Casimir effect and vacuum energy and a bit of relativity....

[asimov42]

Hi all,

Silly question perhaps: I had understood that the energy density of the vacuum is constant throughout spacetime. But, with the Casimir effect, for example, the geometry of (real) matter (i.e., parallel plates), changes the vacuum energy density in between the plates - is this correct? (since only certain modes of the EM field are allowed between the plates)

So, if the vacuum energy density of various regions of space is influenced in various ways by the configuration of matter in that space, doesn't this wreck Lorentz invariance? I.e., you need to have a very specific vacuum energy spectrum to maintain Lorentz invariance, and if the spectrum between e.g., the plates is different, wouldn't this have implications for relativity if the plates were, e.g., accelerated?

Thanks.

J.

 

[Mentz114]

There is no vacuum energy. The vacuum expectation of all fields is zero, nihil, rien, zut, nothing.

There is no vacuum in the plates experiment because the plates are there.

 

[asimov42]

Um, then what produces the Casimir force?

See, e.g., the article here: http://www.scientificamerican.com/article/what-is-the-casimir-effec/

From the article, "the result being that the total amount of energy in the vacuum between the plates will be a bit less than the amount elsewhere in the vacuum."

 

[Mentz114]

Um, then what produces the Casimir force?

See, e.g., the article here: http://www.scientificamerican.com/article/what-is-the-casimir-effec/

From the article, "the result being that the total amount of energy in the vacuum between the plates will be a bit less than the amount elsewhere in the vacuum."

That language has sacrified accuracy for a simple explanation. If you scroll to the end of this page you will see links to other threads about the Casimir effect.I should add that the vacuum in general relativity, transformed to a local frame is always Lorentz invariant. See 'Lambda Vacuum' .

 

[asimov42]

Hi Mentz114,

Thanks for your responses - as a novice / layperson (in physics, anyway), I'm somewhat unclear on what you've said above. A number of the other threads mention vacuum energy.

I can understand that the expectation value (mean) for the EM field is zero - but I'm not sure how this plays into the idea that the vacuum energy density (see e.g., wikipedia article on vacuum energy, which also mentions the Casimir effect) should be non-zero (i.e., should in fact be enormous, but isn't)... why should the the ground state for the EM field have non-zero energy? I think I'm getting off topic and further into areas I'm unlikely to really grasp...

Anyhow, to the original question - I'm assuming then that regardless of the configuration of the plates and their velocity, Lorentz invariance holds (unless it doesn't hold more generally for some reason)?

 

[Demystifier]

Hi all,

Silly question perhaps: I had understood that the energy density of the vacuum is constant throughout spacetime. But, with the Casimir effect, for example, the geometry of (real) matter (i.e., parallel plates), changes the vacuum energy density in between the plates - is this correct? (since only certain modes of the EM field are allowed between the plates)

So, if the vacuum energy density of various regions of space is influenced in various ways by the configuration of matter in that space, doesn't this wreck Lorentz invariance? I.e., you need to have a very specific vacuum energy spectrum to maintain Lorentz invariance, and if the spectrum between e.g., the plates is different, wouldn't this have implications for relativity if the plates were, e.g., accelerated?

The vacuum is defined as the state of lowest energy. Energy can be defined as an eigenvalue of Hamiltonian. But which Hamiltonian? There are many different Hamiltonians used in physics, and hence there are many different "vacuums".

Some Hamiltonians are meant to be fundamental, describing "all" physics, or at least a large part of physics. Other Hamiltonians are merely effective Hamiltonians, describing only a small subset of all physical phenomena. Consequently, some vacuums are supposed to be fundamental, while other vacuums are merely effective vacuums.

The Casimir vacuum is one such effective vacuum. It is the lowest energy state for the system with Casimir plates at a fixed distance y. Clearly, such a vacuum is not fundamental, because in a fundamental vacuum there are no Casimir plates at all. The Casimir vacuum is the vacuum for the effective Hamiltonian, not for the fundamental Hamiltonian.

In what sense is this Hamiltonian not fundamental? In several senses. In this Hamiltonian
1. Only EM fields are quantized, while charged matter is treated as classical. (Technically, dielectric function $\epsilon({\bf x},\omega)$ is not a quantum operator.)
2. Only EM fields are dynamical, while charged matter is treated as a fixed background. (Technically, dielectric function $\epsilon({\bf x},\omega)$ is a fixed function.)
3. Even the distance y between the plates is treated as a fixed parameter. (Technically, it is a consequence of 2.)

