Negative energy/imaginary momentum

[wofsy]

I could use an explanation of negative energy and imaginary momentum.

The situation I read about was the ionized hydrogen molecule, two protons and one electron.

The electron has negative energy.

If the two protons are far apart then the potential is small in most of the space between them so the electron approximates a free particle. With negative energy this implies that the momentum is imaginary.

 

[Redbelly98]

Potential energy can be negative, and does not imply an imaginary momentum.

Imaginary momentum would be implied by negative kinetic energy, which is not implied by negative energy.

Hope that helps. If you're not familiar with potential energy or why it can be negative, post again.

 

[wofsy]

Hi Redbelly

thanks for the reply.

But ... if the potential is nearly zero i.e. the electron is a nearly free particle then its energy is kinetic (ignoring intrinsic energy). This means that the kinetic energy is negative. That is what I was saying.

But I don't get what negative kinetic energy really means and for that matter what imaginary momentum really means.

I guess what is going on in the case of the ionized hydrogen molecule is that the binding energy of the molecule is transformed into negative kinetic energy in regions where the potential is small - presumably it is small because the two protons cancel each other's potentials at large distances.

wofsy

 

[Vanadium 50]

But ... if the potential is nearly zero i.e. the electron is a nearly free particle then its energy is kinetic (ignoring intrinsic energy). This means that the kinetic energy is negative. That is what I was saying.

No, if the particle is bound, it means the kinetic energy is less than the (negative of) the potential energy. If you make the potential nearly zero and have the particle still bound, that means the kinetic energy is smaller still, but not negative.

 

[wofsy]

No, if the particle is bound, it means the kinetic energy is less than the (negative of) the potential energy. If you make the potential nearly zero and have the particle still bound, that means the kinetic energy is smaller still, but not negative.

Ok let me tell you why I asked. In the case of the ionized hydrogen molelcule, if the protons are far apart then the electrostatic potential of the electron is nearly zero over large parts of the region between the protons. In this area the electron is like a free particle and so has a definite energy. But its energy is negative. Since it has definite energy and the potential is zero the de Broglie relations demands that the square of the momentum is also negative and so the momentum is imaginary.

It follows that the amplitude for the electron to flip from one proton to the other is a real number and may be thought of as a potential whose gradient is effectively an attractive force between the two protons.

 

[Vanadium 50]

You are still confusing kinetic energy, which is always positive, with total energy, which can be negative.

 

[Redbelly98]

wofsy, where are you reading this example, can you give a reference?

Right now I can think of 2 possible reasons for the confusion:

• You read "the potential is nearly zero", and then assume you can safely say it is zero. But if the kinetic energy is even less than the (absolute value of) potential energy, it is still positive. We might as well assume potential, kinetic, and total energy are all zero, if we are going to assume potential is zero.
• The energy is appreciably negative, so although classically the electron cannot be very far from either proton, its wavefunction is nonzero in the region you're talking about. There is a small, but nonzero, probability of finding the electron there, where
  
  (total energy) < (potential energy)​
  But I'm pretty sure the correct interpretation is not one of negative kinetic energy ... though I'm not sure off the top of my head what the correct interpretation is.

 

[wofsy]

wofsy, where are you reading this example, can you give a reference?

This example is in Feynmann's Lecture's on Physics, the third book, chapter 10, section 10-5.

The only thing I can think of is a thought experiment like this. Imagine moving an electron towards one of the protons until it binds and suppose that just for an instant it is stationary so its energy is just the negative energy of the electrostatic potential. Now the wave function begins to evolve and there is an amplitude that it will be found in the near zero potential region between the two protons. But here its energy which we know already is negative must be almost all kinetic because the electrostatic potential is nearly zero.

 

[Alewhey]

This example is in Feynmann's Lecture's on Physics, the third book, chapter 10, section 10-5.

