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1. (delta-function potential, g.s.)
Assume that there is a negative-energy solution for a particle of mass m under under the influence of an attractive delta function potential; using only exponentials that diminish as "you" move away from the origin, determine the exact, normalized ground state. (Ground state generally means lowest energy state. In this case it is also the symmetric, node-free state centered at the mirror symmetry point of the potential).
2. (delta-function potential, size) Calculate the expectation value of x^2 for a particle in the state from problem 1. Take its square root; it should have units of length, right? Give this characteristic length scale (which should depend on alpha, hbar and m, I think) a name*, and then rewrite your g.s. using this length scale parameter in both the exponent and the normalization pre-factor. (Your state should now look really simple.)
*may I suggest a short simple name like a_delta (a subscript delta)?
3. (delta-function potential, calculating "V") Calculate the expectation value of the potential energy for a particle in the ground state of an attractive delta-function potential. What is the difference between potential energy as a function of x and the expectation value of the potential energy? What is the meaning of each? [Computationally, this problem is very easy once you know what to do. It is really about understanding the concept of expectation value and what that means.]
PS. (added Sunday 4-5, 7:40 PM) For K.E. of a particle in this state, see problem 9. If 1-3 were too easy, you may like 9 better.
4. (infinite square well) a) What is the (normalized) g.s. wave-function and energy for a particle of mass m experiencing an infinite square well potential of width L?
b) Calculate the expectation value of p for a particle in this state.
c) Calculate the expectation value of p^2 for a particle in this state.
d) Using your result from c), what is the expectation value of the kinetic energy for a particle in the g.s. of an infinite square well.
*** The above problems, 1, 2, 3 and 4, are especially important problems. I would recommend doing them first and trying to finish them by Tuesday, if possible. They help establish some basic concepts that we can discuss and explore. Also, plan to allow enough time for 5- 9, as they may be somewhat computationally difficult. 5a and 8b are especially important to understand. Also, maybe 5b.
I know we haven't actually covered expectation value calculations in this class, but I am hoping you know what to do? If not, please feel free to post your questions here. Also, this is really a peer-to-peer format and anyone can answer anyone's questions.
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5. (finite square well) Consider a particle of mass m in a finite square well. Suppose that someone has numerically determined the parameters of the ground state and that thus you are given k, alpha, B and A (as well as m, hbar and L). (Treat them all as known".) Write down and evaluate the expressions (integrals) for the expectation value of:
a) potential energy
b) kinetic energy
c) momentum
6. With reference to the preceding problem:
a) How does the expectation value of the potential energy compare with that of a particle of the same mass in the g.s. of an infinite square well of the same width?
b) What is the sign of the expectation value of the kinetic energy contribution from the region outside the well?
How does the expectation value of the kinetic energy contribution from inside the well compare with that of a particle of the same mass in the g.s. of an infinite square well of the same width?
7. Momentum expectation values don't seem to depend explicitly on the potential function, V(x), so let's try working with a state without thinking about potentials for a moment.
a) Find the normalization factor for the quantum state:
e{ikx} e{-(x^2/2a^2)
[ I think it is 1/(pi^(1/4) sqrt(a)]
Perhaps someone can check that and post here? Thanks.]
b) Calculate the expectation value of the momentum of a particle in this state.
(Funny how that doesn't have an explicit m in it? I suppose m must have an influence on the momentum in some way?)
8. For a particle in the state: e{-(x^2/2a^2)
find the normalization factor (same as above i think) and then
a) calculate the expectation value of the momentum.
b) calculate the expectation value of the momentum squared.
c) calculate the expectation value of the x^2.
d) What are delta x and delta p for this state? (You know, like x^2-x^2 and p^2 - p^2..., but with the brackets in the right place.)
9. (extra credit-difficult: delta-function potential, K.E.)
Calculate the expectation value of the kinetic energy for a particle of mass m in the g.s of an attractive delta-function potential (as in problem 1-3). Show that the K.E. is indeed positive, as it must be, despite any indications to the contrary. [Be careful around x=0]
10. (extra-credit) Consider an attractive delta function potential in the center of an infinite square well. Assume that L, the size of the infinite sq well is finite, but very large compared to the characteristic size of the ground state of the delta-function potential by itself. Solve for the ground state for this "confined problem", where you can assume the g.s. wave function goes to zero at the boundaries of the well. Look at the nature of the solution in the limit L-->infinity. (I think to solve this you need both exponentials. The growing one will have a small (i think negligible) effect at x=0, but will be essential to making the w-f go to zero at the boundary (x=L/2). This gives us a sense of what the growing exponential can do and what happens to it as L gets really large...
If you're like me and need a refresher on probability, the book I used for CE107 (Stochastic Processes) is quite good as a reference. It's Fundamentals of Probability with Stochastic Processes by Saeed Ghahramani.