The fact 3. is particularly important for the physical interpretation of Casimir effect. As long as y is fixed, there is no Casimir effect at all. The Casimir effect is the force in the y-direction, and the existence of such a force requires y to be a dynamical variable, not a fixed parameter. To make y dynamical, one must add a new kinetic term to the Hamiltonian. But with this new Hamiltonian, the Casimir energy is no longer the lowest possible energy. Instead, the energy can be further lowered by decreasing y. So to describe Casimir effect by a Hamiltonian, Casimir energy cannot be a vacuum energy for that Hamiltonian.

The preceding paragraph can also be rephrased as follows. Parameter y cannot simultaneously be fixed and non-fixed. If it is fixed then Casimir energy can be interpreted as effective-vacuum energy, but in that case there is no Casimir effect. If it is not fixed then there is Casimir effect, but in that case Casimir energy cannot be interpreted as effective-vacuum energy. So Casimir effect cannot be consistently interpreted as being due to effective-vacuum energy.

Finally, a note on Lorentz invariance: The fundamental Hamiltonian is supposed to be Lorentz invariant (at least in the sense that the corresponding Lagrangian is Lorentz invariant). But effective Hamiltonian need not be Lorentz invariant. Consequently, effective vacuum also does not need to be Lorentz invariant.

For related issues at a more technical level see also
http://arxiv.org/abs/hep-th/0503158
http://arxiv.org/abs/1605.04143

 

[Demystifier]

Um, then what produces the Casimir force?

At the fundamental microscopic level of description, Casimir force is produced by van der Waals forces between microscopic charged constituents of metal plates. But for practical calculations such a description turns out to be too complicated (except in a diluted approximation), so in practice one achieves better agreement with measurements when effective theories are used in which only EM fields are dynamical and quantized. When one interprets successful effective theory as "true" theory, then one may have a reason to say that Casimir force is "due to vacuum fluctuations". But strictly speaking, it is not.

 

[bhobba]

From the article, "the result being that the total amount of energy in the vacuum between the plates will be a bit less than the amount elsewhere in the vacuum."

You need to look into something called normal ordering:
https://en.wikipedia.org/wiki/Normal_order

Vacuum energy by definition is a big fat zero.

Thanks
Bill

 

[vanhees71]

There is no vacuum energy. The vacuum expectation of all fields is zero, nihil, rien, zut, nothing.

There is no vacuum in the plates experiment because the plates are there.

The VEV of the Higgs fiels is non-zero :-).

 

[vanhees71]

That language has sacrified accuracy for a simple explanation. If you scroll to the end of this page you will see links to other threads about the Casimir effect.I should add that the vacuum in general relativity, transformed to a local frame is always Lorentz invariant. See 'Lambda Vacuum' .

Well, I guess many physicists just have read about the Casimir effect from introductory chapters of some QED textbooks that unfortunately treat it in this way. In fact it's a limit where the em. coupling constant is sent to $\infty$.

https://arxiv.org/abs/hep-th/0503158

 

[Mentz114]

Well, I guess many physicists just have read about the Casimir effect from introductory chapters of some QED textbooks that unfortunately treat it in this way. In fact it's a limit where the em. coupling constant is sent to $\infty$.

https://arxiv.org/abs/hep-th/0503158

Thanks for the reference to that excellent paper. My own suspicion and mistrust of the 'standard' calculation is vindicated.

 

[bhobba]

Thanks for the reference to that excellent paper. My own suspicion and mistrust of the 'standard' calculation is vindicated.

It's the typical QFT stuff.

To get to grips with the formalism all sorts of intuitive 'views' such as virtual particles are actual particles are promulgated.

Here, when we try to correct misconceptions, its easy to be dismissive its value. Developing intuition is in fact VERY important.

After all its why Feynman displaced Schwinger - he was in possession of much more powerful methods that solved problems in a few hours it took others years.