The only thing I can think of is a thought experiment like this. Imagine moving an electron towards one of the protons until it binds and suppose that just for an instant it is stationary so its energy is just the negative energy of the electrostatic potential. Now the wave function begins to evolve and there is an amplitude that it will be found in the near zero potential region between the two protons. But here its energy which we know already is negative must be almost all kinetic because the electrostatic potential is nearly zero.

So you are saying that in some bound orbital, there is a point at which the potential is almost zero (e.g. a point equidistant between two H+ ions). But that since the energy of the orbital is a negative constant, this appears to imply that the KE is negative, since total E is simply the sum of KE and PE.

I have dug up your reference and you are indeed correct, Feynman does state that in such a situation the electron has imaginary momentum (though I found it in section 10-1, page 10-5). I don't know what to make of this and have not come across it before. I'm tempted to say it is merely a curiosity caused by the general 'weirdness' of QM. In some situations one can consider the KE to be negative. What this 'really' means it to some extent irrelevant, when intuition fails we trust in the mathematics.

 

[wofsy]

So you are saying that in some bound orbital, there is a point at which the potential is almost zero (e.g. a point equidistant between two H+ ions). But that since the energy of the orbital is a negative constant, this appears to imply that the KE is negative, since total E is simply the sum of KE and PE.

I have dug up your reference and you are indeed correct, Feynman does state that in such a situation the electron has imaginary momentum (though I found it in section 10-1, page 10-5). I don't know what to make of this and have not come across it before. I'm tempted to say it is merely a curiosity caused by the general 'weirdness' of QM. In some situations one can consider the KE to be negative. What this 'really' means it to some extent irrelevant, when intuition fails we trust in the mathematics.

Reading more of Chapter 10 makes this idea of imaginary momentum interesting.
It occurs in many particle systems such as neutron proton pairs that are bound through the exchange of a pi meson or electron pairs that exchange a photon.

the amplitude of a particle to be found a distance R in a given direction away from its current position is exp(-i/h_p.R)/R. If R is the distance between the two protons then this is just the amplitude that the electron will flip from one proton to the other. It and its conjugate are the off diagonal terms of the Hamiltonian if one looks at the system as having only two states. If p is imaginary then this term is a real number.

What I don't get is why this means that the off diagonal term is a potential whose gradient is a force attracting the two protons - or the proton and the neutron - or whatever particle pair. Feynman also says that for the exchange of a photon between two electrons one gets a similar formula except that because the rest mass of the photon is zero, the numerator of the exponential is constantly equal to 1 so one just gets Coulomb's Law.

So negative momentum seems to be the cause of the potentials between particles even the Classical ones such as Coulomb forces. This seems profound.

 

[Vanadium 50]

I think you're confusing several things here.

Let's get the basic problem straight before jumping into nuclear physics.

Kinetic energy is always positive, and it is equal to $p^2/2m$. Total energy of a bound state is negative, and it is equal to kinetic energy plus potential energy (which is negative).

 

[wofsy]

I think you're confusing several things here.

Let's get the basic problem straight before jumping into nuclear physics.

Kinetic energy is always positive, and it is equal to $p^2/2m$. Total energy of a bound state is negative, and it is equal to kinetic energy plus potential energy (which is negative).

Well Ok - but if you look through the letters in the thread you will see that the kinetic energy is not always positive and is in fact negative in many situations. You can not forget nuclear physics since this the a quantum mechanical effect of particle interactions.

The arguments we have put forth are rigorous but if if you would like to refer to an authority look at Feynmann's Lectures on Physics Book 3 chapter 10 page 10-5 which discusses the hydrogen molecular ion and there are several other examples in that chapter.

I agree with you that this seems like a weird even senseless idea though mathematically possible and have struggled to accept it.

But fantastically it seems to explain potentials like Coulomb potential and Yukawa potential as quantum mechanical effects. I would love to spend time with you and anyone else really understanding this.

