Why is the square momentum of a virtual photon negative?

probably You need more info what is q and what is q^2 ;) 

Because a virtual photon isn't a physical particle. That q² < 0  implies that the 4-vector q doesn't describe a physical particle for, if it did, in the rest frame, such a particle would have had an imaginary mass.

And that should be expected. Explaining that is the topic of the exercise.

To add: Virtual particles are associated with evanescent waves and having this imaginary component, they fall away with distance. Examples are quantum tunnelling or the near field of an EM transmitter or transformer induction. Virtual photons also have mass.

good day ,i agree with stam and remi  about to explaining   

Evanescent waves are classical objects, virtual particles are not. Virtual particles don't describe any decay process,they describe real interactions. So the two notions don't have anything to do with each other.

The imaginary part of virtual particles isn' t in their momentum-it's in their mass, which doesn't have any real part. 

Sociology is different from physics-so how a word is used doesn't mean anything, beyond the sociology of the community. Sociological considerations don't affect the fact that when the squared 4-momentum is negative the particle isn't physical. Nor that it isn't a ghost, however. That's what matters. 

The tunnel effect in quantum mechanics is just a misnomer, because the quantum particle doesn't tunnel through any barrier: it just sees a different potential than the classical particle. If one wants to keep the classical potential, then one *says* that the quantum particle tunnels through a classical barrier. But it's the assumption that's wrong-the classical potential is the classical limit of the quantum potential. And virtual particles don't have anything to do with the tunnel effect. When two electrons interact through photon exchanges, there's no tunneling involved.

The error is that, in the dispersion relation, evanescent waves have imaginary momentum, whereas virtual particles (would) have imaginary mass. 

E² - p²c² = m²c⁴

On shell: all real quantities

Off shell: some quantities may be imaginary

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Off shell => virtual

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Term evanescent used here:

Quantum tunnel and Virtual particles discussed here:

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Probability for particle to go from one place to another neglecting time (THIS CONSIDERATION IS ONE EXAMPLE SHOWING VIRTUAL PARTICLES. IT SHOWS THE BOUND STATIONARY STATE, SO ENERGY WOULD BE CONSTANT).

<r₂|r₁> = e^(ik,r₁₂/h) / r₁₂

Use dispersal relation p = 1/c.sqrt(E² - m²c⁴)

Imaginary momentum results in an evanescent wave (decaying). This comes about from a large m - either +ve or -ve

Instead of writing meaningless expressions, it's better to actually solve a real problem: compute the interaction of two electrons, to lowest order-one photon exchange-to see that the, virtual, photon doesn't have anything to do with what's claimed. There it will be found, in particular, that the photon's 3-momentum isn't imaginary. By energy-momentum conservation it's equal to the difference between the real  3-momenta of one of the initial and final electron states, that are both on-shell. The rest is algebra and the result is that, if the two electrons are on-shell, the photon can't be.

For any particle, energy and 3-momentum are related by the expression E^2-|p|^2c^2=(mc^2)^2. It's the mass of the particle  that's Lorentz invariant, i.e. independent of the frame, not its energy or its 3-momentum.  So it's wrong to state that a massive particle can have imaginary 3-momentum. 

I guess you are referring to the four momentum square q² of the virtual photon. Because it is off-shell q² can be either positive or negative but not zero. In a head on collision experiment (such as LEP) a virtual photon which is produced has a lot of energy but little 3-momentum then q² is negative, on the other hand in deep inelastic scattering experiments we find that a virtual photon may have a lot of 3-momentum but less energy then q² is positive. I have taken Minkowski signature -1,+1,+1,+1 to evaluate the dot product of four momentum with itself.

PS: Note that four momentum is defined as (E/c, p_x, p_y, p_z)

The signature of the metric doesn't matter. Virtual particles have negative squared 4-momentum, because they're not real particles-and it's the sign of the squared 4-momentum that makes this statement meaningful. The reason is that the squared 4-momentum is a Lorentz invariant. It means that in the rest frame such particles would have had imaginary mass. Were the squared 4-momentum positive, in the rest frame this would be a physical, massive, particle. 

Since this is an exercise I should not tell you too much.  But it generally is a good idea to choose a good viewpoint when one wants to understand the landscape. In this case, a good viewpoint means choosing a good coordinate system (i.e, Lorentz frame).

I should add that although the statement appears to be true (experimentally and model-wise, and within a reasonable interpretation) in QED, is not generally true for gauge bosons in the standard (GWS) model for weak interactions. And one may easily imagine theoretical scenarios (which has not been confirmed experimentally)  where it is not true.

In general, virtual (off-shell) particles may have either sign of its squared four-momentum. 

The statement that it amounts to is that a uniformly moving charge can't radiate. 

For the QED vertex it amounts to showing that, since the photon is massless, not all three particles can be on their mass-shell (in Compton scattering one of the electrons is off-shell, while the photon and the other electron are on-shell). 

And, more generally, if one does have a three particle vertex, where one of the particles is massless, the statement still holds, since a massless particle can't ``decay'', since it doesn't have a rest frame (another way of expressing the same thing). The quotes are to stress that this is a unitary transformation in any consistent field theory, it's a reversible, not an irreversible, process. Calling it a decay simply indicates that one is interested more in one of the two reactions than the other. If  all three particles in the vertex are massive, it is possible to go the rest frame of any one of them. 

