Responsible party: Spherical Chicken, Anaximenes

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Monday, Sept. 28

The equation we use for our basis is

<math>\frac{d^2\psi}{d\xi^2} = \left(\xi^2 - K\right)\psi\left(\xi\right)</math> … (A)

From the note, I am afraid that some points I tried to make were not clear, so I have tried to clarify them in this note. – Yuichi

This can be used to figure out solutions by knowing the value of a solution at one point, and deduce its value at slightly larger/smaller value to <math>\xi</math>.

Since we have learned a while ago (where was it?) that when the potential energy is even (symmetric around zero), the stationary-state solutions are either even OR odd, we can use this to “know” the value of the wave function at <math>\xi=0</math>. For an even solution, the slope: <math>\frac{d\psi}{d\xi}</math> has to be zero (why?), while modulo the normalization of the solution, <math>\psi(0)</math> can be any value (meaning that once an un-normalized solution is found, one can always scale it so that it is normalized). Using the same reasoning, one can argue that for an odd solutions, <math>\frac{d\psi}{d\xi}</math> can be any value, while <math>\psi(0)</math> must be zero.

Let's try to reconstruct a solution for <math>K=1</math> case, which we already know should give an even solution. Starting at some value for <math>\psi(0)</math>, and <math>\frac{d\psi}{d\xi}=0</math>, using the idea of Taylor expansion, <math>\psi(\Delta\xi)\sim\psi(0)+\frac{d\psi(0)}{d\xi}\Delta\xi</math>. At the same time since we need <math>\frac{d\psi(\Delta\xi)}{d\xi}</math> in the next step, let's figure it out. For this we say (also Taylor expansion), <math>\frac{d\psi(\Delta\xi)}{d\xi}=\frac{d\psi(0)}{d\xi}+\frac{d^2\psi(0)}{d\xi^2}\Delta\xi=\frac{d^2\psi(0)}{d\xi^2}\Delta\xi</math>. Here, the Schrodinger equation (A) can be used to evaluate the 2nd derivative from the the wave function and its first derivative!

This process can be repeated millions of time to figure out the values of <math>\psi(\xi)</math> at many values.

Now, let's step back, and look at the bigger picture to see how the wave function will look like. We focus on the sign of the 2nd derivative of the wave function. Knowing that when the 2nd derivative is zero (at some value of <math>\xi</math>), the wave function has an inflection point there, while if it is positive, the wave function is concave up, and if it is negative, it is concave down.

Suppose the K takes the smallest value, 1. Then when <math>0<\xi<1</math>, the term <math>\xi^2-K</math> is negative. So the sign of <math>\psi</math> and <math>\frac{d^2\psi}{d\xi^2}</math> is opposite. For simplicity, let's assume that <math>\psi(0)>0</math>. Then <math>\frac{d^2\psi}{d\xi^2}</math> will be negative (meaning that the wave function is concave down) until either <math>\psi</math> reaches zero and going negative or <math>\xi=1</math> is reached. It turns out that since <math>\xi^2-K</math> is not a big negative (K is small in the current example) so that the wave function never reaches zero before <math>\xi=1</math> causes the inflection, where the concave up region starts. When <math>K</math> is just right (at 1), the downward slope of the wave function at <math>\xi=1</math> is just right so that the wave function will slowly approach zero while the slope also approach zero. If <math>K</math> is slightly less than 1, the curvature of the concave is looser, and the slope of the wave function is less negative. With the concave up region above <math>\xi=1</math> will gradually increase the wave function's slope to a positive value before <math>\xi=\infty</math>, and the wave function diverge to the <math>\infty</math> from there. On the other hand, if <math>K</math> is slightly more than 1, the curvature of the concave is tighter, and the slope of the wave function is more negative. With the concave up region above <math>\xi=1</math> will gradually increase the wave function's slope to a less negative value but before <math>\xi=\infty</math>, the wave function will reach zero. At the point, the sign of the 2nd derivative changes its sign! and the slope starts to become more negative, and the wave function will diverge to the <math>-\infty</math> from there. End of Yuichi's clarification

Essentially, we started with the zero slope point, and then move a little right (for ground state, this is the x=0 point, top of concave down parabola). We start at <math>\frac{d\psi}{d\xi} = 0</math> for <math>\xi = 0 </math>

Then if K is large, the slope starts going down.

When

<math>\frac{d^2\psi}{d\xi^2} < 0 </math> the function will be concave down.

When <math>\frac{d^2\psi}{d\xi^2} > 0 </math> the function will be concave up.

When

<math>\frac{d^2\psi}{d\xi^2} = 0 </math> We have an inflection point.

