Does uncertainty principle imply non-conservation of energy?

[koantum]

Hi Vanesch, in my previous post I just collected my thoughts on the subject; I didn’t mean they are all equally relevant to your challenge, though they certainly have a bearing on it.

I just had a look at what must be your bible (I'm kidding) - The many-worlds interpretation of quantum mechanics edited by DeWitt and Graham. I actually own the book, which isn't half bad. It contains an article by DeWitt "The many-universes interpretation of quantum mechanics", which contains a discussion of a system-apparatus interaction in the Heisenberg picture. DeWitt shows that the apparatus operator after the measurement depends on the undisturbed system operator (which was measured). Since in the Heisenberg picture the system operator depends on time (qua "c-number"), this means that the system's energy (like any other system operator) can be measured with unlimited precision at a precise time!

There is an ambiguity here which is often overlooked: "time of measurement" can mean (i) time at which the outcome becomes "available" or (ii) time at which the system observable possessed the indicated value. We seem to agree now that what I just said is true of (ii). You still deny that it is true of (i).

It seems to me, though, that, although the interaction time with the system is indeed arbitrary short, that the result is not available in this time, because what happened is that there was a transfer of energy from the particle to the first (and the second) condenser, and now we displaced the problem to the energy measurement of the condensers. So in order to measure THIS energy accurately enough, we will need again a time T (or an ingenious scheme which will, in its turn, transfer the knowledge to a further system)... What I learned... is that the system must not remain "available" during the entire measurement time T (the time between the "availability" of the system state, and the "presentation" of the result, which corresponds here with the completion of the energy measurement of the condensors).

What do you mean by "available", "transfer of knowledge", and "presentation of the result"? You are playing the old game of the "shifty cut" (as Bell called it), agreeing that the buck must stop somewhere (such as when the outcome is "available" or "presented"). Available to whom? Presented to whom? I'm echoing Bell's famous question: whose knowledge? Those who use evolution speak (which I try to avoid) call this buck stopper "collapse" (I wouldn’t know of what) or "world branching" (which isn't any better), and they either endorse the slogan "quantum states are states of knowledge" or invoke observers to account for the real or apparent collapses or for the real or apparent world branchings.

In MWI speak... the final evolution into clearly distinguished pointerstates...

This is not specifically MWI speak. So when are pointer states clearly distinguished? When they are distinguishable according to the neurobiology of human, or primate, or mammalian, or vertebrate... perception? If this is your buck stopper, then you are absolutely right because perception (even human) is a notoriously slow process, as psychologists and neurobiologists will confirm.

 

[vanesch]

There is an ambiguity here which is often overlooked: "time of measurement" can mean (i) time at which the outcome becomes "available" or (ii) time at which the system observable possessed the indicated value. We seem to agree now that what I just said is true of (ii). You still deny that it is true of (i).

Well, yes, but I admit having been puzzled by the Bohm example, and that it indicated something important. I think that there's now a flaw in the reasoning which tells us that dT.dE > hbar because of the interaction term. It is true that the MAGNITUDE of this interaction term is correctly described that way, but the magnitude is too coarse a measure to say that it is the measurement error, which can indeed be made smaller, by introducing a reverse evolution. So it is not because you introduce an interaction term with magnitude > dE, that you've introduced an irreducible error on the potential error measurement (as is often claimed, and as I believed). So this kills (ii), and this is what I learned.

However, (i) finds its origin elsewhere. It follows from the claim that the expectation values of all quantities of states which differ by dE only start to show significantly different time evolutions after a time dt such that dt.dE>hbar.

And in as much as, as of Bohm, a very fast interaction can now transfer the energy value to another system, this other system's state will now have to evolve, such that, whatever quantity is going to be measured (position for instance), it is clearly different for E and for E+dE (first of the system, and then of the "helper system" such as the condensers).
Now, to clearly distinguish between a system in a state E and a state E+dE, one must measure something of which the expectation value is different, meaning that we will have to let it evolve for at least the time dt for the expectation values of these states to be clearly distinguishable.

