Negative Energy in Quantum Theory: A Puzzling Problem

[Thomas Ruedel]

In the transition from "basic" quantum theory to quantum field theory, both in the derivation of the Gordon-Klein and of the Dirac-equations, there is a big problem with solutions with negative energies. Can anyone explain me why this is being considered a big problem in the first place (it seems obvious to textbook authors). After all, energy is anyhow only determined up to a constant. And electrons within atoms are generally assigned negative energies.
This is probably one of those stupid questions with obvious answers which people just forget to mention. The more grateful I would be for an answer.

 

[marlon]

In the transition from "basic" quantum theory to quantum field theory, both in the derivation of the Gordon-Klein and of the Dirac-equations, there is a big problem with solutions with negative energies.

There is no problem with that. But just to be clear, negative energy solutions really means the negative square root coming from the Einstein energy relation : E² = p²c² + m'²c^4 where m' is the restmass.

These negative solutions correspond to anti-particles. For example positrons.

In the case of electrons, you are mixing this with potential energy. Indeed the latter is negative (Coulombic potential). However, electrons obey the positive square root relation for energy.

Keep in mind that the Einstein energy formula does not account for the interactions that particles can undergo. It just expresses the energy of a particle with certain rest energy and certain momentum p. This, however does NOT change the negative Coulombic potential between charged particles

marlon

 

[dextercioby]

Einstein's formula is valid for particles which are on their mass sheet.For example,free particles are always on theor mass sheet.Virtual particles,not.

Daniel.

 

[marlon]

As a matter of fact, there are two (connected) ways to determin whether a particle is virtual or not.

1) if a particle does NOT obey the Einstein-energy relation
2) if in a Feynman diagramma, the lines of a particle start and end at the same vertex, the particle is virtual. Though, it can be promoted to a real particle for a short time, thanks to HUP

marlon

 

[Thomas Ruedel]

In the case of electrons, you are mixing this with potential energy. Indeed the latter is negative (Coulombic potential). However, electrons obey the positive square root relation for energy.

marlon[/QUOTE]

Thanks a lot. My problem is really quite basic, I´m afraid. Are you saying something like the following?
" Traditionally one would have thought that potential energy can be negative but kinetic energy cannot. And indeed kinetic energy cannot be negative for the electron. However, the fact that there are solutions for free particles displaying negative (kinetic) energies has led to the discovery of anti-particles, such as the positron. And they can have negative kinetic energies."
Is this a correct understanding?

 

[marlon]

The E in the Einstein relation is NOT kinetic energy. So no, you are wrong. Besides in QM electrons can have negative kinetic energy. This happens with QM-tunnelling effects

marlon

 

[Sterj]

"Einstein's formula is valid for particles which are on their mass sheet.For example,free particles are always on theor mass sheet.Virtual particles,not."

Why aren't virtual particles on their mass sheets? Can you please explain me that?

 

[dextercioby]

Because you can't cave both energy-momentum conservation and ONLY real particles in interaction.Take for example the vertex in QED and try to impose at the same time that the 2 fermions and the photon to be on their mass sheets and 4momentum to be conserved in vertex.

Tell me what u get.

Daniel.

 

[Sterj]

@dexter: I Don't really know what you want (what shall I try to calculate?). Can you give me a beginning of what I should calculate?

 

[dextercioby]

Yes,I've told you what to do.I think it was clear enough.Consider the vertex in QED and try to see what happens if you combine 4momentum conservation+the condition that all 3 particles are real/on their mass sheets...

Daniel.

 

[Sterj]

Ok, I'll try it, I'll give you a solution tomorrow.

 

[dextercioby]

As you wish.I'm not asking you for anything.It was just an example to show u what happens when we have virtual particles and the connection to "mass sheet",the latter being a key concept in SR.

Daniel.

 

[SymmetryBreaking]

I don't know if its related to the topic but does this negative energy really cancels out the positive one? Like for example, we are experimentally getting less of the expected total vacuum energy because of this effect(pls correct me if I'm wrong).

 

[Sterj]

@dexter: yeah, It's clear now, why virtual particles aren't in their mass sheet. (Because they receive their energy from the Heisenberg's uncertainty principle. And this energy most disappear after a short time).

