Saturday, February 2, 2013

Homework 5: momentum, confinement and kinetic energy.

due Friday, Feb 8.
(corrections: problems 3 & 7, Tuesday evening (see comments))
(edited: Sunday Feb 3, intro and problem 8)
(Please report any typos you find and ask about anything that is unclear, strange, etc.)

This is probably our most important homework so far in this class. I think it will be pivotal in helping develop your understanding of key aspects of quantum physics which will allow us to make the transition from describing atoms and other single-well systems to the description of molecules,  semiconductors and other quantum materials. Feel free to ask questions and discuss the problems. These are hard problems and I think it is helpful to work with actual groups and people, and to engage in questions and discussion here.

Problems 1-4 are related to confinement, momentum and kinetic energy. As you know, the Heisenberg relation is often expressed as the inequality $\Delta x \Delta p \ge \hbar / 2$. In the first 4 problems we will do some calculations of $\Delta x$ and $\Delta p$ (and some closely related quantities) for specific systems and states. This will provide a chance to examine the Heisenberg inequality from a physics point-of-view, which is somewhat different from perspectives that are pervasive in popular culture (which tend to fixate on the word "uncertainty"). Uncertainty is a complex word with many possible meanings. Inequalities, such as the one above, refer to the precision with which one can localize a wave-packet, and the need for short wave-lengths in doing so. Such relationships are present in many wave theories, string, E&M, water ..., in which one can envision deconstructing a localized packet into components of precise (well-defined) wavelength.

Problems 5-7 all focus on kinetic energy integrands, especially distinguishing regions of space that contribute negatively to the kinetic energy from the regions which contribute positively to KE. This is a novel and important aspect of quantum mechanics. Another perhaps novel thing is that all the kinetic energies in 5-7 are independent of time.  However, this is not generally the case for KE in quantum physics.

Problem 8 provides a relatively simple example of a state (wave-function) in which KE and PE of an electron vary (in a complementary way) as a function of time. I think if you take the time to do 8, it provides a perspective that allows the whole thing to make more sense. Feel free to start with part b), which is mathematically a little easier than a).

1. a) Calculate $\Delta x$ and  $\Delta p$ for an electron in the ground state of a harmonic oscillator.
    b) What is the product $\Delta x \Delta p$ for an electron in this state? Does it depend on time?

2. a) Calculate $\Delta x$ and  $\Delta p$ for an electron in the ground state of an infinite square well (of width L).
    b) What is the product $\Delta x \Delta p$ for an electron in this state?

3. Show that the Heisenberg relation can be recast as a relationship between kinetic energy and confinement, namely,
$\bar{T} \geq \hbar^2/(4 m (\Delta x)^2$, aaarrrrg, this is incorrect.
$\bar{T} \geq \hbar^2/(8 m (\Delta x)^2$, <---- this one is corrrect i think.
where $\bar{T}$ is an expectation value of kinetic energy and $\Delta x = \sqrt{\bar{x^2}-\bar{x}^2}$ is a length scale which provides a measure of the degree of confinement of an electron.

4. Write $\bar{T}$ in terms of $(\Delta x)^2$, m and $\hbar$ for:
a) an electron in the ground state of a harmonic oscillator.
b) an electron in the ground state of an infinite square well.
(these are different, right?)

5. a) Graph the integrand from the integral that arises in the calculation of $\bar{T}$ for an electron in the ground state of a harmonic oscillator.
 b) Although  $\bar{T}$ is positive, there are regions of x that contribute negatively to the integral (regions that diminish $\bar{T}$).  Where are there? (That is, over what range(s) of x is the integrand negative?)
c) Indicate those regions in your graph. Do you notice anything about the nature of the curvature of $\psi (x)$ for those regions?