Thanks
Bill

 

[vanhees71]

Indeed, the diagrammatic technique is of utmost importance as a calculational tool. However, one should be careful with building too classical intuitions on them. Indeed they are just a very clever notation for evaluating the perturbative Dyson series for S-matrix elements and many other things, also going beyond perturbation theory. If any intuition is at place, it's usually more save to think in terms of fields than particles. The fundamental building block of relativistic QT are fields, and that's reflected by the meaning of the diagrammatical elements: external lines in Feynman diagrams symbolizing S-matrix elements stand for the asymtptotic free wave functions of the initial an final state of the considered scattering event, internal lines stand for propagators, and vertices for interactions in the Lagrangian/Hamiltonian of the field theory. Then a lot of confusion can be avoided and there is indeed a quite intuitive picture developing in ones mind when becoming familiar with the math that is well hidden in the elegance of Feynman's ingenious notation, which is indeed at the basis of all modern applications of QFT, including the systematic solution of the renormalization problem by Bogoliubov, Parasiuk, Hepp, and Zimmermann in the 60ies, which is entirely based on the diagram formalism, including the sometimes cumbersome treatment of the appropriate counter terms for subdivergences and the solution of the problem of overlapping divergences, all culminating in Zimmermann's forest formula, which is entirely formulated in the language of diagrams and their topology. The applications today reach much farther: Feynman diagrams are not only used in relativistic vacuum QFT to evaluate S-matrix elements from the Standard Model and beyond but also in the entire range of many-body theory, reaching from the non-relativistic realm condensed-matter physics (including funny exotic materials like graphene or ultracold Bose and Fermi gases showing all kinds of fascinating QFT objects like anyons, fractional charges/magnetic moments, magnetic monopoles, Weyl fermions,...) to the relativistic thermal QFT heavily used in heavy-ion physics and stellar astrophysics (including the structure of neutron stars, neutron-star mergers, neutron-star collisions...) to learn about the state of strongly interacting matter with its yet not too well understood phase diagram of QCD.

 

[Demystifier]

The vacuum is defined as the state of lowest energy. Energy can be defined as an eigenvalue of Hamiltonian. But which Hamiltonian? There are many different Hamiltonians used in physics, and hence there are many different "vacuums".

Some Hamiltonians are meant to be fundamental, describing "all" physics, or at least a large part of physics. Other Hamiltonians are merely effective Hamiltonians, describing only a small subset of all physical phenomena. Consequently, some vacuums are supposed to be fundamental, while other vacuums are merely effective vacuums.

The Casimir vacuum is one such effective vacuum. It is the lowest energy state for the system with Casimir plates at a fixed distance y. Clearly, such a vacuum is not fundamental, because in a fundamental vacuum there are no Casimir plates at all. The Casimir vacuum is the vacuum for the effective Hamiltonian, not for the fundamental Hamiltonian.

In what sense is this Hamiltonian not fundamental? In several senses. In this Hamiltonian
1. Only EM fields are quantized, while charged matter is treated as classical. (Technically, dielectric function $\epsilon({\bf x},\omega)$ is not a quantum operator.)
2. Only EM fields are dynamical, while charged matter is treated as a fixed background. (Technically, dielectric function $\epsilon({\bf x},\omega)$ is a fixed function.)
3. Even the distance y between the plates is treated as a fixed parameter. (Technically, it is a consequence of 2.)

The fact 3. is particularly important for the physical interpretation of Casimir effect. As long as y is fixed, there is no Casimir effect at all. The Casimir effect is the force in the y-direction, and the existence of such a force requires y to be a dynamical variable, not a fixed parameter. To make y dynamical, one must add a new kinetic term to the Hamiltonian. But with this new Hamiltonian, the Casimir energy is no longer the lowest possible energy. Instead, the energy can be further lowered by decreasing y. So to describe Casimir effect by a Hamiltonian, Casimir energy cannot be a vacuum energy for that Hamiltonian.

The preceding paragraph can also be rephrased as follows. Parameter y cannot simultaneously be fixed and non-fixed. If it is fixed then Casimir energy can be interpreted as effective-vacuum energy, but in that case there is no Casimir effect. If it is not fixed then there is Casimir effect, but in that case Casimir energy cannot be interpreted as effective-vacuum energy. So Casimir effect cannot be consistently interpreted as being due to effective-vacuum energy.

Now a more elaborated version of this is available:
https://arxiv.org/abs/1702.03291

 

[A. Neumaier]

nfluenced in various ways by the configuration of matter in that space, doesn't this wreck Lorentz invariance?

