Lecture 3
The Development of Quantum
Mechanics
Waves reminder
Schrödinger wave equation
The wave function
Particle in a box
Tunneling
Heisenberg’s uncertainty principle
Fyu02- Kvantfysik
David Milstead
Reminder: Standing Waves
Concept from classical physics
Restrict the wave to be zero
everywhere except over a
confined length L.
length = integer x wavelength
i.e. L=nλ/2 (3.1)
L
Easy to do with a string
Essential for understanding quantum mechanics!
Fyu02- Kvantfysik
David Milstead
Animation of rope generated on a string from
http://www.ngsir.netfirms.com/englishhtm/StateWave.htm
Fyu02- Kvantfysik
David Milstead
Αnimation of incident and reflecting waves constructively interfering
to produce a standing wave.
http://www.walter-rendt.du/ph14e/stwaverefl.htm
Fyu02- Kvantfysik
David Milstead
Electron in a Bohr Orbit
nh
Angular momentum L = mvr = pr =
2π
h
Using p = (2.19)
(2.9) e-
v
r
λ
nλ = 2πr = circle circumfere nce (3.2)
Standing wave condition!
Electron forms standing
wave around Bohr orbit
nλ = 2πr
constructive interference
nλ ≠ 2πr
no constructi ve interferen ce
Bohr model and de-Broglie picture consistent.
But what does it mean that the electron is standing waveFyu02? Kvantfysik
David Milstead
Animation of fitting standing waves to Bohr orbits.
http://www.walter-fendt.de/ph11e/bohrh.htm
Fyu02- Kvantfysik
David Milstead
Schrödinger’s Wave Equation
Classical wave equation
∂2 y
1 ∂2y
= 2
2
v ∂t 2
∂x
(3.3) v = speed
Standing waves in one-dimension
y ( x , t ) = ψ ( x ) sin( ω t ) (3.4) ω = angular frequency
Substitute (3.4) into (3.3)
d 2ψ
ω2
=− 2ψ
2
dx
v
Total energy E =
(3.5) and
2
p
+U
2m
v
=k=
2
(3.6)
p2 1 2
where
= mv = kinetic energy
2m 2
U = potential energy
Fyu02- Kvantfysik
David Milstead
p 2 = 2m(E − U )
ω
2m (E − U )
=
(3.7)
2
2
v
hd2
2m
+ - 2 ( E − U )ψ = 0 (3.9)
2
dx
h
2
Schrödinger’s Time Independent Wave
Equation
Equation represents stationary states of an atomic system
for which E is constant with time.
Apply to the Bohr model (coming lecture) and obtain
Fyu02- Kvantfysik
observed energy levels!
David Milstead
Interpretation of the Wave
Function
A physical particle, such as an electron, has an
associated wavefunction (ψ )
ψ
can be a complex number and has no physical interpretation
|ψ |
2
= probability density
| ψ | dV =
2
probability of finding the particle within (3.10)
a volume dV
Quantum mechanics offers no certainties, only
probabilities!
Fyu02- Kvantfysik
David Milstead
Normalising the Wave function
Consider a particle moving along the x-axis.
The particle must exist somewhere in space
+∞
ψ dx = 1
2
(3.11)
−∞
A wave function satisifying 3.11 is normalised.
Fyu02- Kvantfysik
David Milstead
What does probability mean ?
Electron 2-slit diffraction.
It is impossible to predict the final
position of an individual electron.
electron beam
Only a probability is given and for
A large sample the pattern is built up.
Considering the electron as a particle
will not explain this pattern.
Trying to measure which slit an electron passes through
destroys the pattern (see later)!
Fyu02- Kvantfysik
David Milstead
Applications of Schrödinger’s
Equation
A particle of mass m is in a box of
side length L. The potential energy
is zero within the box and infinite
at the walls (infinite potential well).
Particle bounces back and forward in
one dimension.
Classically the particle can be
anywhere in the box 0 ≤ x ≤ L
From quantum mechanics, the
particle has a wave function ψ
ψ=0 outside the box since
the particle is confined.
