Download 3 - Sezione di Fisica

L1 – The Schroedinger
equation
• A particle of mass m on the x-axis is subject to
a force F(x, t)
• The program of classical mechanics is to
determine the position of the particle at any
given time: x(t). Once we know that, we can
figure out the velocity v = dx/dt, the momentum
p = mv, the kinetic energy T = (1/2)mv2, or any
other dynamical variable of interest.
• How to determine x(t) ? Newton's second law:
F = ma.
– For conservative systems - the only kind that occur
at microscopic level - the force can be expressed as
the derivative of a potential energy function, F = -dV/
dx, and Newton's law reads m d2x/dt2 = -dV/ dx
– This, together with appropriate initial conditions
(typically the position and velocity at t = 0),
determines x(t).
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• A particle of mass m, moving along the
x axis, is subject to a force
F(x, t) = -dV/ dx
• Quantum mechanics approaches this
same problem quite differently. In this
case what we're looking for is the wave
function, Y (x, t), of the particle, and we
get it by solving the Schroedinger
equation:
ħ~ 10-34 Js
• In 3 dimensions,
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The statistical interpretation
• What is this "wave function", and what can it tell you?
After all, a particle, by its nature, is localized at a point,
whereas the wave function is spread out in space (it's
a function of x, for any given time t). How can such an
object be said to describe the state of a particle?
• Born's statistical interpretation:
Quite likely to find the particle near A, and relatively
unlikely near B.
• The statistical interpretation introduces a kind of
indeterminacy into quantum mechanics, for even if you
know everything the theory has to tell you about the
particle (its wave function), you cannot predict with
certainty the outcome of a simple experiment to
measure its position
– all quantum mechanics gives is statistical information
about the possible results
• This indeterminacy has been profoundly disturbing
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Realism, ortodoxy,
agnosticism - 1
• Suppose I measure the position of the particle,
and I find C. Question: Where was the particle
just before I made the measurement?
There seem to be three plausible answers to
this question…
1. The realist position: The particle was at C.
This seems a sensible response, and it is the
one Einstein advocated. However, if this is
true QM is an incomplete theory, since the
particle really was at C, and yet QM was
unable to tell us so. The position of the particle
was never indeterminate, but was merely
unknown to the experimenter. Evidently Y is
not the whole story: some additional
information (a hidden variable) is needed to
provide a complete description of the particle
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Realism, ortodoxy,
agnosticism - 2
• Suppose I measure the position of the particle,
and I find C. Question: Where was the particle
just before I made the measurement?
2. The orthodox position: The particle wasn't
really anywhere. It was the act of
measurement that forced the particle to "take a
stand“. Observations not only disturb what is
to be measured, they produce it .... We compel
the particle to assume a definite position. This
view (the so-called Copenhagen interpretation)
is associated with Bohr and his followers.
Among physicists it has always been the most
widely accepted position. Note, however, that
if it is correct there is something very peculiar
about the act of measurement - something that
over half a century of debate has done
precious little to illuminate.
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Realism, ortodoxy,
agnosticism - 3
• Suppose I measure the position of the particle,
and I find C. Question: Where was the particle
just before I made the measurement?
3. The agnostic position: Refuse to answer. This
is not as silly as it sounds - after all, what
sense can there be in making assertions about
the status of a particle before a measurement,
when the only way of knowing whether you
were right is precisely to conduct a
measurement, in which case what you get is
no longer "before the measurement"? It is
metaphysics to worry about something that
cannot, by its nature, be tested. One should
not think about the problem of whether
something one cannot know anything about
exists
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Realism, ortodoxy,
or agnosticism?
• Suppose I measure the position of the particle,
and I find C. Question: Where was the particle
just before I made the measurement?
• Until recently, all three positions had their
partisans. But in 1964 John Bell demonstrated
that it makes an observable difference if the
particle had a precise (though unknown)
position prior to the measurement. Bell's
theorem made it an experimental question
whether 1 or 2 is correct. The experiments
have confirmed the orthodox interpretation: a
particle does not have a precise position prior
to measurement; it is the measurement that
insists on one particular number, and in a
sense creates the specific result, statistically
guided by the wave function.
