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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

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Mr a square unbound

Mr. A. Square Unbound

Continuum States in 1-D Quantum Mechanics


With apologies to shelley
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
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
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
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
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
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
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
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 d1
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
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 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


The step potential

V(x)=V0

Region 2

Region 1

x=0

The Step Potential


Region 1
Region 1

  • “A” is the amplitude of the incident wave

  • “B” is the amplitude of the reflected wave


Region 2
Region 2

  • “C” is the amplitude of the transmitted wave


Matching boundary conditions
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
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


Graphically

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
Reflection and Transmission Coefficients

  • If k2 is imaginary, T=0

  • If k2 is real, then


In terms of energy
In terms of Energy,

  • If E<V0 then R=1 and T=0

  • If E>V0 then


The step potential1
The Step Potential

V(x)=V0

Region 2

Region 1

Region 3

x=0

x=a







Some consequences
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 v 0
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
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
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
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




Applying the boundary conditions
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
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


A matrix approach to scattering

V(x) singularity is given by:

A Matrix Approach to Scattering

Consider a general, localized scattering problem

Region 1

Region 3

Region 2


Wavefunctions
Wavefunctions singularity is given by:


Boundary conditions1
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
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 kik and look for the energies where the S-matrix blows up.


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