Mr a square unbound
Sponsored Links
This presentation is the property of its rightful owner.
1 / 40

Mr. A. Square Unbound PowerPoint PPT Presentation


  • 50 Views
  • Uploaded on
  • Presentation posted in: General

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

Download Presentation

Mr. A. Square Unbound

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


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:


The Free Particle


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


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<V0

    • 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<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


In terms of Energy,

  • If E<V0 then R=1 and T=0

  • If E>V0 then


The Step Potential

V(x)=V0

Region 2

Region 1

Region 3

x=0

x=a


The Wave Function


Boundary Conditions


Apply Boundary Conditions


Solving


Reflection and Transmission Coefficients


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

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


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 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


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=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.


  • Login