(Continued from previous section)
4. The Wave-Particle Duality & Heisenberg Microscope
We
now look in somewhat more detail at wave-particle duality as it arises in
quantum mechanics. In the particle
interpretation, electrons fired from a
device such as an electron gun would not all follow the same path since the
trajectory of an electron – unlike a missile- can’t be predicted from its
initial state. We consider here the case of electron diffraction, whereby
(based on Fig. 7) electrons are emitted from an electron gun and pass through a
slit toward a detector or photographic plate onto which a diffraction pattern
appears. This pattern will also coincide with an intensity distribution such as
shown in Fig. 10.
In
effect, the intensity distribution basically describes the probability for an
individual particle (electron) to strike each of several areas designated on
the photographic film. This discloses a fundamental indeterminacy that has no
counterpart in Newtonian mechanics. Now, consider an electron striking at some
angle q, such as indicated:
Fig. 10: Showing
electron diffraction and intensity pattern on screen.
p
y/ p x = tan q
or p y = p x q (in limit of small q)
Therefore,
the y-component of momentum can be as large as:
p
y = p x (l/ a)
Where a denotes the slit width. The narrower the
dimension of a the broader the diffraction pattern, and the greater D p. From Louis de
Broglie’s matter wave hypothesis (already introduced into the Bohr atom, as we
saw, cf. Fig. 7) : lD = h/ p x
Therefore:
p
y = p x (h/ p x a) =
h/ a or: p y a = h
But ‘a’ represents uncertainty in electron
position vertically (D
y), i.e. as it passes through the slit. We can reduce D py only by narrowing the slit width a and vice
versa. Thus we get:
p
y a = D py D y » h
Which is one form of the Heisenberg
Uncertainty Principle which states that the momentum of a quantum particle
and its positions cannot simultaneously be known to the same arbitrary
precision. One corollary is that to detect a particle any given detector must
interact with it thereby altering the motion of the particle.
This
view is no longer taken in any literal way because we understand that the
quantum measurements are statistical in nature and hence a particular
measurement is the result of a vast statistical assembly. Paul Dirac, in his book Quantum Mechanics, defined the “principle of superposition” thusly[1]:
“A state of a system may be defined as a state of undisturbed
motion that is restricted by as many conditions or data as are theoretically
possible without mutual interference or contradiction"
But let’s examine this in more detail. By “undisturbed motion” Dirac
meant the state is pure and hence no extraneous
observations are being made such that the state experiences interference
effects to displace or disturb it. In the Copenhagen Interpretation,
“disturbance” of mutually defined variables, say x, p or position and momentum,
occurs only if:
[x,
p] = -i h = -i h/ 2p
If it were the case that [x, p] = 0, one would
say the variables “commute” and hence there’s
no interference. If the condition doesn’t hold, then interference exists.
Hence, Dirac’s setting of an upper limit in the last portion of his definition,
specifying as many conditions as theoretically possible “without mutually interfering interference.” This state is
undisturbed. We have a statistical
perspective!
Fig. 11:
Sketch of Heisenberg Microscope and key parameters.
It is
important to see from the preceding, how the Heisenberg Uncertainty Principle
arises not just from an ad hoc assumption, but from the limits (or “tolerance
thresholds”) of explicit quantities (e.g. p, x), when considered in the quantum
limit. Hence, the model of the Heisenberg “microscope” provides a useful
(although not practical, since it can’t actually be constructed) means of
deriving the statistical principle of superposition based on an observational
ansatz.
Consider a measurement made to determine
the instantaneous position of an electron
by means of a microscope. In such a measurement the electron must be
illuminated, because it is actually the light quanta (photon) scattered by the
electron that the observer sees. The resolving power of the microscope
determines the ultimate accuracy with which the electron can be located. This resolving power is known to be
approximately:
l/ 2 sin q
Where l
is the wavelength of the scattered light and q is the half-angle subtended by the
objective lens of the microscope. Then:
Δx = l/
2 sin q
In
order to be collected by the lens, the photon must be scattered through any
range of angle from -q to q. In effect, the electron’s momentum values range from:
+ h
sin q/ l to - h sin
q/
l
Then
the uncertainty in the momentum is given by:
D px = [ h sin q/ l - (- h sin q/ l)] = 2 h sin q/
l
Then the Heisenberg Uncertainty Principle product is:
D px Δx = (2 h sin q/
l )
l/ 2 sin q = h
4. Probability density and Expectation Values
P = ½y (1s) y (1s) *½
Which is the probability
density and a quantity we can actually measure, e.g. for the 1s state of
hydrogen. Then this needs to be generalized to apply to more than one case.
