- #1

geoduck

- 258

- 2

(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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- Thread starter geoduck
- Start date

In summary: The uncertainty principle is a relation between two functions that are correlated but have different standard deviations. In the example given, it states that the two functions have different r.m.s. deviations.

- #1

geoduck

- 258

- 2

(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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- #2

Jano L.

Gold Member

- 1,333

- 75

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

[tex]

\Delta p_x \Delta x >= \hbar/2 *

[/tex]

would be

[tex]

\Delta E \Delta t >= \hbar/2, **

[/tex]

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

[tex]

\Delta E \Delta t = 0,

[/tex]

because

[tex]

\Delta t = 0.

[/tex]

Some people like to use relations like [itex]\Delta x \Delta p \approx h [/itex] 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.

- #3

geoduck

- 258

- 2

Isn't what ψ(t)=a

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

- #4

Jano L.

Gold Member

- 1,333

- 75

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 [itex]\psi(x,t)[/itex] of coordinate(s) x of a particle(s) and time:

[tex]

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

[/tex]

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 [itex]\omega[/itex] 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 [itex] r_x [/itex]. On the other hand, the symbol t means " general coordinate in time ".

http://en.wikipedia.org/wiki/Fourier_Transform

(in 1/3 of the article)

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

[tex]

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

[/tex]

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 [itex]\omega[/itex] 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 [itex] r_x [/itex]. On the other hand, the symbol t means " general coordinate in time ".

Last edited:

The time-energy uncertainty principle is a fundamental principle in quantum mechanics that states that it is impossible to know the exact energy of a quantum system at a given time. This is because the act of measuring the energy of a system will disturb it, making it impossible to simultaneously measure both the energy and the time at which the measurement was taken.

The time-energy uncertainty principle allows for the creation of virtual particles, which are particles that appear and disappear in very short time intervals. This is due to the uncertainty in energy levels, which allows for particles to temporarily borrow energy from the vacuum in order to exist. Virtual particles play a crucial role in many physical phenomena, such as the Casimir effect and Hawking radiation.

No, virtual particles cannot be directly observed because they exist for such a short amount of time and do not leave a lasting trace. However, their effects can be observed and measured indirectly through various physical phenomena.

Virtual particles play a significant role in modern physics and our understanding of the universe. They help explain various phenomena that cannot be explained by classical physics, such as the stability of atoms, the behavior of subatomic particles, and the behavior of the vacuum itself. Additionally, virtual particles are important in the development of theories such as quantum field theory and the Standard Model.

The time-energy uncertainty principle is a fundamental principle of quantum mechanics and has been extensively tested and verified through experiments. It is considered a fundamental law of nature and cannot be violated. However, there are some theories that suggest modifications to the principle in certain extreme conditions, but these are still under debate and require further research and evidence.

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