This is not specifically MWI speak. So when are pointer states clearly distinguished?

When their in product is neglegible.

 

[koantum]

When their in product is neglegible.

I expected this answer. So what are the conditions of negligibility?

 

[marlon]

However (although this is not really about physics, but about the calculational procedure), consider now that you draw an arbitrary cut across the diagram, which is now split into two disjunct pieces A and B, in such a way that all of the In lines are in A, and all of the Out lines are in B.
You cut through a set of virtual particle lines in doing so. Now, my claim is that if you add up all the energy values for all these virtual particles (by picking just any value for all the free loop momenta over which you have to integrate to calculate the Feynman graph), that this algebraic sum of energies "flowing out of A" (and hence "flowing into B") is exactly equal to the energy value of the In states (= energy of the Out states).

Well, to make my point more clear. Let's look at the Feynman diagram of neutron beta decay : A neutron (two down quarks and one up) disappears and is replaced by a proton (two up quarks and one down), an electron, and an anti-electron neutrino.

The Standard Model will tell us that one down quark disappears in this process while an up quark and a VIRTUAL W boson is produced. The W boson then decays to produce an electron and an anti-electron type neutrino. So this is the Feynman diagram i want to talk about.

Now, let's look at the virtual W boson. The mass-energy difference between a neutron and a proton is very much less than the mass-energy of a W boson. This would imply that the W boson cannot exist. Ofcourse, it does...So what do we do with this apparent contradiction ? Well, the answer is in the fact that the W boson is VIRTUAL. Now, if you would compare the energy of the W boson to that of the initial and final states (neutron and proton), energyconservation is NOT conserved. This is what i wanted to say. Ofcourse, when comparing initial and final energies, there is no problem. The violation only appear when we comape initial states with intermediate states or intermediate states with final states. That's what i meant by saying "in between vertex points"

regards
marlon

 

[vanesch]

Now, let's look at the virtual W boson. The mass-energy difference between a neutron and a proton is very much less than the mass-energy of a W boson. This would imply that the W boson cannot exist. Ofcourse, it does...So what do we do with this apparent contradiction ? Well, the answer is in the fact that the W boson is VIRTUAL. Now, if you would compare the energy of the W boson to that of the initial and final states (neutron and proton), energyconservation is NOT conserved. This is what i wanted to say.

But the "mass-energy" of your *virtual* W boson is NOT c^2 times the mass listed in the PDG files for the (real) W boson. That's exactly what it means to be off shell, that E^2 - p^2 must NOT be equal to m^2 (for c=1).
If, in your diagram, you calculate the 4-th component of the energy-momentum vector (E) of your virtual W boson, you will find that it is exactly equal to the sum of the energy of your up and down quark. As such, the total energy of this virtual W boson is MUCH LOWER than m c^2.
No REAL W boson can exist with such a low total energy, because for a REAL W boson, we have that E^2 = p^2 + m^2 (the on-shell condition), so we have that E > m. But this condition is not valid for a virtual boson, and E can have any value, also much lower than m. And E will have exactly the right value for energy to be conserved:

4th component of (real) quark u + (real) quark d = 4th component of (virtual) W boson = 4th component of final states. So there IS conservation of energy, no ?

EDIT: anti-up quark instead of up quark...

 

[George Jones]

As vanesch has emphasized, we are just bantering words back and forth about a perturbation calculation. Sometimes it helps if the words give us a physical picture in our minds.

In your example, another way of thinking about what happens between the vertices is to say that energy is conserved, which means that the "virtual" W does not have the same (rest) mass as a "real" W. To obsreve directly "real" W's requires large amounts of energy.

I think this is the viewpoint of Patrick^2, and, though I am no expert, it is a viewpoint that I think is becoming quite common these days. For example, take this quote form Ticiatti's field theory text:

"The activity of the \psi field represented by this line is not a particle state since q^2 is not required to be m^2. It is called a virtual particle, though it really has nothing to do with particles. It represents unobservable, transient behavior of the field which gives rise to scattering and is the quantum field theory version of a force."