ReplyDeleteAlso with the linear algebra, working through the first chapter of Shankar's book Principles of Quantum Mechanics has proven useful.
I think the library probably has both, either on reserve or for general checkout.
Any chance we can switch back to Kappa for our separation constants? I'm confused about the alpha we used to SCALE the dirac delta potential, and the alpha that we used in the EXPONENTIAL solution... In any case, I feel that alpha was a particularly silly choice because we wanted to avoid confusion, yet alpha is Kappa without a vertical line through it's fishy part.
ReplyDeleteGood point. What if we used q, instead of alpha for outside the well, as in e{-qx}. Then q would have units of inverse length, like k, be related to E and k via the S.Wave eq, as we discussed in class.
ReplyDeleteFor problem 2, I get a scale factor of (h-bar)^5/[2(alpha*m)^(5/2)] with alpha being the "strength" parameter of our delta function potential. I got this by solving for < x^2 > over all space. Assuming I did this calculation right, I don't see how this makes our wave function any simpler. I'm not understanding the motivation for the problem.
ReplyDeleteJerome
Me and my fellow QP superheros have become puzzled over the first portion of our HW:
ReplyDeletei) We end up needing a real and positive k, but should we insert a complex k into an already imaginary exponential (a travelling wave)?
ii) Why does Griffiths solve for the bound state energy outside of the well... in the region x < 0 for the potential V = 0?
iii) When finding the expectation value for the potential energy, shouldn't we split the integral (with limits from -infinity to infinity) so that we can use the appropriate Y(x) for the region? ... Since the delta function only “activates” if the limits encompass x' (from d( x – x' )), then will d( x ) activate if 0 is the lower limit?
Unless we get some help here, we're going to try and set up an appointment with John K on Monday (around 2pm?) - anyone else care to come?
Dear Jerome,
ReplyDeleteI don't think your result has units of length, so it can't be an appropriate scale factor. The thing that puzzles me though, is I am not sure what state you used? 2 says, "Calculate the expectation value of x^2 for a particle in the state from problem 1." What state did you get in problem 1? What is your wave function?
Eepter: i) no, definitely not. That would not be normalizable, and it would not be the ground state.
ii) I am not sure, but E is a "global parameter" and is the same in every region. (This comes from separation of variables.) I think that in a V=0 region with wave-funct. proportional to e^(-kx), the S equation tells us that (hbar k/2m)^2=-E. Does that seem right? If so, then for E negative one would get a real k and a nice, normalizable wave-function, yes?
iii) What you suggest seems reasonable. Let's see -- there are two ways to approach this: One is to integrate from -8 to 8 and see what you get. psi(0) is well defined (i.e., it is the same from either side.) If this seems unsatisfying and you want to get technical, you can break the integral into two parts, as you suggest, but then you would need to give the delta function finite width and carefully take the limit as that width goes to zero (while keeping the area of the d-f fixed at -alpha). Either way, you will get the same thing. Feel free to post what you get.
PS. The short answer to you iii) is: no (don't split it), and yes (it will activate).
Thanks for your response Zack. I corrected a slight problem and posted my results at http://mathbin.net/7605
ReplyDeleteI'm still unclear about the motivation for this problem. I'm not sure what I'm supposed to get from this exercise.
Jerome
Well, I think what you just calculated could be called the "size" of the ground state of the delta-function "atom". It is pretty analogous to what is called the Bohr radius for the H atom in 3 dimensions.
ReplyDeleteHow does this characteristic size depend on hbar, m and alpha? What is the significance of those relations?
Further, will there be a relationship between this size and the potential and kinetic energy? What sort of relationship? Why?
I'll try to keep an eye on this for the next hour or two and respond quickly if anyone has questions or comments.
ReplyDeleteI got meters squared for the root of the expectation value of x^2 which does not give me the proper units for the wave function, it leaves me with a 1/m in the exponential. My integration seems solid though, what are the units of alpha, I thought they were simply Joules?
ReplyDeleteYou raise a good point. Actually the delta function has units; when you integrate it with respect to x you get something unit-less. So that implies that it must have units of inverse length which, in turn, effects the units of alpha. To put it another way, alpha is the integral of V ...
ReplyDeleteI've been working on #5, and when calculating the kinetic energy, got a result that depended only on the component within the well--the contributions from the two regions outside the well canceled each other out. I suspect I messed up one of the integrals, but I can't find a mistake. Anyone else get the same result?
ReplyDeletemight in the limits of integration. (reversing the order creates a sign change)
ReplyDeleteI'm missing something--why am I reversing the order of the limits of integration?
