Classically a particle cant exist in a region where V>E

[suku]

classically a particle can't exist in a region where V>E, v=pot energy, E=total energy, but quantum mechanically its possible. We can explain it(or draw an analogy) with concepts of optics,when light falls from a denser medium to rarer medium.but that wave is a real one,the quantum mechanical wave is a probabilistic one which relates to finding of particle.
If particle can't be found how can its probability wave be found?

thanks for any answer

 

[intervoxel]

Wherever the wave function is different from zero, there is some probability the particle can be found there. So it can be found in a classically forbidden zone. One evidence of this behavior is the existence of transistors where electrons can be found beyond a potential barrier.

 

[SpectraCat]

classically a particle can't exist in a region where V>E, v=pot energy, E=total energy, but quantum mechanically its possible. We can explain it(or draw an analogy) with concepts of optics,when light falls from a denser medium to rarer medium.but that wave is a real one,the quantum mechanical wave is a probabilistic one which relates to finding of particle.
If particle can't be found how can its probability wave be found?

thanks for any answer

Even though there is a finite probability density inside the classically forbidden region (CFR), it can be shown that the particle cannot ever be measured inside that region. Since the momentum is imaginary inside the classical region, that would correspond to a negative kinetic energy, which is nonsensical. What saves this is the uncertainty principle, using which one can demonstrate that the uncertainty in position is always larger than the penetration depth at which one tries to measure, so no measurement of the position could ever confirm that the particle is inside the CFR. A nice workup of this can be found here:
http://books.google.com/books?id=3r...lity in classically forbidden region&f=false"

 

[suku]

Even though there is a finite probability density inside the classically forbidden region (CFR), it can be shown that the particle cannot ever be measured inside that region. Since the momentum is imaginary inside the classical region, that would correspond to a negative kinetic energy, which is nonsensical. What saves this is the uncertainty principle, using which one can demonstrate that the uncertainty in position is always larger than the penetration depth at which one tries to measure, so no measurement of the position could ever confirm that the particle is inside the CFR. A nice workup of this can be found here:
http://books.google.com/books?id=3r...lity in classically forbidden region&f=false"

alpha particles show tunelling phenomenon... what about it??

 

[SpectraCat]

alpha particles show tunelling phenomenon... what about it??

Nothing in what I wrote rules out tunneling, which involves detecting a particle in another classically allowed region of space. My post was about why the particle can never be detected *inside* a classically forbidden region. Did you read the link I posted?

Take the example of a particle tunneling through a 1-D potential. It is possible to observe the particle in the classically allowed region on either side of the barrier, but it is impossible to observe the particle in the CFR represented by the inside of the potential barrier. In other words, the probabilities of observing a reflection event or a transmission event sum to one. This is true even though the particle has a non-zero probability amplitude inside the CFR. You can find a description of this in any elementary Q.M. text.

 

[peteratcam]

Even though there is a finite probability density inside the classically forbidden region (CFR), it can be shown that the particle cannot ever be measured inside that region. Since the momentum is imaginary inside the classical region, that would correspond to a negative kinetic energy, which is nonsensical.

I don't believe this. The whole probability interpretation of QM would be wrong if particles can never be measured in a classically forbidden region where the wavefn is nonzero.

 

[SpectraCat]

I don't believe this. The whole probability interpretation of QM would be wrong if particles can never be measured in a classically forbidden region where the wavefn is nonzero.

I don't think this is true. In any case, there are no known experiments (at least as far as I am aware), that measure a particle while it is inside a CFR. Did you look at the reference I posted? Basically, the exponentially decaying wavefunction inside the CFR ensures that the uncertainty in position is always greater than the penetration depth into the CFR. This means that one can never conduct an observation that conclusively shows a particle located inside the CFR.

 

[peteratcam]

Well I don't know about experiments, but it is a theoretical objection. Assume a particle is in an energy eigenstate of a potential well. The integral of the squared wavefunction within the classically allowed region of the well is always a bit less than 1. But you are saying that a measurement of position will find it inside the classically allowed region with probability 1, which sounds like you are denying the probability interpretation of the squared wavefunction.