Feymann says that the electron in the ionized hydrogen molecule has negative energy. Classically it would be bound to one of the protons with no chance of flipping over to the other. In QM though there is an amplitude that this flip will happen. It is like a tunneling effect.

But there are places between the two protons where the potential is nearly zero because of the cancellation of the potentials of the two protons. So if the electron quantum mechanically ends up in this region its energy will be kinetic not potential and will be negative.

 

[jambaugh]

But there are places between the two protons where the potential is nearly zero because of the cancellation of the potentials of the two protons. So if the electron quantum mechanically ends up in this region its energy will be kinetic not potential and will be negative.

You are confusing force with potential. The forces cancel the potentials add! The potential is the energy required to pull the electron out to infinity. In the region you're thinking this potential is negative but nearly flat so the force which is the gradient of the potential will be small. However the value of the potential is the sum of the potentials you'd get from each proton alone. You add two negatives to get a bigger negative.

Think "escape velocity" as the case where kinetic energy and potential energy exactly equal (in magnitude) and so sum to 0 total energy.

 

[Alewhey]

You are confusing force with potential. The forces cancel the potentials add! The potential is the energy required to pull the electron out to infinity. In the region you're thinking this potential is negative but nearly flat so the force which is the gradient of the potential will be small. However the value of the potential is the sum of the potentials you'd get from each proton alone. You add two negatives to get a bigger negative.

Think "escape velocity" as the case where kinetic energy and potential energy exactly equal (in magnitude) and so sum to 0 total energy.

Well OK, so forget the hydrogen molecule and just consider a single hydrogen ion. The s-orbital, for example, has an infinite extent (form exp[-r/a]). Yes, the probability of finding far from the atom is almost zero, but that is irrelevant; the point is that the PE in this distant region is (also) essentially zero. However, the region is still part of the orbital and hence must still have the energy of that orbital - hence the necessity of an electron 'found' in that region to have negative KE. It does not mean that we could somehow observe the negative KE (whatever that means), it is simply part of the formalism.

The more I think about it the more I'm convinced it holds together... where is the mistake?

 

[wofsy]

Well OK, so forget the hydrogen molecule and just consider a single hydrogen ion. The s-orbital, for example, has an infinite extent (form exp[-r/a]). Yes, the probability of finding far from the atom is almost zero, but that is irrelevant; the point is that the PE in this distant region is (also) essentially zero. However, the region is still part of the orbital and hence must still have the energy of that orbital - hence the necessity of an electron 'found' in that region to have negative KE. It does not mean that we could somehow observe the negative KE (whatever that means), it is simply part of the formalism.

The more I think about it the more I'm convinced it holds together... where is the mistake?

Yes my mistake - its not that the potentials cancel its just that they are small because the region is far away from the protons.

 

[muppet]

Well OK, so forget the hydrogen molecule and just consider a single hydrogen ion. The s-orbital, for example, has an infinite extent (form exp[-r/a]). Yes, the probability of finding far from the atom is almost zero, but that is irrelevant; the point is that the PE in this distant region is (also) essentially zero. However, the region is still part of the orbital and hence must still have the energy of that orbital - hence the necessity of an electron 'found' in that region to have negative KE. It does not mean that we could somehow observe the negative KE (whatever that means), it is simply part of the formalism.

The more I think about it the more I'm convinced it holds together... where is the mistake?

I think the mistake is that you're thinking semi-classically. The wavefunction has infinite extent, but that doesn't mean that the electron is actually there in the classically excluded region (that region which would have negative KE). What I would find interesting is if measuring the electron to be that far out actually removed it from the atom without supplying the "necessary" energy... my best guess at the moment is that it would (in a manner exactly analagous to alpha decay), but I should perhaps think about this tomorrow and go to bed between now and then

 

[Vanadium 50]

What Feynman is trying to say is far from clear in his writing.

