Energy at classically forbiden positions

[DaTario]

TL;DR Summary

Since QM says one can measure a particle's position and find it at places which are forbiden by classicla physics, as in one-dimensional step potential case, which would be the kinetic energy of the particle measured at such positions?

Hi All,

Since QM says one can measure a particle's position and find it at places which are forbiden by classical physics, as in one-dimensional step potential case, with the particle having less energy than the potential heigth, which would be the kinetic energy of the particle measured at such positions?
Put differently: what would happen if one tryed to measure the particle's energy at such position?

Best wishes,

DaTario

 

[BvU]

The operator value for the kinetic energy yields a negative result -- as you expect. How to perform an actual measurement is a different story.

For what it's worth:

https://www.tau.ac.il/~yakir/yahp/yh55.pdf

Pinging @Demystifier to shed some light on this ...

 

[Demystifier]

The operator value for the kinetic energy yields a negative result -- as you expect. How to perform an actual measurement is a different story.

For what it's worth:

https://www.tau.ac.il/~yakir/yahp/yh55.pdf

Pinging @Demystifier to shed some light on this ...

The kinetic energy operator is a positive operator, all its eigenvalues are positive (or zero). Hence a usual projective measurement of kinetic energy cannot give a negative number, not even in the classically forbidden region. The paper you linked, however, is about weak measurement of kinetic energy, which is an entirely different thing. For instance, the first paper on weak measurements pointed out that a weak measurement of spin of a spin-1/2 particle can give the value 100. My opinion is that those weird weak values associated with weak measurements do not have any fundamental meaning.

 

[Demystifier]

what would happen if one tryed to measure the particle's energy at such position?

You cannot measure energy at position, because energy eigenstates are not position eigenstates. The energy operator and the position operator do not commute.

 

[DaTario]

Ok, I guess I (almost - as usually in QM-) fully understand the commutation argument, but let me try rephrasing the question. Let an ensemble of particles, subjected to the step potential, be prepared to be in an eigenstate of energy $E_0$, which is less than $V_0$. We are going to proceed to kinetic energy measurements at a given time $t_0$ after the preparation time of these systems. We know that if these measurements were position measurements, some fraction of the results would be at the classically forbiden region. Wouldn't it be reasonable to expect that the same fraction of the kinetic energy measurements yielded negative results?

Wouldn't it look like contradictory if only positive kinetic energy results were obtained?

 

[Demystifier]

Wouldn't it be reasonable to expect that the same fraction of the kinetic energy measurements yielded negative results?

No.

Wouldn't it look like contradictory if only positive kinetic energy results were obtained?

You are missing quantum contextuality. The act of measurement changes the properties of the system, which is why your arguments are invalid in the quantum realm. See also https://www.physicsforums.com/threa...-tunneling-of-a-particle.934849/#post-5906088

 

[hilbert2]

Ok, I guess I (almost - as usually in QM-) fully understand the commutation argument, but let me try rephrasing the question. Let an ensemble of particles, subjected to the step potential, be prepared to be in an eigenstate of energy $E_0$, which is less than $V_0$. We are going to proceed to kinetic energy measurements at a given time $t_0$ after the preparation time of these systems. We know that if these measurements were position measurements, some fraction of the results would be at the classically forbiden region. Wouldn't it be reasonable to expect that the same fraction of the kinetic energy measurements yielded negative results?

Wouldn't it look like contradictory if only positive kinetic energy results were obtained?

If you measure the position of a particle in such a potential, and find it in a classically forbidden region, the state of the particle after the measurement is such that its energy is sufficient for it to be there (or more precisely, it's a superposition of energy eigenstates that mainly correspond to eigenvalues higher than the height of the potential barrier).

The particle can obtain energy from whatever you use for measuring its position, hence the higher energy state after the measurement.

 

[DaTario]

You are missing quantum contextuality. The act of measurement changes the properties of the system, which is why your arguments are invalid in the quantum realm. See also https://www.physicsforums.com/threa...-tunneling-of-a-particle.934849/#post-5906088

I saw. In fact it would also seems to be contradictory if, in an eigenstate of energy $E_0$, one could obtain a measure of another energy value. Such states are expected to be dispersion free, theoretically. But regarding your suggestion for me to read the post, I must say that I was not proposing the measurement of more than one observable. I proposed the measurement of kinetic energy, only.

Now, consider that we know the particle has energy $E_1 > V_0$. So it classically can overcome the step potential. So, still classically speaking, the particle will have, at different times, two different values for its kinetic energy, namely: $E_1$ and $E_1-V_0$. Are we to suppose that these would be the only possible values obtainable from kinetic energy measurements on the particle in this state?

