#### [SOLVED] Do virtual particles actually physically exist?

I have heard virtual particles pop in and out of existence all the time, most notable being the pairs that pop out beside black holes and while one gets pulled away. But wouldn't this actually violate the conservation of energy? #### @anna v 2015-05-21 04:11:21

Ever since Newton and the use of mathematics in physics, physics can be defined as a discipline where nature is modeled by mathematics. One should have clear in mind what nature means and what mathematics is.

Nature we know by measurements and observations. Mathematics is a self consistent discipline with axioms, theorems and statements having absolute proofs, mathematically deduced from the axioms. "Existence" for physics means "measurable", for mathematics "possible to be included in the self consistent theory.

Modern physics has used mathematical models to describe the measurements and observations in the microcosm of atoms, molecules, elementary particles, adding postulates that connect the mathematical calculations with the physical observables

The dominant mathematical model is the field theoretical model that simplifies the mathematics using Feynman diagrams

These diagrams represent terms in an expansion of the desired solution, each term has a diminishing contribution to the cross section of the interaction. The diagram below would be the dominant term, as the next one would be more complicated and therefore smaller by orders of magnitude. To each component of the diagram there corresponds one to one a mathematical formula twhich integrated properly will give a prediction for a measurable quantity. In this case the probability of repulsion when one electron scatters on another.

This diagram for example, has as measurable quantities the incoming energy and momentum of the electrons ( four vectors) and of outgoing four vectors . The line in between is not measurable, because it represents a mathematical term that is integrated over the limits of integration, and within the integral energy and momentum are independent variables. The line has the quantum numbers of the photon though not its mass , and so it is called a "virtual photon". It does not obey the energy momentum rule which says that :

$$\sqrt{P\cdot P} = \sqrt{E^2 - (pc)^2} = m_0 c^2$$

The photon has mass zero.

Through the above relation which connects energy and momentum through the rest mass, the un-physical mass of the virtual line depends on one variable, which will be integrated over the diagram; it is often taken as the momentum transfer.

Quantum number conservation is a strong rule and is the only rule that virtual particles have to obey.

There are innumerable Feynman diagrams one can write, and the internal lines considered as particles would not conserve energy and momentum rules if they were on mass shell. These diagrams include vacuum fluctuations that you are asking about, where by construction there are no outgoing measurable lines in the Feynman diagrams describing them. They are useful/necessary in summing up higher order calculations in order to get the final numbers that will predict a measurable value for some interaction.

Thus virtual particles exist only in the mathematics of the model used to describe the measurements of real particles . To coin a word virtual particles are particlemorphic ( :) ), having a form like particle but not a particle. #### @Paul 2015-05-21 13:22:48

I haven't read particle physics yet,but Eugene Hecht(optics) says electrons exchange virtual photons when interacting and via virtual photons they can exchange momentum which we call as force.so how this is possible if they exist only in mathematics? #### @anna v 2015-05-21 14:52:52

@Paul It is the Feynman diagrams where the exchange of quantum numbers identifies, photons, gluons , W and Z by their quantum numbers, not by their mass. The experimental fact is that an electron transfers momentum to an electron, in the diagram above. #### @anna v 2015-05-21 14:53:00

continued. It is a simple picture, but the true mathematics has many complicated higher order exchanges, as it is a perturbative expansion. It is simple to think as if "virtual" is like the real, except that one falls into contradictions like energy conservation, and that one can never do an experimental measurement with a virtual photon. Virtual is not real, that is why the adjective is necessary. #### @innisfree 2015-07-01 17:01:56

this isn't correct - energy-momentum is conserved at every vertex of a Feynman diagram #### @anna v 2015-07-01 17:15:07

@innisfree the virtual particle is the one that does not obey it, because it is within an integral. After the integration the input and output real particles obey the conservation rules. If the virtual obeyed it, it would not be off mass shell. #### @Solomon Slow 2015-07-01 17:37:17

@Paul, A more down-to-Earth answer from a non-physicist: A mathematical model is our effort to explain phenomena that we observe in a simple, satisfying way. The best, model that we have so far includes features (e.g., virtual particles) that, according to the theory, we should not ever be able to observe. Are they real? We don't know, but until somebody proves the model wrong, or until somebody finds a better model that does not need virtual particles, then we must accept that either the VP are real, or else there must be some isomorphism between the VP and whatever the real truth is. #### @Ian 2015-07-01 17:54:40

