Transcript DeBroglie Hypothesis

DeBroglie Hypothesis
Problem with Bohr Theory: WHY L = n ?
• have integers with standing waves: n(/2) = L
consider circular path for standing wave:
n = 2r
• from Bohr theory: L = mvr = nh/2
Re-arrange to get 2r = nh/mv = n
Therefore, n = 2r = nh/mv, which leads to
 = h/mv = h/p .
DeBroglie Hypothesis
DeBroglie = h/mv = h/p
In this case, we are considering the
electron to be a WAVE, and
the electron wave will “fit” around the orbit
if the momentum (and energy) is just right
(as in the above relation). But this will happen
only for specific cases - and those are the
specific allowed orbits (rn) and energies (En)
that are allowed in the Bohr Theory!
DeBroglie Hypothesis
The Introduction to Computer Homework
on the Hydrogen Atom (Vol. 5, number 5)
shows this electron wave fitting around the
orbit for n=1 and n=2.
What we now have is a wave/particle duality
for light (E&M vs photon), AND a
wave/particle duality for electrons!
DeBroglie Hypothesis
If the electron behaves as a wave, with
 = h/mv, then we should be able to test this
wave behavior via interference and
diffraction.
In fact, experiments show that electrons DO
EXHIBIT INTERFERENCE when they
go through multiple slits, just as the
DeBroglie Hypothesis indicates.
DeBroglie Hypothesis
Even neutrons have shown interference
phenomena when they are diffracted from a
crystal structure according to the DeBroglie
Hypothesis:  = h/p .
Note that h is very small, so that normally 
will also be very small (unless the mv is
also very small). A small  means very
little diffraction effects [1.22  = D sin()].
Quantum Theory
What we are now dealing with is the
Quantum Theory:
• atoms are quantized (you can have 2 or 3,
but not 2.5 atoms)
• light is quantized (you can have 2 or 3
photons, but not 2.5)
• in addition, we have quantum numbers
(L = n , where n is an integer)
Heisenberg Uncertainty Principle
There is a major problem with the wave/particle
duality:
a) a wave with a definite frequency and
wavelength (e.g., a nice sine wave) does not
have a definite location.
[At a definite location at a specific time the wave
would have a definite phase, but the wave
would not be said to be located there.]
[ a nice travelling sine wave = A sin(kx-t) ]
Heisenberg Uncertainty Principle
b) A particle does have a definite location at a
specific time, but it does not have a
frequency or wavelength.
c) Inbetween case: a group of sine waves can
add together (via Fourier analysis) to give a
semi-definite location: a result of Fourier
analysis is this: the more the group shows
up as a spike, the more waves it takes to
make the group.
Heisenberg Uncertainty Principle
A rough drawing of a sample inbetween case,
where the wave is somewhat localized, and
made up of several frequencies.
Heisenberg Uncertainty Principle
A formal statement of this (from Fourier
analysis) is: x * k
(where k = 2/, and  indicates the
uncertainty in the value).
But from the DeBroglie Hypothesis,  = h/p,
this uncertainty relation becomes:
x * (2/) = x * (2p/h) = 1/2 , or
x * p = /2.
Heisenberg Uncertainty Principle
x * p = /2
The above is the BEST we can do, since there
is always some experimental uncertainty.
Thus the Heisenberg Uncertainty Principle
says: x * p > /2 .
Heisenberg Uncertainty Principle
A similar relation from Fourier analysis for
time and frequency: t *  = 1/2 leads to
another part of the Uncertainty Principle
(using E = hf):
t * E > /2 .
There is a third part: * L > /2 (where L
is the angular momentum value).
All of this is a direct result of the
wave/particle duality of light and matter.
Heisenberg Uncertainty Principle
Let’s look at how this works in practice.
Consider trying to locate an electron
somewhere in space. You might try to
“see” the electron by hitting it with a
photon. The next slide will show an
idealized diagram, that is, it will show a
diagram assuming a definite position for the
electron.
