Analyzing 1-D Quantum States: From Bound to Unbound Continuum Dynamics

1. Mr. A. Square Unbound Continuum States in 1-D Quantum Mechanics

2. With Apologies to Shelley • In the previous section, we assumed • That a particle exists in a 1-d space • That it experiences a real potential, V(x) • That its wavefunction is a solution of the TISE or TDSE • That at infinity, its wavefunction is zero. • In this section, those are removed

3. The consequences • If the boundary condition at infinity is removed, • Then a quantum system is not limited to a discrete set of states but • A continuum of energies is allowed.

4. Normalizing Infinity • One problem if y(x)∞, how do you normalize it? • Well, Postulate 7 (wherein we discuss normalization) is based on the proviso that it mainly applies to bound states. • Mathematically, if we have to find a matrix element, we perform the following operation:

5. The Free Particle

6. Assume k>0 & real, and B(k)=0, then Y describes a wave moving from –x to +x • Obviously, <p2>=2k2 • So • Dp=<p2>-<p>2 =0 • There is no variance in momentum, thus the free particle has mixed momentum • This is in agreement with Newton’s 1st Law

7. Assume k<0 & real, and A(k)=0, then Y describes a wave moving from +x to -x • Obviously, <p2>=2k2 • So • Dp=<p2>-<p>2 =0 • There is no variance in momentum, thus the free particle has mixed momentum • This is in agreement with Newton’s 1st Law

8. Obviously • eikx represents a particle moving from right to left • e-ikx represents a particle moving from left to right

9. The Wave Packet as a solution • Another solution to the TDSE is a “wave packet” • As an example, let B(k)=0 and the solution is in the form of the integral: • Note that this is the inverse Fourier transform • A complication arises in that w is not really independent of k

10. The Wave Packet cont’d • Typically, the form of A(k) is chosen to be a Gaussian • We also assume that w(k) can be expanded in a Taylor series about a specific value of k

11. The Wave Packet cont’d • The packet consists of “ripples” contained within an “envelope” • “the phase velocity” is the velocity of the ripples • “the group velocity” is the velocity of the envelope • In the earlier expansion, the group velocity is dw/dk

12. The phase velocity • So the ripple travels at ½ the speed of the particle • Also, note if <p2>=2k2 then I can find a “quantum velocity”= <p2> /m2 • 2k2/m2= E/2m=vq • So vq is the phase velocity or the quantum mechanical wave function travels at the phase speed

13. The Group Velocity • The group velocity (the velocity of the envelope) is velocity of the particle and is twice the ripple velocity. • BTW the formula for w in terms of k is called the dispersion relation

14. V(x)=V0 Region 2 Region 1 x=0 The Step Potential

15. Region 1 • “A” is the amplitude of the incident wave • “B” is the amplitude of the reflected wave

16. Region 2 • “C” is the amplitude of the transmitted wave

17. Matching Boundary Conditions • The problem is that we have 2 equations and 3 unknowns. • “A” is controlled by the experimenter so we will always solve ALL equations in terms of the amplitude of the incident wave

18. Applying some algebra • If E>V0 then E-V0>0 or “+” • Then k2 is real and y2 is an oscillator propagation • If E<V0 • Classically, the particle is repelled • In QM, k2 is imaginary and y2 describes an attenuating wave

19. V(x)=V0 V(x)=V0 Region 2 Region 2 Region 1 Region 1 x=0 x=0 Graphically • If E>V0 then E-V0>0 or “+” • Then k2 is real and y2 is an oscillator propagation • If E<V0 • Classically, the particle is repelled • In QM, k2 is imaginary and y2 describes an attenuating wave

20. Reflection and Transmission Coefficients • If k2 is imaginary, T=0 • If k2 is real, then

21. In terms of Energy, • If E<V0 then R=1 and T=0 • If E>V0 then

22. The Step Potential V(x)=V0 Region 2 Region 1 Region 3 x=0 x=a

23. The Wave Function

24. Boundary Conditions

25. Apply Boundary Conditions

26. Solving

27. Reflection and Transmission Coefficients

28. Some Consequences • When ka=n*p, n=integer, implies T=1 and R=0 • This happens because there are 2 edges where reflection occur and these components can add destructively • Called “Ramsauer-Townsend” effect

29. For E<V0 • Classically, the particle must always be reflected • QM says that there is a nonvanishing T • In region 2, k is imaginary • Since cos(iz)=cosh(z) • sin(iz)=isinh(z) • Since cosh2z-sinh2z=1 • T cannot be unity so there is no Ramsauer-Townsend effect

30. What happens if the barrier height is high and the length is long? • Consequence: T is very small; barrier is nearly opaque. • What if V0<0? Then the problem reduces to the finite box • Poles (or infinities) in T correspond to discrete states

31. An Alternate Method We could have skipped over the Mr. A Square Bound and gone straight to Mr. A Square Unbound. We would identify poles in the scattering amplitude as bound states. This approach is difficult to carry out in practice

32. The Dirac Delta Potential • The delta barrier can either be treated as a bound state problem or considered as a scattering problem. • The potential is given by V(x)=-ad(x-x0) x=x0 Region 1 Region 2

33. Wavefunctions and Boundary Conditions

34. From the previous lecture, the discontinuity at the singularity is given by:

35. Applying the boundary conditions • R cannot vanish or only vanishes if k is very large so there is always some reflection

36. Solving for k and E • This is in agreement with the result of the previous section. • If a is negative, then the spike is repulsive and there are no bound states

37. V(x) A Matrix Approach to Scattering Consider a general, localized scattering problem Region 1 Region 3 Region 2

38. Wavefunctions

39. Boundary Conditions • There are four boundary conditions in this problem and we can use them to solve for “B” and “F” in terms of “A” and “G”. • B=S11A+S12G F=S21A+S22G • Sij are the various coefficients which depend on k. They seem to form a 2 x 2 matrix • Called the scattering matrix (s-matrix for short)

40. Consequences • The case of scattering from the left, G=0 so RL=|S11|2 and TL=|S21|2 • The case of scattering from the right, F=0 so RR=|S22|2 and TR=|S12|2 • The S-matrix tells you everything that you need to know about scattering from a localized potential. • It also contains information about the bound states • If you have the S-matrix and you want to locate bound states, let kik and look for the energies where the S-matrix blows up.