@jameslarge There is no reason that we should accept that virtual particles are real. Quantum field theory never asserts that virtual particles are real. To put it simply, virtual particles are just factors that look mathematically like particles but don't behave quite like particles. But, the theory does not make the logical jump to state that VPs actually exist. These VPs are a mathematical way of quantifying the interaction between real particles. In that sense, your idea of an "isomorphism" is more correct since the diagrammatic calculation keeps track of interactions through VPs. #### @innisfree 2015-07-01 17:58:27

@annav have you ever derived an amplitude the long way, without the Feynman rules? If so, you'd see that there is a Dirac function at every vertex imposing momentum conservation. Once integrated over, it gives a Dirac function for overall energy-momentum conservation #### @anna v 2015-07-01 18:11:59

@innisfree Too long ago, at some graduate level course. I am not talking of vertices but of the propagator which has the pole and is the source of the virtual particle meme. #### @innisfree 2015-07-01 18:14:01

The internal particle corresponding to a propagator with pole at $m^2$ won't always satisfy $p^2 = m^2$ - it is off shell - but energy-momentum is still conserved. #### @Ooker 2015-11-22 07:48:18

if so, are photons real? We get photon after quantizing the EM field, and quantizing is a mathematical process. #### @anna v 2015-11-22 08:38:13

@Ooker we get photons from the photoelectric effect, a measurement. and absorption and emission lines, other measurement. These measurements are external lines in the appropriate feynman diagrams. Yes photons are real, because they are measured. The quantization is done to model the measurements #### @reductionista 2015-12-04 15:48:38

@annav If you still believe this (about energy not being conserved for virtual particles), could you explain more how this works on my question here? physics.stackexchange.com/questions/221842/… #### @Andrei Geanta 2017-11-04 07:04:14 #### @anna v 2017-11-04 07:45:25

As my answer says, no I would not rely on it. As an experimentalist I consider "real" synonymous with "measurable" and one cannot set up an experiment to measure the (x,y,z,t) or (p_x,p_yp_z,E) of a virtual particle. #### @Dr. Ikjyot Singh Kohli 2015-07-01 17:24:39

I think one has to be very careful when talking about "particles popping in and out of existence".

This interpretation is only sort of fine in flat-spacetime QFT, where the Minkowski metric is time-invariant, so has a global timeline Killing vector. The definition of a particle depends on the notion of there existing time invariance! Since black hole solutions are static and asymptotically flat, "particles popping in and out" are sort of okay there too.

BUT, quantum field theory is not a theory of particles, it is a theory of fields. So, "particles popping in and out of existence" is based on a naive "particle-interpretation" of QFT, which is not quite accurate for the following reasons (see also the book by Wald, QFT in Curved Spacetime)

Consider a two-level quantum mechanical system which is coupled to a Klein-Gordon field, $\phi$ in a Minkowski spacetime, for simplicity. The combined system will have a total Hamiltonian of the form

$\mathcal{H} = \mathcal{H}_{\phi} + \mathcal{H}_{q} + \mathcal{H}_{int}$,

where $\mathcal{H}_{\phi}$ is the Hamiltonian of the free Klein-Gordon field. We will consider the quantum mechanical system to be an unperturbed two-level system with energy eigenstates $| x_{o} \rangle$ and $|x_{1} \rangle$, with energies $0$ and $\epsilon$ respectively, so we can define

$\mathcal{H}_{q} = \epsilon \hat{A}^{\dagger} \hat{A}$,

where we define

$\hat{A} |x_{0} \rangle = 0, \quad \hat{A} |x_{1} \rangle = |x_{0} \rangle$.

The interaction Hamiltonian is defined as

$\mathcal{H}_{int} = e(t) \int \hat{\psi}(\mathbf{x}) \left(F(\mathbf{x}) \hat{A} + o\right) d^{3}x$,

where $F(\mathbf{x})$ is a spatial function that is continuously differentiable on $\mathbb{R}^{3}$ and $o$ denotes the Hermitian conjugate. One then calculates to lowest order in $e$, the transitions of a two-level system. In the interaction picture, denoting $\hat{A}_{s}$ as the Schrodinger picture operator, one obtains

$\hat{A}_{I}(t) = \exp(-i \epsilon t) \hat{A}_{s}$.

Therefore, we have that

$(\mathcal{H}_{int})_{I} = \int \left(e(t) \exp(-i \epsilon t) F(\mathbf{x}) \psi_{I}(t,\mathbf{x}) \hat{A}_{s} + o\right) d^{3}x$.

