Transcript Quantum2
Quantum mechanical versionthe particle is confined by an infinite potential on either side. The boundary condition- the probability of finding the particle outside of the box is ZERO! For the quantum mechanical case: here, we assumed the walls were infinite—there was no possibility the particle could escape. at x 0 1( x, t ) 2( x, t ) A1ei ( kx t ) A2 ei ( kx t ) A1ei ( 2 mE x / Et / ) A2 ei ( 2 mE x / Et / ) A1ei ( Et / ) A2 ei ( Et / ) 0 A1 A2 at xL A1ei ( 2 mE L / Et / ) Ae iEt / e i A2 ei ( 2 mE L / ei 2 mE L / Et / ) 2 mE L / 2iA sin( 2mE L / ) 0 (0, t ) 0 at x 0 ( L, t ) 0 at x L 2mE L / n n 2 2 2 En 2mL2 n 1, 2 ,3... 0 0 I II III U E d 2 ( x) 2m ( E U ) ( x) 2 2 dx Since E<U: d 2 ( x ) 2m 2 (U E ) ( x) 2 dx Generally, solutions are then: 0 Setting 2m 2 (U E ) 2 L This begins to look like the familiar undamped sho equation: d 2 ( x) 2 ( x) 2 dx ( x) C1ex C2ex C1e But remember the conditions imposed on wavefunctions so that they make physical sense: 2 m(U E ) x / C2e 2 m(U E ) x / I II ( x) C1ex C2ex C1e III C2e 2 m(U E ) x / C1 must be 0 in Region III and C2 must be zero in Region I, otherwise, the probabilities would be infinite in those regions. U E -L/2 2 m(U E ) x / L/2 Note that the wavefunction is not necessarily 0 in Regions I and II. (It is 0 in the limit of an infinite well.) How is this possible when U>E?? The uncertainty principle. Now we must use the condition of continuity (the wavefunction must be continuous at the boundary, and so must its first derivative). Suppose we had a discontinuous function… y(x) L/2 x Here, the acceleration would be infinite. Uh-oh! dy/dx L/2 x Even solutions Odd solutions Note that the particle has negative kinetic energy outside of the well: K=E-U The wavefunction decreases rapidly- 1/e in a space of 1/ 1/ is a penetration depth. To observe the particle in this region, you must measure the the position with an accuracy of less than 1/ 2 x 1 / 2m(U E ) From the uncertainty principle: p h / x h 2 p K 2m 2m(U E ) 2 8 2 m(U E ) The uncertainty in the measurement is larger than the negative kinetic energy. An “real world example” of a finite box would be a neutron in a nucleus. Note that the allowed energies are inversely proportional to the length of the box. The energy levels of a finite box are lower than for an infinite well because the box is effectively larger. There is less confinement energy. an application As close as one gets to a potential well in real life. CCD must perform 4 tasks to generate an image: * Generate Charge --> Photoelectric Effect * Collect Charge --> pixels: an array of electrodes (called gates) * Transfer Charge --> Apply a differential voltage across gates. Signal electrons move down vertical registers (columns) to horizontal register. Each line is serially read out by an on-chip amplifier. * Detect Charge --> individual charge packets are converted to an output voltage and then digitally encoded. Low noise * high efficiency for weak signals * uniform response * large dynamic range At x=0, all of a particle’s energy is potential energy, as it approaches the boundary, its kinetic energy becomes less and less until it all of the particles energy is potential energyit stops and is reflected back. An example would be a vibrating diatomic molecule. This is analogous to a classical system, such as a spring, where potential energy is being exchanged for kinetic energy. In contrast to the square well, where the particle moves with constant kinetic energy until it hits a wall and is reflected back, in the parabolic potential well of the harmonic oscillator, the kinetic energy decreases (wavelength increases) as the boundary is approached. p2 E U p 2 m( E U ) Kinetic energy: K 2m p h 2 m( E U ) The wavelength is position dependent: Making an approximationassuming small penetration depth and high frequency, the condition for an infinite number of half wavelengths as in an infinite well must be recast as an integral to account for a variable wavelength: b a Boundary conditions: 0 at x 1 3 2m[ E U ( x)]dx ,1, ... 2 2 Note that the width of the well is greater for higher energies. As the energy increases, the “confinement energy” decreases. The levels are evenly spaced. We have Planck’s quantization condition! 1 E ( n ) 2 n 1,2,3,... In classical physics, the “block on a spring” has the greatest probability of being observed near the endpoints of its motion where it has the least kinetic energy. (It is moving slowly here.) This is in sharp contrast to the quantum case for small n. In the limit of large n, the probabilities start to resemble each other more closely. •We can find allowed wavefunctions. •We can find allowed energy levels by plugging those wavefunctions into the Schrodinger equation and solving for the energy. •We know that the particle’s position cannot be determined precisely, but that the probability of a particle being found at a particular point can be calculated from the wavefunction. •Okay, we can’t calculate the position (or other position dependent variables) precisely but given a large number of events, can we predict what the average value will be? (If you roll a dice once, you can only guess that the number rolled will be between 1 and 6, but if you roll a dice many times, you can say with certainty that the fraction of times you rolled a three will converge on 1 in 6…) If you roll a dice 600 times, you can average the results as follows: 1 2 4 6 5 5 3 4 ... 600 Alternatively, you can count the number of times you rolled a particular number and weight each number by the the number of times it was rolled, divided by the total number of rolls of the dice: 99 97 104 (1) (2) (3) ... 600 600 600 After a large number of rolls, these ratios converge on the probability for rolling a particular value, and the average value can then be written: x xPx This works any time you have discreet values. What do you do if you have a continuous variable, such as the probability density for you particle? 2 x xPx dx x ( x, t ) dx It becomes an integral…. The expectation value can be interpreted as the average value of x that we would expect to obtain from a large number of measurements. Alternatively it could be viewed as the average value of position for a large number of particles which are described by the same wavefunction. We have calculated the expectation value for the position x, but this can be extended to any function of positions, f(x). For example, if the potential is a function of x, then: U U ( x) ( x, t ) dx 2 expression for kinetic energy p2 KE ; 2m the potential kinetic plus potential energy gives the total energy p k position momentum x p x potential energy U U(x) kinetic energy K 2 2 2m x 2 total energy E i t operator observable i x In general to calaculate the expectation value of some observable quantity: Q * Qdx We’ve learned how to calculate the observable of a value that is simply a function of x: U U dx U ( x)dx U ( x) dx * * 2 But in general, the operator “operates on” the wavefunction and the exact order of the expression becomes important: 2 2 dx K K dx 2 2m x * *