Regards,
George

 

[vanesch]

As vanesch has emphasized, we are just bantering words back and forth about a perturbation calculation. Sometimes it helps if the words give us a physical picture in our minds.

In your example, another way of thinking about what happens between the vertices is to say that energy is conserved, which means that the "virtual" W does not have the same (rest) mass as a "real" W. To obsreve directly "real" W's requires large amounts of energy.

I think this is the viewpoint of Patrick^2, and, though I am no expert, it is a viewpoint that I think is becoming quite common these days. For example, take this quote form Ticiatti's field theory text:

"The activity of the \psi field represented by this line is not a particle state since q^2 is not required to be m^2. It is called a virtual particle, though it really has nothing to do with particles. It represents unobservable, transient behavior of the field which gives rise to scattering and is the quantum field theory version of a force."

Regards,
George

Indeed, that's exactly what I meant.

 

[marlon]

In your example, another way of thinking about what happens between the vertices is to say that energy is conserved, which means that the "virtual" W does not have the same (rest) mass as a "real" W. To obsreve directly "real" W's requires large amounts of energy.

Yes, i indeed agree with this and i do get the point. I also realize that both Vanesch as well as nrqed are saying this. My point is however that total energyconservation is defined based upon final and initial states. Thus, when one wants to talk about W boson in this context, one automatically refers to the real W bosons. It is this point that i was trying to outline : the (virtual) W bosons in beta decay are off mass shell just because the real W bosons could never be used in this example of beta decay.

However, i do realize that this is getting a semantical discussion and in the long run i very well understand what Vanesch is trying to say. So as a summary let's conclude this : "the (virtual) W bosons in beta decay are off mass shell just because the real W bosons could never be used in this example of beta decay. These virtual W bosons can exist thanks to the HUP and therefore they are short-lived just to do their interaction."


marlon

 

[vanesch]

Yes, i indeed agree with this and i do get the point. I also realize that both Vanesch as well as nrqed are saying this. My point is however that total energyconservation is defined based upon final and initial states.

Sure, I agree with this, and I'd even say that *it doesn't matter* whether, in a Feynman graph, there is "calculational conservation of energy" in the intermediate lines, because it is a *calculational tool* in order to calculate a series expansion approximation and not something physical. Of course, conserving energy in each step of the calculation will guarantee you conservation between initial and final states, so it is *useful* to require this.

Only, in the standard way of calculating a Feynman graph, it happens to be so that there is this conservation (and if you happen to find your own different way of calculating such a diagram, where it is not conserved, then that wouldn't even be a problem).

 

[reilly]

Yes, i indeed agree with this and i do get the point. I also realize that both Vanesch as well as nrqed are saying this. My point is however that total energyconservation is defined based upon final and initial states. Thus, when one wants to talk about W boson in this context, one automatically refers to the real W bosons. It is this point that i was trying to outline : the (virtual) W bosons in beta decay are off mass shell just because the real W bosons could never be used in this example of beta decay.

However, i do realize that this is getting a semantical discussion and in the long run i very well understand what Vanesch is trying to say. So as a summary let's conclude this : "the (virtual) W bosons in beta decay are off mass shell just because the real W bosons could never be used in this example of beta decay. These virtual W bosons can exist thanks to the HUP and therefore they are short-lived just to do their interaction."


marlon

First, energy conservation, when valid, applies to a system throughout it's lifetime, not just for beginning and final states. (See, for example, Noether's Thrm.)

Second: Feynman diagrams represent terms in a infinite perturbation series. It is the series sum (if it exists) that preserves unitarity, and thereby forces the requirement of all possible itermediate states -- sometime, conveniently or not, termed virtual states or virtual particles -- just as is the case with, say NR Coulomb scattering, the exact scattering states are superpositions of plane waves of many different free Hamiltonian energies. But, of course, a free particle is not an eigenstate of the full Hamiltonian, hence said free particle is not in a well defined energy state.

With my very cynical hat on, I'll say, once again, 'virtual' particles are a convenient conceptual fiction -- a perusal of the history of QFT will flesh-out this notion. They are useful to the theoretician, as long as she does not take them too seriously.

