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 Aug 15 comment Determining the Wave Function From Initial Conditions Oh, I see. You are using terminology that I am unfamiliar with. I still do not see how this relates to the wave function being only a mixture of the first two stationary states. Aug 15 comment Determining the Wave Function From Initial Conditions Well, I did not define them, but the textbook I am using does, which is Griffith's Introduction To Quantum Mechanics. Here is what he says regarding stationary states: "Although the wave function itself does (obviously) depend on $t$, the probability density, $|\Psi(x,t)|^2 = \Psi^* \Psi = \psi e^{i Et/\hbar} \psi e^{-i Et/\hbar} = |\psi(x)|^2$ does not. Aug 15 comment Determining the Wave Function From Initial Conditions So, what other justification could be used? The answer key does not use this operator, nor is it spoken of in the chapter from which this problem comes from. Aug 15 comment Determining the Wave Function From Initial Conditions @ACuriousMind I do not know of this time evolution operator of which you speak. Aug 15 comment Determining the Wave Function From Initial Conditions Yes, exactly. As far as I understand, the most general solution of the infinite well is $\Psi (x,t) = \sum_{n=1}^{\infty} c_n \psi_n(x) \phi_n(t)$, where the coefficients are $c_n = \sqrt{\frac{2}{a}} \int_{0}^{a} \sin(\frac{n \pi}{a} x) \Psi(x,t)$. Why wouldn't I use these two equations to calculate the wavefunction for all future times? Jan 2 comment Collision Between Two Particles: Writing the Mass As A Function of The Angle I don't believe this is a duplicate: I am asking for advice as to how I might write $\displaystyle \frac{m_1}{m_2}$ as a function of the angle (some angle); but the other question asks to verify unrelated equations. Jan 2 comment Collision Between Two Particles: Writing the Mass As A Function of The Angle Well, I was defining $\theta$ as the angle between the final velocity vectors. Would this not be helpful? Feb 19 comment Finding the Electric Field (and other information, besides) @MichaelBrown Will it still be true that I need to multiply the magnitude of the electric field at that point by two, because I am only applying Gauss's law to one plate? Feb 19 comment Finding the Electric Field (and other information, besides) @MichaelBrown Yes, I have actually; but I wasn't sure if I need it. So, should I generate a Gaussian plate, infinite in size, and place it between the two sheets? Then, find the magnitude of the electric field from sheet, and multiply it by two? Feb 19 comment Finding the Electric Field (and other information, besides) To be frank, the textbook you have provided appears to be a little advanced. Seeing as I am only in a undergraduate Physics II course, I am not sure it will be very helpful at the moment. Notwithstanding, I still glad you did provide me with a link, because the book does look interesting, and I'd like to read it someday. Feb 16 comment Higher To Lower Electric Potential @MichaelBrown Actually, I have one more question. I was able to answer the question I originally posed, except for part (b). I thought, because the electric potential energy was increasing at that point, that the electric potential at that point would be higher. After all, electric potential is defined one way as, "the amount of electric potential energy that a unitary point charge at that location would have." So, if the electron has more PE at the final position, wouldn't that mean the electric potential was greater at that final position too? Feb 16 comment Higher To Lower Electric Potential Thank you for the explanation. I think that was the root of my confusion; what they said in the orignal problem I posted seemed odd, that potential energy is increasing, but electric potential was becoming more negative. Feb 16 comment Higher To Lower Electric Potential @MichaelBrown So, even though we have a positive increase in potential energy for a negative test charge, the change in electric potential could be negative because of the sign negative test charge? Feb 16 comment Higher To Lower Electric Potential Yes, I believe so. Just one question, if I push a negative test charge near a negative source charge, the test charge will be gaining potential energy, and that potential energy will be positive, right? Alternatively, if I move a negative test charge near a positive source charge, making sure it doesn't accelerate towards it, it will loss potential energy, showing up as a negative potential energy, right? And the mathematics will show that these are true? I guess the root of my problem is seeing a generalized situation, and the mathematics that accompany it. Feb 16 comment Higher To Lower Electric Potential Sorry for the inundation of questions. I've heard electric potential being described as being "the measure of potential energy per unit charge." From this, I have developed the understanding that electric potential is a scalar that is assigned a different value at every point in an electric field.It almost seems like a conversion factor. It's how much potential energy a charged particle could possession if it were there. It's a potential of a potential. Again, sorry for all of the comments. I hope you can help me. Feb 16 comment Higher To Lower Electric Potential Also, is it true that if a charge has negative electric potential energy, then it is not necessary that it has lost that energy. In other words, negative electric potential energy can mean a gain in that sort of energy? Feb 16 comment Higher To Lower Electric Potential I understand it, for the most part; one bit I don't quite understand, however, is this, "negative charge that rolls 'down this potential hill'." I keep reading allusions about this "potential hill" analogy, but I can't find an actual source of it. Do you know of any online articles that treat this analogy? Feb 11 comment Finding the electric field Actually, I am having difficulty applying Gauss's law to a cylinder. Feb 2 comment Electric Fields How do I color parts of my text? Feb 2 comment Electric Fields @Crazybuddy I needed those selections to be colored.