Quantum Mechanics
Towards the end of the nineteenth century it became clear that there were
limitations in the ability to explain key observations in physics with only
classical mechanics. Two experiments were key to this finding.
The Photoelectric e↵ect
Incident monochromatic light on certain materials was expected to eject
electrons with an energy dependent on the intensity of the incident light.
Rather constant intensity with varying frequency determined the kinetic
energy of the ejected electrons.
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E
= hf
is the work function which is the minimum energy required to liberate an
electron. h = 6.625 ⇥ 10 34 Js is Plank’s constant. Resulting in one of the
key equations is QM:
E = hf
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Wave-Particle Duality
Is an electron a particle or a wave?
The above experiment shows that it is indeed both. This led to Louis De
Broglie to postulate that all matter has wave-like qualities, resulting in
another key equation in QM:
p=
h
where p is momentum of the particle , and
wave representing the particle.
is the wavelength of the
TRY: calculate your own wavelength if you know your mass and assume
you are in car traveling at 100 kmh 1 .
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The Schrödinger Equation aka the Wave Equation
Erwin Schrödinger, an Austrian physicist, developed the equation below
which formalised the wave-like-nature of matter:
~2 r2
(r) + V (r) (r) = E (r)
2me
h
This is the time independent Schrödinger equation. Here ~ = 2⇡
(hbar),
m is the mass of the particle, V the potential of the system and E the
total energy, is known as the wave function, all meaningful parameters
can be determined from it. This equation states that the total energy of
the system is the sum of the kinetic and potential energy.
Important One of the key axioms of quantum mechanics is that the wave
function of physical states admits a probabilistic interpretation.
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Finding the Wave Function
Let us consider a simple case of Schrödinger Equation in one dimension
(r = x). This would be the case of a free particle with V (x) = 0.
~2 r2
(x) = E (x)
2me
In on-dimension r2 =
d2
dx 2
We can now rewrite the equation as:
00
(x) + k 2 (x) = 0 with k 2 =
2mE
~2
So we have a second order di↵erential equation. What are the possible
solutions?
8
>
<sin(kx)
(x) = const cos(kx)
>
: ±ikx
e
TRY: Make sure you are happy that the above are solutions. What about
their sum?
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With the above solutions and given that there are no boundary
conditions to the problem, the allowed values of E are:
~2 k 2
0
2m
k, can take on any value, so the energy values are not quantised and
exhibit classical behaviour.
E=
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How do we use these solutions, to determine something of value? Let’s
talk probability and probability density!
Probability:
Example : 10 coins in a bag, 6 silver, 4 gold, what is the probability of
picking a gold from the bag versus a silver? What should your answers
sum to?
The above example is discrete , how would we deal with calculating the
probability of finding a particle we know exists in an infinite universe at a
particular position?.....make sure you understand the issue!!!!
Probability Density:
This will save us. How can we talk about a probability spread over an
infinite domain?
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Let us consider a very small region in a one dimensional infinite space.
This interval of this region is given by [x, x + dx]. If we assume that the
probability of finding a particle in this infinite domain is spread smoothly
over the entire domain, we can state that in a small region of this infinite
domain [dx] there should be a small probability [dP] of finding the
particle. We can therefore state that the probability will be proportional to
the width of interval such that:
dP = p(x)dx
p(x) is a proportionality constant know as the probability density function.
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More on Wave Functions
We have seen in the free particle quantum system what the solutions to
the wave equation look like, but what do they actually mean?
These wave functions do not provide direct information about the
physical observable (position, momentum etc). The square of the wave
function 2 is however a probability density.
The value 2 at a particular point represents the probability density of
finding the particle at that point. Integrating 2 over a specific region
yields the probability of finding the particle within that region.
⇤ where ⇤
Mathematically, the probability density is given by 2 =
denotes the complex conjugate of . Squaring the wave function ensures
that the resulting probability density is always positive or zero, allowing us
to interpret it as a genuine probability.
