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### 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 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:

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

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

Obviously

- eikx 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 w 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 w(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 dw/dk

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

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

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>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

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

Reflection and Transmission Coefficients

- If k2 is imaginary, T=0
- If k2 is real, then

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

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

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

An Alternate Method long?

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 long?

- 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

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

Applying the boundary conditions singularity is given by:

- R cannot vanish or only vanishes if k is very large so there is always some reflection

Solving for k and E singularity is given by:

- 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

V(x) singularity is given by:

A Matrix Approach to ScatteringConsider a general, localized scattering problem

Region 1

Region 3

Region 2

Wavefunctions singularity is given by:

Boundary Conditions singularity is given by:

- 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)

Consequences singularity is given by:

- 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 kik and look for the energies where the S-matrix blows up.

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