While the sign of the squared 4-momentum depends on the convention, what is true is that virtual particles have negative mass squared. (However, once the convention is fixed, it isn't true that the  sign of the squared 4-momentum of virtual particles can be arbitrary. It's always consistent with the property that the 4-momentum, for virtual particles, is always space-like, whereas it's time-like or null, for on-shell, physical, particles.)  That does not depend on the convention (where one must be careful to keep track of minus signs).  So the invariant formulation is:  why is the squared mass of a virtual photon negative?  

see this

I didn't find Baez's summary, which has been around for years, much use.

Quite simply, the definition of a virtual particle is that it is off-shell.

Stam dismissed one of my response where I talked about the stationary bound state/tunnelling where the particle is off-shell. Look at the maths I put up. An off-shell POSITIVELY massive particle in that case will have imaginary momentum. It is a simple matter to see that its wavefunction decays exponentially with distance.

Near field photons of a transmitter also have mass (and a longitudinal polarisation state) and this mass is the explanation for the falling off (decay, not decay process as Stam incorrectly read me) of the near fields. Nuclear binding forces are also short ranged because the particles are so massive too.

Stam and Biswajoy and Kare are talking about another scenario, electron-photon scattering.

Come on Stam; if this was true LEP would never have worked (or been funded) :-)

I interpret Elias' problem to be restricted to the context of electron-photon scattering, but all my qualifications are related to more general possibilities.

With neutrino oscillations the process mu -> e + gamma is in principle possible, although with an unobservable small rate. That would violate the statement to be proven (but it can also be said that then the photon no longer couple to a QED vertex). And in the dream of the next generation of particle experimentalists, the International Linear Collider, the virtual photon emerging from the QED vertex is planned to have a positive mass squared of up to (1 TeV/c^2)^2.

Of course virtual particles have negative squared mass-that's what virtual means. The photon exchanged in deep inelastic scattering, for example,  has space-like 4-momentum; that's why people got interested in the ``deep Euclidian region'' in the first place. The center of mass energy is different from the momentum transfer.

Once more: The virtual photons created by electron-positron annihilation has a positive mass squared > (4 m_e²). In LEP as large as about (115 GeV/c²)². To be considerably increased in ILC, if it is ever built. (This at least in circles where the notions of "virtual" and "off-shell" are used interchangeably).

No! There's a confusion between center of mass energy and momentum transfer. There are two independent 4-vectors, in a 2-particle collision (the third is related to the two others), p1+p2, that's time-like, its square is the center of mass energy; and p1-p2=q, that expresses the 4-momentum transfer.  It isn't the virtual photons that had positive energy squared (115 GeV/c^2)^2 at LEP:  the, on-shell electrons and positrons that were colliding, were carrying it (each half of it).  And same holds for the LHC, the ILC and so on. Virtual particles aren't emitted on-shell-and they're described by a space-like 4-momentum, that's why.The colliding particles are on-shell.

It's a standard exercise, once more, to check that a uniformly moving massive particle can't emit a massless paticle: energy-momentum conservation implies that not all three particles in the vertex can be on the respective mass shells-the part that's not mentioned that much is that  the one that isn't has a 4-momentum vector that's space-like. Proving that is the subject of the exercise under discussion.

Consider e⁺e^- -> μ⁺μ^- to lowest order in QED, and calculate q² for the single virtual photon involved in that process, where q is it's 4-momentum.

Simples, there are no virtual photons- they are an invention of modern physics to get over some knotty problems in their maths

Andrew, no. The bound state in a chemical bond requires imaginary momentum from the dispersion relation. Why else would the shape of the orbitals fall off with distance than to have some function with e^ip/r as the envelope?

@ Remi,,

Actually the solution is, the vacuum is not empty but full of dark energy. The originators of quantum physics did not know this, but you can do away with virtual particles and virtual photons knowing this. Its not at all complicated to do, without inventing particles and photons that do not exist.

After all, all baryonic matter makes up only 5% of the Universe, dark energy makes up 70% of the Universe.

Oh yeah!? Please show us how you in an uncomplicated way can calculate the electron magnetic moment to fifth order in the fine structure constant without using the concept of virtual particle.

On second thought, maybe you can start by calculating the first order (Schwinger) term for us. That should not take more than an afternoon, since it is not at all complicated. 

(Kåre Olaussen):  ``Consider e+e- -> μ+μ- to lowest order in QED, and calculate q2 for the single virtual photon involved in that process, where q is it's 4-momentum.'' 

Indeed, something more is required to be specified, namely the channel: In the s-channel, the virtual exchanged particle  has time-like 4-momentum; it's in the t-channel that it has space-like 4-momentum.  

Agree. But to higher orders, inside diagram loops, we are forced to consider virtual particles with all possible real values of q².

The reason is that, beyond tree level, the global conservation laws aren't sufficient and there are integrations involved-which means that  the amplitude can't depend on q itself, anyway,  but on two of the three invariants, s, t and u. These, though, are functions of the external momenta only so the channel, still, matters. The singularities are related but not identical. And that can lead to confusion, that ought to be avoided, by stating all assumptions. No need to play games.

WHY?