An inflection point is the point in the function when the slope changes from concave up to concave down (or vice versa).

When we start the solving, we start with <math>\psi\left(\xi\right) = 0</math>.

However when <math>\psi = 0</math>, there is an inflection point as well. Also when the wave function = 0 there is an inflection point. (when the function drops below the x axis).

When <math>\xi^2 >K </math> then <math>\left(\xi^2 - K\right)</math> is positive. Then <math>\psi < 0 </math> .

Suppose K is slightly larger than the solved for value. At this point, the convexity is steeper and the function will not converge.

If the wave function touches the x axis, (when say K is slightly larger and raises the function up) then the function will continue to cross, be concave up, and grow to infinity. Of course then, the function cannot converge and hence is not normalizable.

The same is true if the K is just slightly smaller. The function's slope will go to zero, and then drop off to negative infinity. This as well cannot be normalized. Thus there are very special solutions of K. However we should not forget that there are multiple values for K. But for each K, the value is unstable and will go off to negative or positive infinity, and to a non-normalizable value.

Remember, <math>K = 2n + 1 </math> thus for n = 0,1,2… , K is 1, 3, 5….

For the SHO, as K becomes larger, the inflection points will be spread out (<math>\xi=1, \sqrt{2}, \sqrt{3}</math>, … for the first, second and third energy states. See the graph above.) The wiggly region of the wave function slowly grows longer). The final inflection point is where <math>\left(\xi^2 - k \right) = 0 </math>.

We remember that we got <math>\xi^2</math> from potential energy, <math>\frac{1}{2}kd^2 </math> and the K from total energy. thus <math> \frac{1}{2}kx^2 - E = 0 </math> Is the inflection point. So classically, in the case of the SHO, this is where the particle on the spring starts turning back from its expansion (or compression).

There was a question prof. Kubota addressed, from an email or online, that asked how for the infinite square well, <math>E_n \prop n^2</math> However for the SHO <math>E \prop n</math> So – Why do we lose a factor of n for the infinite square well vs. the SHO?

The answer lies in remembering the equations for energy. <math> E \prop \frac{k^2 \hbar^2}{2m} </math> and since <math>k \prop n {\rm then } E \prop n^2</math>

However in an infinite square well, we lose the power of n. To understand how this works. imagine the physical representations of these two systems. for the infinite square well, the energy levels are raised up the ladder of a square box. However for the SHO, as the energy levels go up, they also get more room, because they are ladders in a parabola. Thus as E increases, the wave spreads out. So <math>\lambda \prop \Delta x </math> where <math>\Delta x</math> is the region between the two inflections, and <math>\lambda \prop \frac{\Delta x}{n} \prop \frac{1}{\sqrt n} </math>

then <math>k \prop \sqrt n </math>

so when you square k, it turns out that

<math> E \prop n </math>

The key to this explanation being that in the SHO, the waves are not confined to the same space, but are allowed to spread out as the energy increases.

• We are supposed to know the formulas in a general form, as well as to know that certain formulas exist and may be useful for certain cases. If we have a problem remembering exact numbers or details, we are encouraged to ask the proctor for assistance, in which case he will help fill in what you already have for the equation.
• This is a no calculator quiz, as well. No extravagant calculations should be necessary.

For the Free Particle, we have the equation

<math>\psi(x) = e^{ikx} </math>

(Which can also be written cos (kx) + isin(kx) )

With the time dependance it is written

<math>\Psi(x,t) = e^{i(kx-\omega t)}</math>

where omega is <math>\omega(k) = \frac{E}{\hbar} = \frac{k^2 h^2}{\hbar2m}</math>

(Note that these are not normalized waves.)

The wave packet can be written as * a wave packet *linear combination of plane waves

Which means that if we can understand the second representation, we can understand the first better. (It is much easier to compute the second).

From 2601 last year we remember representing a particle wave hitting a finite potential. Part of the wave bounces back, part and part of it goes through. In this way we can calculate the wave forms from the boundary conditions.

<math> \left|\frac{B}{A}\right |^2 = Reflection </math>

A is the particles going in, and B is the particles going out.

<math>\left | \frac{C}{A} \right| ^2 \frac{K'}{K} = Transmission </math>

A is the number of particles / unit length, and C is number of particles transmitted right.

Thus if the potential goes up, the speed of particles going through is slower. <math> Transmission = \frac{# Particles/second}{# Particles / Second}</math>

and <math>\frac{K'}{K} <math> is the conversion of particles/meter to particles to second. (these are the velocity ratios of the two areas).

Now we've done this in terms of the plane wave. Now we'll do it in terms of the wave packet. The reason we do it for the plane wave is because it's much easier than the wave packet.

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