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The probability of finding the particle inside the wall is non-zero, What happen if we do find the particle is inside the wall, would that violate energy conservation? since total energy is smaller than the potential energy of the wall
The probability of finding the particle inside the wall is non-zero, What happen if we do find the particle is inside the wall, would that violate energy conservation? since total energy is smaller than the potential energy of the wall
I'm not sure I understand the problem. 'Particle' has energy set by the initial conditions of the particle, say E. Then, one is to solve the problem of finite barrier and determine the wave-function psi(x,t) and (if that's of interest) in particular the psi(x=x_V) i.e. at the position of the barrier. Now, if you were to do a kind of measurement that answers: is the particle here or not, after successive number of measurements you would reconstruct |psi(x)|^2, and the probability of particle being inside the V would then be |psi(x=x_V)|^2.
Or, did I miss the point of the question?
But if E, the energy of the particle, is smaller than V, then how can it be inside the wall?
To be inside the wall, the particle must have energy larger than V, right?
But if E, the energy of the particle, is smaller than V, then how can it be inside the wall?
To be inside the wall, the particle must have energy larger than V, right?
not in quantum mechanics, that is the point of tunneling.
I have no problem with quantum tunneling. It's fine if the particle is found on either side of the wall, but not inside the wall, since the energy of the wall is V, so the particle can be on either side of it, but not inside it
not in quantum mechanics, that is the point of tunneling.
I suppose a similar question is, if a single particle from a collimated beam passes a single slit, and is detected at a high angle of diffraction, how is momentum conserved? (And does the aperture just zero out part of the wave-function for particles that get through, or strictly does the aperture entangle with every particle that it changes the wave-function of?)
But I was speaking of just a single individual particle. We agree that energy and momentum are conserved on average, I guess the question is whether they are conserved only on average? (I'm guessing not, since otherwise shouldn't independence of previous outcomes send the average on a random walk?)the momentum is conserved of course, because for the high angle you mention there's equal and opposite contribution on the other side, so the transverse (to slit) momentum cancels out [..] When you speak of waves, conservation of energy is a non-local concept.
But I was speaking of just one solitary individual quanta alone. We agree that energy and momentum are conserved on average, I guess the question is whether they are conserved only on average? (I'm guessing not, since otherwise I would expect independence of previous outcomes to result in a random walk.)
so here is a good question, can we measure energy and position at the same time in QM?
sure, as [x,p] .ne. 0, and so is [x,p^2] .ne. 0H and x does not commute
sure, as [x,p] .ne. 0, and so is [x,p^2] .ne. 0
you can use that mathematical construct to give the answer to this problem - i agree. there are multiple ways of saying the same thing, so perhapsa this ill work better?
i guess i saw where you were going with this :) I'm new to this forum, but would guess that this response shouldn't really go into thread discussion... so, i'll put some relevancy to the theme of the thread below (not necessarily in response to your comments):hehe you answered really quick, I changed my answer, I wanted to take that point later ;-)
I think it is really important to be consistent with QM if one are to discuss QM. When we talk about measurements, we must do in the realm of QM. And so on, otherwise it is not strange that paradoxes will arise.
It is quite meaningless to ask "what is the energy at position x_1?"