Quantum Mechanics for Scientists and Engineers. David Miller

Transcription

1 Quantum Mechanics for Scientists and Engineers David Miller

2 Particles in potential wells The finite potential well

3 Insert video here (split screen) Finite potential well Lesson 7 Particles in potential wells Insert number

4 Particle in a finite potential well We will choose the height of the potential barriers as V o with potential energy at the bottom of the well The thickness of the well is L Now we will choose the position origin in the center of the well V o / L / L

5 Particle in a finite potential well If there is an eigenenergy E for which there is a solution then we already know what form the solution has to take sinusoidal in the middle exponentially decaying on either side V o / L / L

6 Particle in a finite potential well For some eigenenergy E with k me/ and mv o E/ for L / Gexp for L / L / Asin k Bcos k for L / Fexp L / with constants A, B, F, and G L / V o

7 Particle in a finite potential well Now we need to apply the boundary conditions to solve for the unknown coefficients constants A, B, F, and G Gexp L Asin kbcos k Fexp L or at least three of them the fourth could be found by normaliation V o / L / L / / / L / L

8 Particle in a finite potential well From continuity of the wavefunction at L Writing X gives / V o L / Fexp L / Asin kl / Bcos kl / S C L L L exp L / sin kl / cos kl / FX AS BC L L L / L / L

9 Particle in a finite potential well Similarly at L / V o GX AS BC L L L Continuity of the derivative gives at L / GX AC BS k at L / FX AC BS k L L L L L L / L / L

10 Particle in a finite potential well So we have four relations GX AS BC L L L FX L ASL BCL GX L ACL BSL k FX L ACL BSL k Now we need to find what solutions are compatible with these V o / L / L

11 Particle in a finite potential well Adding gives GX AS BC L L L FX AS BC L L L Subtracting FX L ACL BSL k from GX L ACL BSL k gives BS F GX k BCL F G XL L L V o / L / L

12 Particle in a finite potential well As long as F G we can divide BS F GX by k BC F G X to obtain L L L L tan kl / / k V o This relation is effectively a condition for eigenvalues / L / L

13 Particle in a finite potential well Subtracting from gives GX AS BC L L L FX AS BC L L L ASL F G XL V o Adding and gives FX L ACL BSL k GX L ACL BSL k AC F GX k L L / L / L

14 Particle in a finite potential well Similarly, as long as F G we can divide AC F GX k by AS F G X to obtain This relation is also effectively a condition for eigenvalues L L V o L L cot kl / / k / L / L

15 Particle in a finite potential well For any case other than F G which leaves tan kl / / k but not cot kl / / k or F G which leaves cot kl / / k but not tan kl / / k V o then the solutions tan kl / / k and cot kl / / k are contradictory / L / L

16 Particle in a finite potential well So the only possibilities are V o 1 - F G and tan kl / / k F G and cot kl / / k / L / L

17 Particle in a finite potential well V o 1 - F G and tan kl / / k Note from ASL F G XL and AC F GX k S L and C L cannot both be so A Hence in the well we have cos k which is an even function L L / L / L

18 Particle in a finite potential well 1 - F G and cot kl / / k Note from BCL F G XL and BS F GX k S L and C L cannot both be so B Hence in the well we have sin k L L which is an odd function V o / L / L

19 Particle in a finite potential well Though we have found the nature of the solutions we have not yet formally solved for the eigenenergies E and hence for k and We do this by solving tan kl / / k and cot kl / / k V o / L / L

20 Solving for the eigenenergies Change to dimensionless units Use the energy of the first level in the infinite potential well width L leading to a dimensionless eigenenergy and a dimensionless barrier height Also E 1 m L / / / 1 / E/ E 1 v V / E k me L E E L mve / / L VE / E / L o o 1 o o o 1

21 Solving for the eigenenergies Vo E vo Consequently k E kl E L Vo E and v o E1 E1 So tan kl / / k becomes tan / v o / or tan / vo and cot kl / / k becomes cot / v o / or cot / vo

22 Graphical solution Choose a specific well depth and plot the curve v o o 4 6 8

23 Graphical solution Choose a specific well depth and plot the curve v o o

24 Graphical solution Choose a specific well depth and plot the curve v o o

25 Graphical solution Choose a specific well depth and plot the curve v o Now add the curves tan cot o

26 Graphical solution For a specific the solutions are the values of at the intersections of v o and tan or cot o 8

27 Solutions These are the solutions for a well depth V o of 8E 1 Note that they are all lower energies than the corresponding solutions for the infinitely deep well of the same width Vo 8 E n 3 n n 1

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29 Particles in potential wells The harmonic oscillator

30 Mass on a spring A simple spring will have a restoring force F acting on the mass M Mass M S

31 Mass on a spring A simple spring will have a restoring force F acting on the mass M proportional to the amount y by which it is stretched y For some spring constant K F Ky The minus sign is because this is restoring it is trying to pull y back towards ero This gives a simple harmonic oscillator S Force F Mass M

32 Mass on a spring From Newton s second law d y F Ma M Ky dt i.e., d y K y y dt M where we define K / M we have oscillatory solutions of angular frequency K / M e.g., y sint S

33 Potential energy 1 K V Force F Mass m The potential from the restoring force F is 1 V F d o K o do K S1 m

34 Harmonic oscillator Schrödinger equation 1 With this potential energy V m the Schrödinger equation is d 1 m E md For convenience, we define a dimensionless distance unit m so the Schrödinger equation becomes 1 d E d

35 Harmonic oscillator Schrödinger equation One specific solution to this equation is 1 d E d exp( / ) with a corresponding energy E / This suggests we look for solutions of the form A exp / H n n n where H n is some set of functions still to be determined

36 Harmonic oscillator Schrödinger equation Substituting n Anexp / Hn into the Schrödinger equation 1 d E d gives d Hn dhn E 1 H n d d This is the defining differential equation for the Hermite polynomials

37 Harmonic oscillator Schrödinger equation Solutions to d Hn dhn E 1 H n d d exist provided E 1 n n, 1,, that is, 1 En n n, 1,,

38 Harmonic oscillator Schrödinger equation The allowed energy levels are equally spaced separated by an amount where is the classical oscillation frequency Like the potential well there is a ero point energy here / 1 En n n, 1,,

39 Hermite polynomials The first Hermite polynomials are Note they are either odd or even i.e., they have a definite parity They satisfy a recurrence relation 1 H H n H n n1 n successive Hermite polynomials can be calculated from the previous two H 1 H H 4 H3 8 1 H

40 Harmonic oscillator solutions Normaliing gives A exp / H n n n 1 A n n n! m 1 En n, 1,, n

41 Harmonic oscillator solutions Normaliing gives or n A exp / H n n n 1 A n n n! m 1 En n, 1,, n 1 m m m H n n n! exp

42 Harmonic oscillator eigensolutions A 4 4 exp / Energy 1 / 9 / A exp / / A exp / 4 5 / A exp / 1 3 / A exp / / 4 4

43 Classical turning points The intersections of the parabola and the dashed lines give the classical turning points where a classical mass of that energy turns round and goes back downhill Energy / / 7 / 5 / 3 / /

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