What he is saying is that if one has a sufficiently flat and shallow potential, one can approximate the correct solution by plane waves. When you do this, you get unphysical values for energy and momentum, but if you blindly plug them in, you can reproduce certain features of the system.

He is not saying that kinetic energy is negative. He is saying that in certain cases, one can lump kinetic and potential energy together, and pretend that kinetic energy is negative - i.e. treat total energy as if it were purely kinetic. And sometimes the answer that you get by doing the wrong thing turns out to be close to right.

 

[jambaugh]

Well OK, so forget the hydrogen molecule and just consider a single hydrogen ion. The s-orbital, for example, has an infinite extent (form exp[-r/a]). Yes, the probability of finding far from the atom is almost zero, but that is irrelevant; the point is that the PE in this distant region is (also) essentially zero. However, the region is still part of the orbital and hence must still have the energy of that orbital - hence the necessity of an electron 'found' in that region to have negative KE. It does not mean that we could somehow observe the negative KE (whatever that means), it is simply part of the formalism.

The more I think about it the more I'm convinced it holds together... where is the mistake?

We are diverging from the OP which I believe was expressing a misunderstanding about the relationships between Classical PE, KE and total energy.

You bring up an interesting example.

Consider what it means to observe the electron in this far region. You must at the minimum throw in a test particle to interact with it. In so doing I think you will see that in an actual case you must create a potential via the observation process sufficient for the electron when observed to have a positive KE. In essence you must create the possibility of a tunneling event and your position reduces to the question of interpretation of the energy when a particle tunnels.

There are various ways to look at tunneling, borrowed energy from vacuum fluctuations being one though not my preference. See some of the threads on tunneling.

But all I can say at this point is that if you are invoking negative KE you are redefining the meaning of KE i.e. that KE is total energy minus PE. Fine but we then need to parse what PE and total energy mean exactly before discussing KE.

 

[wofsy]

We are diverging from the OP which I believe was expressing a misunderstanding about the relationships between Classical PE, KE and total energy.

You bring up an interesting example.

Consider what it means to observe the electron in this far region. You must at the minimum throw in a test particle to interact with it. In so doing I think you will see that in an actual case you must create a potential via the observation process sufficient for the electron when observed to have a positive KE. In essence you must create the possibility of a tunneling event and your position reduces to the question of interpretation of the energy when a particle tunnels.

There are various ways to look at tunneling, borrowed energy from vacuum fluctuations being one though not my preference. See some of the threads on tunneling.

But all I can say at this point is that if you are invoking negative KE you are redefining the meaning of KE i.e. that KE is total energy minus PE. Fine but we then need to parse what PE and total energy mean exactly before discussing KE.

perhaps you could elaborate a little more. I am still confused. I guess Feynman is just saying the KE is negative for pedagogical reasons though usually when he does something like that he has a disclaimer in the footnotes.

I still come down to his original argument which is 1) the bound electron in the hydrogen molecule has negative energy. 2) In regions of space where the electrostatic potential is low it is essentially a free particle. 3) from this it follows that p^2/2m is negative and p is imaginary ( de Broglie relations for a free particle). 4) the reason the electron can be found in these regions is because the quantum mechanical amplitude for the electron to flip from one proton to the other is non-zero.

 

[wofsy]

What Feynman is trying to say is far from clear in his writing.

What he is saying is that if one has a sufficiently flat and shallow potential, one can approximate the correct solution by plane waves. When you do this, you get unphysical values for energy and momentum, but if you blindly plug them in, you can reproduce certain features of the system.

He is not saying that kinetic energy is negative. He is saying that in certain cases, one can lump kinetic and potential energy together, and pretend that kinetic energy is negative - i.e. treat total energy as if it were purely kinetic. And sometimes the answer that you get by doing the wrong thing turns out to be close to right.

but actually he does not say that we can pretend.

further there are other example where he refers to negative kinetic energy

 

[Redbelly98]

It's one thing to imagine, as Feynman was apparently doing, that a wavefunction being non-zero in certain regtions suggests a negative kinetic energy. But, quantum mechanics works by dealing with what is observable and measurable. So we have to know what it entails to measure the particle's position and kinetic energy.

