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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:

• In 3 dimensions,

ħ~ 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, <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 Ab, 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, ph 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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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

3 comments on the stationary solutions: 1

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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

3 comments on the stationary solutions: 2

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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

3 comments on the stationary solutions: 3

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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)