Скачать презентацию

Идет загрузка презентации. Пожалуйста, подождите

Презентация была опубликована 2 года назад пользователемМария Вишневская

1 The Harmonic Oscillator Turning Points, A, -A no friction Stretch spring, let go. Mass, m, oscillates back and forth. m Hooke's Law linear restoring force spring constant force amplitude mass Harmonic oscillator - oscillates sinusoidally. A is how far the spring is stretched initially. At the turning points, A, -A, motion stops. All energy is potential energy. Copyright – Michael D. Fayer, 2007

2 Potential is Parabolic oscillator frequency, Hz oscillator frequency, rad/s Energy of oscillator is A can take on any value. Energy is continuous, continuous range of values. A - classical turning point. A Copyright – Michael D. Fayer, 2007

3 Quantum Harmonic Oscillator Simplest model of molecular vibrations Bond dissociation energy Molecular potential energy as a function of atomic separation. Bonds between atoms act as "springs". Near bottom of molecular potential well, Molecular potential approximately parabolic Harmonic Oscillator. Copyright – Michael D. Fayer, 2007

4 V x Potential Turning point Kinetic energy zero; potential energy max. Turning point Classical particle can never be past turning point. This can't happen for Q.M. harmonic oscillator. Uncertainty Principle indicates that minimum Q.M. H.O. energy Particle can be stationary at bottom of well, know position, x = 0; know momentum, p = 0. Copyright – Michael D. Fayer, 2007

5 One Dimensional Quantum Harmonic Oscillator in the Schrödinger Representation Schrödinger Representation kinetic energypotential energy Define Substitute H and definition of k. Mult. by -2m/ 2. Copyright – Michael D. Fayer, 2007

6 Find Must obey Born Conditions 1. finite everywhere 2. single valued 3. continuous 4. first derivative continuous Use polynomial method 1. Determine for 2. Introduce power series to make the large x solution correct for all x. Copyright – Michael D. Fayer, 2007

7 For very large x Therefore, can be dropped. Try Then, This is negligible compared to the first term as x goes to infinity. Copyright – Michael D. Fayer, 2007

8 Two solutions This is O.K. at This blows up at Not finite everywhere. Therefore, large x solution is For all x Must find this. Copyright – Michael D. Fayer, 2007

9 Need second derivative in Schrödinger equation With and Substitute into the original equation and divide by gives Equation only in f. Solve for f and have. Copyright – Michael D. Fayer, 2007

10 divide by substitute Gives Hermite's equation Substitute series expansion for H( ) Copyright – Michael D. Fayer, 2007

11 substitute in series The sum of these infinite number of terms in all powers of equals 0. In order for the sum of all the terms in this expression to vanish identically for any, the coefficients of the individual powers of must vanish separately. To see this consider an unrelated simpler equation. Fifth degree equation. For a given set of the a i, there will be 5 values of x for which this is true. However, if you know this is true for any value of x, then the a i all must be zero. Copyright – Michael D. Fayer, 2007

12 In general Even and odd series. Pick a 0 (a 1 = 0), get all even coefficients. Pick a 1 (a 0 = 0), get all odd coefficients. Recursion Formula Coefficients of like powers of. Copyright – Michael D. Fayer, 2007

13 Have expression in terms of series that satisfy the diff. eq. But not good wavefunction. Blows up for large |x| if infinite number of terms. (See book for proof.) For infinite number of terms and large |x|. blows up Unacceptable as a wavefunction. Copyright – Michael D. Fayer, 2007

14 Quantization of Energy If there are a finite number of terms in the series for H( ), wavefunction does not blow up. Goes to zero at infinity. The exponential goes to zero faster than n blows up. Then, because if a 0 or a 1 is set equal to zero (odd or even series) series terminates after = n a finite number of terms. = (2n + 1) To make series finite, truncate by choice of. n is an integer. Copyright – Michael D. Fayer, 2007

15 Any value of with = (2n + 1) is O.K. Any other value of is no good. Therefore, definition of definition of Solving for E n is the quantum number Lowest energy, not zero. Energy levels equally spaced by h. Copyright – Michael D. Fayer, 2007

16 Energy Levels Wavefunctions normalization constant Hermite Polynomials Copyright – Michael D. Fayer, 2007

17 Lowest staten = 0 Classical turning points potential total energy energy classical turning points - wavefunction extends into classically forbidden region. Copyright – Michael D. Fayer, 2007

18 More wavefunctions - larger n, more nodes Copyright – Michael D. Fayer, 2007

19 Probability for n = 10 Copyright – Michael D. Fayer, 2007 Looks increasingly classical. For large object, nodes so closely spaced because n very large that can't detect nodes. Classical turning points ~ = 4.6 Time oscillator spends as a function of position.

