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There really is a test this friday? That's not a typo on the syllabus? Is this basically the finite square well, delta function and the new vector notation we've been learning? Anyone else as excited about this as I am? :p
I am! Just don't forget how complicated scattering states can be… and that the transmission coefficient isn't always |F/A|^2. Otherwise, I think this should be a very manageable test. If it's anything like the sample exam I'll be just fine ;)
Depending on your degree of sarcasm, i am either just as excited as you or much less excited than you. :)
I'm pumped!
At the bottom of page 97 Griffiths shows that the momentum operator is hermitian (for functions in Hilbert space). However, example 3.2 shows that there are no eigenfunctions for the momentum operator in Hilbert space. What are we supposed to make of this?
I guess that's why physicists want to expand the Hilbert space to include some non-normalizable functions - plane waves which correspond to momentum eigenstates which reside in a sort of pseudo-Hilbert Space.
What is the difference between the operator Q with a hat, and just Q? I keep seeing them in different places, and some of the problems I've seen seem to imply that they are not exactly the same, like I was assuming.
My understanding is that <math>\hat{Q}</math> is the operator and <math>Q</math> is the observable quantity.
I apologize that I have not been consistent, but as Chavez says, the intention is that <math>\hat{Q}</math> is the operator, though sometimes, I use just <math>Q</math> as an operator.
Griffiths used <math>\hat{Q}</math> explicitly for an operator that is related to some function Q(x, p) that is the observable, on p. 96. So on 97 he says that <Q> = <Q>* is the observable, and < ψ | <math>\hat{Q}</math> ψ > is the operator and associated wavefunction.
I'm a little confused about the chapter 3.1 footnote 6 which states that even the finite integral from 'a' to 'b' of a function that is zero everywhere except a few isolated points (say it equals 50 at x=2) still vanishes. How is this? I know he says it doesn't occur in physics, but it's still a little disconcerting.
My impression is that if you have a function that has a value of 50 at x=2, then taking an integral from a<2<b would mean finding the area under f(x) from a to b. Since it has a single value at x=2 then this spike has no area; there is no width to a spike that is 50 units tall, so the area contribution to the integral is zero, and the total integral is 0 assuming there is no other contributions to the total area from a to b. The Dirac delta function is defined to have an area of 1, which seems like an ad-hoc bypass to this problem.
Yep, these isolated points jumping out of line mean the function is discontinuous. And the reason why we don't have to worry about that is because real world physical functions are continuous. Or at least they usually are. I guess I don't know for sure but I can't think of any off hand that aren't.
In this same footnote, what does Griffiths mean when he says that “technically, vectors in Hilbert space represent equivalence classes of functions.” What is an equivalence class?
The equivalence class of a vector v in Hilbert space consists of all vectors in Hilbert space that are equivalent to v. For any v, there is thus an equivalence relation satisfying transitive, symmetric, and reflexive properties for each vector in its equivalence class. In the footnote, he is basically saying that Hilbert space is partitioned into equivalence classes representing sets of functions having the same square integral.
Will the homework solutions be available some time before the test? We're covering the finite square well, transmission/reflection, hilbert space, dirac and double dirac…. anything else anyone can think of? Proof of hermitianness?
I have just posted all solutions except Quiz 1, which will be posted tomorrow. If you find anything missing, let me know.
Since we have been doing so many problems about boundary condition, every time I look at a problem I thought I understood it, actually not. I am still confused under which case the negative or positive infinity term would vanish to zero, when should we have the reflection term?
Infinity square well,
left side <math>\phi=Bexp(i\kappa x)</math>,
right side <math>\phi=Fexp(-i\kappa x)</math>
Infinity step well,
left side <math>\phi=Bexp(\kappa x)</math>,
right side <math>\phi=Fexp(-\kappa x)</math>
for potential well, we have i in the exponential term , for step well no imaginary term in the exponential, is that right ?