The plates already break Lorentz invariance. The Casimir effect is caused by the electromagnetic field emanating from the plates.

 

[Demystifier]

The Casimir effect is caused by the electromagnetic field emanating from the plates.

With the important caveat that the average electromagnetic field is zero
$$\langle\psi |{\bf E}|\psi\rangle = \langle\psi |{\bf B}|\psi\rangle =0$$

 

[vanhees71]

That's why I'd put it as: "The Casimir effect is caused by the fluctuations of the electromagnetic field due to the presence of electric charge within the plates." In our context it's important to emphasize that there are no vacuum fluctuations but fluctuations of quantum fields (in this case the electron and photon fields in "minimal QED").

 

[Demystifier]

That's why I'd put it as: "The Casimir effect is caused by the fluctuations of the electromagnetic field due to the presence of electric charge within the plates." In our context it's important to emphasize that there are no vacuum fluctuations but fluctuations of quantum fields (in this case the electron and photon fields in "minimal QED").

Sure, but as I stress in the paper above, it depends on what exactly one means by "vacuum". In particular, the relevant state can be called vacuum because it is annihilated by certain lowering operators. However those are not photon lowering operators. Those are operators for polaritons - quasi-particles which can be thought of as mixtures of photons and quanta of polarization field.

 

[vanhees71]

What's a "polarization field". Are you referring to the problem to define asymptotic free states in models containing massless excitations a la Kulish&Faddeev?

 

[Demystifier]

What's a "polarization field".

It's electric dipole moment per unit volume, ${\bf P}({\bf x},t)$, appearing in electrodynamics of continuous media (see e.g. Jackson) in the equation
$${\bf D}({\bf x},t)={\bf E}({\bf x},t)+{\bf P}({\bf x},t)$$

 

[stevendaryl]

Vacuum energy by definition is a big fat zero.

The analogy is made between QFT and a simple Harmonic oscillator: In the quantum SHO, the zero-point energy is not zero, but is $\frac{1}{2} \hbar \omega$. If you naively quantize the field modes of QFT, you get a similar contribution for every field mode, so you get a total zero-point energy that is divergent (since there are infinitely many field modes). The normal-ordering prescription for QFT is analogous to redefining the Hamiltonian of the simple harmonic oscillator: Instead of

$H = \frac{1}{2m} p^2 + \frac{m \omega^2}{2} x^2$

You use an alternative Hamiltonian

$\tilde{H} = A^\dagger A \hbar \omega$

where $A$ and $A^\dagger$ are combinations of $p$ and $x$ satisfying $A |\psi_0\rangle = 0$, when $|\psi_0\rangle$ is the ground state. The Hamiltonian $\tilde{H}$ produces the same physics as $H$ without the annoying zero-point energy.

However, in the case of the SHO, the ground state is still nontrivial, because even though the energy is redefined to be zero, and the expectation values of $x$ and $p$ are both zero, the expectation values for $x^2$ and $p^2$ are both nonzero. The analogy with QFT holds true: even though normal ordering makes the expectation value for the energy density of the vacuum rigorously zero, the expectation value for quadratic terms in the field operators is nonzero. When people loosely say that something is an effect of the zero-point energy, I think they might mean that it is an effect of the zero-point variance of the field operators, which is nonzero.

 

[Demystifier]

However, in the case of the SHO, the ground state is still nontrivial, because even though the energy is redefined to be zero, and the expectation values of xx and pp are both zero, the expectation values for $x^2$ and $p^2$ are both nonzero. The analogy with QFT holds true: even though normal ordering makes the expectation value for the energy density of the vacuum rigorously zero, the expectation value for quadratic terms in the field operators is nonzero. When people loosely say that something is an effect of the zero-point energy, I think they might mean that it is an effect of the zero-point variance of the field operators, which is nonzero.

Good point! Indeed, Eqs. (51) and (52) in my paper above show explicitly how expectation value of $x^2$ leads to a Casimir-like force.

 

[_PJ_]

Sorry if this is a hijack of the thread, it's relevance should be obvious.

With respect to the OP and initial question, then -

How is the notion of "spontaneous virtual pair generation" to be considered?
I understand even with such a situation, there is a net result of ZERO - In the "popular science" description, these virtual pairs are providing pressure against the plates - which would mean they are imparting energies, but that energy is required to be 'paid back' to the "vacuum" - so... a deficit?