I'm too many years removed from my QM classes to think out the details on this issue. However I am confident that it is not possible to observe a negative kinetic energy or an imaginary momentum.

 

[wofsy]

It's one thing to imagine, as Feynman was apparently doing, that a wavefunction being non-zero in certain regtions suggests a negative kinetic energy. But, quantum mechanics works by dealing with what is observable and measurable. So we have to know what it entails to measure the particle's position and kinetic energy.

I'm too many years removed from my QM classes to think out the details on this issue. However I am confident that it is not possible to observe a negative kinetic energy or an imaginary momentum.

I hear what you are saying.
I am very new to this stuff.

I would not dogmatically say that one doesn't need to make the measurement because integrating the Shroedinger equation tells you what the amplitudes to have transitioned to any region are. In the region of low potential it is the equation for free particle.

 

[muppet]

Let's think about the maths. Suppose it were possible to have an "imaginary momentum", or negative K.E. The common eigenfunctions of the KE and momentum operators are complex exponentials; so eigenfunctions corresponding to imaginary momentum would be real, decaying exponentials. So hydrogenic wave functions bear a superficial resemblance to such eigenfunctions, but they aren't actually eigenfunctions of those operators. So you can't claim that an atomic electron has negative K.E., because it doesn't actually have any definite value for K.E. What's interesting is that such states would exist in the classically excluded region of a finite square well potential.

What I'd like to do next is work out the probability of finding a particle in such a state given that prior to measurement the atom is in the ground state, but unfortunately I can't evaluate the integral. (Any takers? Me, Mathematica, wikipedia and my textbook all draw a blank; short of some tricks from complex analysis I can't think of anything else).

 

[jambaugh]

perhaps you could elaborate a little more. I am still confused. I guess Feynman is just saying the KE is negative for pedagogical reasons though usually when he does something like that he has a disclaimer in the footnotes.

I still come down to his original argument which is 1) the bound electron in the hydrogen molecule has negative energy. 2) In regions of space where the electrostatic potential is low it is essentially a free particle. 3) from this it follows that p^2/2m is negative and p is imaginary ( de Broglie relations for a free particle). 4) the reason the electron can be found in these regions is because the quantum mechanical amplitude for the electron to flip from one proton to the other is non-zero.

1.) Right but that's total energy PE+KE and both classically and quantum mechanically that's no problem PE+KE< 0 <---> bound particle. (remember PE < 0)

2.) "Essentially" being the loaded word. Classically and quantum mechanically that's fine provided the KE is still less than the PE in magnitude. Be careful here to distinguish the cases of
I. > An excited but still bound particle far from the potential center (positive KE slightly less than PE in magnitude so TE < 0 but very close to 0 and momentum real.)
vs
II. >the nearly zero component of a ground-state (or nearly so) particle far from the center.

It is the first case applies both classically and quantum mechanically. No negative KE here.

I think it is in this case where in QM we may compare the bound particle's wave-function locally with that of a plane-wave. But remember the plane-wave solution is assuming the potential has been removed.

It is no different in essence than comparing the classical bound particle at a point in its orbit to the classical particle which is not bound (given you remove the potential) having the same momentum and moving in a straight line (tangent to the first's orbit at this point).

In the second case one is trying to make sense of observing a particle at a given position which has a kinetic energy less in magnitude than the potential energy at that position. This may be puzzling at first but you must remember that the act of measurement which gives this probability meaning will necessarily perturb both the potential and the total energy. In the specific case of say a ground state H atom the act of measuring position doesn't commute with the projection onto the s-orbital. Thus the fact that the electron is in the s-orbital is invalidated once the measurement of position is made (said measurement being what defines this probability far from the center.)


