L4

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L4 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
• In 3 dimensions,

? 10-34 Js
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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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Collapse of the wavefunction

• Suppose I measure the position of the particle,
  and I find C. Question Where will be the
  particle immediately 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, ltxgt 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 ltxgt is
  the average of them.

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Momentum
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More on operators

• One could also simply observe that Schroedingers
  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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The uncertainty principle (qualitative)
x(m)

• Imagine that you're holding one end of a long
  rope, and you generate a wave by shaking it up
  and down rhythmically.
• Where is that wave? Nowhere, precisely - spread
  out over 50 m or so.
• What is its wavelength? It looks like 6 m
• By contrast, if you gave the rope a sudden jerk
  you'd get a relatively narrow bump traveling down
  the line. This time the first question (Where
  precisely is the wave?) is a sensible one, and
  the second (What is its wavelength?) seems
  difficult - it isn't even vaguely periodic, so
  how can you assign a wavelength to it?

x(m)
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The uncertainty principle (qualitative, II)
x(m)
x(m)

• The more precise a wave's x is, the less precise
  is l , and vice versa. A theorem in Fourier
  analysis makes this rigorous
• This applies to any wave, and in particular to
  the QM wave function. l is related to p by the de
  Broglie formula
• Thus a spread in l corresponds to a spread in p,
  and our observation says that the more precisely
  determined a particle's position is, the less
  precisely is p
• This is Heisenberg's famous uncertainty
  principle. (we'll prove it later, but I want to
  anticipate it now)

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Exercise

• A particle is represented at t0 by the
  wavefunction
• Y (x, 0) A(a2-x2) x lt a (agt0).
• 0 elsewhere
• a Determine the normalization constant A
• b, c What is the expectation value for x and for
  p at t0?

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Exercise (cont.)

• A particle is represented at t0 by the
  wavefunction
• Y (x, 0) A(a2-x2) x lt a (agt0).
• 0 elsewhere
• d, e Compute ltx2gt, ltp2gt
• f, g Compute the uncertainty on x, p
• h Verify the uncertainty principle in this case

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L5 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) mv2/2 V(x).
• The corresponding Hamiltonian operator, obtained
  by the substitution p -gt -i ? ?/ ?x, is

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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 (V0, everywhere)
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(however, for any finite volume V, however large,
y is normalizable)
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• Let us assume that f(k) is narrowly peaked about
  some particular value k0. Since the integrand is
  negligible except in the vicinity of k0, we may
  Taylor-expand the function w(k) about that point
  and keep only the leading terms

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L6 - Gradino di potenziale

1. E lt U0
2. E gt U0

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a. E lt U0
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b. E gt U0
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Barriera di potenziale
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Finite squarewell
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L7 Loscillatore armonico 1-d
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Operatori di creazione e distruzione
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