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I feel like I have a pretty good understanding of the uncertainty principle in quantum mechanics (the idea being that uncertainty arises from the failure of two operators to commute). however, energy-time uncertaintly doesn't seem to fit that framework, since time is not an operator.

does energy-time uncertainty exist in quantum mechanics, or does it only arise in quantum field theory, where it's really energy/momentum-spacetime uncertainty?

and what does energy-time uncertainty mean for practical purposes? something like "the better we know the energy, the less well we know when it happened"??

2007-05-16 05:11:12 · 4 answers · asked by Anonymous in Science & Mathematics Physics

4 answers

You're right to be a little edgy. Any time a physicist invokes the uncertainty principle in an argument, keep a hand on your wallet.

For practical purposes, Energy/time uncertainty means that you can borrow energy from the uncertainty principle to create a virtual particle, but only for a short period of time (given by the uncertainty relationship). So virtual particles that are almost on-shell (don't need to borrow much energy) can stick around for a while. Particles that are way off shell (don't have nearly enough real energy to exist) are very short-lived, so the forces they mediate have short ranges.

2007-05-16 05:26:00 · answer #1 · answered by Anonymous · 0 1

energy- time uncertainty, I have that every day. I am never quite sure that I will have the energy to complete the task in the allotted time.
So here is the short version of that equation:
em/t = a bm about every three or four days,
conversely: bm/too much time+ an EMT - a lot of cash=lol.

2007-05-17 00:01:19 · answer #2 · answered by Wayne B 1 · 0 0

Energy-time uncertainty follows from momentum-position uncertainty, since the product of momentum and position has the same dimensions as the product of energy and time.

2007-05-16 12:43:09 · answer #3 · answered by Anonymous · 0 1

Time _is operator. More acuurately time t is variable
and operator t-hat is operator.

Operator t-hat acts exactly as x-hat, i.e. if you have ψ(x,t), then
x-hat(ψ(x,t)) = x ψ(x,t), and
t-hat(ψ(x,t)) = t ψ(x,t).

Commutator is also easy:
i h-bar ∂ψ/∂t = H ψ(x,t),

[t, H] = [t, i h-bar ∂/∂t] = i hbar

2007-05-16 12:34:33 · answer #4 · answered by Alexander 6 · 0 0

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