6. a) Graph the integrand from the integral that arises in the calculation of $\bar{T}$ for an electron in the ground state of a finite square well. (Note: For this problem you are not expected to calculate T or to know the exact state for this problem. Please use the form of the ground state that was provided in problem 7 of HW 4. I think you can do this entire problem using that form (and without evaluating A or B).)
 b) Although  $\bar{T}$ is positive, there are regions of x that contribute negatively to the integral (regions that diminish $\bar{T}$).  Where are there? (That is, over what range(s) of x is the integrand negative?)
c) Indicate those regions in your graph. Do you notice anything about the nature of the curvature of $\psi (x)$ for those regions?

7.  In this problem we would like to sketch the integrand associated with the $\bar{T}$ calculation for a double square-well focusing in particular on the nature of the contribution to $\bar{T}$ that arises from the region between the two wells. A double square-well refers to two finite square wells. In this case, let's put the origin of coordinates between the wells, define a center-to-center distance of b, and the width of the wells as L. In equations:
$U(x) = -V$, for (-b/2) - (L/2) $\leq x \leq$ (-b/2) + (L/2),
$U(x) = -V$, for (b/2) - (L/2) $\leq x \leq$ (b/2) + (L/2),
$U(x) = 0$, elsewhere.
(In this case, the two wells are identical in depth (measured in energy) and width (L). We assume that b is greater than L, so that the wells are not overlapping and that b is not too large,  so that the two wells are close to each other. (We will learn how to define what "close" means. That is an important question! (thanks for asking).

Additionally let's use the following notation for the ground state:
$\psi(x) =  A cos(k(x-b/2))$, for (b/2) - (L/2) $\leq x \leq$ (b/2) + (L/2),
$\psi(x) =  B e^{-\gamma (x-(b+L)/2)}$, for $x \geq (b+L)/2)$,
$\psi(x) =  C cosh(\gamma x)$, for (-b/2) + (L/2) $\leq x \leq$ (b/2) - (L/2),
Where the A cos( ) term is for the wave-function within the region of the right-hand-well, which is centered at b/2, I believe; the $B e^{-( )}$ term is for the region beyond the right-hand-well, and the very important C cosh( ) term is for the region between the two wells

 a) graph the potential U(x) as a function of x (for b = 2L approximately).
 b) graph the ground state wave-function $\psi (x)$ as a function of x. (I would suggest using an assumed $\gamma$ of roughly 3/b.)
 c) calculate and then graph the second derivative of the wave function for these regions (leaving A, B and C undetermined)
 d) graph the integrand of the kinetic energy calculation for this double-well system,
$\psi (x) (-d^2 \psi (x) / dx^2)$ for these regions.
 e) What is the sign of the contribution to $\bar{T}$ from the region between the wells? Why might this be relevant to understanding molecular bonding? Thinking about K.E. and confinement, what does it signify?

8.  Advanced. (I would recommend reading the whole problem and then starting with the calculation of part b. I think that is a lot less difficult than calculating $\bar{T}$ for this mixed state.) All of the kinetic energy expectation values we have calculated so far have been "constants of motion", that is, not dependent on time. This is a different from what you may be used to, since in a classical harmonic oscillator, or a wave on a string, etc., we expect the kinetic and potential energies to oscillate with time. This can happen in QM also, however, just not for the pure states (energy eigenstates) that we have been studying. 
a) Show that for an electron in a harmonic oscillator in the mixed state that is an equal mix of the ground state and 2nd excited state, the kinetic and potential energy each oscillate as a function of time. What is the period of the oscillation? How does it relate to an oscillation in the size of the wave-function?
e.g., at t=0, $\psi(x) = (\psi_o (x) - \psi_2 (x))/\sqrt{2}$
Start here ==>  b) Calculate $\bar{x^2}$ as a function of time for an electron in this mixed state. What is its smallest value? What is is largest value? Graph it (vs t). How is $\bar{U}$ related to $\bar{x^2}$? How would you guess that $\bar{T}$ might be related to $\bar{U}$?  (Feel free to discuss here. Is there a simple relationship between $\bar{T}$ and $\bar{U}$? This is not obvious so anything you say or ask will be good.)

