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I'm sure trivial and I can figure out by looking closely, but what is the expectation value of the partial derivative, with respect to time, of momentum operator?
This is the problem that was in the homework and the exam. <math>\frac{\partial <p>}{\partial t} = \left< - \frac{\partial V}{\partial x} \right></math>. The way you worded your question, however, seems to be asking about <math>\left< \frac{\partial p}{\partial t} \right></math>, which I assume would be the same, but I don't know.
Yes, the second way is what I have in mind. I think it is different though, because the partial is already inside the integrand and does not operate on conjugate wave function. I will look, but I think it is zero?
Hmmm… If you look at the definition of <math>\hat{p}</math>, we get
<math>\hat{p} = \frac{\hbar}{i}\frac{d}{dx},</math>
<math>\left< \frac{d \hat{p}}{dt} \right> = \left< \frac{d}{dt} \left[ \frac{\hbar}{i} \frac{d}{dx} \right] \right></math>
<math> = \left< \frac{\hbar}{i} \frac{d^2}{dtdx} \right></math>
<math> = \frac{\hbar}{i} \int_{-\infty}^{\infty} \Psi^* \frac{d^2}{dtdx} \Psi dx</math>
<math> = 0? </math>
because <math>\frac{d \Psi}{dx}</math> always goes to zero at infinity, and the derivative with respect to time of that would be zero? I assume operators can operate on other operators, too, right?
Why do you want to know, anyway?
I realize I forked my own thread the other day, and I am still quite curious about this - what intuitive methods, general techniques, and approximations could be used to finish the first exam in one hour or less? My re-write took 13 pages and 6 hours or something like that. Yet somebody got a 98 the first time through. Perhaps that person would like to share their thoughts? Broadly, I think this is a very important question for the class. The mean was about 42. I'm happy to say I got a 43. And I was writing pretty much the whole time. If I recall, no one left early. I invite anyone who got a 98, or close to it, to help the rest of us out. Thanks and please, East End
I agree, the test was much too long for many people to finish. I also wrote as fast as I could the entire time. What gave me trouble was the fact that we had a “think outside the box” type of problem - problem 3. Although it closely paralleled the homework/book examples, the fact that it was slightly different required a little “thinking” time, which I had no time for, since I didn't have enough time to write the easier problems too which I definitely knew the answers. When correcting the exam on my own time, I found the problem not difficult, but just required extra thought. Also, a suggestion: allow us to have trig identities. Even if we memorize them, we may not have time to check to see we didn't make a mistake. Suggestion to East End: reviewing the book, homeworks, and discussion problems helped me. Just make sure you know the basic concepts of each significant topic we cover.
You were definitely missing some shortcuts if you took 13 pages. In problem 1, the key was the trigonometric identity <math>sin^2(x)=\frac{1-cos(2x)}{2}</math>. Then all you have to do is integrate the RHS of that equation for <math><p^2></math>. <math><x^2></math> is a little trickier. Multiply RHS of the identity by <math>x^2</math>. The first term is trivial. The second term is a classic application of integration by parts, which needs to be done twice. A lot of the terms in the sum you end up with will vanish due to the limits of integration - they will be products of x and a sine or cosine of 2x. The lower limit will be 0, causing any terms containing x to vanish, and the trig functions will be 1 or 0, at one or both limits.
To make it easier to read (and write) I dropped the constant coefficients.
This would be a fantastic homework problem. It's actually in the book (2.11). It would be a good test problem too, if we were allowed to spend more time on it, AND we were given that trigonometric identity. There are countless tricks in calculus, trigonometry, and algebra that we have all learned, and we can't possibly have all of them on tap at all times. I am not a physics student, so perhaps my experience is not on par with the class, but I certainly have not had to integrate enough squares of trig functions for a double-angle formula to pop into my head as soon as I see it. Maybe it's a very common thing in QM? Us students don't know that yet. If we did SEVERAL problems using this trick for homework/discussion, we would wise up and memorize it. For example, problem 4 was pretty tricky, but we did it as homework, as well as covering it in discussion (for my section at least), so I think it's perfectly reasonable to expect us to know the tricks for that (again, it's a somewhat long derivation - time was still tight). Having notecards will not alleviate this problem, as we can't write down the tricks if we don't know they're important.
I think the best improvement that can be made on the quiz is to specifically give us the tricks we need to use to solve each problem, if they have not been thoroughly covered in class. Seeing a double-angle formula printed on a piece of paper is not going to hurt our intuitive understanding of quantum mechanics.