 

[Demystifier]

Are we to suppose that these would be the only possible values obtainable from kinetic energy measurements on the particle in this state?

No. The wave function is a wave packet represented by a Fourier transform containing many different values of the wave vector $k$. Each wave vector corresponds to another possible kinetic energy $\hbar^2k^2/2m$.

 

[DaTario]

This is a simple problem that I have never got into such level of detail.

 

[Vanadium 50]

Let an ensemble of particles, subjected to the step potential, be prepared to be in an eigenstate of energy $E_0$... Wouldn't it be reasonable to expect that the same fraction of the kinetic energy measurements yielded negative results?

No.

If you are in an energy eigenstate, every time you measure the energy you get the same number, in this case $E_0$.

If you then say "but I only want kinetic energy", I would ask you how you intend to measure it. If you say "it's total energy minus potential energy", if you write that down, it's $E_0 - V(x)$. If you are in an energy eigenstate, you're not in a position eigenstate and vice versa. So this is not well-defined.

If you have a way to measure kinetic energy, write it down, and we can tell you how it behaves. But what will surely happen is that if kinetic energy is well defined, position will not be, so "kinetic energy when the particle is in the classically forbidden region" is undefined.

 

[DaTario]

But what will surely happen is that if kinetic energy is well defined, position will not be, so "kinetic energy when the particle is in the classically forbidden region" is undefined.

Thank you, Vanadium 50, I guess now I got a nice road to understand a little better this issue. Both energy concepts, total and kinetic, depend on linear momentum and that seems to be the reason why position states and energy states cannot be well defined simultaneously. But, still considering the particle in the step potential with energy $E_0$ larger than the heigth $V_0$ (it is in the eigenstate of energy $E_0$), does the momentum spectrum consists of two values, namely:
$$ p_I =\sqrt{2mE_0}$$
and
$$ p_{II} = \sqrt{2m(E_0-V_0)}$$
?
or the transition (discontinuity) in $x=0$ introduces many other momentum components in this state?

One last question: is there a way to find experimentally in the energy context at least a small signature of the "strangeness" of this evanescent wave in the position representation, that extends into the classically forbiden region? I am still considering the particle in the step potential with energy.

 

[vanhees71]

Of course, in the momentum eigenstate momentum is not determined either, since including a potential the energy eigenstate is (usually) not a momentum eigenstate.

The tunnel effect itself has been observed. One of the earliest examples is Gamov's theory of $\alpha$ decay. In a nucleus that is instable against $\alpha$ decay among the nucleons making up the nucleus an $\alpha$ particle is "preformed", and then tunnels out of the binding potential.

Today there are plenty of more examples, even used in semiconductor technology:

https://en.wikipedia.org/wiki/Quantum_tunnelling

 

[Vanadium 50]

does the momentum spectrum consists of two values

No.

In QM, we can only talk about expectation values, and <p> = 0; the particle moving to the left and the particle moving to the right are equiprobable.

We can talk about <p²> (or it's square root), but that is essentially kinetic energy, so we're right back where we were a few messages back.

 

[DaTario]

But when we consider the spetrum of an observable (in the present case, linear momentum), the set of eingenvalues represents the set of possible outcomes in direct measurent of this observable. So, besides expectation values we can discuss issues involving specific measurement results.
If we consider the particle to be incident from the left in a step potential that goes like $V(x) = V_0 \Theta(x)$, where $\Theta(x)$ is the Heaviside function, is the symmetry you referred to still present in the expectation value?

 

[Vanadium 50]

Making this increasingly Rube-Goldbergy is not going to help you.

If you have a one-dimensional step function, and the energy is greater than the barrier you have no classically forbidden region. If you make it less than the barrier, if you are in an energy eigenstate, you are not in a position eigenstate and cannot talk about a particle's energy in the classically forbidden region. If you constrain the particle to be in the classically forbidden region, it is not only not in an energy eigenstate, any measurement of the energy will be greater than the barrier.

This has been explained in messages 4, 7 and 11.

 

[DaTario]

Ok, Thank you, Vanadium, for your kind contribution. But I don't think it is fair to say that I am trying a Rube-Goldberg way in this discussion. My appealling to basic notions such as momentum to talk about measurements of kinetic energy doesn't seem to be absurd or too complex at all.
Correct me please if I am wrong, but in message 7 there is a connection between the quantum dispersion in momentum produced after a measurement of position and physical actions on the particle due to the measurement process. It seems that there is some debate (see for instance this Nature paper by G. Rempe) on this point, i.e., that uncertainty principle has to do with direct influences of the apparatus on the system (particle).