41 comments:

  1. I'm a bit confused about how to approach number 3. Using the uncertainty equation to solve for delta-p and the writing the kinetic energy operator, but replacing the p operator with delta-p I was able to get an answer which is half of the answer we are trying to obtain.

    When I got to number 4, however, it made me question my approach to number 3. Number 3 does not specify a wave function with respect to which to solve for T-bar. Number 3 asks to solve for T-bar using two different wave functions. Are the answers to all three (3, 4(a) and 4(b)) supposed to be the same?

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    1. 1st. Thanks. There is an error in 3. It should be an inequality. Thanks for catching that.

      "I was able to get an answer which is half of the answer we are trying to obtain."
      what did you get? I better check what I have in the problem.

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    2. I fixed it. I think you should get 5 pts extra credit on the HW for calling our attention to that error in the problem before it got too late in the week. If it had stayed as it was, it would have made problems 1-4 very confusing for everyone.

      I do think it is correct now (as an inequality) with the 4 in the denominator. What are other people getting.

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    3. I'm still confused about number 4. It seems that (a) and (b) are only different in-so-far as Δx is different, but if we're expressing T-bar in terms of Δx then how would they be different?

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    4. Try writing L/pi in terms of delta x and you'll see that the number in the denominator is different. It is not 8. It is smaller.

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  2. Should number 3 be T = hbar / 8*(dx)^2 ?

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  3. Rather ... T >= hbar / 8*(dx)^2

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  4. So it makes more sense now that the inequality is in there, but I still get what Brent got T>=hbar/(8*(delta-x)^2). Is substituting the value of delta-p from the uncertainty equation into the kinetic energy operator equation the right approach to this problem?

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  5. Seems like it should be T>=hbar/(8*(delta-x)^2).

    That's the relation that I'm seeing elsewhere (and getting in the calculation).

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    1. you guys are totally right. I fixed it!

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    2. I just wanted to clarify, is substituting the value of delta-p from the uncertainty inequality into the T-operator equation an acceptable way of solving for ? It seems somewhat iffy, but it yields the right result.

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    3. I don't believe it is iffy. I think it is a true inequality since $\bar{p^2} is greater than or equal to $(\Delta p)^2$. What do other people think?

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  6. For 7 I think the region of the $Acos(k(x-b/2))$ equation should be $b/2-L/2\leq x \leq b/2+L/2$ since the function has a maximum at $x=+b/2$. The limits there are for the well on the left.

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    1. Additionally, I think the cosh term should be $Ccosh(\gamma x)$. This would be centered at x=0 but the function given is centered at x=b/2 which is the center of the well on the right.

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  7. For numbers 6 and 7 I'm having some trouble sorting out where the integrands will be negative. If the wave function is Acos(kx) inside the well and Be^-gamma(x-L/2) outside the well ... the integrands for each should be (A^2)(k^2)cos^2(kx) and (gamma^2)(B^2)e^-gamma(x-L/2) respectively ... I'm not seeing where those could be negative.

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    1. For the exponential term don't forget to multiply by $-\hbar/2m$ from the T operator. That gives you the negative sign.

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    2. the $\hbar$ should be squared.

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    3. I agree with what Dylan said. Also, let me add that taken two derivatives of cos(kx) produces -cos(kx). On the other hand, taking two derivatives of an exponential produces a positive sign, So they have different signs, One will contribute a positive contribution and the other a negative contribution. Does that make sense?

      Or, to make this comment shorter, two derivatives of a cos function yields a negative sign, two derivatives of an exponential (or cosh) does not.

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    4. In number 6 is it Ok for there to be discontinuities in the graph of the integrands?

      Also, (and this applies to problem 5 as well) shouldn't part (c) really ask about the curvature of the integrand, not Ψ? The integrand is Ψ ^2 not just Ψ.

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  8. I made some videos about momentum and momentum calculations. They are rendering now. (The compression takes awhile.) I will post them tomorrow morning. It is more than an hour of video (related to today's class...)

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  9. In number 7, would you like three different graphs for each region, or should we assume continuity and put them all together?

