There's indeed such a problem. In regions where energy of particle is smaller than the minumum of particle, the solutions are exponential rather than oscialating, and kinetic energy operator becomes negative. For instance, (in nuclear physics) in a spherical potential well, the solution in the...
Ouch! Sorry! I'm not a native English speaker, that's probaby why an quite-ambioguous sentence turned into disaster in my hands!
Hmmm. Right. But then, the width of slit isn't measure of it's position at the instant of momentum measurement. Let's see what I've understood from your...
Wow, that's whom I'd call a braveheart.
Fine, but which is the last word?
BTW, there should be limitations on your CCD (and our technology) by uncertanity principle, so I suppose it's impossible to make a device that measures momentum of a particle with perfect accuracy. Err.. something...
Though I'm aware of HUP is talking about standard deviation, I'm curious about one thing. So I'd like to raise this question: in what precision can we measure position and momentum of one, single electron. Feel free to talk in terms of eigenstates and wavefunction collapse.
For someone who's bored with wave mechanics, would you suggest studying Heisenberg's matrix mechanics (which was the first formulation)?
Are there any major/conceptual differences? (except one talks about waves and the other doesn't!)
And for someone who wants to study it, what books/online...
Horray! :approve: Huge thanks, finally Schrödinger equation did come out!
One last question, how could you (or Feynman) see which term you had to expand? I rather tried expanding e^{iLdt/\hbar} directly, without expanding the potential term around a fixed point, and without leaving a...
Ups. I missed that bit. So my A became
A \sqrt{\frac{\pi}{-im/\hbar \epsilon}} = 1. V(x') \frac{\partial^2 \varphi}{\partial x^2}... term is dropped since it has an second order \epsilon. OK, by doing so, I had the potential term nicely, but I still have a weird extra coefficent in the same...
Hmm. How about this:
Following (Fourier transform) holds for any wave packet
\psi(x,t) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \phi(k) e^{i(kx-wt)} dk
Let us assume that matter is a kind of wave, so obeys this equation. Now, to relate the wave with quantities we all know and love, we...
Thanks for the reply!
I didn't know the bounds are infinities... Also, I couldn't get \lambda to the 3/2th power, for -iV'(x') y \lambda, but as you did, I used only the first term for potential.
Anyway, I couldn't see how Schrödinger equation comes out from what you've written, so
I've tried...
What i mean is this: if you have a "piece of wave" that is too short -such as a sudden small bump-, would you call it a wave? To talk about a wave, you'll need a wavelength, roughly. (Of course, if you're pedantic, you actually need something that is periodic, but one-wavelength piece is fairly...
I think your teacher's statement was rather "intutive". (Note that he seems to be talking about a precise wavelength, no \Delta \lambda is involved. That'd mean a precise momentum).
More likeky, his pointly was this: if you have a minimal thing that would represent a wave, maybe a pulse or a...
Yup.
p = \frac{h}{\lambda}
then, by differentiating both sides
\Delta p = -\Delta \lambda \frac{h}{\lambda^2}
But I see no point in messing with \Delta \lambda. Your teacher is talking about \lambda.
Also note that these deltas can be quite misleading. What Heisenberg means with "uncertainty...