 

[SpectraCat]

Well I don't know about experiments, but it is a theoretical objection. Assume a particle is in an energy eigenstate of a potential well. The integral of the squared wavefunction within the classically allowed region of the well is always a bit less than 1. But you are saying that a measurement of position will find it inside the classically allowed region with probability 1, which sounds like you are denying the probability interpretation of the squared wavefunction.

I am certainly not denying the Born interpretation of the squared wf. What I am saying is that any measurement of the position will correspond to one of the following cases:

1) it will return a mean value that lies within the classically allowed region (CAR),

2) it will return a mean value that lies within the CFR for a given quantum state, but with an uncertainty so large that one cannot exclude the possibility that the act of measurement excited the particle to a higher lying state where that measured value *does* correspond to the CAR.

So, from a certain point of view, I guess that one could say that the probability of *measuring* the position (or any other physical property) in the CAR is indeed one, but that says nothing about the probability density of the wavefunction. I don't think there is such a thing as a "position space probability operator", so I don't think it is possible to measure the probability density directly.

Can you concoct a theoretical experiment whereby the a quantum particle is measured inside a CFR without violating the HUP? I have tried to do this, but I have not been able to come up with anything.

One mitigating factor here that is getting glossed over is that what I have been talking about so far is formulated from the point of view of stationary states, whereas any meaningful measurement is necessarily time-dependent (there must be a "before" and "after"). Thus, one should take into account that any real measurement will amount to a time-dependent perturbation of the quantum state of the system, I have never worked out such a treatment, nor have I seen it done.

 

[Truecrimson]

First of all, thanks for the discussion that got me thinking about this problem.

Now, my point is: can't we find a particle in a classically forbidden region? I'm not really convinced by the argument in the book. Actually I perfer this explanation http://www.chem.uci.edu/undergrad/applets/dwell/uncertainty.htm

To summarize: we cannot simultaneous know that a region is classically forbidden and that a particle is really located there. In a sense we can't "catch the particle in the act" of being quantum mechanical!

Another mathematical point is that if the momentum is imaginary in the forbidden region, that means that the momentum operator is not hermitian there. Then how can we use the uncertainty relation?

 

[suku]

Take the example of a particle tunneling through a 1-D potential. It is possible to observe the particle in the classically allowed region on either side of the barrier, but it is impossible to observe the particle in the CFR represented by the inside of the potential barrier. In other words, the probabilities of observing a reflection event or a transmission event sum to one.

if a particle can't be located inside a CFR how can reflection and transmission coefficient add to 1? that is the particle would only be reflected and not transmitted...

 

[Frame Dragger]

First of all, thanks for the discussion that got me thinking about this problem.

Now, my point is: can't we find a particle in a classically forbidden region? I'm not really convinced by the argument in the book. Actually I perfer this explanation http://www.chem.uci.edu/undergrad/applets/dwell/uncertainty.htm



Another mathematical point is that if the momentum is imaginary in the forbidden region, that means that the momentum operator is not hermitian there. Then how can we use the uncertainty relation?

Now THAT makes sense.

 

[zenith8]

As is so often the case, the explanation is clear using the http://www.tcm.phy.cam.ac.uk/~mdt26/pilot_waves.html" of QM. There, the wave is not a 'wave of probability' as you state - it is an objectively existing wave which exists in addition to the electrons and pushes them around (this picture logically follows from the ordinary equations of QM - the only assumption that differs from normal is that particles continue to exist when you don't look at them, i.e. when you don't measure where they are)..

Consider tunneling through a square barrier - the case where the initial energy of the particle E is less than the barrier height V. Initially the particle - traveling along with a \Psi-wave packet - moves towards the barrier. The interaction of the packet with the barrier leads to the formation of reflected and transmitted packets of diminished amplitude, perhaps together with a small packet persisting inside the barrier. The particle ends up in one of these (which one depends upon its initial starting position within the wave packet)..

