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Quantum Mechanics 102 Tunneling and its Applications
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Interference of Waves and the Double Slit Experiment Waves spreading out from two points, such as waves passing through two slits, will interfere d Wave crest Wave trough Spot of constructive interference Spot of destructive interference
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Interpretation The probability of finding a particle in a particular region within a particular time interval is found by integrating the square of the wave function: P (x,t) = | (x,t)| 2 dx = | (x)| 2 dx | (x)| 2 dx is called the “probability density; the area under a curve of probability density yields the probability the particle is in that region When a measurement is made, we say the wave function “collapses” to a point, and a particle is detected at some particular location
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Particle in a box (x) = B sin (n x/a) (x) | (x)| 2 n=2 n=3 Only certain wavelengths = 2a/n are allowed Only certain momenta p = h/ = hn/2a are allowed Only certain energies E = p 2 /2m = h 2 n 2 /8ma 2 are allowed - energy is QUANTIZED Allowed energies depend on well width
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What about the real world? Solution has non-trivial form, but only certain states (integer n) are solutions Each state has one allowed energy, so energy is again quantized Energy depends on well width a Can pick energies for electron by adjusting a | (x)| 2 n=1 n=2 x
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Putting Several Wells Together How does the number of energy bands compare with the number of energy levels in a single well? As atom spacing decreases, what happens to energy bands? What happens when impurities are added?
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Quantum wells An electron is trapped since no empty energy states exist on either side of the well
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Escaping quantum wells Classically, an electron could gain thermal energy and escape For a deep well, this is not very probable
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Escaping quantum wells Thanks to quantum mechanics, an electron has a non-zero probability of appearing outside of the well This happens more often than thermal escape
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What if free electron encounters barrier? Do Today’s Activity
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What Have You Seen? What happens when electron energy is less than barrier height? What happens when electron energy is greater than barrier height? What affects tunneling probability? T e –2kL k = [8 2m(E pot – E)] ½ /h
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A classical diode According to classical physics, to get to the holes on the other side of the junction, the conduction electrons must first gain enough energy to get to the conduction band on the p-side This does not happen often once the energy barrier gets large Applying a bias increases the current by decreasing the barrier
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A tunnel diode According to quantum physics, electrons could tunnel through to holes on the other side of the junction with comparable energy to the electron This happens fairly often Applying a bias moves the electrons out of the p-side so more can tunnel in
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Negative resistance As the bias is increased, however, the energy of the empty states in the p-side decreases A tunneling electron would then end up in the band gap - no allowed energy So as the potential difference is increased, the current actually decreases = negative R
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No more negative resistance As bias continues to increase, it becomes easier for conduction electrons on the n-side to surmount the energy barrier with thermal energy So resistance becomes positive again
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The tunneling transistor Only electrons with energies equal to the energy state in the well will get through
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The tunneling transistor As the potential difference increases, the energy levels on the positive side are lowered toward the electron’s energy Once the energy state in the well equals the electron’s energy, the electron can go through, and the current increases.
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The tunneling transistor The current through the transistor increases as each successive energy level reaches the electron’s energy, then decreases as the energy level sinks below the electron’s energy
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Randomness Consider photons going through beam splitters NO way to predict whether photon will be reflected or transmitted! (Color of line is NOT related to actual color of laser; all beams have same wavelength!)
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Randomness Revisited If particle/probabilistic theory correct, half the intensity always arrives in top detector, half in bottom BUT, can move mirror so no light in bottom! (Color of line is NOT related to actual color of laser; all beams have same wavelength!)
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Interference effects Laser light taking different paths interferes, causing zero intensity at bottom detector EVEN IF INTENSITY SO LOW THAT ONE PHOTON TRAVELS THROUGH AT A TIME What happens if I detect path with bomb? No interference, even if bomb does not detonate!
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Interpretation Wave theory does not explain why bomb detonates half the time Particle probability theory does not explain why changing position of mirrors affects detection Neither explains why presence of bomb destroys interference Quantum theory explains both! Amplitudes, not probabilities add - interference Measurement yields probability, not amplitude - bomb detonates half the time Once path determined, wavefunction reflects only that possibility - presence of bomb destroys interference
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Quantum Theory meets Bomb Four possible paths: RR and TT hit upper detector, TR and RT hit lower detector (R=reflected, T=transmitted) Classically, 4 equally-likely paths, so prob of each is 1/4, so prob at each detector is 1/4 + 1/4 = 1/2 Quantum mechanically, square of amplitudes must each be 1/4 (prob for particular path), but amplitudes can be imaginary or complex! e.g.,
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Adding amplitudes Lower detector: Upper detector:
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What wave function would give 50% at each detector? Must have |a| = |b| = |c| = |d| = 1/4 Need |a + b| 2 = |c+d| 2 = 1/2
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