• Still some agnosticism is tolerated…
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• Suppose I measure the
Collapse of position of the particle, and I
find C. Question: Where will
the
be the particle immediately
wavefunction after?
• Of course in C. How does the
orthodox interpretation
explain that the second
measurement is bound to give
the value C? Evidently the
first measurement radically
alters the wave function, so
that it is now sharply peaked
about C. The wave function
collapses upon measurement
(but soon spreads out again,
following the Schroedinger
equation, so the second
measurement must be made
quickly). There are, then, two
entirely distinct kinds of
physical processes:
"ordinary", in which Y evolves
under the Schroedinger
equation, and
"measurements", in which Y
suddenly collapses.
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Normalization
•
|Y (x, t)|2 is the probability density for finding
the particle at point x at time t.
The integral of |Y (x, t)|2 over space must be 1
(the particle has to be somewhere).
• The wave function is supposed to be
determined by the Schroedinger equation--we
can't impose an extraneous condition on Y
without checking that the two are consistent.
• Fortunately, the Schroedinger equation is
linear: if Y is a solution, so too is A Y , where
A is any (complex) constant. What we must
do, then, is pick this undetermined
multiplicative factor so that The integral of |Y
(x, t)|2 over space must be 1 This process is
called normalizing the wave function.
• Physically realizable states correspond to the
"square-integrable" solutions to Schroedinger's
equation.
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Will a normalized function stay as such?
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Expectation values
• For a particle in state Y, the expectation value
of x is
• It does not mean that if you measure the
position of one particle over and over again,
this is the average of the results
– On the contrary, the first measurement (whose
outcome is indeterminate) will collapse the wave
function to a spike at the value obtained, and the
subsequent measurements (if they're performed
quickly) will simply repeat that same result.
• Rather, <x> is the average of measurements
performed on particles all in the state Y, which
means that either you must find some way of
returning the particle to its original state after
each measurement, or you prepare an
ensemble of particles, each in the same state
Y, and measure the positions of all of them:
<x> is the average of them.
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Momentum
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More on operators
• One could also simply observe that
Schroedinger’s equations works as if
(exercise: apply on the plane wave). In
3 dimensions,
Compound operators
• Kinetic energy is
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Angular momentum
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Exercise
•
A particle is represented at t=0 by the wavefunction
Y (x, 0)
= A(a2-x2)
|x| < a (a>0).
=0
elsewhere
a
Determine the normalization constant A
b, c What is the expectation value for x and for p at t=0?
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Exercise (cont.)
•
A particle is represented at t=0 by the wavefunction
Y (x, 0)
= A(a2-x2)
|x| < a (a>0).
=0
elsewhere
d, e Compute <x2>, <p2>
f, g Compute the uncertainty on x, p
h
Verify the uncertainty principle in this case
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L2 – The time-independent
Schroedinger equation
• Supponiamo che il potenziale U sia
indipendente dal tempo
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y, soluzione della prima equazione (eq.agli autovalori detta
equazione di S. stazionaria), e’ detta autofunzione
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3 comments on the stationary solutions: 1
1. They are stationary states. Although
the wave function itself
does (obviously) depend on t, the
probability density does not - the time
dependence cancels out. The same
thing happens in calculating the
expectation value of any dynamical
variable
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3 comments on the stationary solutions: 2
2. They are states of definite energy. In
mechanics, the total energy is called
the Hamiltonian:
H(x, p) = p2/2m + V(x).
The corresponding Hamiltonian
operator, obtained by the substitution p
-> p operator
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3 comments on the stationary solutions: 3
3. They are a basis. The general solution
is a linear combination of separable
solutions. The time-independent
Schroedinger equation might yield an
infinite collection of solutions, each with
its associated value of the separation
constant; thus there is a different wave
function for each allowed energy.
The S. equation is linear: a linear
combination of solutions is itself a
solution.
It so happens that every solution to the
(time-dependent) S. equation can be
written as a linear combination of
stationary solutions.
To really play the game, mow we must
input some values for V
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The infinite square well
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Infinite square well, 2
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Infinite square well, 3
But…
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Infinite square well, 4
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Free particle (V=0, everywhere)
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