Tunneling arises from the modification of the total energy of the particle due to the rapid spacetime fluctuations of the \Psi-wave in the vicinity of the barrier (the force -\nabla Q exerted by the wave on the particle turns out to be proportional to the curvature of the wave). The effective 'barrier' encountered by the particle is not V but V+Q - where Q is the potential energy function of the wave field. This may be higher or lower than V and may vary outside the 'true' barrier. For tunneling you require only that the energy of the particle \geq V + Q then the particle may enter or cross the barrier region.

In effect, the additional 'force' exerted by the wave field on the particle just shoves the particle over the barrier.

As far as I know it is impossible to explain this effect consistently using an interpretation of QM involving the wave function alone (all you can do is predict the probabilities of the final outcomes). Cue much wailing and gnashing of teeth and people telling you that I'm not allowed to explain this to you because the interpretation is non-standard.

 

[Frame Dragger]

As is so often the case, the explanation is clear using the http://www.tcm.phy.cam.ac.uk/~mdt26/pilot_waves.html" of QM.

Now why doesn't that surprise me? Of your explanation of tunneling is correct however, why wouldn't say, an electron tunnel EVERY time? If the energy is there in the pilot wave to "push" it over the energy differential, why doesn't it happen every time?

That said... as far as I know you're correct, and while its necessity is explained and predicted by QM, there is no single explanation involving the \Psi alone. Bah.

 

[zenith8]

Of your explanation of tunneling is correct however, why wouldn't say, an electron tunnel EVERY time? If the energy is there in the pilot wave to "push" it over the energy differential, why doesn't it happen every time?

Hidden variables, mate. Repeat the experiment 1 million times with the same state preparation procedure (i.e. the same starting wave function). However, the individual members of the ensemble are not the same because of the initial positions of the electrons (which are distributed according to the square of the wave field). As the system time evolves, different starting points evolve to different finishing points i.e. the electrons follow a different trajectory each time. If the probability of tunneling through the barrier is 0.3, then 30 per cent of initial trajectories will end up on the other side.

See the attached picture.

That said... as far as I know you're correct, and while its necessity is explained and predicted by QM, there is no single explanation involving the \Psi alone. Bah.

Really annoying isn't it..

 

[Frame Dragger]

Hidden variables, mate. Repeat the experiment 1 million times with the same state preparation procedure (i.e. the same starting wave function). However, the individual members of the ensemble are not the same because of the initial positions of the electrons (which are distributed according to the square of the wave field). As the system time evolves, different starting points evolve to different finishing points i.e. the electrons follow a different trajectory each time. If the probability of tunneling through the barrier is 0.3, then 30 per cent of initial trajectories will end up on the other side.

See the attached picture.


Really annoying isn't it..

Ahhhh... I see.

and... YES! lol.

 

[SpectraCat]

First of all, thanks for the discussion that got me thinking about this problem.

Now, my point is: can't we find a particle in a classically forbidden region? I'm not really convinced by the argument in the book. Actually I perfer this explanation http://www.chem.uci.edu/undergrad/applets/dwell/uncertainty.htm

ok ... but that is just your preference. The two explanations are formally identical.

Another mathematical point is that if the momentum is imaginary in the forbidden region, that means that the momentum operator is not hermitian there. Then how can we use the uncertainty relation?

This is the same argument as in the book you like, except I didn't put the word "logically" negative in there. IOW, you can describe a CFR as a region where the kinetic energy is "logically" negative, or as a region where the momentum is "logically" imaginary. Either concept is non-sensical, and I think they are identical, since a negative KE implies an imaginary momentum.

However, I do see how the non-Hermetian point is a little more serious ... but isn't that a general problem for solving the Schrödinger equation inside a CFR as well? If one of the postulates of Q.M. is that the eigenvalues of physical observables is real, then how can we use the mechanics of Q.M. to calculate a probability density in a region where a relevant observable is imaginary?