3.) Not necessarily as I've qualified 1.) and 2.)

4.) That "flipping" is a at worst a tunneling event, you never observe the particle where you would define its KE < 0. Only near one of the two protons where its KE > 0. So the issue becomes what do you mean by KE of the particle in a quantum setting between acts of measurement.

Also again you are deriving KE = TE-PE and assuming TE remains unchange and PE remains defined by the potential in the absence of external influences. When discussing the probability of measuring the position you are invalidating the "no external influences" assumption so it is not proper to hold TE constant while talking about the probability of measuring the particle at a given position.

Say you try to measure the position of the electron by focusing a pulse light at a region and seeing if the electron scatters light from a point in that region. (i.e. you try to "photograph" the electron.) Then the light being an electromagnetic wave will alter the electromagnetic potential and the scattering event will alter the total energy of the electron.

Now if you want to use negative KE and imaginary PE as a mathematical tool, fine. But you can't measure a particle with these properties and so shouldn't give them meaning in a physical context.

One example is where you resolve a physical particle's wave-function into components which do not all represent physical particles but rather differences in wave-functions of physical particles. This is sometimes necessary and we get "ghosts" or "virtual particles".

But this is no different in essence than say in a classical setting resolving a relativistic time-like particle motion (as seen by some observer) into a non-physical instantaneous spatial jump $x\to x+\Delta x$ and a stationary physical wait $t\to t+\Delta t$. These are but components of the physical displacement vector $(x,t)\to(x,t)+(\Delta x,\Delta t)$.

We can write
$(x+\Delta x,t + \Delta t) = (x,t) + (\Delta x,0) + (0,\Delta t)$
but we don't take the $(\Delta x, 0)$ seriously as a physical motion. Its meaning is at best as the difference in two physical motions: $(\Delta x,\Delta t)-(0,\Delta t)$.

So too we may sometimes write $\psi = \psi_1 + \psi_2$ where say $\psi_1$ by itself doesn't represent a physical mode.

Finally if you are trying to find say an imaginary eigen-value to the momentum operator you are choosing a representation where the momentum operator is no longer Hermitian and it becomes problematic to give it meaning as an observable. The eigen-function may be useful mathematically but saying it corresponds to a particle with imaginary momentum is stepping outside the domain of momentum as a physical observable.

 

[wofsy]

1.) Right but that's total energy PE+KE and both classically and quantum mechanically that's no problem PE+KE< 0 <---> bound particle. (remember PE < 0)

2.) "Essentially" being the loaded word. Classically and quantum mechanically that's fine provided the KE is still less than the PE in magnitude. Be careful here to distinguish the cases of
I. > An excited but still bound particle far from the potential center (positive KE slightly less than PE in magnitude so TE < 0 but very close to 0 and momentum real.)
vs
II. >the nearly zero component of a ground-state (or nearly so) particle far from the center.

It is the first case applies both classically and quantum mechanically. No negative KE here.

I think it is in this case where in QM we may compare the bound particle's wave-function locally with that of a plane-wave. But remember the plane-wave solution is assuming the potential has been removed.

It is no different in essence than comparing the classical bound particle at a point in its orbit to the classical particle which is not bound (given you remove the potential) having the same momentum and moving in a straight line (tangent to the first's orbit at this point).

In the second case one is trying to make sense of observing a particle at a given position which has a kinetic energy less in magnitude than the potential energy at that position. This may be puzzling at first but you must remember that the act of measurement which gives this probability meaning will necessarily perturb both the potential and the total energy. In the specific case of say a ground state H atom the act of measuring position doesn't commute with the projection onto the s-orbital. Thus the fact that the electron is in the s-orbital is invalidated once the measurement of position is made (said measurement being what defines this probability far from the center.)


3.) Not necessarily as I've qualified 1.) and 2.)