 

[stevendaryl]

Sorry if this is a hijack of the thread, it's relevance should be obvious.

With respect to the OP and initial question, then -

How is the notion of "spontaneous virtual pair generation" to be considered?
I understand even with such a situation, there is a net result of ZERO - In the "popular science" description, these virtual pairs are providing pressure against the plates - which would mean they are imparting energies, but that energy is required to be 'paid back' to the "vacuum" - so... a deficit?

As @A. Neumaier has pointed out on numerous occasions, the language of virtual particles and borrowing energy and so forth is just a failed pop-science attempt to explain quantum field theory to people without using the actual mathematics. So don't spend too much time trying to make sense of it.

 

[_PJ_]

As @A. Neumaier has pointed out on numerous occasions, the language of virtual particles and borrowing energy and so forth is just a failed pop-science attempt to explain quantum field theory to people without using the actual mathematics. So don't spend too much time trying to make sense of it.

I'm trying to ask what is the QFT explanation then? All I see from Neumaier is:

"All I see from The plates already break Lorentz invariance. The Casimir effect is caused by the electromagnetic field emanating from the plates."

 

[A. Neumaier]

this is a hijack of the thread

... because what you say is not connected to the Casimir effect.

As @A. Neumaier has pointed out on numerous occasions

and in particular here, with the connection to the Casimir effect pointed out here.

 

[_PJ_]

... because what you say is not connected to the Casimir effect.

and in particular here, with the connection to the Casimir effect pointed out here.

Thank you!

 

[haushofer]

I've printed the papers by Jaffe and Demystifier, since I'm very interested in this Casimir-description. I think I get the reasoning, but I'm still wondering why the QFT 'vacuum fluctuation' calculation yields the same answer as the Van Der Waals calculation for the Casimir force. Is there any simple/intuitive way of seeing why these two seemingly completely different approaches lead to the same answer?

 

[A. Neumaier]

why the QFT 'vacuum fluctuation' calculation yields the same answer as the Van Der Waals calculation for the Casimir force.

Perturbation theory in quantum field theory always produces results expressed in terms of vacuum expectation values, no matter which problem is treated. Thus the coincidence of the results is not surprising. What is surprising is only that many people think something significant is expressed by talking about these vacuum expectation values in terms of vacuum fluctuations.

 

[Vanadium 50]

why these two seemingly completely different approaches

Why do you think they are completely different? In both cases, the force arises through the boundary conditions. The issue comes about when one tries to paint a mental picture of the results of this calculation.

 

[stevendaryl]

Why do you think they are completely different? In both cases, the force arises through the boundary conditions. The issue comes about when one tries to paint a mental picture of the results of this calculation.

If I understand what the two approaches to deriving the Casimir force are, they sure seem very different:

1. Treat the electromagnetic field classically, but assume that the charges inside the metal plates are undergoing random internal motion. Then you get a Van der Waals type force between the plates.
2. Treat the plates classically (as just boundary conditions for the electromagnetic field), and treat the field modes of the E&M field quantum mechanically.

The two approaches seem to be focusing on completely different subsystems, and are making completely different approximations about what to treat classically and what to treat quantum mechanically.

 

[haushofer]

Why do you think they are completely different? In both cases, the force arises through the boundary conditions. The issue comes about when one tries to paint a mental picture of the results of this calculation.

Well, I have to think about that, but the VdW-force I associate with dipole-interactions, giving a 1/r^6 law. For me that's really different from 'vacuum interactions' as described by Feynman diagrams.

-edit Steven's post above me states my issue a lot clearer, I guess.

 

[Haelfix]

Just so people are clear, this business is deeply controversial in the literature and not settled. There are quite a few subtleties involved, and really requires going through the entire analysis with a scalpel.

 

[haushofer]

Just to make sure i get it: in Jaffe's paper fig.3, those external legs are the conducting charges of the plates, right?

What I still can't reconcile is the dependancy of the Casimir force on the distance between the plates, i.e. eqn.(3) v.s. the VdW expression between eqn.5 and 6 (top of page 5). Shouldn't one expect F~1/d^6? What am I missing?

 

[Vanadium 50]

Shouldn't one expect F~1/d^6? What am I missing?

The field from a charged plane is not the same as a field from a charged point.