 

[SpectraCat]

As is so often the case, the explanation is clear using the http://www.tcm.phy.cam.ac.uk/~mdt26/pilot_waves.html" of QM. There, the wave is not a 'wave of probability' as you state - it is an objectively existing wave which exists in addition to the electrons and pushes them around (this picture logically follows from the ordinary equations of QM - the only assumption that differs from normal is that particles continue to exist when you don't look at them, i.e. when you don't measure where they are)..

Consider tunneling through a square barrier - the case where the initial energy of the particle E is less than the barrier height V. Initially the particle - traveling along with a \Psi-wave packet - moves towards the barrier. The interaction of the packet with the barrier leads to the formation of reflected and transmitted packets of diminished amplitude, perhaps together with a small packet persisting inside the barrier. The particle ends up in one of these (which one depends upon its initial starting position within the wave packet)..

Tunneling arises from the modification of the total energy of the particle due to the rapid spacetime fluctuations of the \Psi-wave in the vicinity of the barrier (the force -\nabla Q exerted by the wave on the particle turns out to be proportional to the curvature of the wave). The effective 'barrier' encountered by the particle is not V but V+Q - where Q is the potential energy function of the wave field. This may be higher or lower than V and may vary outside the 'true' barrier. For tunneling you require only that the energy of the particle \geq V + Q then the particle may enter or cross the barrier region.

In effect, the additional 'force' exerted by the wave field on the particle just shoves the particle over the barrier.

As far as I know it is impossible to explain this effect consistently using an interpretation of QM involving the wave function alone (all you can do is predict the probabilities of the final outcomes). Cue much wailing and gnashing of teeth and people telling you that I'm not allowed to explain this to you because the interpretation is non-standard.

Maybe I am being dense, but how does this explain the original question about detecting the particle inside a CFR?

Also, while I like dBB and I am impressed by it every time I learn more, I don't really find this explanation any more clear than the TCI-probability based one, which I think I understand (uh oh .. that means I don't, according to Feynman).

 

[peteratcam]

I am certainly not denying the Born interpretation of the squared wf. What I am saying is that any measurement of the position will correspond to one of the following cases:

1) it will return a mean value that lies within the classically allowed region (CAR),

2) it will return a mean value that lies within the CFR for a given quantum state, but with an uncertainty so large that one cannot exclude the possibility that the act of measurement excited the particle to a higher lying state where that measured value *does* correspond to the CAR.

Talking about the mean of a set of measurements is quite different from the result of a single measurement. Furthermore, we can assume we have very good experimentalists so there is no uncertainty in any direct measurements they make.

Can you concoct a theoretical experiment whereby the a quantum particle is measured inside a CFR without violating the HUP? I have tried to do this, but I have not been able to come up with anything.

Put an electron in an energy eigenstate of a potential well.
Turn off the potential well and simultaneously turn on a huge electric field which accelerates the electron downwards into a piece of photographic paper. Sometimes, the dot on the paper will be found outside of the classically allowed areas on the paper. The HUP is built into the formalism of quantum mechanics, so it can't be violated by anything I do.

Also, I don't know what it means to say momentum is imaginary in the CFR. Momentum is a hermitian operator, or a real number.

 

[SpectraCat]

Talking about the mean of a set of measurements is quite different from the result of a single measurement. Furthermore, we can assume we have very good experimentalists so there is no uncertainty in any direct measurements they make.

That doesn't change the situation at all, but perhaps my choice of terms was poor. The upshot of this is that any time that you measure a position observable that corresponds to the CFR for a particular quantum state, the conditions of the measurement will be such that you cannot be certain that the act of conducting the measurement did not perturb the system so that it was a higher lying state, where the measured position corresponds to a CAR.

Put an electron in an energy eigenstate of a potential well.
Turn off the potential well and simultaneously turn on a huge electric field which accelerates the electron downwards into a piece of photographic paper. Sometimes, the dot on the paper will be found outside of the classically allowed areas on the paper. The HUP is built into the formalism of quantum mechanics, so it can't be violated by anything I do.