ReplyDeleteno, don't reverse them. I am just saying i think that is where your sign error might be. You are getting contributions that cancel, you say, and intuitively, by symmetry, they should be identical in both sign and magnitude. I was just trying to help you find your sign error.
ReplyDeleteOkay, thanks.
ReplyDeletelooking at number 5...I cant figure out the wave equation for a finite square well. i havent even attempted the actual "problem", since i dont know what the wave-function is. inside the well, i think it is Acos(kx/L), where k= {2m(V-E)/h-bar^2}. outside the well, the solution is Bexp(alpha*x).
ReplyDeleteIF my solutions are right, im not sure how to solve for A and B. also, i have alpha and k are related by the energy. im confused. if someone can tell me the solution for the wave-equation for a fintie square well, and how they got the normalization constants, i'd appreciate it. then i can actually do the expectation values :)
Suppose that someone has numerically determined the parameters of the ground state and that thus you are given k, alpha, B and A (as well as m, hbar and L). (Treat them all as "known".)
ReplyDeleteAlso, just a point on language. The wave equation is the Schrodinger eqn. The things your refer to are solutions to the wave equation (states, or wave functions). We don't want to confuse those.
Just as a general comment to everybody. I think that our square well oonventions are as follows: inside the well use k (as in A cos(kx), outside the well q qould be preferable to alpha (this happened in an earlier comment), no big deal either way, but anyway that means that outside you can use B exp{-qx). And we also assume V=0 inside the well and V_0 outside, though if you prefer you can use -V_0 inside and 0 outside. What I would discourage is calling the inside V_0 and treating it as negative. That is confusing (to me) and makes communication difficult.
ok. i used wave eqn simply because its shorter than shrodinger, but ill use schrodinger from now on. im still unsure how to solve for the normalization constants A and B. set the integral from -infinity to infinity of psi-star times psi equal to 1, but in an infinite square well, we only solved for one constant, where as here, we have two. we also have to apply the smoothness and continuous boundary conditions at the walls of the well, but all i have (from lecture notes) is that q and k are related to energy from the boundary condition analysis/calculations. this still leaves me puzzled as to what the normalization constants A and B are in the shrodinger eqn solutions.
ReplyDeleteSorry about the confusion, but that is not the distinction I was trying to make. Saying "wave equation" is fine and strongly encouraged. I was concerned with confusing the equation with its solutions. Sorry I wasn't more clear about that.
ReplyDeleteRegarding your main point, you were asked to treat A,B, k and q as given! You DO NOT need to calculate them! Otherwise this problem would be intractably difficult (as you have pointed out). You are supposed to start with A, B, k and q; and then find expressions for V and T etc, in terms of them.
ahhhhhhhhhhhhhhhh. thank you. i can sleep in peace now.
ReplyDeleteThis may make me sound like an asshole, but in a problem like the delta potential, when we solve it for both sides in the general form, with exp(-) and exp(+), and then throw one of them out becuase otherwise the function would blow up, does that bother anyone else? If the heart of the quantum system is the schrodinger equation, and the schrodinger equation tells us that a solution is that the function goes up and up until it reaches infinity at a certain point, why isn't the particle just definately found there? Is there a larger idea behind the equation that makes this obvious?
ReplyDeleteYou don't sound like and asshole at all. I think this is a good question. We should think about it carefully,and see what we come up with. Maybe it has to do with limits of some kind?
ReplyDeleteWhen I first read your comment I just thought, "well that is just the way things work," but as I am writing this i am thinking maybe one can say something more appealing. Does anyone have any ideas about this?
I think maybe we throw out the infinite solution for a good reason. The heart of the quantum system isn't just in the schrodinger equation, it's in the schrodinger equation + boundary conditions.
ReplyDeleteIf I were to make a guess at the answer to Hefeweizen 's question, it would be that maybe there is a boundary condition that we don't state--that the wavefunction of a quantum system does not approach infinity at any boundary (?). That would make sense to me, because we wouldn't expect a wave to ever go all the way to infinity, because it has to come back down without discontinuity, right?
The first reason that pops in mind is that we are solving for a bound state, so perhaps it's just that the relevant boundary condition is that the wavefunction does not blow up at +/- infinity.
ReplyDelete-- John (the TA)
PS: I'll try to follow the blog and comments, commenting when I can.
I can take a guess of why we throw out the solutions do not tend to zero at infinity. If we allow them, the interpretation of quantum mechanics says the the square of the wave function in the x-basis is the probability density of finding the particle in a given interval x and x+dx. If our function do not tend to zero there can be infinite probability in finding a particle at plus and minus infinity. Mathematically our state of our system is represented by a state vector in Hilbert space. If our solutions do not tend zero they leave the Hilbert space which is the space that represents the system. A little more mathematical our Hamiltonian to be hermitian also depends on the functions subjected to it. If they do not converge to zero our Hamiltonian may not be hermitian, meaning that the eigenvalues which our the allowed energies can be complex. Complex eigenvalues can lead to violation of conservation of probability.
