Does energy time uncertainty violate conservation of energy for a small time delta t?

Tanjore

Hi folks, I am new to physics forums , an engineer by profession, I have newly developed some interest for pure physics, can someone please enlighten me on the violation of energy conservation by the time energy UP? The way I understand it the principle only puts a LOWER bound.

Is conservation of energy indeed violated for a short time, taking the example of
electron-electron interaction?
for a short time that the 'virtual' photon exists is the conservation of energy violated?
what about the conservation of momentum?

The_Duck

The energy-time uncertainty principles basically says that sometimes a physical state has an uncertain energy. For instance, an excited state of an atom has a finite lifetime before it decays to the ground state, which translates into a slight uncertainty in the energy of the excited state. A practical result is that the light emitted by atoms falling down from that excited state can have a range of frequencies (= photon energies) instead of all having the same exact frequency.

However, this is not the same thing as saying that energy is not conserved. In quantum mechanics energy is still conserved. If you measure the energy of a system, leave it alone for any amount of time, and then measure the energy again, you will get the same value. This is because measuring the energy forces the system into a state of definite energy.

In Feynman diagrams energy and momentum are conserved even when virtual particles are present. For example in electron scattering you have an electron which gives up some of its energy and momentum to a virtual photon, which then deposits that energy and momentum in the other electron.

tom.stoer

Unfortunately popular books quite frequently use statements regarding "virtual particles borrowing energy for a short time from the vacuum" in order to explain the time-energy uncertainty relation. This is missleading for a couple of reasons.

First the time-energy uncertainty relation is very different from the position-momentum uncertainty relation as tzhe latter one can be derived strictly based on a mathematical (geomezrical) structure of the Hilbert space plus certain operators acting on it. In contradistinction the time-energy uncertainty relation is more a "visualization"; there is no time operator in QM.

Second the term "virtual particles" is missleading as they are only mathematical artefacts arising in a certain approximatin, perturbation theory; they do not have properties of particles as used in quantum field theory, there are no associated states in an Hilbert space.

Third, if one looks at virtual particles in perturbation theory, they are represented by inner lines in the famous Feynman diagrams. Every vertex in such a diagramm (where several lines meet = where the particles interact) carry an algebraic structure descring the interaction plus a so-called delta function which guarantuees in the incoming and the outgoing momentum and energy are equal; that means at such a vertex no momentum or energy are destroyed, created, borrowed o whatever. Energy-momentum-conservation holds strictly.

Edit: this agrees with Duck's explanation

Tanjore

Hmm. I could not understand the technical things that you mentioned but thanks a lot for mentioning about the vertices of the feynman diagram, if energy and momentum are conserved at the vertices then they ought to be conserved all along the internal as well as the external lines. Is this right?

tom.stoer

yes!

Tanjore

Thanks a lot both of you!

DJsTeLF

Just though you might appreciate this extra 'nugget' to continue to stimulate your new-found interest in particle physics. It's an explanation (that I've copy and paste from here: http://www.physicsforums.com/showthread.php?t=32834) of what is meant by the terms - "on mass shell" and "off mass shell":

The 'mass shell' for a particle of mass m is the surface in the E - p1- p2 - p3 space satisfying the equation

E^2 - p1^2 - p2^2 - p3^2 = m^2 (factors of c omitted)

It's kind of a hyperboloid in 4 dimensional space.

Or if P is the energy-momentum 4-vector (E, p1, p2, p3),P.P = m^2

(You may see this with the opposite sign of P.P depending on the
metric.)

In classical special relativity all particles are 'on mass shell'. For
LaTeX Code: p=0 we recover the infamous

E^2 = m^2

or with factors of c replaced

E^2 = m^2 c^4

In quantum field theory particles are allowed to propagate even if
they do not satisfy this equation. In that case they are said to be
'off mass shell'. These are also called 'virtual particles'. However,
there is a penalty to pay for this, because in that case the amplitude
of interaction goes as

1/(P.P - m^2)

Particles that can propagate to infinity ('real particles') are always
'on mass shell'. In a collider experiment, a few centimetres is a
reasonable approximation to infinity, so one takes the incoming and
outgoing particles 'on shell'. Intermediate states are 'off shell' or
'virtual'.

tom.stoer

The main point is that virtual particles are allowed to be of-shell i.e. to violate E² - p² = m².

But (!) a virtual particle is not just "one off-shell particle", but it is "an integral over all off-shell particles integrated of the complete (E,p) i.e. 4-momentum space - restricted by delta-functions at the vertices. That's the reason why a virtual particle is not a particle at all but a description for a mathematical calculation.