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    1. you can definitely assume continuity. and put them all adjacent.

      You can assume continuity and smoothness (1st derivative continuous), and that can be derived and proven directly from the Schrodinger equation! But all bets are off when you are graphing the 2nd derivative, which does not have to be continuous. Does that seem right?

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    2. Yes that seems correct, and very interesting! It does make sense to require the first derivatives to be continuous, and that the second does not require that. Then this way at changes in potential, the wave function can immediately change in it's second derivative, which ultimately decides where it's "going". Does that seem right?

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    3. hmm. That's interesting. I think about it a little differently. The difference between inside and outside is that:
      1) inside, the form of the wave is a propagating wave (cos(kx), sin(kx), $e^{ikx}$ are propagating waves),
      2) outside, it is an evanescent wave.
      The second derivative of the wave-function:
      1) has the opposite sign as the wave-function for a propagating wave,
      2) has the same sign as the wave-function for an evanescent wave.
      You get the same relationships and the same thing in E&M.

      In this case nothing is going anywhere:
      1) inside, because there are two propagating waves combining to make a "stationary" standing wave,
      2) outside, because it is evanescent.

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  11. I believe that in number 6 $-Be^{−\gamma(x−L/2)}$ can also take the form $-Be^{\gamma(x−L/2)}$ (no negative). This would improve the graph greatly and make much more sense for the negative region of x. Otherwise we would be dealing with a negative infinity, which doesn't converge, and this is bad right?

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    1. Yes, in the negative region (to the left of the wells) it can only take the form Beγ(x−L/2) (no negative). Otherwise we would be dealing with a negative infinity, which is very bad! (as in not normalizable, hence not an allowed solution to the S equation

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  12. By the way, for 6 and 7, the focus is entirely on the graph. Please don't spend your time matching boundary conditions.

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  13. I'm getting weird answers for numbers 1 & 2. I used the wave function Ψ=1/(sqrt(pi)*a)*e^(-x^2/2a^2) for number 1 and Ψ=sqrt(2/L)*sin(pi*x/L) for number 2. For number 1 I got x-bar^2=1/(4*pi^2) and x^2-bar=a/(4*sqrt(pi)) and Δx=sqrt(a*pi^1.5-1)/(2pi). I got p-bar^2=-h-bar^2/(4*pi^2*a^4), p^2-bar=h-bar^2*(pi^1.5*a^5-2)/(2*pi^2*a^8) and Δp was messier than that. Is anyone else getting really messy answers like this?

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    1. Check your wave function for #1--I believe it should be $$\Psi(x)=\left(\frac{1}{a\sqrt{\pi}}\right)^{1/2}e^{-x^2/2a^2}$$.

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  14. Oh. I see. I was missing the square root. I'll try that. Thanks.

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    1. No problem. The values I'm getting for $\Delta p\Delta x$ for the infinite square well are still ugly, but better than they'd be without the square root.

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  15. So did you use this wave function for the infinite square well? Ψ=sqrt(2/L)*sin(pi*x/L)

    What did you get for Δx? I got L((1/2-3/pi^2)/6)^.5).

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    1. I would suggest just turning that into a decimal number and then using that to establish a relation between $\Delta x$ and $L/\pi$ for this particular case. I got a $\pi^2/12 -.5$ and turned that into a decimal number...

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  16. i got (((L^2)/2)((1/6)-(1/pi^2)))^(1/2) sorry for the messy equation. me and a few friends got this but at first out answers looked very different. it can look very different as you rearrange it

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    1. I would calculate L^2/pi^2 and turn the 1/6-1/pi^2 into an actual number. i think it is about .07 or so? My recommendation is to keep a pi^2 with the L^2. you will need it later since T is proportional to (L/pi)^-2.

      In the end, you just want a number to go in the denominator instead of the 8. An actual number like 5.1 or 7.4 or whatever you actually get. (The grader will not be able to evaluate an expression to see if your answer is not wrong. he will just look at the number...)

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