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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, =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
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
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 =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 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
    • 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
    • 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
  • 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
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

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
  • 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
  • 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
  • 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)A Matrix Approach to Scattering

Consider a general, localized scattering problem

Region 1

Region 3

Region 2

boundary conditions1
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
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.
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