Are you sure about this? Has the experiment been done? What are the characteristics of the trap? What are the characteristics of the "huge" field? In order for this to work, you would have to be sure that the transverse distance dx traveled by the electron during its acceleration by the huge field is smaller than the penetration depth of the wf into the CFR, let's see some numbers for those conditions. Note also that it would need to be a 3D-trap, which means that the electronic wavefunction would have non-zero components along the direction of the accelerating field, which is a serious complication in this case, because it means that the wavefunction will couple to your "huge" field, and therefore have the chance to be perturbed by it. This then means that you are in the same boat as above, you cannot ever be sure that the measured position is in the CFR for the quantum state of the system, because you cannot independently know the quantum state of the system from the same single measurement. That is the consequence of the HUP.

Also, I don't know what it means to say momentum is imaginary in the CFR. Momentum is a hermitian operator, or a real number.

I don't know either ... but isn't that a problem for any measurement in the CFR? But I guess you agree that the K.E. would have to be negative in the CFR, and the K.E. is given by p^2/2m, so what other conclusion could one draw than that the momentum is imaginary? Did you see my earlier post on this question?

 

[SpectraCat]

Tunneling arises from the modification of the total energy of the particle due to the rapid spacetime fluctuations of the \Psi-wave in the vicinity of the barrier (the force -\nabla Q exerted by the wave on the particle turns out to be proportional to the curvature of the wave). The effective 'barrier' encountered by the particle is not V but V+Q - where Q is the potential energy function of the wave field. This may be higher or lower than V and may vary outside the 'true' barrier. For tunneling you require only that the energy of the particle \geq V + Q then the particle may enter or cross the barrier region.

In effect, the additional 'force' exerted by the wave field on the particle just shoves the particle over the barrier.

After some more thought, I am not sure I completely understand this explanation. It sounds like you are saying that there are trajectories where the particle physically travels over the barrier, extracting the necessary "boost" from the quantum potential. If this were true, then one should be able to detect the particle in the classically forbidden region, since it actually passes through that space, right? How is this dealt with in the framework of dBB?

As far as I know it is impossible to explain this effect consistently using an interpretation of QM involving the wave function alone (all you can do is predict the probabilities of the final outcomes). Cue much wailing and gnashing of teeth and people telling you that I'm not allowed to explain this to you because the interpretation is non-standard.

Well, no wailing or gnashing, but a little head-scratching ... I guess it is fair to say that the TCI avoids this by saying "you aren't allowed to ask that", but can you answer my question above about how the dBB picture can be consistent with never observing the particle in the CFR?

 

[zenith8]

It sounds like you are saying that there are trajectories where the particle physically travels over the barrier, extracting the necessary "boost" from the quantum potential. If this were true, then one should be able to detect the particle in the classically forbidden region, since it actually passes through that space, right? How is this dealt with in the framework of dBB?

Yes - that's correct. See the picture in my last post.

Would you mind explaining to me why you think this is a problem?

 

[SpectraCat]

Yes - that's correct. See the picture in my last post.

Would you mind explaining to me why you think this is a problem?

I am not sure if it is a problem, I may be thinking too linearly. However it sounds like this picture could be consistent with both knowing the quantum state of the particle and measuring its position as being inside the CFR, since it seems to involve the particle having classical trajectories where it travels over the barrier (as opposed to through it).

Also, I am a bit unclear on how these putative trajectories conserve energy. After all, conservation of energy is how we "know" that the particle cannot penetrate or cross the barrier in the classical case.

 

[zenith8]

I am not sure if it is a problem, I may be thinking too linearly. However it sounds like this picture could be consistent with both knowing the quantum state of the particle and measuring its position as being inside the CFR, since it seems to involve the particle having classical trajectories where it travels over the barrier (as opposed to through it).

Apart from the fact that I'm not sure what the difference is between traveling over the barrier or through it, then sure.. so what's your point?

Also, I am a bit unclear on how these putative trajectories conserve energy. After all, conservation of energy is how we "know" that the particle cannot penetrate or cross the barrier in the classical case.

They do conserve energy, but only if you include the quantum potential energy as well.