And, as can be seen from many books, cf. Goldberger&Watson Collision Theory, Blatt and Weiskopf's Nuclear Theory -- talks about a virtual n-p state, and the loose nature of the use of "virtual" in physics language, similar discussions about "virtual" can be found in Mott and Massey's Theory of Atomic Collisions (1933). The idea of "virtual" has been around for a very long time. To be fair, I'll note that in the early days, virtual states often meant resonant or unstable states. Somehow the idea of virtual has, in my opinion. probably outrun its usefulness -- too many folks are too confused about the idea to make it safe to use. Better to talk about internal line in a diagram, period.

The problem is strictly one of language not of physics. People sometimes take figurative language too seriously, and then spend tortuous energy trying to understand that which cannot be well understood. (How high is up?)

Regards,
Reilly

 

[marlon]

First, energy conservation, when valid, applies to a system throughout it's lifetime, not just for beginning and final states. (See, for example, Noether's Thrm.)

Yep that is true. I don't think anybody tries to counter that. The question is ofcourse about the "when valid"-part.

With my very cynical hat on, I'll say, once again, 'virtual' particles are a convenient conceptual fiction -- a perusal of the history of QFT will flesh-out this notion. They are useful to the theoretician, as long as she does not take them too seriously.

Well, obviously this is your opinion on this matter and i respect it. I just do not agree with what you are saying but there is no point in starting a debate on this.

regards
marlon

 

[nrqed]

Yep that is true. I don't think anybody tries to counter that. The question is ofcourse about the "when valid"-part.


Well, obviously this is your opinion on this matter and i respect it. I just do not agree with what you are saying but there is no point in starting a debate on this.

regards
marlon

I am still hoping that you will pick a specific Feynman diagram (a weak interaction at tree level with a virtual W boson in the intermediate state if you want, or anything else) so that we can write down the expression and show that the total energy is conserved even in the intermediate state.

Let me consider the electron-positron scattering diagram in the s channel at tree level. The photon in the intermediate state is off-shell of course. But the energy (as well as the three-momentum) of the off-shell photon is euqal to the energy (and three-momentum) of the initial electron and positron, q_{\gamma}= P_{electron} + P_{positron} (where the P's are the physical four-momenta of the electron and positron). Therefore energy is conserved. QED (no pun intended)



Regards

Patrick

 

[marlon]

I am still hoping that you will pick a specific Feynman diagram (a weak interaction at tree level with a virtual W boson in the intermediate state if you want, or anything else) so that we can write down the expression and show that the total energy is conserved even in the intermediate state.

Ok. How about we look at the Feynman diagram of the neutron beta decay. This is the example that i have used several times to illustrate my point. The W boson NEEDS to be off mass shell because a real W boson is far to heavy. As we have agreed before, energyconservation is defined for initial and final states, for real particles, etc etc... For energyconservation to be respected, the W boson should be real yet this is impossible. Hence the apparent violation. Now ofcourse, in the beta decay energy is indeed conserved throughout the entire diagram (thanks to the W boson being virtual, not real). I know that that is exactly what you want to say and it has been my fault not to agree with you on that directly. I should have been more clear on that.

Let me consider the electron-positron scattering diagram in the s channel at tree level. The photon in the intermediate state is off-shell of course. But the energy (as well as the three-momentum) of the off-shell photon is euqal to the energy (and three-momentum) of the initial electron and positron, q_{\gamma}= P_{electron} + P_{positron} (where the P's are the physical four-momenta of the electron and positron). Therefore energy is conserved. QED (no pun intended)



Regards

Patrick

Yes that is true. I do see your point and i hope that i have made my point more clear. Again, i think most of the confusion/discussion was actually caused by the fact that i did not state my point clear enough. I hope that's solved now.

regards
marlon

 

[nrqed]

Ok. How about we look at the Feynman diagram of the neutron beta decay. This is the example that i have used several times to illustrate my point. The W boson NEEDS to be off mass shell because a real W boson is far to heavy. As we have agreed before, energyconservation is defined for initial and final states, for real particles, etc etc... For energyconservation to be respected, the W boson should be real yet this is impossible. Hence the apparent violation. Now ofcourse, in the beta decay energy is indeed conserved throughout the entire diagram (thanks to the W boson being virtual, not real). I know that that is exactly what you want to say and it has been my fault not to agree with you on that directly. I should have been more clear on that.

