# Negative energy/imaginary momentum

1. Jan 19, 2009

### 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.

2. Jan 19, 2009

### Redbelly98

Staff Emeritus
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.

3. Jan 19, 2009

### 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

4. Jan 19, 2009

### Vanadium 50

Staff Emeritus
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.

5. Jan 19, 2009

### wofsy

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.

6. Jan 19, 2009

### Vanadium 50

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

7. Jan 19, 2009

### Redbelly98

Staff Emeritus
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.

8. Jan 19, 2009

### wofsy

9. Jan 19, 2009

### Alewhey

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, Feynmann 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.

10. Jan 19, 2009

### wofsy

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. Feynmann 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.

11. Jan 20, 2009

### Vanadium 50

Staff Emeritus
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).

12. Jan 20, 2009

### wofsy

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.

13. Jan 20, 2009

### jambaugh

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.

14. Jan 20, 2009

### Alewhey

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?

15. Jan 20, 2009

### wofsy

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

16. Jan 20, 2009

### muppet

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

17. Jan 21, 2009

### Vanadium 50

Staff Emeritus
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.

18. Jan 22, 2009

### jambaugh

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.

19. Jan 22, 2009

### wofsy

perhaps you could elaborate a little more. I am still confused. I guess Feynmann 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.

Last edited: Jan 22, 2009
20. Jan 22, 2009

### wofsy

but actually he does not say that we can pretend.

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

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?