4.) That "flipping" is a at worst a tunneling event, you never observe the particle where you would define its KE < 0. Only near one of the two protons where its KE > 0. So the issue becomes what do you mean by KE of the particle in a quantum setting between acts of measurement.

Also again you are deriving KE = TE-PE and assuming TE remains unchange and PE remains defined by the potential in the absence of external influences. When discussing the probability of measuring the position you are invalidating the "no external influences" assumption so it is not proper to hold TE constant while talking about the probability of measuring the particle at a given position.

Say you try to measure the position of the electron by focusing a pulse light at a region and seeing if the electron scatters light from a point in that region. (i.e. you try to "photograph" the electron.) Then the light being an electromagnetic wave will alter the electromagnetic potential and the scattering event will alter the total energy of the electron.

Now if you want to use negative KE and imaginary PE as a mathematical tool, fine. But you can't measure a particle with these properties and so shouldn't give them meaning in a physical context.

One example is where you resolve a physical particle's wave-function into components which do not all represent physical particles but rather differences in wave-functions of physical particles. This is sometimes necessary and we get "ghosts" or "virtual particles".

But this is no different in essence than say in a classical setting resolving a relativistic time-like particle motion (as seen by some observer) into a non-physical instantaneous spatial jump $x\to x+\Delta x$ and a stationary physical wait $t\to t+\Delta t$. These are but components of the physical displacement vector $(x,t)\to(x,t)+(\Delta x,\Delta t)$.

We can write
$(x+\Delta x,t + \Delta t) = (x,t) + (\Delta x,0) + (0,\Delta t)$
but we don't take the $(\Delta x, 0)$ seriously as a physical motion. Its meaning is at best as the difference in two physical motions: $(\Delta x,\Delta t)-(0,\Delta t)$.

So too we may sometimes write $\psi = \psi_1 + \psi_2$ where say $\psi_1$ by itself doesn't represent a physical mode.

Finally if you are trying to find say an imaginary eigen-value to the momentum operator you are choosing a representation where the momentum operator is no longer Hermitian and it becomes problematic to give it meaning as an observable. The eigen-function may be useful mathematically but saying it corresponds to a particle with imaginary momentum is stepping outside the domain of momentum as a physical observable.

You are right - the momentum operator is positive definite Hermitian and so has all positive eigen values. This means that any observation of momentum must be positive.

Perhaps you can explain a lingering question about Feynmann's exposition.
He uses the imaginary momentum to show that the amplitude of the electron to switch protons is a potential whose gradient defines an attraction between the protons.

He then uses the same type of argument to derive the Coulomb potential for two electrons and the Yukawa potential for a neutron and a proton.

For the two electrons the potential is the amplitude for the electrons to exchange a photon. Since the electron has zero mass the numerator of the amplitude, the de Broglie term, is zero and one just gets 1/r. For the Yukawa potential the numerator is complex and is the Broglie term for a pi+ meson.

regards

wofsy

P.S. I did find a paper in Physics Review that discusses measurements of negative kinetic energy in the "forbidden domain" i.e. the region where the classical particle can not tread. The authors try to reconcile this with the positive definiteness of the momentum operator and I think try to explain their "error" in measurement in terms of underlying assumptions about how measurements are made.

If you would like to read it together I will send you the link. There is stuff in there that I don't know anything about and would not be able to grok without a serious project.

 

[wofsy]

OK

I guess since momentum can not be imaginary I need to know how to interpret some basic examples and could use your help.

Let's take the case of a particle in a jump potential. The potential is zero to the left of some point on the x-axis then jumps to some constant postive value after that to the right.

Assume that the energy of the particle is less than the potential.

The stationay solution of the Shroedinger equation is a simple wave function of the form exp(ik + wt) in the region where the potential is not zero. The wave number,k, is purley imaginary.

This a case where if one verbatim applies the de Broglie relations one gets an imaginary momentum.

So what is the momentum of this particle and how do we understand it?