ReplyDeleteSometimes taking understanding things that become infinite can be addressed by thinking about limits. In this case the infinity occurs as x--> infinity, right? If there were a way to control the large x region and then take some sort of limit, perhaps that might help us understand this better.
ReplyDeleteIf we look at what the solution to the Schrodinger equation represents, then we have a possible answer. Why is there a normalization condition? We were told that the integral for the ground state of one electron/atom/whatever over all space (in our case one-dimension) equals one. Why one? Because we know for a fact that there's one electron/atom/whatever somewhere in that infinite space, so we will eventually find it if we look everywhere.
ReplyDeleteIf the solution diverges, the integral too diverges. This would imply an infinite number of particles in the space, right?
Thinking about normalizing the wavefunction is also good, as we require (postulate, or via probability theory/unitarity) that it normalizes to 1 in our space. Including both positive and negative powers for the exponential solutions outside the potential violates this.
ReplyDelete-- John
In regards to the last part of #8, "delta x and delta p." Do you mean SIGMA x and p? It seems like a natural progression from earlier calculations to end up calculating the standard deviation. If this isn't the case, what do DELTA x and DELTA p mean?
ReplyDeleteIntersting discussion. In addition, I think Griffiths says that complex eigenvalues do not correspond to real observables. Therefore it sounds like a wave function that diverged to infinity would never be observable. Is this correct?
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteAre most people getting a negative kinetic energy for problem 5(b)?
ReplyDeleteDelta x and Delta p are also labeled with sigma (different notations). What we mean is the variance, or square of the standard deviation (see what Zack wrote after he said delta x...).
ReplyDeleteRegarding diverging wavefunctions, I think the general stance is that only normalizable wavefunctions are physical states. Sort of what you said, but a bit more than that; it's not a physically meaningful object at all (by itself). Of course, physical wavefunctions can be constructed out of pieces that are not normalizable (and not physical) by themselves, but combined can be a "real" state.
-- John
For the kinetic energy, I would check your signs and i's. Also note that p^2 does not mean |p|^2 (not norm squared), so do not change any signs when doing p^2 (= p*p).
ReplyDeleteI'm going to add an extra-credit problem that I think is related to this issue. I will edit into the end of the assignment post. (problem 10)
ReplyDeleteJerome, I got a length scale (for problem 2) that is slightly different from yours. I got h-bar^2/(m*alpha), times 1/sqrt(2). The units work, using Zack's assertion that kappa has units of inverse length, and that kappa=m*alpha/h-bar^2. Please let me know if my notation is indecipherable.
ReplyDeleteMe too, Heather! cool.
ReplyDeleteHas anyone gotten an answer for #9? I got a positive number, but it's too simple and elegant to be a correct answer.
Thanks for the comments on my question everyone. I did #10, and yes it's incredibly cool how as the walls go out to infinity the exp(-) term takes over and the exp(+) disappears. So is the idea that we're all in a potential well? Does this have something to do with some sort of repulsion from the outer galaxies or something? Perhaps just that in any test area there are actual material walls and that those walls are effectively infinite potential walls? If that's it, then cool. Amazingly cool.
ReplyDeleteIn response to Jerome, on number 5b, I got a term with a minus sign for the exponential part, but also a term that has a plus sign for the part inside the well. Whether the whole kinetic energy is negative would depend on the values of the constants.
ReplyDeleteKind of a last minute post, but did anyone else get an imaginary value for the momentum on problem 5?
ReplyDeletethis thing is crazy. "blog" ? what does that stand for, anyway?
ReplyDeletefar too many posts to follow. great idea in theory, but bad for those who dont think to check it often.
-Subdued by technology
Two things: just to clarify above, I meant p*p as in p times p, NO complex conjugation. But when you normalize wavefunctions, you do conjugate (i.e. psi-star times psi).
ReplyDeleteRegarding #9, for those of you at section, yes taking the limit of a finite square well is one way to do it. You must know which constants depend on the potential depth, and take the limit as the width goes to zero and the depth to infinity, but width times depth is constant (same as "strength" of delta function potential).
Seems like you can do it directly from the delta potential solution, as the second derivative looks continuous to me (easy to see why only odd derivatives have the discontinuity). Looks like I get the same answer both ways (at least schematically).
-- John
Heather, I got an imaginary solution for 5C as well.
ReplyDeleteAs a correction (for completeness), what I said above, that the second derivative looks continuous, is incomplete. As you see in the solutions, you need to reproduce the discontinuity that is in the first derivative. Alternatively, the S-Equation says explicitly that the second derivative has a piece proportional to a delta function.
ReplyDelete-- John