0
L
x
Fyu02- Kvantfysik
David Milstead
d
Boundary conditions ψ and
continous over a boundary.
dx
ψ = 0 at x = 0 and L = 0
Take U = 0 and Schrödinge r's Eqn. becomes
d 2ψ
2
k
(3.12)
ψ =0
+
2
dx
where k = 2mE - (3.13)
h
Solution is
ψ = A sin( kx + φ ) (3.14)
where φ is a constant.
Fyu02- Kvantfysik
David Milstead
Apply boundary conditions
ψ = 0 at x = 0
ψ = 0 at x = L
Hence
φ =0
(3.15)
kL = n π (3.16)
nπx
ψ ( x ) = A sin
L
; n = 1, 2 ,3 ,... (3.17)
where A is a constant.
2π
nπ
Since k =
=
(3.18)
λ
L
2L
h
= de Broglie wavelength =
λ=
n
mv
nh
(3.19)
v=
2mL
Fyu02- Kvantfysik
David Milstead
1
2
Total energy E n = kinetic energy = mv
2
n2h2
En =
; n = 1,2,3, . . . (3.20)
2
8mL
A confined particle has discrete energy levels.
Energy is never zero.
Lowest energy at n=1 is zero-point energy and is not
zero!
E(h2/8mL2
9
4
1
Classically, particle has zero potential energy at 0K.
Quantum mechanics disagrees !
Fyu02- Kvantfysik
David Milstead
Where is the particle ?
Probability P of a particle at between x and x+dx is
P = ψ dx (3.21)
2
ψ1
ψ12
ψ2
ψ
ψ3
ψ32
0
x
L
2
2
particle is
never there
Standing
waves!
Fyu02- Kvantfysik
David Milstead
How do we think about the particle in the box ?
Classically, the particles
bounces back and forward and
there is no preferred position
for it.
0≤ x≤L
Quantum mechanically
the wave function is
a standing wave made up
of the superposition of waves
travelling in opposite directions
There are regions in which the
particle can never be found!
0
L
standing wave
Fyu02- Kvantfysik
David Milstead
Question
Consider a 10-7 kg dust particle confined to a box of side length 1cm.
(a) What is its minimum possible speed ?
(b) What is the quantum number if the particle’s speed is 10-3 mm/s
(a) From (3.20), minimum allowed energy is E1
h2
1 2
E1 =
mv (remember the particle has no potential energy)
=
2
8mL
2
h
6.63 × 10 −34
v=
=
2mL 2 × 10 − 7 × 10 − 2
= 3.32 × 10 - 25 m/s.
Quantum mechanical effects on a particle are only observable
at the atomic level !
Fyu02- Kvantfysik
David Milstead
n2h2 1
2
mv
=
(b)
8mL2 2
2 mvL 2 × 10 −7 × 10 − 6 × 10 − 2
18
10
≈
n=
=
6.63 × 10 −34
h
A dust particle is a classical object obeying
classical laws of physics. Observable s of the particle
should have very large quantum numbers according to
Bohr's correspond ence principle.
Fyu02- Kvantfysik
David Milstead
Finite Potential Well
I
Consider particle trapped in
a well of potential energy depth U.
U=0 at the well bottom.
Clasically, if the particle energy
E < U , it cannot escape!
Schrödinger equation in region II
d 2ψ II
2
k
+
ψ = 0 (3.22)
2
dx
where k = 2 mE h
Solution is ψ II = C sin( kx ) (3.23)
where C is a constant.
II
III
U
m
0
L
x
Fyu02- Kvantfysik
David Milstead
At x=0 and x=L, ψ ≠ 0
In regions I and III U>E and the Schrödinger eqn. is
d 2ψ
= K 2ψ
2
dx
(3.24)
where K 2 = 2 m (U − E )
h-2
(3.25)
General solution :
= Ae Kx + Be − Kx (3.26)
In region III,
III
= Be − Kx (3.27)
In region II,
II
Match
→ 0, as x → ∞
→ 0, as x → −∞
= Ae Kx (3.28)
I
and
II
at x = 0 and
Use boundary conditions
I (0) =
II ( 0 ) and
II ( L ) =
II
and
III
III
at x = L.
( L ) (3.29)
and
d
( 0 ) d II ( 0 )
d II ( L ) d III ( L )
=
=
and
dx
dx
dx
dx
I
(3.30)
Fyu02- Kvantfysik
David Milstead
Wave functions of particle in finite potential well.