Heisenberg Uncertainty Principle
We fire an incoming photon at the electron,
have the photon hit and bounce, then trace
the path of the outgoing photon back to see
where the electron was.
incoming
photon
electron
Heisenberg Uncertainty Principle
screen
slit so we can
determine direction
of the outgoing
outgoing
photon
photon
electron
Heisenberg Uncertainty Principle
Here the wave-particle duality creates a
problem in determining where the electron
was.
photon hits here
slit so we can
determine direction
of the outgoing
photon
electron
Heisenberg Uncertainty Principle
If we make the slit narrower to better
determine the direction of the photon (and
hence the location of the electron, the wave
nature of light will cause the light to be
diffracted. This diffraction pattern will cause
some uncertainty in where the photon actually
came from, and hence some uncertainty in
where the electron was .
Heisenberg Uncertainty Principle
We can reduce the diffraction angle if we
reduce the wavelength (and hence increase
the frequency and the energy of the photon).
But if we do increase the energy of the
photon, the photon will hit the electron
harder and make it move more from its
location, which will increase the uncertainty
in the momentum of the electron.
Heisenberg Uncertainty Principle
Thus, we can decrease the x of the electron
only at the expense of increasing the
uncertainty in p of the electron.
Heisenberg Uncertainty Principle
Let’s consider a second example: trying to
locate an electron’s y position by making it
go through a narrow slit: only electrons that
make it through the narrow slit will have the
y value determined within the uncertainty of
the slit width.
Heisenberg Uncertainty Principle
But the more we narrow the slit (decrease y),
the more the diffraction effects (wave aspect),
and the more we are uncertain of the y motion
(increase py) of the electron.
Heisenberg Uncertainty Principle
Let’s take a look at how much uncertainty
there is: x * p > /2 .
Note that /2 is a very small number
(5.3 x 10-35 J-sec).
Heisenberg Uncertainty Principle
If we were to apply this to a steel ball of mass
.002 kg +/- .00002 kg, rolling at a speed of
2 m/s +/- .02 m/s, the uncertainty in
momentum would be 4 x 10-7 kg*m/s .
From the H.U.P, then, the best we could be
sure of the position of the steel ball would
be: x = 5.3 x 10-35 J*s / 4 x 10-7 kg*m/s
= 1.3 x 10-28 m !
Heisenberg Uncertainty Principle
As we have just demonstrated, the H.U.P.
comes into play only when we are dealing
with very small particles (like individual
electrons or photons), not when we are
dealing with normal size objects!
Heisenberg Uncertainty Principle
If we apply this principle to the electron going
around the atom, then we know the electron
is somewhere near the atom,
(x = 2r = 1 x 10-10 m)
then there should be at least some uncertainty
in the momentum of the atom:
px > 5 x 10-35 J*s / 1 x 10-10 m = 5 x 10-25 m/s
Heisenberg Uncertainty Principle
Solving for p = mv from the Bohr theory
[KE + PE = Etotal, (1/2)mv2 - ke2/r = -13.6 eV
gives v = 2.2 x 106 m/s ] gives
p = (9.1 x 10-31 kg) * (2.2 x 106 m/s)
= 2 x 10-24 kg*m/s;
this means px is between -2 x 10-24 kg*m/s and
2 x 10-24 kg*m/s, with the minimum px
being 5 x 10-25 kg*m/s, or 25% of p.
Heisenberg Uncertainty Principle
Thus the H.U.P. says that we cannot really
know exactly where and how fast the
electron is going around the atom at any
particular time.
This is consistent with the idea that the
electron is actually a wave as it moves
around the electron.
Quantum Theory
But if an electron acts as a wave when it is
moving, WHAT IS WAVING?
When light acts as a wave when it is moving,
we have identified the
ELECTROMAGNETIC FIELD as waving.
But try to recall: what is the electric field?
Can we directly measure it?
Quantum Theory
Recall that by definition, E = F/q. We can
only determine that a field exists by
measuring an electric force! We have
become so used to working with the electric
and magnetic fields, that we tend to take
their existence for granted. They certainly
are a useful construct even if they don’t
exist.
Quantum Theory
We have four LAWS governing the electric
and magnetic fields: MAXWELL’S
EQUATIONS. By combining these laws
we can get a WAVE EQUATION for
E&M fields, and from this wave equation
we can get the speed of the E&M wave and
even more (such as reflection coefficients,
etc.).