Actually, the way you describe is exactly how I solved the problem (the identity, the integration by parts, the terms going to zero, you name it). That isn't exactly using intuition and approximations based on what's going on, which is what the exam hinted at. I am curious about these intuitive methods and approximations. Anybody else?
(Also, I disagree about the notecards. I am pretty bad at memorizing mechanically usually, and think being able to write down things like the definition of a+/-, [a+,a-], etc, would be a nice, small, and reasonable help. I personally think this is exactly what reference materials are for in the “real world,” and think that we should be allowed a full-sized crib sheet. For me, historically, creating a crib sheet is excellent review time, forcing me to review everything, and giving me the chance to realize what topics I don't understand fully and spend more time on those topics in my studying.)
In the problem with the two finite wells (2.47), the system could model a diatomic molecule sharing an electron. In the ground state of the wave function, it is energetically favorable to have the wells closer together. This means there is an attractive force while the electron is in the ground state. Is there a repulsive force when the electron is in the first excited state?
Oddly, yes, that's exactly right. We know that Energy is proportional to frequency, which is inversely proportional to wavelength, which in discussion we determined is proportional to b for the first excited state; increasing the width between the wave crests increased its effective wavelength, decreasing its frequency and therefore reducing its energy, so Energy is inversely proportional to b for the first excited state. Since E minimizes as b→ infinity, the nuclei are pushed apart. For the even waveform, the ground state, the energy is lowest for b=0. The frequency is essentially zero and the wavelength is essentially infinite, making E essentially zero. For b>0, the 'dip' between the wave crests can be considered the valley of oscillation, resulting in an apparent wavelength and an increased frequency. In this case, as frequency increases Energy does as well, meaning that E increases with b. Since E is minimized at b=0, the ground state electron strives to pull the nuclei together. This is a very interesting problem conceptually, and serves as an introduction to the important anti intuitive results we obtain in studying Quantum Mechanics. I hope that explanation was clear and correct.
I am a bit perplexed with the free particle and other such problem where the wavefunction isn't constructable or says Griffths. Griffths later on finds the wavefunction of such a free particle. What happened here and why do we have a wavefunction for something that he said didn't exist? Is he just saying that the wavefunction isn't normalizable or just fails some condition that I don't see?
The free particle solution (plane wave) does not exist in Hilbert space because it is not square integrable, hence it is not normalizable. States that are not bound in quantum mechanics are not normalizable since their spectra are continuous, not discrete like bound states. This can be interpreted by the fact that plane waves have sharply defined momenta, so their energies cannot be sharply defined at the same time whilst obeying the uncertainty principle. The remedy is given as a wave packet, a linear superposition of plane waves in which the momentum, energy and position of the particle are no longer known exactly. We know from ordinary differential equations that a superposition of linearly independent solutions is also a solution, so this works well for us.
Oh, ok! This makes perfect sense! Thanks!
Later in chapter 3, Griffths talks about how the free particle has a continuous spectrum k and how infinite square well has a discrete spectrum n, but says that wavefunction of a continuous spectrum sometimes lie in Hilbert space, which makes our wavefunction unphysical. Could this be a reason why he said we didn't have a wavefunction which worked for the free particle in chapter 2?
When dealing with all of these matrices and eigenvectors it is easy to lose track of what physical importance any of this is. To break it down some, what exactly would it mean for something like problem 3.37, when “The Hamiltonian for a certain three-level system is represented by the matrix…”? Is a three level system in this sense a system that has three possible states and we try to find them or is it a three particle system that has these particles in each of these three states?
Also does anyone have any quick hints on how to conceptualize the rest of this linear algebra better in terms of physics?
Can someone explain the significance of the Axiom on page 102?
I believe this is similar to saying that any wavefunction can be described by the linear combination of certain other wavefunctions, as described by Fourier's theorem. In this axiom, the eigenfunctions of an observable correspond to the wavefunctions of that observable, which as we have seen in the previous chapters can indeed be described by a combination of wavefunctions.
This question is not entirely pertinent to the current subject, yet it crossed my mind nonetheless. When working in the simple harmonic oscillator potential scenario, we found that energy values are increments of 1/2. Does this mean that only Fermions can be modeled with SHO?
The simple harmonic oscillator has energy level increments of 1. <math>E_{n} = (n + \frac{1}{2})\hbar\omega</math>
I thought I understood SHO,and just like chavaz said <math>E_{n} = (n + \frac{1}{2})\hbar\omega</math>, but now I am confused, how do you make the jump from SHO to fermions?