20 Dirac Approach to Q.M. Harmonic Oscillator Very important in theories of vibrations, solids, radiation Want to solve eigenkets, normalized We know commutator relation To save a lot of writing, pick units such that In terms of these units identity operator Copyright – Michael D. Fayer, 2007

21 Define operators is the complex conjugate (adjoint) of a since P and x are Hermitian. Then Hamiltonian commutator Copyright – Michael D. Fayer, 2007

22 Similarly and Can also show Therefore Very different looking from Schrödinger Hamiltonian. Copyright – Michael D. Fayer, 2007

23 Consider E; eigenket of H. scalar product of vector with itself only if We have Then normalized, equals 1 Therefore, Copyright – Michael D. Fayer, 2007

24 Now consider eigenket of H commutator rearrange transpose factor these are same Operate H on ket, get same ket back times number. is eigenket with eigenvalue, E - 1. eigenvalueeigenket Maybe number multiplying. Direction defines state, not length. Then, Copyright – Michael D. Fayer, 2007

25 a is a lowering operator. It gives a new eigenvector of H with one unit lower energy. Each application gives new ket one with one unit lower energy. Could keep doing this indefinitely, but Therefore, at some point we have a value of E, call it E 0, such that if we subtract 1 from it But E can't be < 1/2. Therefore For eigenvector not zero in conventional units Copyright – Michael D. Fayer, 2007

26 Raising Operator using the commutator rearranging, operating, and factoring as before These are the same. Therefore, is an eigenket of H with eigenvalue E + 1. takes state into new state, one unit higher in energy. It is a raising operator. number, but direction defines state Copyright – Michael D. Fayer, 2007

27 is the state of lowest energy with eigenvalue (energy) 1/2. Apply raising operator repeatedly. Each application gives state higher in energy by one unit. eigenvalue, one unit higher in energy With normal units Same result as with Schrödinger Eq. Copyright – Michael D. Fayer, 2007

28 is the state of lowest energy with eigenvalue (energy) 1/2. Apply raising operator repeatedly. Each application gives state higher in energy by one unit. eigenvalue, one unit higher in energy With normal units Same result as with Schrödinger Eq. Copyright – Michael D. Fayer, 2007

29 Eigenketslabeled with energy Can relabel kets with quantum number Take to be normalized. Raising and Lowering operators numbers multiply ket when raise or lower Will derive these below. Copyright – Michael D. Fayer, 2007

30 Consider operator operating on Therefore is an eigenket of operator with eigenvalue n. number operator. Eigenvalue – quantum number Important in Quantum Theory of Radiation and Solids and called creation and annihilation operators. Number operator gives number of photons in radiation field or number of phonons (quantized vibrations of solids) in crystal. Copyright – Michael D. Fayer, 2007

31 To find Take complex conjugate Now Work out from here Copyright – Michael D. Fayer, 2007

32 But Then and Therefore, Copyright – Michael D. Fayer, 2007 True if

33 Using the occupation number representation with normal units Consider Therefore, are eigenkets of H with eigenvalues. Copyright – Michael D. Fayer, 2007

34 Units in the raising and lowering operators Many constants. This is the reason why derivation was done in units such that. Need constants and units to work problems. Add operators, P cancels. x in terms of raising and lowering operators. Subtract operators, get P in terms of raising and lowering operators. Copyright – Michael D. Fayer, 2007

35 Can use the raising and lowering operator representation to calculate any Q.M. properties of the H. O. Example for ground state, average value of x 4 In Schrödinger Representation Copyright – Michael D. Fayer, 2007

36 constant - C Many terms. Must keep order correct. Operators dont commute. Copyright – Michael D. Fayer, 2007

37 Could write out all of the terms, but easier way. Any term that doesnt have same number of as and a + = 0 Example orthogonal = 0 Any operator that starts with a is zero. Can't lower past lowest state. Terms with are also zero because Copyright – Michael D. Fayer, 2007

38 Only terms left are No integrals.Must be able to count. Copyright – Michael D. Fayer, 2007

39 Vibrational Wave Packet ground electronic state excited electronic state vibrational levels short pulse optical excitation pulse bandwidth A short optical pulse will excite many vibrational levels of the excited state potential surface. Launches vibrational wave packet Copyright – Michael D. Fayer, 2007

40 Time dependent H. O. ket Model Excited State Vibrational Wave Packet with H. O. States Superposition representing wave packet on excited surface Calculate position expectation value - average position - center of packet. Copyright – Michael D. Fayer, 2007

41 only non-zero if Then But and This expression shows that time dependent. Time dependence is determined by superposition of vibrational states produced by radiation field. Copyright – Michael D. Fayer, 2007

42 Simplify Take n large so n >1 Also, i = Otherwise j = 0 Each state same amplitude in superposition for some limited set of states. Using these Position oscillates as cos( t). Copyright – Michael D. Fayer, 2007

43 Wave packet on harmonic potential surface. Packet moves back and forth. I 2 example Ground state excited to B state ~ 565 nm 20 fs pulseband width ~700 cm -1 Level spacing at this energy~69 cm -1 Take pulse spectrum to be rectangle and all excited same within bandwidth. States n = 15 to n = 24 excited (Could be rectangle) Copyright – Michael D. Fayer, 2007

44 Cos+1 to -1 distance traveled twice coefficient of Cos 10 equal amplitude states. Distance traveled = 1.06 Å. Comparable to bond length. Copyright – Michael D. Fayer, 2007

Еще похожие презентации в нашем архиве:

Готово:

AVL-Trees COMP171 Fall 2005. AVL Trees / Slide 2 Balanced binary tree * The disadvantage of a binary search tree is that its height can be as large as.

AVL-Trees COMP171 Fall 2005. AVL Trees / Slide 2 Balanced binary tree * The disadvantage of a binary search tree is that its height can be as large as.

© 2018 MyShared Inc.

All rights reserved.