It still seems to me that we are not saying the same thing. Well, it seems like some statements agree with me and then other statements are in disagreement. I agree that the energy is conserved throughout the entire diagram thanks to the W being off-shell, but then you seem to say the opposite in the next sentences so I am a bit confused.

I am saying that even for the virtual (i.e. off-shell) states energy is conserved. My example of the s-channel of the electron-positron scattering is such an example. Energy of the initial state = energy of the virtual photon in the intermediate state = energy of the final state. Don't we agree on this?

This is just saying that the total energy of the initial state (electron + positron) = energy of the virtual photon in the intermediate state = total energy of the electron and positron in the final state. This is what I mean by energy conservation in the intermediate state!


Also, you say that for energy to be conserved, the W would have to be real. I would say exactly the opposite. If the W was real, energy would not be conserved. Imposing energy conservation forces the W to be off-shell (i.e. virtual).

If you say that the W is off-shell and that the energy is not conserved, then how do you calculate the energy of the W? There is no way to calculate it so the whole Feynman diagram is undefined! There would be no way to calculate a cross section or anything if there is no rule to get the four-momentum of the virtual W (one cannot use P^2 = M^2 c^4 since it is not on-shell). I am saying that the way the energy of the virtual W is determined is by imposing conservation of energy!


Regards

PAtrick

 

[pmb_phy]

Yes, this is exactly the paper that koantum talked about and which I commented. It is apparently true that the interaction between the measurement apparatus and the system can be arbitrary short, I didn't realize this. But this was not the claim. The claim was that it takes time dT to have the result available.
In the Bohm example, in a very short time, the px momentum (and eventually, the py momentum) energy equivalent has been transferred to the condenser state...
But now we must still measure the energy of the condenser !
While it is true (and I ignored this) that you can now consider the condensers as independent from the system under measurement, you will now have a new energy measurement to perform, so the problem starts all over again.
If you accept to talk about the quantum states of the condensers, after the interaction with the particle, they are now entangled with the particle states, but their own states are not yet sufficiently separated for them to be "pointer states". They will need to evolve during at least a time dT, before they are "grossly orthogonal" for different energy input states which are of the order of dE.

Nevertheless (and that's what I learned), the advantage of this is that you can have rapidly successing energy measurements on a same system.

However, for instance, you do not have the result available in very short time (so that you could take a decision based upon that to act on the system, for instance).

The article I quoted shows, not that you can measure energy in an instant, but that dE*dT > h can be violated.
This expression is not a postulate of quantum mechanics. It comes about in certain systems in which the relation is true, but the relation is not valid in all possible situations and not as interpreted as I believe that you are interpreting it. BTW - I don't know what you're talking about regarding this capacitor example you gavel. Where did it come into play here?

Pete

 

[cgyyh]

I found it somewhere,maybe useful.

Do they violate energy conservation?
We are really using the quantum-mechanical approximation method known as perturbation theory. In perturbation theory, systems can go through intermediate "virtual states" that normally have energies different from that of the initial and final states. This is because of another uncertainty principle, which relates time and energy.
In the pictured example, we 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 (except for recoil) 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.
Some descriptions of this phenomenon instead say that the energy of the system becomes uncertain for a short period of time, that energy is somehow "borrowed" for a brief interval. This is just another way of talking about the same mathematics. However, it obscures the fact that all this talk of virtual states is just an approximation to quantum mechanics, in which energy is conserved at all times. The way I've described it also corresponds to the usual way of talking about Feynman diagrams, in which energy is conserved, but virtual particles can carry amounts of energy not normally allowed by the laws of motion.