In summary, squaring the wave function in quantum mechanics is necessary
to obtain the probability density distribution, which allows us to calculate
the probabilities associated with di↵erent measurement outcomes.
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A Quantum Example - The Potential Barrier
Consider the potential barrier as shown in the image below.
For x < 0, the potential energy is zero, for x
x = 0, there is a step Epot = E0 .
0, Epot = E0 6= 0. at
For x < 0, from the left side free particles with energy E fly into the +x
direction. They therefore have solutions given by:
I (x)
= Ae ikx + Be
ikx
(1)
Why: are there both positive and negative k’s in the above solution. Look
at the image for the solution.
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For x
0 the Schrödinger equation is given by:
d2
2m
+ 2 (E
2
dx
~
E0 ) = 0
which can be simplified to:
d2
dx 2
↵2
=0
where
↵=
the solutions are given by:
p
2m(E0
II (x)
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E )/~
= Ce ↵x + De
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↵x
(2)
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We now apply boundary conditions such that the solutions are continuous
across the boundary at x = 0.
I (x
= 0) = II (x = 0) =) A + B = C + D
✓
◆
✓
◆
d I
d II
|x=0 =
|x=0 =) ik(A B) = +↵(C
dx
dx
D)
We now consider two di↵erent situations where the kinetic energy
Ekin = E of the incident particle is smaller than the potential step E < E0
or greater than the potential step E > E0 .
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E < E0
p
In this case ↵ = 2m(E0 E )/~ > 0, is real and C in equation 2 must be
zero, otherwise II (x) becomes infinite as x ! 1 and can no longer be
normalised, therefore:
ik + ↵
2ik
A and D =
A
ik ↵
ik ↵
The wave function in the region x < 0 is then:
h
i
ikx + ik + ↵ e ikx
I (x) = A e
ik ↵
The real part of the wave function is given in the figure below:
B=
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The fraction R of the reflected particles is:
R=
|Be ikx |2
|B|2
ik + ↵
=
=
2
ikx
2
|A|
ik ↵
|Ae |
2
=1
So, all the particles are reflected from the boundary. Classically you would
expect this as well. Have a closer look at the wave function image on the
previous page, it penetrates somewhat into the barrier before going to zero.
Lets calculate the probability of finding a particle at x
P(x) = |
II (x)|
2
= |De
↵x |2
=
4k 2
|A|2 e
↵2 + k 2
2↵x
0.
=
4k 2 2
|A| e
k02
2↵x
where k02 = 2mE0 /~2 .
After a penetration depth x = 1/(2↵), the probability has dropped to 1/e
of its value (4k 2 /k02 )A2 at x = 0.
Quantum particles can enter regimes a classical particle can not.
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E > E0
Here the kinetic energy Ekin = E of the incident particles is greater than
the potential barrier E0 . Classically all particles will travel into the region
x 0, while their kinetic energy Ekin (x 0) = E E0 will become smaller.
↵ is now purely imaginary and we replace it by the real quantity:
0
k = i↵ =
The solutions for
I
p
E0 )/~
are as given in equation 1, for
II (x)
= Ce
For x > 0 no particles travel in the
be zero and we obtain:
II (x)
M. D. Blumenthal
2m(E
0
ik x
+ De ik
0
(3)
II
we replace 2 with:
x
(4)
x direction, the amplitude C has to
= De ik
0
x
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(5)
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Apply the same boundary conditions as before:
0
k k
2k
B=
A
0 A and D =
k +k
k + k0
The wave functions then become:
0
⇣
k
ikx + k
e
(x)
=
A
e
I
k + k0
ikx
and
II (x)
=A
⌘
for x
0
2k ik 0 x
e
for x < 0
k + k0
The reflection coefficient is:
0
|B|2
k k
R=
=
2
|A|
k + k0
2
The transmission coefficient is:
0
|D|2
4kk
T =
=
2
|A|
|k + k 0 |2
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Transmission and reflection for Ekin > E0
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