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Almost all detection of visible light is by the “photoelectric effect” (broadly defined.) There is always a threshold photon energy for detection, even in film.

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Blackbody radiation

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A

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A particle in a box of width 2L has the wavefunction shown. By what factor is the particle more likely to be in the right side of the box, compared with the left side? A] equally likely B] twice as likely C] four times as likely D] eight times as likely C

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A particle in a box of width 2L has the wavefunction shown. What is the probability of the particle being found in the right side of the box? A] 50% B] 66.6% C] 75% D] 80% D

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A particle in a box of width 2L has the wavefunction shown. What is the constant ‘a’? (Use pencil & paper.) A] 1 B] C] D] D

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Note: Measuring the particle’s position may (in fact does) change the wavefunction. More on this later…

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For a free particle with wave function What is the direction of wave motion? A] -x B] +x C] no direction - it’s a stationary state A

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For a free particle with wave function What is the sign of the momentum? A] - B] + C] the momentum is zero A

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For a free particle with wave function Where is the particle? A] at the origin (x=0) B] not necessarily at, but near the origin C] somewhere, but we don’t know exactly where D] everywhere at once D (not C!)

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For a free particle with wave function Is this a stationary state? A] yes B] no C] can’t tell A

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Wave packets

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B

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Summary (so far) What is complicated about quantum mechanics: Particles must have (or be) waves. Equations that describe waves are more mathematically complicated than F=ma. But how waves behave is “classical”. If you can understand waves in water or sound, you can understand QM waves. (For example, classical waves are quantized… like waves on a violin string.) What is simple and beautiful in QM: Matter and light have the same description, a hybrid of wave and particle. What is unfathomable in QM: How do we know when to “switch” between a particle description and a wave description? This switch seems to occur when we measure the position of a photon or electron… then, it acts like a particle, and the wave associated with it collapses to its measured location.

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What is n for the state shown? A] 1 B] 2 C] 4 D] 8 E] 0 C

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If a measurement of the particle’s position is made, what are the relative probabilities that it will be found in regions A, B, and C? Note: the wavefunction is drawn offset from zero. We often draw the zero of the wavefunction at the energy of the state. A] all regions are equally probable B] region A is most probable, then B, then C C] region B is most probable, then A& C tie D] region B is least probable, A&C tie D

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At a later time, the probability of finding the particle at B will be larger than finding it at A or C. A] true B] false False.

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After measuring the position of the particle, it will still be in this wavefunction or “state”. A] true B] false False.

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After measuring the energy of the particle, it will still be in this wavefunction or “state”. A] true B] false True.

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A measurement is made of the momentum of the particle. What are the results? A] 0 B] ± with equal probability B

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Correspondence principle: How could this EVER LOOK LIKE a classical particle, bouncing around inside a box? Answer(s) We need to consider a wavepacket, And… Macroscopic systems are never in pure energy states. (Pure states are called “eigenstates”) http://groups.physics.umn.edu/demo/applets/qm1d/index.html

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All things are waves, obeying wave equations. “Measurement” (which is NOT precisely defined in QM!) changes the wave. It “collapses” it to a wave appropriate to the measured value. The best possible measurement of energy puts a system in an energy “eigenstate”, which we also call a “stationary state” When in a stationary state, the probability density is stationary. Absorption and Emission of light by atoms is (apparently) such an energy measurement. Some folks think “measurement” means “a conscious observer observes”, but QM is NOT precise about what a measurement is. If measurement requires a conscious observer, you get odd things, like Schroedinger’s Cat. There are almost no restrictions on what an arbitrary wavefunction looks like. (Must be finite, continuous, and have a continuous derivative if U<infinity.) In general, wavefunctions are not stationary states (or any special state, like a state of precise p.)

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Suppose we have a “superposition” wavefunction The probability of measuring an “eigenvalue” is proportional to the Magnitude squared of the coefficient of that eigenfunction. Correspondence principle: The behavior of wave packets with high quantum numbers looks classical.

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Harmonic potential. (A spring.) The effect of the widening of the potential at higher energies is to make the energy eigenstates equally spaced.

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Energy eigenstates of the harmonic potential.

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Which is the n=3 energy eigenstate? B

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A particle is in a bound energy eigenstate in a finite square well. A measurement of the particle position finds it in the “classically” forbidden region. In this region, the potential energy is higher than the particle’s (original) total energy. Explanation?? A] energy (and momentum) are not precisely conserved in QM B] the particle was in a classically allowed spot, but the measurement perturbed it C] the particle is still in the same energy state, so energy is conserved D] the measurement adds energy to the particle in an unspecified way D

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A particle is in a bound energy eigenstate in a finite square well. A measurement of the particle position finds it in the “classically” forbidden region. After the measurement: A] the particle will be in the same bound energy eigenstate B] the particle will be in a different bound energy eigenstate C] the particle will be in superposition of bound states D] the particle will be unbound C or D

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A particle is in a bound energy eigenstate in a finite square well. A measurement of the particle position finds it in the “classically” allowed region. After the measurement: A] the particle will be in the same bound energy eigenstate B] the particle might still be bound C] the particle must be unbound B

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Tunneling: In the barrier, the wavefunction decays exponentially. How should I draw the wave on the other side of the barrier? A] There is no wave on the other side B] The wave on the other side should have the same wavelength, but a smaller amplitude C] The wave on the other side should have the same amplitude, but a smaller wavelength D] The wave on the other side should have a smaller amplitude and a larger wavelength E] None of these is correct B

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In this simulation, Why is the energy of the wavepacket shown as a green schmear? A] The exact kinetic energy of the wavepacket is unknown. B] In order for tunneling to occur, the wavepacket must have some energies (or energy components) higher than the barrier C] Wavepackets intrinsically contain a spread of energies. C

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In this simulation, If I make the wavepacket bigger, what do you think will happen to the energy schmear? A] It will get narrower B] It will stay the same C] It will get broader A

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In this simulation, What will happen to the wavelength on the right if I lower the potential energy there? A] It will be shorter than the wavelength on the left B] it will remain equal to the wavelength on the left C] it will become longer than the wavelength on the left A.

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Radioactive decay is tunneling

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