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562. Chem Lecture 6,7 Simple systems by Quantum Mechanics. I September 21, 2007 1 Quantization and boundary condition: one-dimensional "particle" in the box The problem is very similar to the free "particle" case - except that the potential is zero only within a region [0, L] but outside of this region. 1.1 A heuristic solution The intimate relationship between boundary condition and energy quantization can be illustrated by an heuristic derivation based, once again, on the de Broglie relation. If we divide the space into regions I ([0, L]) and II (x < 0 and x > L), we realize that the wavefunction in region II has to be zero II (x) = 0; x < 0; x > L (1) In region I, the wave function is finite but also has to be continuous at the boundary x = 0, L. In other words, we should have I (x) = 0; x = 0, L (2) What effect does this have on the energy of the system? Since p = h/, and E = p2 /2m, we expect that E = h2 /(2m2 ). Is arbitrary? No! For a wave to satisfy the boundary condition at x = 0 and x = L, the wave length () has to be such that an integral number of half-period (/2) fits in region I (otherwise the value of wave does not vanish at x = 0, L). Therefore, = 2L/n, where n is an non-zero integer. Substituting this into the energy expression, we immediately get quantized energy levels: E = n2 h2 /8mL2 . 1 1.2 Solving the Schr dinger Equation o As mentioned above, the wavefunction in region II has to be zero II (x) = 0; x < 0; x > L (3) In region I, the equation has exactly the same form as the free "particle" case and therefore has exactly the same general solutions: I (x) = Aeikx + Be-ikx and E= k 2m 2 2 (4) (5) But we know the result has to be different from the free "particle" - but how? It turns out that now the coefficients A, B and k are not longer arbitrary but fixed through boundary conditions - i.e., the wavefunction has to be continuous across the region boundaries (x = 0,x = L): Aeik0 + Be-ik0 = A + B = 0 (6) and AeikL + Be-ikL = 0 (7) Solving the two equations, we reach the important result that only specific values of k are allowed: n (8) k= L where n is an non-zero integer - which is also referred to as the quantum number. The quantization in k immediately implies quantization of energy through Eq.5: E= n2 h2 8mL2 (9) What about the wavefunction? With the boundary condition Eq.6, we have I (x) = Csin n nx L (10) where the constant C(= 2A) can be specified if we further impose the normalization condition for the wavefunction (as stated earlier, the physics doesn't change with different values of C). It's straightforward to show that the normalized wavefunction is: I (x) = n 2 nx sin L L (11) See ppt for the illustration of wavefunctions with different quantum numbers. 2 1.3 Understanding trends in the eigenvalue and eigenfunctions: general for any 1D potential 1. Curvature in the wavefunction increases as the energy increases 2 2 ^ Recall that the kinetic operator energy is a second-order differential operator ( - x2 ). 2m Therefore, curvature of the wavefunction reflects kinetic energy. 2. Lowest energy boundary states always have finite energy - "Zero-point" energy Note that n = 1 for the ground state of the particle in the box problem. Two ways of understanding this: Even the lowest energy state needs finite curvature to satisfy the boundary condition - which immediately implies finite kinetic energy. Uncertainty principle - the system is confined in a finite space, thus the uncertainty in x has to be finite. Due to the uncertainty principle, the uncertainty in momentum has to be finite - which means that the momentum (thus kinetic energy) can not be zero. 3. The nth eigenstate always has (n-1) node (zero-crossing) Mathematically, this trend can be thought as the way to satisfy the orthonormal condition for eigenfunction of Hermitian operators (e.g., the Hamiltonian). (x)m (x)dx = mn n (12) Physically, this trend can be understood as the consequence of increased kinetic energy (thus increased curvature) as n increases. Exercise: see ppt. The node in the wavefunction implies that there is also node in the probability density, |n (x)|2 = 2 nx sin2 L L (13) This is clearly a phenomena not observed in classical mechanics. In fact, since the probabilities of finding the system on the left and right of the node are finite, if the system behaves classically, then there must be finite probability at the node (as the system travels from the left to the right of the node). Apparently, for microscopic systems, classical thinking is not very productive. The existence of node in probability density is due to the interference effect between components in the wavefunction that have different momenta (i.e. e ikx where k = n/L). Indeed, since n (x) eikx - e-ikx , the probability density has cross terms between e ikx : |n (x)|2 (e-ikx - eikx )(eikx - e-ikx ) = 2 - 2cos2kx = 2 - 2(1 - 2sin2 kx) = 4sin2 kx (14) 3 4. Correspondence principle: as the quantum number (n) or the size of the system increases, the result approaches that of classical mechanics 2 As n increases, the oscillation in |n (x)|2 = L sin2 nx is so fast that the distribution L approaches uniform - which is the classical result. The interval between adjacent energy levels, E = n2 h2 (2n + 1)h2 (n + 1)2 h2 - = 8mL2 8mL2 8mL2 (15) As L , E 0 - i.e., energy becomes continuous - which is again the classical result. Another use of Eq.15 is the rational of fluorescence change in quantum dots - which are small semiconductor particles (nm range). As the size of the dots changes, the fluorescence wavelength (thus color) changes. Question: what is the expected trend - i.e., as the size increases, does the fluorescence shift to the red (long wavelength) or blue (short wavelength)? Example: What can we learn about the behavior of the system given a specific wavefunction. See ppt. 4

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