Quantum Theory
But what do we do for electron waves?
What laws or new law can we find that will
work to give us the wealth of predictive
power that MAXWELL’S EQUATIONS
have given us?
Quantum Theory
The way you get laws is try to explain
something you already know about, and
then see if you can generalize. A successful
law will explain what you already know
about, and predict things to look for that
you may not know about. This is where the
validity (or at least usefulness) of the law
can be confirmed.
Quantum Theory
Schrodinger started with the idea of
Conservation of Energy: KE + PE = Etotal .
He noted that
• KE = (1/2)mv2 = p2/2m, and that =h/p, so
that p = h/ = (h/2)*(2/) = k = p, so
KE = 2k2/2m
• Etotal = hf = (h/2)*(2f) = .
Quantum Theory
He then took a nice sine wave, and called
whatever was waving, :
(x,t) = A sin(kx-t) = Aei(kx-t) .
He noted that both k and  were in the
exponent, and could be gotten down by
differentiating. So he tried operators:
Quantum Theory
(x,t) = A sin(kx-t) = Aei(kx-t) .
pop = i[d/dx] = i[-ikAe-i(kx-t)] = k
= (h/2)*(2/)* = (h/ = p .
similary:
Eop= i[d/dt] = i[-iAei(kx-t)] = 
= ((h/2)*(2f)* = (hf = E .
Quantum Theory
Conservation of Energy: KE + PE = Etotal
becomes with the momentum and energy
operators:
-(2/2m)*(d2/dx2) + PE* = i(d/dt)
which is called SCHRODINGER’S
EQUATION. If it works for more than the
free electron, then we can call it a LAW.
Quantum Theory
What is waving here? 
What do we call ? the wavefunction
Schrodinger’s Equation allows us to solve
for the wavefunction. The operators then
allow us to find out information about the
electron, such as its energy and its
momentum.
Quantum Theory
To get a better handle on , let’s consider
light: how did the E&M wave relate to the
photon?
Quantum Theory
The photon was the basic unit of energy for
the light. The energy in the wave depended
on the field strength squared.
[Recall energy in capacitor, Energy = (1/2)CV2,
where for parallel plates, Efield = V/d and
C// = KoA/d, so that
Energy = (1/2)*(KoA/d)*(Efieldd)2
= KoEfield2 * Vol, or Energy Efield2.]
Quantum Theory
Since Energy is proportional to field strength
squared, AND energy is proportional to the
number of photons, THEN that implies that
the number of photons is proportional to the
square of the field strength.
This then can be interpreted to mean that the
square of the field strength is related to
the probability of finding a photon.
Quantum Theory
In the same way, the square of the
wavefunction is related to the probability of
find the electron!
Since the wavefunction is a function of both x
and t, then the probability of finding the
electron is also a function of x and t!
Prob(x,t) = (x,t)2
Quantum Theory
Different situations for the electron, like being
in the hydrogen atom, will show up in
Schrodinger’s Equation in the PE part.
Different PE functions (like PE = -ke2/r for
the hydrogen atom) will cause the solution
to Schrodinger’s equation to be different,
just like different PE functions in the
normal Conservation of Energy will cause
different speeds to result for the particles.
Schrodinger’s Equation
Let’s look at the case of an electron confined
to a length, L, but otherwise free. We will
first consider the 1-D case. Since the
electron is free, PE=0. However, since it is
confined to the length, L, we have
boundary conditions:
x=0,t) = 0, and (x=L,t) = 0.
(   / 2m)(  /  x )  0  i(  /  t )
2
2
2
Schrodinger’s Equation
(   / 2m)(  /  x )  0  i(  /  t )
2
2
2
To solve this differential equation, we use the
technique of separation of variables:
(x,t) = X(x)*T(t) .
The partial derivative with respect to x does
not affect T(t), and the partial derivative
with respect to t does not affect X(x).