I
II
I
ψ5
ψ4
Non-zero wavefunction
outside of the box.
Wave does not fall to zero
at x=0, x=L
ψ3
ψ2
Possibility to find the
particle in regions I and III
ψ1
L
0
x
Fyu02- Kvantfysik
David Milstead
Barrier Penetration
A particle with energy E approaches
a region of potential energy U where E<U
potential
barrier
Clasically the particle bounces back.
(think of a car at constant speed approaching
a hill. If the driver turns off the engine, the car
Will roll back if the speed is too small.)
Quantum mechanically, the particle
can ’tunnel’ through the barrier.
x
This is the basis of α−decay and is used in superconducting
and scanning tunneling electron microscopes.
Can be understood in terms of borrowing an energy ∆E
for a short enough time ∆t to overcome the barrier (Heisenberg’s
Fyu02- Kvantfysik
uncertainity principle)
David Milstead
Animation of barrier tunneling with variable parameters
http://phys.educ.ksu.edu/vqm/html/qtunneling.html
Fyu02- Kvantfysik
David Milstead
Heisenberg’s Uncertainity
Principle
The most beautiful and mysterious part of quantum mechanics???
The act of measurement will always disturb the measured object.
It is impossible to simultaneously know the
position of a particle and its linear momentum
to arbitrary precision (Heisenberg 1927)
However good our apparatus is, nature imposes
this restriction.
∆x∆p > h
(3.31)
∆p = uncertainty of linear momentum
∆x = uncertainty on position
Fyu02- Kvantfysik
David Milstead
Reaching the Uncertainty Limit
Locate an electron by illuminating with light.
Unable to locate it to a precision better than the light wavelength
∆ x ≈ λ (3.32)
Photon can transfer some or all of its linear momentum to
the electron
γin
∆p =
h
λ
(3.33)
x p ≈h
Increasing position resolution means reducing
wavelength and increasing momentum resolution!
The act of measurement disturbs the system !
e-
γout
e-
Fyu02- Kvantfysik
David Milstead
Deriving the Uncertainty Principle
Consider diffraction of
electrons through a single
slit.
From (2.21), position of
first minimum is given by
sin θ =
λ
a
=
λ
∆y
a
θ
(3.34)
e∆y=uncertainty in position (in y-direction) of
Electron position as it passes through slit.
∆py > psinθ (θ=angle of first minimum)
Using de Broglie formula p=h/λ (2.19)
∆ y∆ p y > h
p
θ
Fyu02- Kvantfysik
David Milstead
∆py
What is happening ?
Understand the uncertainty principle in terms of the particle
wave function.
No knowledge of its position.
A waveform with a precise wavelength is
continuous over all space. ∆p=0,∆x=infinity
Α superposition of waves
Leads to a localised wave
Packet with a welldetermined location but
with a less well-defined
wavelength.
Fyu02- Kvantfysik
David Milstead
Question
Are the results obtained for the momentum of the particle in the
infinite potential well consistent with the uncertainty principle ?
From 3.19 and taking n = 1
∆ p = 2 p (particle can go back or forward)
2 mnh h
=
2 mL
L
∆ x ≈ L (particle has to be somewhere in the box! ! )
∆ p = 2 mv =
p x ≈h
Increase n and lose p precision but gain more
knowledge of likely particle position.
0
L
Fyu02- Kvantfysik
David Milstead
Another Uncertainty Relation
Τhe energy of a system can fluctuate from the value set
by the conservation of energy by an amount ∆E for a time
∆t.
∆E∆t > h
(3.35)
’Explains’ Quantum mechanical tunneling if a particle borrows
energy ∆E for a time ∆t to tunnel through a potential barrier.
Zero point energy is explained by the continual creation and
destruction of particles for short time periods according to 3.35.
Alternatively, using 3.31, if the energy of a particle in the box
is zero then the momentum is zero and is precisely known.
∆p = 0
But ∆x ≈ L ≠ ∞
∆x∆p = 0 and the uncertaint y principle is violated.
A particle which is confined must have a non - zero energy!
Fyu02- Kvantfysik
David Milstead