Since the
electron locations can’t be computed from Newtonian mechanics but more
plausibly based on an analogous probability density to what we saw above, then we can generalize and
write:
P
ab = ò ba ‖y(x) ‖2 dx
Where x is the state under consideration and this system
is 1-dimensional with the probability assessed from a to b. Note that we define the normalization condition as:
ò ba ‖y‖2 dx = 1
ò ba ‖y‖2 dx = 0
The
wave function, y(x)
satisfies the Schrodinger equation. For the simple one dimensional case we can
write:
d2 y/dx2 + F(x) y = 0
Though
the wave function y(x)
it self is not a measurable quantity, other measurable quantities such as the
energy E and momentum of the particle can be derived from it. Also, if the wave
function is known it is possible to compute the average position of the
particle, known as the expectation value:
This
expression implies the particle is in a definite state so that the probability
density is time –independent.
Example
Problem: Consider the 1D quantum system shown, and a particle confined therein:
E
n = (h2/ 8m L2) n2
Let
the wave function be:
y(x) = Ö2/ ÖL sin (kx)
Solution:
We
rewrite the wave function as:
y(x) = Ö2/ ÖL [sin (px/L)] where k = p/L
Then:
P
ab = ò ba ‖y(x) ‖2 dx = ò L 0 (Ö2/ ÖL )2 sin2 (px/L) dx
= 2/ L ò L
0 ½
[1 - cos (px/L)]
dx
(Let
q = px/L and use:
sin2 q=
½ (1 – cos 2q))
P
ab = 2/ L [½
ò L 0 dx - ò L 0 cos (px/L)] dx
P
ab = 1 -
1/p sin (2px/L)]
L 0 = 1 -
1/p sin (2p)
But
: sin (2p) = 0 so P ab = 1
The
expectation value is:
Finding
the energy: We have the Schrodinger
equation:
dy2/dx2 + K2
y = 0
where K = Ö [2mE] / ħ
where K = Ö [2mE] / ħ
If
we examine the sketch below:
We see plots of the wave function y(x) vs. position x (far left), and of the probability density (middle) and
the energy levels. Since we have represented the wave function by a sinusoidal
function then it follows that the allowed wavelengths are those for which the
length L is equal to an integral number of half wavelengths, or:
L = nl/ 2
These allowed states are called stationary
states and represent standing waves (analogous to the ones seen earlier for the
Bohr atom). Thus, the wavelengths of the particle are restricted by the
condition:
l = 2L/ n
Then
the magnitude of the momentum p is also restricted to specific values (e.g.
using p = h/ l)
[2]
such that:
p
= h/ l = h/ 2L/ n = nh/ 2L
The
energy associated with the particle is then:
E
= 1/2 mv2 = p2/ 2m = (nh/ 2L)2 / 2m
E=
( h2/ 8mL2) n2
(n=
1, 2, 3 etc.)
Thus
the energy is quantized with the energy of the lowest energy state
corresponding to n =1 so:
E1=
( h2/ 8mL2)
This
least energy that the particle can have is called the “zero point energy” and
means the particle can never be at rest.
Note that the above energy result can also
obtained through the use of differential equations, e.g.
The
probability density can be extended to 3 dimensions by writing:
P
= ò ¥- ¥ ‖y(x) ‖2 dV
The
quantized energy will then be (for a 3D box, for which dV = dx dy dz):
E=
( h2/ 8mL2)[ n x 2 + n y
2 + n z 2 ]
Problems:
1)For
a 1D box, let one electron inside have the wave function:
y(x) = Ö2/ ÖL [sin (2px/L)]
Find
the probability of locating the electron between x = 0 and x = L/4.
2)Use
the uncertainty principle to estimate the uncertainty in momentum for a
particle in a 1D box. Estimate the ground state energy using this means and
compare it to the actual ground state energy.
3)
The wave function for a particle confined to moving in a 1D box is given by:
y(x) = A [sin (npx/L)]
Use
the normalization condition on y(x) to
show the constant A is given by: A = Ö2/
ÖL
4) It is known from quantum mechanics that a
particle in a one dimensional potential well (such as shown in the diagram) can
exist in a number of energy states. Imagine an electron confined between the
boundaries x and x + Dx, where Dx is 0.5 Angstroms.
Approximately,
what is the uncertainty in the x-component of
the momentum of the electron?
5)(a) Consider a free particle confined between
two impenetrable walls at x and x + L.
What is the probability according to classical physics that the particle
will be found between x and x + L/3 if no other information is given?
b)
What is the probability according to quantum mechanics that the particle in its
lowest energy state will be found
between x and x + L/3?
c)
What is the probability according to quantum mechanics that the particle in the
second
lowest energy state will be found between x and x + L/3?
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