Time-energy uncertainty and virtual particles

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SUMMARY

The discussion centers on the time-energy uncertainty relation and its implications for virtual particles. Participants argue that the conventional inequality ΔE Δt ≥ h should not apply to energy and time, as time is not an operator and lacks a conjugate variable. Instead, they propose that ΔE Δt = 0 is more accurate, emphasizing that time is an independent variable unaffected by particle properties. The conversation also critiques the misuse of Fourier transforms in deriving uncertainty relations, clarifying that time should not be treated as a variable subject to statistical scatter.

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  • Understanding of quantum mechanics principles, particularly uncertainty relations.
  • Familiarity with Fourier transforms and their applications in physics.
  • Knowledge of non-relativistic wave mechanics and operator theory.
  • Basic concepts of virtual particles and their role in quantum field theory.
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  • Study the implications of the uncertainty principle in quantum mechanics.
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geoduck
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I have seen an argument that the time-energy uncertainty relation allows heavy particles to exist for short periods of time as virtual particles:

(ΔE Δt)>h

Δt>h/ΔE

However, shouldn't the inequality be the other way:

Δt<h/ΔE

to suggest that a heavy particle with large ΔE can exist for short time?
 
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Hello geoduck,

good question! In fact, a naive attempt at generalization of the relation

<br /> \Delta p_x \Delta x &gt;= \hbar/2 *<br />

would be

<br /> \Delta E \Delta t &gt;= \hbar/2, **<br />

which would be in fact the exact opposite of what one would want if he wanted to explain creation of short-lived particles. You are right that the opposite sign would make much more sense.

However, I would like to say that neither of these relations for energy and time
make much sense. The situation is entirely different from that of momentum and position.

The derivation of * cannot be directly applied to derive ** for energy and time, because the time is not an operator and there is no analogous commutation relation between the energy and time.

This is because in non-relativistic wave mechanics, coordinates and momenta of particles play very different role from that of time variable.

Coordinate and its conjugated momentum have associated non-commuting operators, but there is no such couple of conjugated operators for energy and time variables.

Time is just an independent variable that has nothing to do with any special particle. Although the energy can have nonzero standard deviation for some psi function, the time is the time, always with zero standard deviation. It is common for the whole description and is not influenced by anything. The statistics and averaging are only in particle properties.

So if we wanted to write the relation down anyway, we would have to write

<br /> \Delta E \Delta t = 0,<br />

because

<br /> \Delta t = 0.<br />


Some people like to use relations like \Delta x \Delta p \approx h to find numbers when they cannot or are lazy to really derive or calculate from the theory.
One has to be careful not to believe in such fairy-tales too much.
 
I thought all it took to establish Δa Δb>h for two quantities 'a' and 'b' is that they are Fourier transforms of each other. Then a wide Gaussian in 'a' will correspond to a thin Gaussian in 'b' and vice versa, which gives you the uncertainty principle.

Isn't what ψ(t)=a1 exp(-E1 t)+a2 exp(-E2 t)+...

saying is that E and t are Fourier transforms of each other?

I think you're absolutely right that t commutes with anything, so ΔE Δt=0 , but I'm a bit confused with the Fourier transform argument that I just put up which suggests ΔE Δt > h
 
The connection with Fourier transform is different. The uncertainty principle relates root mean square deviations of a function and its Fourier transform:

http://en.wikipedia.org/wiki/Fourier_Transform
(in 1/3 of the article)

In your example, you have a function \psi(x,t) of coordinate(s) x of a particle(s) and time:

<br /> \psi(x,t) = c_1 \Phi_1(x) e^{iE_1t/\hbar} + ...<br />

You can calculate its Fourier transform with respect to coordinate x (the new function will be a function of k) and then the r. m.s. deviations of both functions fulfil the uncertainty relation. This can be interpreted as relation between statistical scatters - r.m.s deviations of coordinate and momentum.

You can do the same with respect to time to obtain a function of \omega and again these obey the uncertainty inequality. This gives relation between degrees of localizations of both functions (it is impossible to localize both). However, now it does not makes sense to interpret it as statistical scatter of time. Time is parameter, which is always known precisely, not a variable that would have to be measured as a property of the system.

The difference lies in that the symbol x means " coordinate of the particle ", not " general coordinate in space ". It is better to use a different symbol for particle coordinate, for example r_x. On the other hand, the symbol t means " general coordinate in time ".
 
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