Using Fock space index notion, we can then consider for some $\Psi \in \mathbb{H}$, where $\mathbb{H}$ is the associated Hilbert space, and note that the field is in the state

$|n_{\Psi} \rangle = \left(0, \ldots, 0, \Psi^{a_{1}} \ldots \Psi^{a_{n}}, 0, \ldots \right)$.

The initial state of the full system is then given by

$|\Psi_{i} \rangle = | x \rangle |n_{\Psi} \rangle$.

One then obtains the final state of the system as being

$|\Psi_{f} \rangle = |n _{\Psi} \rangle |x \rangle + \sqrt{n+1} \| \lambda \| (\hat{A} |x \rangle) |(n+1)^{'}\rangle - \sqrt{n} (\lambda, \Psi) (\hat{A}^{\dagger} |x\rangle) |(n-1)_{\Psi}\rangle$,

where $| (n+1)^{'} \rangle$ is defined as in Eq. (3.3.18) in Wald, and $\lambda$ is defined as in Eq. (3.3.15) in Wald.

The key point is that if $|x \rangle = |x_{0} \rangle$, that is, the system is in its ground state, the derivation above shows explicitly that this two-level system can make a transition to an excited state, and vice-versa. Note that the probability of making a downward transition is proportional to $(n+1)$, and even when $n = 0$, this probability is non-zero. This in the \emph{particle interpretation} is interpreted as saying that the quantum mechanical system can spontaneously emit a particle. However, the above calculation in deriving explicitly shows that it is the interaction of the quantum mechanical system with the quantum field that is responsible for the so-called spontaneous particle emission. This misleading picture of the vacuum state is precisely promoted by the particle interpretation of quantum field theory. As the work above also shows, this is not spontaneous particle emission from nothing'' in any sense of the word. One must have both a well-defined quantum mechanical system interacting with a well-defined vacuum state for such spontaneous emission to occur, I emphasize that these are not nothing!

The more important point is perhaps that in general curved space times such as the FLRW class of metrics that describe our universe, one can never talk about particles popping in and out of existence, because in general curved space times, there exist no timelike Killing vectors, no Poincare symmetries, no way of defining a covariant ground state, and hence the concept of "particles" has no meaning. #### @Andrii Magalich 2016-06-15 23:13:23

So, I tried to follow your logic and even spent significant time tidying up the formulas... until I understood that this makes no sense. You provide a messy and complicated textbook computation and conclude that this overthrows the QFT. What is this field $\phi$ you invent? How does it fix the fundamental Heisenberg-principle-style consideration that allows to find a particle in a vacuum for a short period of time? Do you imply that interaction with gravitational field or a definition of vacuum in curved space plays a role? How does it work in a low energy/Minkowski limit that we live in? #### @innisfree 2015-07-01 17:00:53

Energy and momentum are conserved at every vertex of a Feynman diagram in quantum field theory. No internal lines in a Feynman diagram associated with a virtual particles violate energy-momentum conservation. It is true, however, that virtual particles are off-shell, that is, they do not satisfy the ordinary equations of motion, such as $$E^2=p^2 + m^2.$$

There is an added complication. A process might have a definite initial and final state, but an "intermediate state" between the two is in a linear superposition of possible states - in this case, a linear superposition of Feynman diagrams - that interfere with each other. We cannot speak of what particles are in this intermediate state, let alone what their momentum is.

But in spite of that complication, I don't think it's ever justifiable to claim that energy-momentum conservation may be violated briefly because of an uncertainty relation. See e.g. this question for a discussion about the interpretation of $\Delta E \Delta t$. #### @William 2015-05-21 02:11:31

To understand this one shall take in quantum-mechanical approximation method namely perturbation theory into account. In perturbation theory, systems can go through intermediate virtual states which often have energies different from that of the initial and final states. This is because of time energy uncertainty principle.

Consider an intermediate state with a virtual photon in it. It isn't classically possible for a charged particle to just emit a photon and remain unchanged itself. The state with the photon in it has too much energy, assuming conservation of momentum. However, since the intermediate state lasts only a short time, the state's energy becomes uncertain, and it can actually have the same energy as the initial and final states. This allows the system to pass through this state with some probability without violating energy conservation. #### @reductionista 2015-12-04 15:43:58

"Consider an intermediate state with a virtual photon in it. It isn't classically possible for a charged particle to just emit a photon and remain unchanged itself. The state with the photon in it has too much energy" But this isn't possible in quantum mechanics either, and doesn't happen that way as far as I understand. When the photon is emitted the electron loses an amount of energy exactly equal to the photon's energy--it does not remain unchanged as you suggest. It regains the energy later when the photon gets absorbed. 