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Quantum Mechanics Lecture Notes 10/9

• Scattering problems using delta functions

We remind ourselves that the delta potential is defined as:

<math>V(x)=\pm \alpha \delta (x)</math>

Unlike in bound state problems, we are able to use <math>\pm</math> the delta function for scattering. We chose to use

<math>V(x)=- \alpha \delta (x)</math>

for this example. Therefore we want solutions to Schrödinger's Equation when E>0.

Schrödinger's Equation states:

<math>-\frac{\hbar^2}{2m}\frac{\partial^2}{{\partial x}^2}\psi (x) + V (x)\psi (x)=E\psi (x)</math>

In lecture it was asked how a particle can pass over the potential without being trapped. Yuichi said that if you started your “experiment” at t=0, the well would be empty and would likely trap particles. However, after some time has passed the potential “fills up” and reaches a steady state, and after this point most particles are just transmitted or reflected.

• Break the delta potential into two regions. One to the left of zero called <math>\psi_I(x)</math>, and one to the right of zero called <math>\psi_{II}(x)</math>
• Solve Schrödinger's equation for both regions and get the resulting wave functions
• Consider boundary conditions to solve unknowns

Yuichi mentioned that there are only a few problems in Quantum Mechanics that can be solved analytically (ISW, SHO, Hydrogen atom). The others we must approximate or solve numerically. For discontinuous potentials we have to solve it in separate parts and then “splice” them together.

1.) In region I & II V(x)=0. From this we know that Schrödinger's Equation is equal to:

<math>-\frac{\hbar^2}{2m}\frac{\partial^2}{{\partial x}^2}\psi (x)=E\psi (x)</math>

dividing out by

<math>-\frac{\hbar^2}{2m}</math>, we get:

<math>\frac{\partial^2}{{\partial x}^2}\psi (x)=(-\frac{2mE}{\hbar^2})\psi (x)</math>.

Now we introduce the constant <math>k^2</math> and set that equal to: <math>k^2=\frac{2mE}{\hbar^2}</math>

The Schrödinger Equation now reads:

<math>\frac{\partial^2}{{\partial x}^2}\psi (x)=-k^2 \psi(x)</math>

Similar to lec_notes_1007, we then know that the solutions to the Schrödinger Equation are:

<math>\psi_I (x)=Ae^{ikx}+Be^{-ikx}</math>

<math>\psi_{II}(x)=Ce^{ikx}+De^{-ikx}</math>

Now consider the difference between the scattering solutions and the bound state solutions. We notice immediately that these solutions contain the imaginary <math>i</math> value. In the bounded case we could say that the B and C values must have been zero because they blow up as they approach<math>\pm \infty</math>. Can we say the same in the scattering state? The answer is no, because with the <math>i</math>, the exponentials behave as sine and cosine functions. So it's not nice and zero, but it is also not <math>\infty</math> and must therefore still be taken into consideration.

2.) Now we look for boundary conditions. When we integrate the Schrödinger Equation from <math>-\varepsilon </math> to <math>\varepsilon</math> we get the first boundary condition:

<math>\Delta\psi\prime(0)=-\frac{2m\alpha}{\hbar^2}\psi(0)</math>

The <math>\psi</math> in the above boundary condition can be either <math>\psi_I</math> or <math>\psi_{II}</math>. This is because of the second boundary condition that states:

<math>\psi_I(0)=\psi_{II}(0)</math>

When we count up the unknowns, we see that we have 'A', 'B', 'C', and 'D'. Is 'k' also an unknown? We might think that it is, because we don't necessarily know E (which is a part of k), but we must remember the big picture. As an experimenter, we do have control over the energy of the particles in this case. With the bound state scenario (previous lecture) we had no control. The particle is just “thrown” and and you wait for it to settle. In the scattered state we “shoot” the particle in and therefore we have control of the energy, and consequently we can know the value of 'k'.

So there are 4 unknowns: 'A', 'B', 'C', 'D' and two equations. We need to consider the physical significance of these constants.

Let's say we shoot the particle from -<math>\infty</math>, then:

*the A term corresponds to the amplitude of a wave coming in from the left (incident)

*the B term corresponds to the amplitude of the wave returning to the left (reflection)

*the C term corresponds to the amplitude of the wave traveling off to the right (transmission)

*the D term represents the particle coming in from the right (<math>+\infty</math>)

Upon consideration, we realize that the D term must be zero. This is because we shot the particle from -<math>\infty</math>, and therefore the particle cannot reasonably come from <math>+\infty</math> for scattering from the left.

With our two boundary conditions

<math>\Delta\psi\prime(0)=-\frac{2m\alpha}{\hbar^2}\psi(0)</math>

and

<math>\psi_I(0)=\psi_{II}(0)</math>

we then use some algebraic manipulation to find B and C in terms of A. First let's use the convenient notation:

<math>\beta = \frac{m\alpha}{k\hbar^2}</math>.

Then,

<math> B=\frac{\i\beta}{1-\i\beta}A</math>

<math> C=\frac{1}{1-\i\beta}A</math>

From this we can derive certain useful ratios to help deal with the unknowns.

<math>\frac{B}{A}=\frac{i\beta}{1-i\beta}</math>

<math>\frac{C}{A}=\frac{1}{1-i\beta}</math>

We then define the Reflection coefficient as:

<math>R=|\frac{B}{A}|^2=\frac{\beta^2}{1+\beta^2}</math>,

and the transmission coefficient as:

<math>T=|\frac{C}{A}|^2=\frac{1}{1+\beta^2}</math>

It was asked in lecture why we must square the ratio in order to obtain R and T. We square the ratio because the ratios alone are complex. Without multiplying by the complex conjugate, they can't have any physical significance.

We approximate or assume what a potential looks like irl, even though we don't really know. However, experimentally we can vary the energy and since we know <math>\beta</math>, we can expect values for R and T. We can tell if the potential is like a <math>\delta</math> function by seeing how well the values match up.

A general point: transmission increases with greater energy, and reflection decreases.

This makes sense, considering that higher energy particle's would be less likely to be affected by the potential well (and thus merely pass over). It follows that particles with energies closer to V(x) are more likely to be affected.

Using plane waves we can form the wave packet:

<math>\Psi(x) = \int \phi(k) e^{-ikx} dx.</math>

Does this satisfy the Schrödinger Equation? That might be a little too optimistic, and in fact it doesn't happen. However, we can say:

<math>\Psi_I(x) = \int \phi(k) [e^{ikx}+\frac{B}{A}e^{-ikx}] dk</math> for x<0

and

<math>\Psi_{II}(x) = \int \phi(k) [\frac{C}{A}e^{ikx}] dk</math> for x>0.

Each k will be slightly different for each equation, but once these two equations are formed into a linear combination they will satisfy the Schrödinger Equation and boundary conditions.

*I understood the description of the real incident wave packet and the imaginary wave packet from the end of lecture but I don't think I can explain it very well on here. If someone wants to give it a shot, feel free.

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