BTW, I have not quite understood the symmetry point in this context. The eigenstate of energy $E_0 > V_0$ of a particle in the step potential does not seem to present symmetry in momentum so that $\langle p \rangle = 0$. Recall that we have three propagating waves: incident, reflected and transmitted. Do they produce $\langle p \rangle = 0$ ?

Best wishes and sorry for my being somewhat awkward.

 

[Vanadium 50]

I really don't think adding complications will be helpful. Surely the problem isn't that this just isn't complicated enough.

Which statement are you troubled by?

1. If you have a one-dimensional step function, and the energy is greater than the barrier you have no classically forbidden region.
2. If you make it less than the barrier, if you are in an energy eigenstate, you are not in a position eigenstate and cannot talk about a particle's energy in the classically forbidden region.
3. If you constrain the particle to be in the classically forbidden region, it is not only not in an energy eigenstate, any measurement of the energy will be greater than the barrier.

 

[DaTario]

When you are locked inside an unconfortable room, I don't think trying to exit through the roof is adding complications (may be too dramatic...). But wrt your question, now I am ok with all of them, and I am grateful to all participants of this debate.

a) I have seen some computer simulations of wave packets incident from the left in the interface of the step potential and some crazy things seems to happen in the interface from the "Fourier point of view". So I tend to think that a particle in a eigenstate of energy $E_0 > V_0$ may not present only two values of momentum (as I pointed out in #12).

b) It is not clear that the momentum expectation value of an eigenstate of energy $E_0 > V_0$ is zero (it is not clear this symmetry).

c) Are issues of contextuality present even when we are considering the measurement of just one observable? (for instance: I prepare the state $10^6$ times and measure position $10^6$ times $\Delta t$ after the preparation is concluded. I do some statistics with the results. Now I prepare the state $10^6$ times and measure momentum $10^6$ times $\Delta t$ after the preparation is concluded, and also do some statistics with these results. Only after these two sessions of preparing+measurement I start doing comparisons.)

 

[Nugatory]

computer simulations of wave packets incident from the left in the interface of the step potential

Of course these wave packets are not energy (or momentum, or position) eigenstates. They are superpositions of momentum eigenstates, and their behavior provides little insight into the behavior of energy eigenstates.

 

[DaTario]

Yes, I agree, but my point has to do with the effects of the discontinuity at $x=0$ on the momentum spectrum of the particle.

 

[vanhees71]

I really don't think adding complications will be helpful. Surely the problem isn't that this just isn't complicated enough.

Which statement are you troubled by?

1. If you have a one-dimensional step function, and the energy is greater than the barrier you have no classically forbidden region.
2. If you make it less than the barrier, if you are in an energy eigenstate, you are not in a position eigenstate and cannot talk about a particle's energy in the classically forbidden region.
3. If you constrain the particle to be in the classically forbidden region, it is not only not in an energy eigenstate, any measurement of the energy will be greater than the barrier.

It doesn't make sense to say "the particle has a certain energy at one point" already in classical mechanics. The energy of a particle is just a quantity, depending on position and momentum.

In QM: If you prepare a particle in an energy eigenstate, it has usually neither a determined momentum nor position, but you have probability distributions for these observables, and then the above quoted items by @Vanadium 50 is all there is to say about it.

 

[vanhees71]

Of course these wave packets are not energy (or momentum, or position) eigenstates. They are superpositions of momentum eigenstates, and their behavior provides little insight into the behavior of energy eigenstates.

Here I have some simulations of wave packets in various 1D simple step potentials. Though the text is in German, maybe it's nice to look at the simultations:

https://itp.uni-frankfurt.de/~hees/qm1-ss09/wavepack/wavepack.html

 

[vanhees71]

Yes, I agree, but my point has to do with the effects of the discontinuity at $x=0$ on the momentum spectrum of the particle.

The momentum spectrum is $\mathbb{R}$, independent of the Hamiltonian (if you don't put the particle in a box with perdiodic boundary conditions). Where are there discontinuities?

 

[DaTario]

I refer to the discontinuity in the step potential at $x=0$.

 

[vanhees71]

Fine, but these are not discontinuities in the momentum spectrum!

 

[DaTario]

I know. I thank you all very much. In some sense I must appologize for having changed the focus of the OP. It seems to me that the main point has been addressed properly already. Some exotic preocupations still come into my mind regarding these quantum departures from classical behavior, but I believe that it will be better to save it for another thread. Thank you all again.