We can also divide both sides of the equation
by X(x)*T(t) to get:
Schrodinger’s Equation
(   / 2m)( X /  x ) / X  E  (i T /  t ) / T
2
2
2
Note that the left side depends solely on x,
and the right side solely on t. This means
that as far as x is concerned, the right side is
a constant, and as far as t is concerned, the
left side is a constant (which turns out to be
the Energy).
Schrodinger’s Equation
We now have two ordinary differential
equations instead of one partial differential
equation:
-(2/2m)d2X/dx2 = E*X for X(x)
i dT/dt = E*T
for T(t).
In fact, any situation where the PE does not
explicitly depend on time will have this
solution.
Schrodinger’s Equation
i dT/dt = E*T
for T(t)
The above 1st order ordinary differential
equation is easily solved if we recall that the
differential of an exponential gives an
exponential back again. So try T(t) = Aeat:
i(aAeat) = E(Aeat) or ia = E, or a = E/i.
But we know that E = , so this means that:
a = /i = -i, and T(t) = A e-it .
Schrodinger’s Equation
• Note that the probability depends on 2
(actually, since  can be a complex
quantity, Prob(x,t) = where  is
the complex conjugate of ).
• Note also that T**T = Ae-it * Ae+it = A2 ,
but A can be incorporated as a constant into
X(x). Thus the T part of  can effectively
be ignored if the PE does not depend on t.
Schrodinger’s Equation
This means that when the PE does not depend
on time, Schrodinger’s Equation can be rewritten as (since iddt = E*) :
(-2/2m)*d2/dx2 + PE* = E* .
Note that  can be complex (just like E could
be written in complex form), but the
Probability must be real (since it is a
measurable quantity).
Now we go on to the X equation.
Schrodinger’s Equation
-(2/2m)d2X/dx2 = E*X for X(x)
This equation can again be solved by inspection
since we know that the second derivative of a sine
(or cosine) function gives itself back again with a
minus sign, so we’ll try X(x) = B sin(kx+φo) :
-(2/2m)(-Bk2 sin(kx+φo)) = E B sin(kx+φo) , or
2k2/2m = E (since p = k, E = p2/2m = KE
but this is correct since PE = 0 in this case!)
Schrodinger’s Equation
X(x) = B sin(kx +φo)
We now need to apply the Boundary
conditions:
X(x=0) = 0 works fine for the sine function,
but not the cosine function so φo= 0.
X(x=L) = 0 demands that kL = n, or that
k = n/L (note that k is quantized, so p is
quantized, and E is quantized!).
Schrodinger’s Equation
Note that Schrodinger’s Equation GIVES US
A QUANTUM NUMBER. It comes when
we apply the boundary conditions! We did
not have to put the quantum number into the
theory as Bohr did.
Note that the Schrodinger’s equation we have
written down is for 1-D only (x). We now
need to see about a 3-D form.
Schrodinger’s Equation
Momentum is a vector, so the DeBroglie
relation: p = k is really a component
relation: px = kx, py = ky, pz = kz ,
where the kx simply describes how the
E&M field varies in the x-direction.
Thus the pop = i d/dx must now become
px op = i  /  x ; p y op = i  /  y;
pz op = i  /  z
Schrodinger’s Equation
KE = p2/2m = (px2 + py2 + pz2)/2m.
This can be used in the Schrodinger’s
equation with the component operators for
p, and using the symbol
 2 =  2 /  x 2 +  2 /  y 2 +  2 /  z2
Schrodinger’s Equation
The three dimensional Scrodinger’s
Equation becomes (for PE not dependent
on time):
-{(h/2π)2 /2m}2Ψ + PE Ψ = E Ψ.
Schrodinger’s Equation
This equation can also usually be solved by
separation of variables, and when applying
the boundary conditions (for x, y and z) we
usually get THREE QUANTUM
NUMBERS (just like we got 1 quantum
number in the 1-D case).
These quantum numbers come out of the
theory rather than being put into the theory
as Bohr did in his.
Quantum Theory
From Chemistry, you may recall that FOUR
quantum numbers were required, instead
of the three that the Schrodinger’s Equation
predicts.
The fourth quantum number (spin) can be
predicted from a relativistic quantum
theory, where the equation is called the
DIRAC EQUATION. (This also predicts
anti-matter as well!)