One more thing just for reference, you can't use plane-wave incident, reflected and transmitted waves to analyze the dynamics of the situation, as is normally done in school, since this in no way corresponds to a situation where a particle is incident on a barrier and may or may not tunnel through it. An infinite plane incident wave means the particle can start anywhere in the universe, even on the other side of the barrier. You need to use a proper traveling time-dependent wave packet, if you want to avoid nonsense.

 

[Frame Dragger]

I'm with Zenith here at least in principle. There is no barrier to go over or through, but rather an apparent requirement for an energy "boost" that occurs despite conditions being to the contrary.

On the other hand, she's so blase abgout the the particle having a classical trajectory because that's what particles HAVE in dBB. From other approaches there would be some violations, but not from the point of view of dBB for which this isn't a special phenomenon at all.

 

[zenith8]

I'm with Zenith here at least in principle. There is no barrier to go over or through, but rather an apparent requirement for an energy "boost" that occurs despite conditions being to the contrary.

On the other hand, she's so blase about the the particle having a classical trajectory because that's what particles HAVE in dBB.

You shouldn't use the word 'classical trajectory' in this context; to many people that implies particles following Newton's laws which is not what's happening here (see figure in my post #15). How about 'quantum trajectory'?

From other approaches there would be some violations.

Example?

 

[Frame Dragger]

You shouldn't use the word 'classical trajectory' in this context; to many people that implies particles following Newton's laws which is not what's happening here (see figure in my post #15). How about 'quantum trajectory'?


Example?

Oh you'd LOVE to see the dBB referred to as the default QM trajectory... veeeerrrry cute. lol. Let's call it the "Bhomian" Trajectory... :)

As for violations, any purely probabilistic system would reject the trajectories entirely. You would be able to measure its state and position at the same time. I think. I'm not sure on this one.

 

[zenith8]

Oh you'd LOVE to see the dBB referred to as the default QM trajectory... veeeerrrry cute. lol. Let's call it the "Bhomian" Trajectory... :)

Well, they're essentially just the streamlines of the Schroedinger probability current, so why don't we call them 'Schroedinger trajectories' - it has less negative connotations since people are taught to despise Bhom, or even Bohm, on principle.

As for violations, any purely probabilistic system would reject the trajectories entirely. You would be able to measure its state and position at the same time. I think. I'm not sure on this one.

Take your time..

 

[Frame Dragger]

Well, they're essentially just the streamlines of the Schroedinger probability current, so why don't we call them 'Schroedinger trajectories' - it has less negative connotations since people are taught to despise Bhom, or even Bohm, on principle.


Take your time..

Ok, I actually find that a perfect terminology. From what I can see based on double-checking my assumptions... I don't see any violations of the QM system like the one SpectraCat describes. :shy: That said, this is an area where I must do more research. I'm familiar with tunneling more in the context of MOSFETs and the like, and less in the context of theory.

Schrödinger Trajectories a good term, I agree... it would be really fantastic if that turns out to be correct in nature too. File it under, "I want to believe."

 

[Truecrimson]

ok ... but that is just your preference. The two explanations are formally identical.

The first time I read the explanation in the book you linked to, I understood it in the same way as when you said

Take the example of a particle tunneling through a 1-D potential. It is possible to observe the particle in the classically allowed region on either side of the barrier, but it is impossible to observe the particle in the CFR represented by the inside of the potential barrier.

But in my preferred explanation, we indeed can observe a particle inside the potential barrier. It may be just a forbidden region "logically" and we are free to say, if we actually found the particle there, that now it's an allowed region because we are unsure about the kinetic energy of the particle, but the potential barrier is always there. Probably it's just your wording that makes me confused. If so, then I'm sorry.

Edit: I think we are in agreement about this now. I'm posting this nitpicking just in case if the OP finds this specific post by you unclear as I did.

Also the sum of the reflection and the transmission coefficients is one because they're properly normalized ratios of probability, not the probability given by the wave function.

About the problem with non-hermitian operator, it may not add much to the discussion. Please feel free to ignore it.