# Zero-point energy and QED vacuum state

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1. Jun 25, 2015

### JG1009

I'm relatively new to QFT and was wondering how the QED vacuum has a dormant zero average-field condition if there is a zero-point energy of the field as well? How is there a zero average and a zero point energy?

2. Jun 25, 2015

### bhobba

The existence of actual zero point energy is open to question. In QFT its usually eliminated by something called normal ordering.

But if you want to read a detailed discussion of it see:
http://users.ox.ac.uk/~lina0174/vacuum.pdf

I don't entirely agree with the above - for example I believe you can explain the Casmir effect without ZPE.

Thanks
Bill

3. Jun 25, 2015

### WannabeNewton

The zero-point energy is due to quantum fluctuations so it doesn't show up in the average energy.

4. Jun 26, 2015

### Zarqon

I'm not an expert on this, but have been reading a bit about it lately, and my conclusion is this: The reason the field (in amplitude) averages to zero is because fluctuations over time are not phase coherent, but random in phase. Of course, if you time average fields over random phases you get zero, but that does not mean that the field is zero for any given moment in time. On the other hand, if you instead look at the average of the energy/intensity you square the field first and then there is no longer any cancellation in the time average. In fact, if you look at the absolute value of the field, you will find that the average value for this is also non-zero.

5. Jun 26, 2015

### Xertese

All you need to know is that stuff can exist in a vacuum so the idea of zero point energy isn't so crazy.

Space is described as a void but it is anything but.

6. Jun 26, 2015

### jerromyjon

Only when there is stuff in it, if you take away the stuff you have no "space" to describe. Space and the zero point vacuum state are a geometry.

7. Jun 26, 2015

### andresB

8. Jun 26, 2015

### stevendaryl

Staff Emeritus
There is a sense in which zero-point energy is only relevant in General Relativity. For quantum mechanics and quantum field theory, energy is unimportant, only energy differences are important. So if you add or subtract a constant to the Hamiltonian, it will make no difference. So we free to choose the zero-point energy to be 0, if we like.

When you go to General Relativity, energy does make a difference, because a constant, nonzero energy density will cause spacetime curvature. But, there, I think that it's true that any constant energy density can be absorbed into the cosmological constant. So the issue really becomes: Is there some deeper theory unifying GR and QM that would allow us to compute the cosmological constant?

9. Jun 26, 2015

### stevendaryl

Staff Emeritus
That paper makes a side-point about the solutions to the Klein-Gordon equation which makes no sense to me. Could somebody explain it?

The author says that the KG equation:

$(\partial_\mu \partial^\mu + \mu^2) \phi= 0$

has 4 different classes of solutions:
1. $\phi = e^{i \omega t - i \vec{k} \cdot \vec{x}}$
2. $\phi = e^{i \omega t + i \vec{k} \cdot \vec{x}}$
3. $\phi = e^{-i \omega t - i \vec{k} \cdot \vec{x}}$
4. $\phi = e^{-i \omega t + i \vec{k} \cdot \vec{x}}$
I cannot understand what could possibly be the difference between 1&2, or between 3&4. Solution 1 implicitly allows $\vec{k}$ to point in any direction. So it includes, for any vector $\vec{k}_0$, both the solution $e^{i \omega t - i \vec{k}_0 \cdot \vec{x}}$ (by letting $\vec{k} = \vec{k}_0$) and the solution $e^{i \omega t + i \vec{k}_0 \cdot \vec{x}}$ (by letting $\vec{k} = - \vec{k}_0$). So how is 2 giving us any additional solutions?

In the appendix, the author suggests that the relationship between velocity $\vec{v}$ and $\vec{k}$ can be either $\vec{v} = \pm \vec{k}/\omega$, but I don't understand that explanation, either.

