# Quantum tunneling and energy conservation

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

## Answers and Replies

Matterwave
Gold Member
I'm not 100% sure on this (like, at all), but I do remember my professor once saying that you can't "really" find a particle with energy lower than the potential wall because when you observe that particle in that region you actually kick the energy of that particle up above the potential 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
The potential energy of the wall does not have the same value everywhere. You only know the average value. The place where you find the particle (at that particular time) could have a much lower potential than other places.

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 thats 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?

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 thats 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?

malawi_glenn
Homework Helper
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.

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?
No, I disagree. For a finite barrier there's a decaying (in space) solution inside the barrier. However, if barrier is infinitely high, the inside solution goes to zero. Afterall, thats tunneling...
BTW, the fact that solution inside is non-zero is just a manifestation of the wave-nature (here the wavefunction). Also, in optics you can have similar behavior with so called evanescent wave.

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. If the particle is inside the wall, its energy must be larger than V, am I right?

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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
The question is of the form of the wavefunction psi(x) that determines whether the particle can be found inside the well. If you believe that particle can be on one side and then on the other, then connecting the solutions (through boundary conditions on psi) - particle can be found inside.

not in quantum mechanics, that is the point of tunneling.
Say a free charged particle is prepared (with specific kinetic energy E) and at some later time it is shown to be inside a wide barrier (of constant potential V>E), by a process of elimination (i.e., by operations/measurements performed only on the regions in front and behind of the barrier, so as to not disturb the particle directly).

If we just wait, then measure the particle's energy after it has left the barrier, will it now have KE V>E (consistent with the knowledge that it was definitely "in" the barrier, and if so then was energy correspondingly subtracted from the mechanism that was used to check that it wasn't outside of the barrier)?

Or, if the particle's total energy at all times remains constant (implying negative KE in the barrier), what speed will we measure for it (say, if we equalise the electric potential outside, to widen the barrier, then measure how far the particle propagates in some time period)?

Mathematically, do we just zero out the reflected/transmitted parts of a wave-packet, renormalise and wait-see what its evolution limits to?

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?)

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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?)
You're right - tunneling is one form of interference. Like in the example of diffraction in the single slit its the average over space that needs to be taken in order to talk about conservation of energy/momentum. Incidentally, 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 leaving only components away from the slit - same as what was before the slit. You can also say, what about conservation of energy? At some fringe I can measure more than came in??? The answer is, if you integrate over all space (i.e. all the points of destructive interference) then overall energy will be conserved (assuming lossless slit). Same way in this problem... integrate over all space and you'll find that $$\int |\psi(x)|^2 dx$$=1, even though at some point (not in this problem but in general) |psi(x=a)|^2 can be greater than 1 (you wouldn't say that you've violated conservation of energy). When you speak of waves, conservation of energy is a non-local concept.

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 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?)

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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.)
If we are talking about quantum mechanics, then we are talking about probabilities. We have to have enough events to build up the probability distribution (i.e. multiple measurements). If I toss a coin 5 times and get heads all the five times does that mean that the random nature of the process is violated? I have to toss the coin N (where N is a very large number) of times before i get 1/2. If i shoot electrons one by one through double slits, eventually i will build up an interference pattern, - that doesn't mean that any particular electron comprising the distribution has violated momentum/energy/whatever. Perhaps this sounds like stating the obvious, but -- that is the explanation. In quantum mechanics we look for expectation values -- statistical representation of the events. You can't draw one event from a pile and infer about the distribution as a whole.

malawi_glenn
Homework Helper
so here is a good question, can we measure energy and position at the same time in QM?

so here is a good question, can we measure energy and position at the same time in QM?
how long is the measurement?

malawi_glenn
Homework Helper
It is quite meaningless to ask "what is the energy at position x_1?"

H 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 perhaps this will work better?

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malawi_glenn
Homework Helper
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?
hehe you answered really quick, I changed my answer, I wanted to take that point later ;-)

hehe you answered really quick, I changed my answer, I wanted to take that point later ;-)
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):

tunneling is not a quantum phenomenon but a wave phenomenon. you can observe tunneling with water waves, electromagnetic waves, etc. inside the barrier all waves will be attenuated in the direction of propagation, but the energy density would still be non-zero, unless at infinity.

malawi_glenn
Homework Helper
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.

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.
agreed about consistency! i also think that its equally important to find analogies in various branches of physics and relate them to the problem at hand. perhaps for some it is a revelation that situation discussed can be viewed totally classically?

i'm new on here and still haven't quite gotten the reply 'etiquette' if such even exists, but I guess i'm viewing the purpose of these discussions as a collective attempt to help each other understand something, and not necessarily demonstrate knowledge of for e.g. jargon (which is of course very useful to know too, don't get me wrong). so if an analogy (or a mathematical statement) can clarify the point or explain something - i don't see anything wrong with either. I can have flour and corn tortillas and switch freely between the two :P

as for paradoxes - they can 'spawn' anywhere - from having incorrect physical picture to not doing the math right...