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

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

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.

Normalizing Infinity One problem if  (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:

The Free Particle

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

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

Obviously e ikx represents a particle moving from right to left e -ikx represents a particle moving from left to right

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  is not really independent of k

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

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 d  /dk

The phase velocity So the ripple travels at ½ the speed of the particle Also, note if =  2 k 2 then I can find a “quantum velocity”= /m 2  2 k 2 /m 2 = E/2m=v q So v q is the phase velocity or the quantum mechanical wave function travels at the phase speed

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  in terms of k is called the dispersion relation

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

“A” is the amplitude of the incident wave “B” is the amplitude of the reflected wave

Region 2 “C” is the amplitude of the transmitted wave

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

Applying some algebra If E>V 0 then E-V 0 >0 or “+” Then k 2 is real and  2 is an oscillator propagation If E<V 0 Classically, the particle is repelled In QM, k 2 is imaginary and  2 describes an attenuating wave

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

Reflection and Transmission Coefficients If k 2 is imaginary, T=0 If k 2 is real, then

In terms of Energy, If E>V 0 then If E<V 0 then R=1 and T=0

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

The Wave Function

Boundary Conditions

Apply Boundary Conditions

Solving

Reflection and Transmission Coefficients

Some Consequences When ka=n* , 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

For E<V 0 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 cosh 2 z-sinh 2 z=1 T cannot be unity so there is no Ramsauer- Townsend effect

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

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

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)=-  (x-x 0 ) x=x 0 Region 1 Region 2

Wavefunctions and Boundary Conditions

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

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

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

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

Wavefunctions

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=S 11 A+S 12 G F=S 21 A+S 22 G S ij 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)

Consequences The case of scattering from the left, G=0 so R L =|S 11 | 2 and T L =|S 21 | 2 The case of scattering from the right, F=0 so R R =|S 22 | 2 and T R =|S 12 | 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  i  and look for the energies where the S-matrix blows up.