10. Jun 27, 2015

### moss

K-G eq is a wave eq. and its solutions must be plane waves. The above sols. are incoming and outgoing plane waves and that is the difference of 1 & 2. say.
Sol 3 & 4 are negative energy sols. they that must not be there but they are the concequence of the K-G eq.
but your text is correct as it is the short coming of K-G theory that it accepts the negative energy sol.

Note that the +/- signs in the exponent depends on the metric (-+++) or (+- - -). From 4-vector k goto 4-vector p then you will see the negative energy easily.

11. Aug 28, 2015

### Bob108

I am the author of the cited paper http://arXiv.org/abs/astro-ph/0309679.

Regarding the question of Science Author’s solutions 1 & 2 seeming to be the same thing, since 3-momentum k can have plus or minus sign, this is the most common initial reaction to the paper. But, note that in ey , y can have plus or minus sign. Yet, that does not mean ey and e-y are the same thing. They are not. Nor are eiy and e-iy , or ei(x+y) and ei(x-y) , or the above referenced solutions 1 & 2. Similar logic holds for solutions 3 & 4.

There is a difference between the sign before a quantity in an algebraic relation and the sign that quantity itself may have.

If one does the traditional QFT analysis (see Chap. 3 of my text Student Friendly QFT at www.quantumfieldtheory.info, pgs.53-56) one finds the infinite vacuum energy arises directly. With the usual ad hoc, sleight-of-hand, normal ordering (see same Chap. 3, Sect. 3.6.6, pgs 60-61), one can make it go away. But, using the more natural approach, outlined in the cited article, of including all the possible solutions to the field equations in one’s analysis, the huge vacuum energy term (which is not observed) is cancelled by an equal and opposite negative energy term.

The negative energy particles, if they exist, do not appear in our universe as real particles, but they might well play a role as virtual particles.

As an aside, I devote all of Chap. 10 in the book to the vacuum.

Bob Klauber

12. Sep 22, 2015

### andresB

I'm not sure about it, but assuming for the moment that the procedure of hte article is correct one have to wonder how that affect some of the calculation of QED. For example the derivation of the Euler-Heisenberg lagrangian (in for example, Landau and Lifshitz) is based int he shift of the energy of the infinite negative energy constant.

13. Sep 22, 2015

### Bob108

In my book referenced above, I derive QED in detail, and note at certain points along the way, that the results for interactions are the same in both approaches (provided the additional solutions I suggest are not coupled to the traditional solutions except via gravity). I do have a lot more work to do developing the theory, which I hope to accomplish in the coming year. It may, ultimately, not hold water, yet I submit the field equations of QFT do admit solutions that have not been incorporated into the traditional theory and should be investigated.

The Euler-Heisenberg effective Lagrangian has not been used much for decades (no modern QFT texts I am aware of discuss it), and frankly, I know little about it. I do believe, however, that it pertains to pair production in an electromagnetic field, rather than a vacuum, and that the infinite quantity is of the sort one finds before renormalization in usual QFT (i.e., arising from divergent integrals that need to be regularized in the process of renormalization.) See Itzykson and Zuber, Quantum Field Theory (McGraw-Hill 1980), pgs. 195-197.

14. Sep 25, 2015

### vanhees71

It's not clear to me from your paper, how you discuss away the supplemental modes. As treated in your article, I'd expect they occur in the physical spectrum, but having negative probabilities, which is a contardictio in adjecto, because then the theory becomes inconsistent, not providing a unitary S-matrix. Usually one simply discards these solutions, and they are not needed anywhere. This argument about "natural" cancellation of the infinite vacuum-energy term is also not too convincing to me since you have an operator-ordering problem anyway, and you have to define the proper order of operators in any case. Normal ordering is one prescription to save a little work when renormalizing the theory only in the sense that certain tadpole terms (i.e., a propagator loop beginning and ending at the same space-time point in the Feynman diagram). You can as well keep all the terms in some other operator ordering and renormalize the corresponding divergences as all other UV divergences.