RedX

I remember that an electron in an excited state decays very quickly (E&M processes are very fast compared to strong or weak processes), so that should correspond to a huge spread in energy by the time-energy uncertainty relationship. However, you do see spectral lines rather than a rainbow smear, so the uncertainty can't be that bad?

Anyways, does this mean that you can never excite hydrogen into the n=2 state? Because for it to be in an n=2 state it has to stay there for infinite time, since if it only lasts between say t=-10 seconds and t=10 seconds, then it was really a wavepacket composed of states of all n?

Is this the spectral density form of the propagator? I remember that not only do you have to integrate over the complete (E,p) space, but also sum over the many-particle states which are part of the Hilbert space in the interaction picture. However, for this part, you only integrate over space for which E^2-p^2>=4m^2 since there must be at least two particles. But it seems that this cloud of virtual particles are real states in the full Hilbert space.

I read somewhere that the reason that weak interactions have short range is that the W and Z are very heavy, so if your initial particle (say a neutron) doesn't have the energy equal to the mass of the W, then it has to borrow heavily to create a W, which means the W can only exist for a short time by the time-energy uncertainty principle, which means it can only travel a short distance before it turns into an electron and antineutrino: in fact this distance is so short that the interaction can be approximated as a 4-point interaction - i.e., an instantaneous interaction. This is what Fermi calculated in the old days before the W was even known about.

tom.stoer

Unfortunately things like that have been written in popular books, but that's confusing and missleading. The particle doesn't borrow energy E to create the W; instead the W violates the mass-shell condition E²-p²=m²; E and p are conserved, nothing is borrowed, not even for a very short time. Yes, there may be not enough energy E (and momentum p) to create a real W; instead the process uses the available energy E (and momentum p) and creates a W of some mass m=m(E,P); m becomes a function of E and p! That's what is meant be "off-shell". If you look at an off-shell photon that it has non-zero mass.

RedX

I didn't get it from a popular book. The last paragraph of page 13 of a review article written for Annual Reviews of Nuclear and Particle Science:

http://arxiv.org/abs/hep-th/0701053

says something like it too.

I agree with you: I can see the delta functions that conserve everything at each vertex. But how is it that if W is really heavy, you can ignore it for low energy processes and say that weak decay occurs by a point interaction? So I feel that it's not just a mistake by the author and there might be something to the energy-time uncertainty relation. I don't know.

tom.stoer

OK, fine.

Omitting loops you immediately have (p²-M²)^-1 = 1/M² + ... for p² << M². So for small p the propagator between two vertices collapses to a single vertex; the coupling is therefore ~ 1/M²

It is not a mistake but a perfectly valid statement; but you need not refer to something like "borrowing energy". It is simply an application of standard Feynman diagram analysis.

Tanjore

@RedX : </i>then it has to borrow heavily to create a W, which means the W can only exist for a short time by the time-energy uncertainty principle, which means it can only travel a short distance before it turns into an electron and antineutrino:</i>

But the time energy uncertainty principle gives a lower limit for how long a particle of energy E can last, it does not put an upper limit.

Tanjore

Also I have another question, in the time energy uncertainty relationship there is an inequality, I mean it is an inequality, so now my question is this , is the inequality just mathematical or does it have any importance in the physics of the process meaning can the inequality be ignored( because I have seen in textbooks that mass of pi meson is calculated simply by ignoring the inequality, can that be done?) can we simply ignore the inequality. My friend said, it could be ignored because the uncertainty in the energy of the particle is at the most equal to the energy of the particle. Is it right? can it be ignored?

QuantumClue

Some of this is horrible to read. I am sure it would make sense, but is horrible to read without proper latex.

Anyway, to the OP, my few cents. The uncertainty between energy and time [tex]\Delta E \Delta t[/tex] does not violate any conservation laws. You cannot extract energy which is already there.

All it means is that a state which exists for a very short time cannot have a well-defined energy. Nothing amazingly weird is going on. The shorter you make a measurement, the more undefined that measurement becomes in relation to the energy of the system. If you watch over a system for a great length of time, you have a well-defined energy in contrast.

tom.stoer

as QuantumClue said: it does not borrow energy! Yes, the W exists for a very short time and travels a short distance. Measuring the lifetime of the W (which is impossible as it is too short) would result in a probability distribution with a width Delta T; measuring the energy of the W results in a distribution with a width Delta E; these two values are related by the time-energy uncertainty. But again: energy is always conserved.

ZapperZ

In addition to all the answers given here, please also read our FAQ in the General Physics forum. And yes, it implies that this is a very frequent question that we get on here.

Zz.