “Expectation value" of position
Expectation value < x > of the position of a particle described
by wave function (x,t), is the value of x we would obtain if
we measured the position of a large number of particles
described by the same wavefunction at some instant t and then
averaged the results.
It is not the most probable value of a measurement.
For a large number of identical particles, distributed along x-axis
so that there are N1 particles at x1, N2 at x2 etc…:
The average position of the identical particles:
N1 x1  N 2 x2  N3 x3  ...  Ni xi
x

N1  N 2  N3  ...
 Ni
If we deal with single particle, we must replace Ni of particles at xi by
the probability Pi, that the particle be found in interval dx at xi.
Pi  i dx
2
where i is the particle wave-function at x=xi.
Nx

x
N
i i
i
Expectation value of the position of a single particle is:

x 
x
2
dx


  dx
2

If  is normalised wave function,

x 
x

dx

2

Evaluate the expectation values: <E> and <p> ?
Problem: A particle limited to the x-axis has the wave function  = ax
between x = 0 and x = 1;  = 0 elsewhere.
(a) Find the probability that the particle can be found between x = 0.45
and x = 0.55.
(b) Find the expectation value <x> of the particle’s position.
Solution:
(a) The probability is:
x2
3 0.55
0.55
x
x  dx  a 0.45x dx  a 3
1
2
2
2
2
 0.0251a 2
0.45
(b) The expectation value is:
1
4 1
1
x
x   x  dx  a  x dx  a
4
0
0
2
2
3
2
0
2
a

4
Consider a free particle wave function:
i ( kx t )
 ( x, t )  Ae
Knowing that
E = ħ and
p = ħk
 ( x, t )  Ae
i
( px Et )

 ( x, t )  Ae
i
( px Et )

Successive differentiation will give

i
 E ( x, t )
t

 i
 p ( x, t )
x
and
 2  2 p 2


 ( x, t )
2
2m x
2m
Now for a free non-relativistic particle
E = p2/2m
Therefore, we have

   ( x, t )
i

2
t
2m x
2
2
This is the one-dimensional time dependent Schrödinger
equation for a free particle.
We also get that E and p can be represented by:

E  i
t

p  i
x
i

 E ( x, t )
t
 i

 p ( x, t )
x
Till now we have assumed the particle to be free. If we now assume the
particle to be in a field characterized by the potential energy function
V(x,t) then according to classical mechanics, the total energy would be
given by:
E= p2/2m + V(x,t)
Since the potential energy function does not depend on p and E, we
have from preceeding two important equations that the wave-function
should satisfy
Where H= p2/2m + V and is known as the
Hamiltonian.
Shrödinger Equation
For conservative systems having time-independent potential energy V(x):
Separation of variables
 2 1  2 ( x)
1

 V ( x) 
H ( x)  E
2
2m  ( x) x
 ( x)
f (t )  Ce
iEt


( x, t )  C ( x)e

n ( x, t )  Cn n ( x)e
iEt


iEn t

Note that t in the above equation is only a phase factor.
It does not effect the probability distribution or expectation value.
As t does not appear in the measurable quantities the eigenstates of
the time-dependent Schrödinger equation of a conservative system
are known as the stationary states.
Time independent Schrödinger eqn.
One dimensional Potential well
V0
Infinite
L
-V0
V(x)= 0 for 0<x<L
=∞ for x<0 and for x>L
The one dimensional Schrödinger equation becomes
 2
2
 k  ( x)  0
2
x
2mE
2
where k  2

The general solution of this equation is:
ψ=A sin kx + B coskx
The boundary condition is ψ(x=0) = ψ(x=L) = 0
Using this condition, you will get that B=0 and A is either 0, or kL= nπ.
The condition that A=0 leads to the trivial solution of ψ
vanishing everywhere, the same is the case for n=0. Thus
the allowed energy levels are given by
2
2
nh
En 
; n  1,2,3,...
2
8mL
For the nth energy level the wavefunction is given by:
nx
 n ( x)  A sin
L
Before finding the probability amplitude, we have to normalize the
wavefunction to unity
nx
2 L
0  ( x) n ( x)dx  1  A 0 sin L dx  A 2
2
A
L
L
L
*
n
2
2
The corresponding eigen-functions are:
Eigenfunctions are the wavefunctions associated with the eigenvalues:
 n ( x) 
2
 n
sin 
L  L

x ; (0  x  L)

The complete time-dependent wave function is:
n ( x, t )   n ( x)e

iEnt

2
nx

sin
e
L
L

iEn t

 n ( x) 
2
 n
sin 
L  L
L
-V0

x ; (0  x  L)

Finite depth well
V0
Finite
L
-V0
One dimensional Potential step
E > V0
V0
E < V0
x
One dimensional Potential barrier
Quantum mechanical tunneling
E < V0
V0
x
More the barrier height less the transmission probability
More the barrier width less the transmission probability
Applications in atom trapping experiments
THE END