So how do you overcome this problem with the "negative-probability states" when keeping the "supplemental modes" as physical modes?

15. Sep 25, 2015

### Bob108

Good points. Several interwoven issues are involved and my responses could be fairly long, but I will try to keep it as short as possible.

The concept clearly does not work if the supplemental particles are coupled to the traditional particles (within the standard model – exclusive of gravity). Two non-interacting (except for gravity) sectors are envisioned, so unitarity within the SM should be OK. Interaction probabilities for the SM traditional particles remain just as they are, positive and of identical magnitudes.

If by probability, you mean, as in NRQM, the square of the absolute value of the state wave function integrated over space (norms), the supplementals have alternating positive and negative norms, as the particle number of the state increases, i.e., an indefinite metric in Fock space. I address, and believe resolve, this in my second paper on the subject (available at www.quantumfieldtheory.info link near bottom of home page).

If by probability, you mean, as in RQM and QFT, the relation derived from manipulating the relativistic wave equation and its conjugate (or equivalently from Noether’s theorem), then the antiparticle expressions for probability of traditional SM antiparticles are negative anyway. This led to interpretation of the probability density as charge density. Exactly parallel reasoning holds for supplemental particles. Relativistic “probability” (charge, really) density is opposite sign for supplemental particles vs supplemental anti-particles.

Yes, I agree normal ordering has advantages when it comes to streamlining the theory development, for tadpoles and elsewhere. But I am not aware of anyone using renormalization to remove the infinite (or at least ginormous) half quanta vacuum contributions that one gets without the ad hoc introduction of normal ordering (that happens to violate the basic commutation principle upon which the entire theory is based). Lots and lots of people talk about why this disagrees with empirical observation (by a factor of 10^120 or so). A number of papers have been published on possible solutions.

I submit the supplemental solution approach is the simplest mathematically of all of them. The troublesome half quanta simply drop out straightforwardly by including all of the mathematically possible solutions to the wave equation in the analysis.

What does this all mean physically? I’m working on it. As Dirac said,

“ .. it [is] an easier matter to discover the mathematical formalism needed for a fundamental physical theory than its interpretation … with the interpretation most unexpected things may turn up.”

16. Sep 26, 2015

### vanhees71

The probabilities in conventional QFT are all positive (i.e., one works in a bona fide Hilbert space with a positive definite scalar product). I don't discuss socalled "relativistic QM", which is not consistent. That's why we use relativistic QFT rather than relativistic QM! Anyway, so you simply don't couple the supplemental particles to the interacting usual particles. This is fine for vacuum QFT, because then you can omit the supplemental particles right away, because they don't contribute to S-matrix elements. But what happens with them at finite temperature? There they contribute to the partition sum, and I'm not sure whether this leads to a consistent theory, because if their energies are negative, you'd never have a converging partition sum, because your energy is not bounded from below. I don't see all troubles resolved yet.

Of course, you have to renormalize tadpoles. Naive normal ordering can be problematic in gauge theories, and thus usually one uses the path-integral formalism, where one needs Weyl ordering and not normal ordering and renormalizes the tadpole diagrams. An example, where this becomes clear in the most easy way, is in scalar QED, where you have a tadpole matter-particle loop in the leading-order self-energy correction of the photon. Only taking it into account together with the usual one-loop diagram you get a transverse regularized photon self-energy when using a regularization scheme that is compatible with gauge invariance (e.g., dimensional regularization). The same holds true for the gluon self-energy in QCD (or pure Yang-Mills). There you also need the gluon-tadpole loop go get a transverse self-energy.

17. Sep 29, 2015

### Bob108

As I said, there are a number of things to be worked out. Hopefully, I can get to them in the coming year. The “not bounded below” issue is one. I’ve thought about it and may have a resolution, but I’m a ways from going public yet. As with most models, odds are it is wrong, but it has chance to answer some perplexing issues in a simple manner.