But I've another question. I asked a physicist, why for example virtual photons annihilate after a short time (I mean the antiparticle collides with the particle but for this the antiparticle has to know the position of the particle and to collide the particle and the antiparticle have to go in the same direction, but what is if they don't go in the same direction?). You know what I mean?

--> He's answers didn't sadisfy me. He said, that it is so, ... (I think he didn't know an answer, perhaps there isn't really an answer for that).

What do you mean?

 

[reilly]

I suggest that a little history is in order. Back in Dirac's time, the existence of negative energy states, down to - infinity, indicated a complete collapse of electrons, protons as wel, etc. unless the negative energy states were filled creating a rather extensive Fermi Sea. The rest is history, positrons were holes in the big Fermi Sea, and finally, in modern QM, particles are anti particles are afforded equal protection under the law. All of this stuff is, or should be, discussed in great detail in any text or treatise on relativistic QM, Weinberg, Schweber's QED and the Men Who Made It -- just a couple of suggestions.
Regards,
Reilly Atkinson

 

[Sterj]

I know that "reilly". But I can't find my answer in any textbook of qm or qed.

again:
"""""""But I've another question. I asked a physicist, why for example virtual photons annihilate after a short time (I mean the antiparticle collides with the particle but for this the antiparticle has to know the position of the particle and to collide the particle and the antiparticle have to go in the same direction, but what is if they don't go in the same direction?). You know what I mean?

--> He's answers didn't sadisfy me. He said, that it is so, ... (I think he didn't know an answer, perhaps there isn't really an answer for that).

What do you mean?""""""

 

[marlon]

I know that "reilly". But I can't find my answer in any textbook of qm or qed.

again:
"""""""But I've another question. I asked a physicist, why for example virtual photons annihilate after a short time (I mean the antiparticle collides with the particle but for this the antiparticle has to know the position of the particle and to collide the particle and the antiparticle have to go in the same direction, but what is if they don't go in the same direction?). You know what I mean?

--> He's answers didn't sadisfy me. He said, that it is so, ... (I think he didn't know an answer, perhaps there isn't really an answer for that).

What do you mean?""""""

Again, if you look at a particle that interacts with an anitiparticle, don't imagine just two particles that are annihilated. In QM you need to look at this as two BEAMS of particles that interact (read : scatter) with each other. The probability of such an interaction is expressed by quentities like the cross section of the beam and the scattering amplitude.

Virtual particles exist for a short while because of the uncertainty principle. Suppose you look at the vacuum (ie the lowest energy state). It has a NON-zero energy value because otherwise you cannot detect any particles in this state (you know, because of :$$\Delta x * \Delta p = constant$$)Now Since the energy value is NONZERO we have mass (because of E=mc²) and therefore the vacuum is filled with particles. These particles (of energy E)exist for a short period of time because of $$\Delta E * \Delta t = constant$$. These particles are the socalled vacuumfluctuations. These particles are indeed virtual because they only exist for an intermediate stage and they can be used in interactions between real particles that are on mass shell

regards
marlon

 

[Sterj]

@Marlon, that's what I know. But it exists only a probability that the particle and antiparticle interact (annihilate). So how can a particle find its "friend" (I know that we can say that particle and antiparticle are a kind of wave with probability to be at a position).

 

[marlon]

@marlon, that's what I know. But it exists only a probability that the particle and antiparticle interact (annihilate). So how can a particle find its "friend" (I know that we can say that particle and antiparticle are a kind of wave with probability to be at a position).

Clearly you did not get my point. A particle does NOT find or recognize another anti-particle. Again i refer to the scattering processes and the probability of interaction (annihilation)

marlon

ps : for all that is interested, i posted a text on virtual particles in my journal. Check it out, it is the 5th entry on the designated page (on vacuum fluctuations and virtual particles)
https://www.physicsforums.com/journal.php?s=&action=view&journalid=13790&perpage=10&page=2

 

[Sterj]

Your text makes it a little bit simpler to understand. With "number operator" you mean this N=ab with a the creation operator and b the annihilation operator, right?

"suppose each such particle is represented by one LOWEST energy quantum of a harmonic oscillator"
You mean each particle has energy of: E=h(bar)w*1/2=h/(4pi)*w=h*f/2. Shouldn't the energy of a photon be: E=hf? Or is the ground state of the electro magnetic field described by the harmonic oscillator?

