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W and Z bosons have large masses, but they can still come into existence for short periods of time by "borrowing" energy from the vacuum.

When this energy is borrowed and the particles come into existence, does that mean there's less vacuum energy available to be borrowed by other temporary particles?

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...they can still come into existence for short periods of time by "borrowing" energy from the vacuum.

You are being misled by a bad science reporting metaphor gone haywire.

The idea of "borrowing" energy from a vacuum reservoir with fluctuating monetary liquidity is bound to steer you into this very misconception. When people discuss such Ws and Zs, what they have in mind is Feynman diagrams, formal mathematical expressions in covariant perturbation theory, a computational approximation scheme for QFT amplitudes. In these diagrams (and the metaphorical imagery accruing to them, when people talk about "the W" or the "the Z") energy and momentum is strictly conserved, so there is no borrowing in a tree diagram where each external line has a well-specified energy and momentum. enter image description here

It is just that the contribution of such a diagram is vastly suppressed at low energies, since the energy-momentum balance of these internal lines is not satisfied for particles, and that "particle" is said to not be "real", but instead "virtual".

At very high energies, the energy-momentum constraint is satisfied, and real Ws and Zs are produced with vastly enhanced probabilities; being so heavy and unstable, they decay almost instantly and are studied in colliders. Nobody "borrowed" any energy or momentum: the projectile and target particles in a collision had enough energy to produce real Ws and Zs.

At higher orders in perturbation theory, one has "loop" diagrams, with closed loops involving Ws and Zs, and there are integrals over all momenta and energies, a feature of sums-over-all-intermediate-states of perturbation theory. Again, at every vertex, energy and momentum is conserved, but you might visualize a closed loop as a time-travel banking system from hell, continuously creating and absorbing circulating funds. This interferes with availability of such integrated variables for other loops or variables in no way -- beyond strict conservation of energy and momentum at every vertex.

These integrals get the lion's share of their contributions at energies and momenta close to such obeying the above reality constraint, "near the mass shell" in technical lingo. You might dream of these internal particles coming into reality for such values of the dummy E,p variables integrated over.

But they have no impact on the independent integrations in other loops, independent internal circulation machines, provided the strict energy momentum conservation laws at every juncture (vertex) are strictly satisfied.

There is never any "borrowing": just partitions of incompressible flows.

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It only looks like borrowed energy because you are using the free particle Hamiltonian. If you use the full Hamiltonian that includes interactions, you would see that energy is always conserved. In the asymptotic limit, $t\to\pm\infty$, the two Hamiltonians coincide, but the virtual particles will disappear before then after a time $\Delta t\sim \hbar/\Delta E$, where $\Delta E$ is how much energy you thought was "borrowed" using the free Hamiltonian.

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  • $\begingroup$ Could you elaborate this perspective / point to somewhere that this is worked out in more detail? This is a nice perspective on the energy-time HUP that I haven't seen before. $\endgroup$
    – jacob1729
    Commented Aug 25, 2020 at 14:56
  • $\begingroup$ It's just time-dependent perturbation theory. The transition matrix element is $-i\int\!dt\,\langle f |V_I(t)|i\rangle$, where $V_I(t)$ is the perturbation to the Hamiltonian in the interaction picture, $e^{iH_0 t}Ve^{-iH_0 t}$. Take $|i\rangle$ and $|f\rangle$ to be energy eigenstates of $H_0$, $V$ constant, and integrate from $-\Delta t/2$ to $\Delta t/2$, and you get the result I said. But it's not really an uncertainty principle because the interpretation is completely different. $\endgroup$ Commented Aug 25, 2020 at 17:32
  • $\begingroup$ In perturbative QFT, all states of definite particle number are eigenstates of $H_0$. $\endgroup$ Commented Aug 25, 2020 at 17:38

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