Mr. A. Square Unbound

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# Mr. A. Square Unbound - PowerPoint PPT Presentation

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

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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, =2k2
• So
• Dp=-

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, =2k2
• So
• Dp=-

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 =2k2 then I can find a “quantum velocity”= /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
V(x)=V0

Region 2

Region 1

x=0

The Step Potential
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
• Classically, the particle is repelled
• In QM, k2 is imaginary and y2 describes an attenuating wave
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
• 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
In terms of Energy,
• If E
• If E>V0 then
The Step Potential

V(x)=V0

Region 2

Region 1

Region 3

x=0

x=a

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
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

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)=-ad(x-x0)

x=x0

Region 1

Region 2

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 a is negative, then the spike is repulsive and there are no bound states
V(x)A Matrix Approach to Scattering

Consider a general, localized scattering problem

Region 1

Region 3

Region 2

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)
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.