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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

• 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

• 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.

• 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

• eikx represents a particle moving from right to left

• e-ikx represents a particle moving from left to right

• 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

• 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 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

• 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 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 2

Region 1

x=0

The Step Potential

• “A” is the amplitude of the incident wave

• “B” is the amplitude of the reflected wave

• “C” is the amplitude of the transmitted wave

• 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

• 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

• If k2 is imaginary, T=0

• If k2 is real, then

• If E<V0 then R=1 and T=0

• If E>V0 then

V(x)=V0

Region 2

Region 1

Region 3

x=0

x=a

• 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

• 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

• 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 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 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 Scattering

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