 

[marlon]

Your text makes it a little bit simpler to understand. With "number operator" you mean this N=ab with a the creation operator and b the annihilation operator, right?

yes indeed

"suppose each such particle is represented by one LOWEST energy quantum of a harmonic oscillator"
You mean each particle has energy of: E=h(bar)w*1/2

Yes

Shouldn't the energy of a photon be: E=hf?

Yes, indeed. Keep in mind that in the case of an harmonic oscillator, the severeal energy modes are the socalled fonons...

Or is the ground state of the electro magnetic field described by the harmonic oscillator?

In QFT, a field needs to be seen as a mattress, built out of many interconnected springs. So a quantumfield is indeed a bunch of connected harmonic oscillators. If you put something on this mattress, it will start to vibrate because the many oscillators will vibrate. Now, each vibration corresponds to a certain energy-level and you can go from one vibration to another by adding energy. You can proove that this added energy corresponds to a certain particle of mass m and certain spin-properties that depend on both the field and the symmetry of the "thing" that initiated the vibration of the field (you know, like the masses we put on the mattress to make it vibrate). This is how photons are excitations of the EM-field. The vibrating EM-mattress really generates the photons.

regards
marlon

 

[Sterj]

Ahhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

Let me proofe if I'm understanding it right now:

Consider a free electro magnetic field. We can describe the energy of the field with phonons. Thus, E(oscillator)=h(bar)w(n+0.5) and then the whole energy of the field is: E(whole)=E(oscillator 1)+...+E(oscillator n)

Now, by adding a photon to the field with the creation operator the energy of the correspondanding oscillator will be for example:
E=h(bar)w*1.5+h*f=h(bar)w*1.5+h(bar)"w=2.5h(bar)*w and so the oscillator is on the next state.

And the velocity of the oscillators has to be the speed of light. How large is such an oscillater? I mean its spreading a long the x,y,z axis? (<--- d(x), d(y), d(z) (position uncertainty) of the photon)?

 

[marlon]

Ahhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

Let me proofe if I'm understanding it right now:

Consider a free electro magnetic field. We can describe the energy of the field with phonons. Thus, E(oscillator)=h(bar)w(n+0.5) and then the whole energy of the field is: E(whole)=E(oscillator 1)+...+E(oscillator n)

Now, by adding a photon to the field with the creation operator the energy of the correspondanding oscillator will be for example:
E=h(bar)w*1.5+h*f=h(bar)w*1.5+h(bar)"w=2.5h(bar)*w and so the oscillator is on the next state.

And the velocity of the oscillators has to be the speed of light. How large is such an oscillater? I mean its spreading a long the x,y,z axis? (<--- d(x), d(y), d(z) (position uncertainty) of the photon)?

Err, let me give you some advice. Forget everything you just wrote in the above post because it is all wrong. This is not an insult, so don't get me wrong. I am just saying that these topics require a good knowledge of introductory QFT. This is not explained in a sequence of two or three posts.

Do you know your QM and your special relativity ? If so, i suggest you by Antony Zee's QFT in a Nutshell. Go see my journal (the info on the web entry) for more links on QFT

regards
marlon

 

[marlon]

On a more intuitive level, you are correct. However the EM-quantumfield has NOTHING to do with fonons. These are energy quanta of waves that express the way atoms in a lattice vibrate around their equilibrium position (you know, their fixed site on the lattice)

marlon

 

[Sterj]

Ok, thanks marlon, I'll first read the pdf file.

 

[Sterj]

Ok, now,
"So a quantumfield is indeed a bunch of connected harmonic oscillators."
So, why isn't then the energy of the field:
E=E(oscillator1)+E(oscillator2)+...?

 

[Sterj]

Mhh, can you say me, what a "four potential" is?

 

[reilly]

strj -- If you know all that, why ask the question? Once again -- negative energy solutions to relativistic QM equations implied that matter had to be unstable. That's by any definition, a problem. I cited Weinberg and Schweber to provide two excellent and detailed discussions of the issue. You might, arguably, profit from reading these two authors.