For a potential well(bound states)E<0, In an order from left to right,
region 1:<math>\phi=Aexp(i\kappa x)+Bexp(i\kappa x)</math> x<0
region 2:<math>\phi=DCos{lx}</math>
region 3:<math>\phi=Fexp(-i\kappa x)</math>
For a step potential(scattering) E>0,
region 1:<math>\phi=Bexp(\kappa x)</math> x<0
region 2:<math>\phi=DCos{lx}</math>
region 3:<math>\phi=Bexp(-\kappa x)</math>
Im in the same boat Can! If same has an awesome way of charactorizing boundary conditions for each of the different cases please share what seems to be top secret info. Question, for a free particle and finding the relationship between quantum velocities and classical velocities on pg 65 how do they go from the wave function at t=0 to “and at later times”. Where ψ(x,t) with an associated shift from x to x-w0't?
All these all correct? I wish some one can do a summary for all these general cases. thx!
In terms of when to drop either the negative or positive infinity term all I can say is this. In these situations we are only dealing with one wave coming in from the left hand side. When it hits a barrier it is going to produce a reflection wave as well as a transmitted wave. Now in the second region, if their is a preceding barrier after the first the we need to include both terms. For instance, in a well their are particles going back and forth between the two barriers due to reflection. Thus we need both <math>Aexp(ikx)</math> and <math>Bexp(-ikx)</math>. However, once the final barrier has been breached then we only need to concern ourselves with <math>Fexp(-ikx)</math> since their are no more reflections and we want the wave to decay to zero.
On top of what Pluto said, it is usually convenient to look at the Schrodinger equation and figure out what your k, <math> \kappa </math>, and l values are, and consider if E > V or V < E. From there you can figure out if you need an imaginary term or not. The easiest way to figure out if the term blows up or becomes zero at infinity is to just plug in infinity for x and disregard the imaginary term, although I think this might be a little naïve.
This is difficult because of the varying characteristics of each potential. It is then further complicated by the energy(is E<0 or E>0).
If E is negative then <math>k=\frac{\sqrt{(-2mE)}}h</math>.
If E is positive then <math>k=\frac{\sqrt{(2mE)}}h</math>.
If <math>V_o</math> and E are both negative then <math>l=\frac{\sqrt{(2m(E+V_0)}}h</math>. This is because you're taking E and subtracting negative <math>V_o</math>.
If <math>V_o</math> and E are both positive then <math>l=\frac{\sqrt{(2m(E-V_0)}}h</math>.
If <math>V_o</math> is positive and E is negative then <math>l=\frac{\sqrt{(-2m(E+V_0)}}h</math>.
This is then complicated further depending on if the potential is a well, hill, step up, step down, monkey, delta-function, free particle, ect. Did I get this correct?
I agree with you. Those expressions look like correct for me, but the fourth one should be assumed with E>Vo.
Monkey?…eh? Anyway, maybe my brain is fried from studying, but doesn't Can have the boundary equations for E<0 and E>0 switched?
In discussion we worked a problem that ended up with a Transmission Coefficient that looked something like this: T= <math>\frac1{1+Xsin(X)}</math>, where <math>X</math> is stuff. I just noticed that it's in a form that is similar to Equation [2.141]. Question one, do we want to simplify our equations so that it looks like that so we can more easily evaluate its properties or will any other form due? Second question, this Transmission Coefficient ended up with a sin in it which, as my TA explained, at certain energy levels the Transmission is 100%, in what “situations” (problems) does this occur? Is it common?
