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• main points understood, and expand them - what is your understanding of what the points were.
  • expand these points by including many of the details the class discussed.
• main points which are not clear. - describe what you have understood and what the remain questions surrounding the point(s).
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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{\delta^2}{{\delta 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. Code Text

• Break the delta potential into two regions. One to the left 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 i 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 i, 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.

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