Quantum Theory
In the hydrogen atom we do not use
rectangular coordinates (x,y,z); instead we
use spherical coordinates (r, because
of the spherical symmetry of the potential
energy (PE = -ke2/r).
The solution of the Schrodinger’s equation
gives us three quantum numbers: n, l, and
ml . The fourth number is due to spin, ms .
Quantum Theory
The n quantum number is related to the r equation
and is related to energy.
The l quantum number is related to the  equation
and is related to angular momentum.
The m quantum number is related to the  equation
and is related to the z-component of angular
momentum, which is related to the magnetic
properties of the state.
Quantum Theory
The fourth quantum number, ms, does not
have a classical explanation - it is a
relativistic quantum phenomenon. It is
related to magnetic behavior, and hence has
the m name. The closest thing classically
we can relate it to is to the case of the
electron “spinning”, so that its spinning
charge creates a magnetic field. But this does
not work out according to classical calculations.
Pauli Exclusion Principle
How do we extend the quantum theory to
systems beyond the hydrogen atom?
For systems of 2 electrons, we simply have a
 that depends on time, and the coordinates
of each of the two electrons:
(x1,y1,z1,x2,y2,z2,t)
and the Schrodinger’s equation has two
kinetic energies instead of one.
Pauli Exclusion Principle
It turns out that the Schodinger’s Equation can
again be separated:
 = Xa(x1,y1,z1) * Xb(x2,y2,z2) * T(t) .
This is like having electron one in state a, and
having electron two in state b.
Pauli Exclusion Principle
However, from the Heisenberg Uncertainty
Principle (from wave/particle duality), we
are not really sure which electron is electron
number 1 and which is number 2. This
means that the wavefunction must also
reflect this uncertainty.
Pauli Exclusion Principle
There are two ways of making the
wavefunction reflect the indistinguishability
of the two electrons:
sym = [Xa(r1)*Xb(r2) + Xb(r1)*Xa(r2) ]* T(t)
and
anti = [Xa(r1)*Xb(r2) - Xb(r1)*Xa(r2) ]* T(t)
.
Pauli Exclusion Principle
sym = [Xa(r1)*Xb(r2) + Xb(r1)*Xa(r2) ]* T(t)
anti = [Xa(r1)*Xb(r2) - Xb(r1)*Xa(r2) ]* T(t)
.
Which (if either) possibility agrees with
experiment?
It turns out that some particles are explained
nicely by the symmetric, and some are
explained by the antisymmetric.
BOSONS
sym = [Xa(r1)*Xb(r2) + Xb(r1)*Xa(r2) ]* T(t)
Those particles that work with the symmetric
form are called BOSONS. All of these
particles have integer spin as well. Note
that if electron 1 and electron 2 both have
the same state,  > 0. This means that both
particles CAN be in the same state at the
same location at the same time.
FERMIONS
anti = [Xa(r1)*Xb(r2) - Xb(r1)*Xa(r2) ]* T(t)
Those particles that work with the anti-symmetric
wavefunction are called FERMIONS. All of these
particles have half-integer spin. Note that if
electron 1 and electron 2 both have the same
state,  = 0. This means that both particles CAN
NOT be in the same state at the same location at
the same time.
Pauli Exclusion Principle
BOSONS. Photons and alpha particles (2
neutrons + 2 protons) are bosons. These
particles can be in the same location with
the same energy state at the same time.
This occcurs in a laser beam, where all the
photons are at the same energy
(monochromatic).
Pauli Exclusion Principle
FERMIONS. Electrons, protons and neutrons
are fermions. These particles can NOT be in
the same location with the same energy state
at the same time.
This means that two electrons going around
the same nucleus CAN NOT both be in the
exact same state at the same time! This is
known as the Pauli Exclusion Principle!
Pauli Exclusion Principle
Since no two electrons can be in the same
energy state in the same atom at the same
time, chemistry is possible (and so is
biology, psychology, sociology, politics,
and religion)!
Thus, the possibility of chemistry is explained
by the wave/particle duality of light and
matter, and electrons acting as fermions!