Good observation and questions. I'll also add that equation [2.169] is in a similar form, albeit inversed. The short answer is yes, your first statement is correct. It looks “nicer” and more importantly we know that if we change any of those variables in the X term the function will drop as <math>\frac{1}{1+X}</math>, and will also oscillate due to the sine-squared term between a maximum of 1 (for perfect transmission) and the <math>\frac{1}{1+X}</math> curve. Go ahead and plug <math>\frac{1}{1+X}</math> into a graphing calculator along with <math>\frac{1}{1+X*sin^2(X)}</math> and you'll see it more clearly. Before we had graphing calculators and software people relied on these familiar functions to roughly sketch out the shape of the curve and then look for implications and smaller particulars without painstakingly computing and plotting dozens of points. For us in the Computer Age, it just looks a little nicer to be simplified; but since we don't plot by hand anymore it's more of a convenience for someone reviewing our work than a necessity. As for your second question, you can kind of think of a water wave against the “potential” barrier that is the side of a pool (this is a watered down description, I know, and no pun was intended). The water wave has a certain amplitude and velocity. When it strikes the wall it can be anywhere in its amplitude oscillation cycle, striking anywhere from the crest to the valley. How perfectly the strike coincides with the crest of the wave determines (more or less) how much of the wave is transmitted beyond the barrier. Notice in figure 2-19 that as the wave gains more and more energy (higher amplitude and speed) it's more likely to cross the wall, and eventually a giant wave won't even notice that the wall exists. Conversely, a tiny wave in a shallow pool with a tall wall will likely just hit the wall and be reflected back entirely, making the transmission close to zero. (Be careful making and interpreting analogies to classical mechanics. There are still some significant differences) In terms of this type of problem's general occurrence in Quantum Mechanics, any time that you have a finite potential you can have scattering (if the energy is high enough). The Infinite Square Well and the Harmonic Oscillator are the only situations we've looked at that don't have scattering states because the particle's energy must be higher than the potential. One of the sample test's questions about modeling fission in a nucleus is a good example of this type of problem. I hope that helps and is accurate.
In regards to the second problem on the practice quiz, how do you determine how many acceptable values of z exist? Do you just count the number of intersections on the plot? And would the asymmetric solution have just one fewer acceptable value of l?
Also, for the third question, is it necessary to use the velocity correction to the transmission coefficient we used in the homework question 2.34?
I think with the second problem it's actually every other intersection that you count since the graph shows solutions for tangent and cotangent. (The odd and the even states)
For the third question, I think you do need to do a velocity correction if you calculate T. But the correction factor is different, I think, than the homework problem. Oh but wait! You can also solve for R and not use the velocity correction. And then T = 1-R.
By the way what did people get for answers to this? I just did it, and got T = 8/9, (and R=1/9), but I have low confidence in my answer. What did other people get?
I did the problem wiht a couple other people, and we all got 8/9. (This problem was pretty much exactly example problem 2.35C and their hint is that you should use T= 1-R
Does anyone know if the free particle will be covered on this test?
If it is then it probably won't be anything to extensive. Just make sure you know how to do the problem if ask. More than likely we will be dealing with wells/steps and Hermitian operators.
I assumed the test was covering everything we did since the last test, which would be from 2.4 to 3.3.1. Maybe Pluto4ever knows something I don't??
The footnote on 95 says that the equation 3.9, or the inner product of a function with itself ( <f(x)|f(x)> ) will vanish and is zero only where f(x) = 0. …. even for a function that is zero everywhere but for a few isolated points….. Does this make sense? What about the delta function? Isn't this the definition of a delta function? Does this equation make sense for a Delta function?
The footnote is not referring to the delta function. The delta function is a magical object that, even though it is infinitesimally thin, has an 'area' of 1.
Maybe, as Griffiths says, I shoulda been a mathematician… I realize he's not explicitly mentioning a delta function, it just sounds like what he refers to “could be” a delta function as well… but ya. It's just not. Is a delta function “square integrable”? – Yes because it has an 'area' yes?
The 'isolated points' the footnote related is talking about points with finite values. Of course delta function is not square integrable, because <math>\int_{-\infty}^{+\infty} \delta(x)^2 dx = \int_{-\infty}^{+\infty} \delta(x) \delta(x) dx =\delta(0) = + \infty </math> (equation [2.113])
So, for the finite square well problem in the sample quiz, we are asked to calculate the first two values of l. Is this doable without a computer? If so, how?
I think this is as simple as looking at the graph provided and looking for where the lines cross. For symmetric z= 2, 4 (approximately) with z=la then l= 2/a, 4/a. For asymmetric z= 3, 5 (approximately) then l= 3/a, 5/a. I think. It's then up to you to figure out if those numbers are sensible.
In the homework we just turned in, question 3.4a doesn't really make sense. A priori it says “<math>\alpha</math> is imaginary, and ^Q is hermitian. Under what conditions is alphaQ hermitian?” – The answer is “Alpha is real”. isn't this contradicting what we were told to assume?