Virtual Particles -- To a certain degree, much ado about nothing. It all started in the ancient beginnings of wave mechanics in which it was found that "particles" could formally have negative kinetic energy -- like a particle in a potential well with finite barriers, a very standard QM problem. That is, in QM a particle could be, probabilistically, where it could not be classically. But, the method of finding such a particle is somewhat indirect. With appropriate hand-waving, this finite barrier potential well model became the standard model for radioative nuclear decay, particularly apha decay - to transverse points "well barrier outside" in sequence, the particle must go through the barrier -- right? As a plausible metaphor, particles in such forbidden regions are called virtual particles.

In QFT, it can be argued that virtual particles are simply an artifact of perturbation theory -- perturbartion in it's most general form. That is, particularly for diagrammatic approaches which support internal lines -- photon exhanges, self energy, box diagrams and so forth. That is, virtual particles help steer us through the maze of QFT mathematics. Interpret them as you like.

Regards,
Reilly Atkinson

 

[Sterj]

@reilly: Yeah, I know that. Mhh, I didn't know that particles that are somewhere where they mustn't be (in classical mechanics) (caused of probability) are also virtual particles. So, particles that use the tunnel effect are virtual particles.

 

[marlon]

So, particles that use the tunnel effect are virtual particles.

Certainly not. Electrons can tunnel through a barrier. Just look at how a STM-microscope works. In QFT the instantons tunnel from one gauge-configuration to another. The alpha decay is an example of a tunneling process.

Tunneling lowers the ground state energy in both QM and QFT. Particles that tunnel through barriers can have "negative mass" though. However they have NOTHING to do with virtual particles, which already have been explained to you

marlon

 

[reilly]

Strj-- Read.

Regards,
Reilly Atkinson

 

[Sterj]

Ok, ok. I read a few pdf. files. There are files that say:
The total energy of an electromagnetic field is E=h(bar)*w/2+h(bar)*w+h(bar)*w+...
and there are some files where it is written:
The total energy of an electromagnetic fiel is E=h(bar)*w+h(bar)*w+...
The second is without the ground oscillation.

Now, what is true?

 

[Edgardo]

Ok, ok. I read a few pdf. files. There are files that say:
The total energy of an electromagnetic field is E=h(bar)*w/2+h(bar)*w+h(bar)*w+...
and there are some files where it is written:
The total energy of an electromagnetic fiel is E=h(bar)*w+h(bar)*w+...
The second is without the ground oscillation.

Now, what is true?

Hello Sterj,

I found this document from http://iftia9.univ.gda.pl/~sjk/skok/om03.pdf
Have a look at page 37, "We renormalize the energy by dropping the term 1/2"

Do you know the book "Quantum Field Theory" by Mandl and Shaw?
In chapter 1.2.3 the text says: "This constant ( $\frac{1}{2} \sum_{k} \sum_{r} \hbar \omega_{k}$) is of no physical significance. Just scale the energy by replacing equation (1.30) by $H_{rad} = \sum_{\vec{k}} \sum_{r} \hbar \omega_{k} a^{\dagger}_{r}(\vec{k}) a_{r}(\vec{k})$, where (1.30) is
$H_{rad} = \sum_{\vec{k}} \sum_{r} \hbar \omega_{k} \left( a^{\dagger}_{r}(\vec{k}) a_{r}(\vec{k}) +\frac{1}{2} \right)$

 

[Sterj]

ok, but in reality the term is there, but we don't use it.
thanks

 

[Cinderella]

Antiparticles have positive kinetic energie

In the case of electrons, you are mixing this with potential energy. Indeed the latter is negative (Coulombic potential). However, electrons obey the positive square root relation for energy.

marlon

Thanks a lot. My problem is really quite basic, I´m afraid. Are you saying something like the following?
" Traditionally one would have thought that potential energy can be negative but kinetic energy cannot. And indeed kinetic energy cannot be negative for the electron. However, the fact that there are solutions for free particles displaying negative (kinetic) energies has led to the discovery of anti-particles, such as the positron. And they can have negative kinetic energies."
Is this a correct understanding?[/QUOTE]


Antiparticles do not have negative kinetic energy, because if a particel with energy E and an antiparticel wirh energy (-)E collide the resulting energy, which may be set free as radaiation, is not zero, its 2E, this is an experimental fact. Out from the relativistic wave equations e.g. the Klein-Gordon equation it does not follow necessarily that that kinetic energy is negative. The energy occures there in terms of the product Et, so that an altermative interpretation of a negaive product -Et is that antiparticles move backwards in time with positive energy.