- #1

- 91

- 14

You should upgrade or use an alternative browser.

- I
- Thread starter hideelo
- Start date

- #1

- 91

- 14

- #2

ShayanJ

Gold Member

- 2,809

- 604

Some say that the uncertainty relation between energy and time is only a myth. And to some extent its true because it can't be thought of in the same way you think about the uncertainty relation for position and momentum. But there are some situations where it is true in a different meaning. You can check "Modern Quantum Mechanics" by Sakurai for such different meanings of this relation but I don't remember which section.

And at the end, there is the famous saying of Griffiths:

"In general, when you hear a physicist invoke the uncertainty principle, keep a hand on your wallet."

- #3

- 91

- 14

I can't speak to the QFT part of your response, because I don't know QFT.

Some say that the uncertainty relation between energy and time is only a myth. And to some extent its true because it can't be thought of in the same way you think about the uncertainty relation for position and momentum. But there are some situations where it is true in a different meaning. You can check "Modern Quantum Mechanics" by Sakurai for such different meanings of this relation but I don't remember which section.

And at the end, there is the famous saying of Griffiths:

"In general, when you hear a physicist invoke the uncertainty principle, keep a hand on your wallet."

I know that the derivation of x and p uncertainties doesn't exactly transfer over identically to the t and E uncertainty. But the pattern is so simple and clean, it seems like there should be some sense in which meaningful.

I see no reason not to have a time operator which would just be the projection into the t (or 0) coordinate in Minkowski space, it's even hermitian. The issue is in considering E its fourier transform.

- #4

ShayanJ

Gold Member

- 2,809

- 604

In non-Relativistic QM, Minkowski spacetime is not used, but a three dimensional space with a Newtonian absolute time.I see no reason not to have a time operator which would just be the projection into the t (or 0) coordinate in Minkowski space, it's even hermitian. The issue is in considering E its fourier transform.

- #5

- 91

- 14

You're correct of course, but am I to conclude then that this is just a coincidence?In non-Relativistic QM, Minkowski spacetime is not used! Its the three dimensional space with a Newtonian absolute time.

- #6

ShayanJ

Gold Member

- 2,809

- 604

The two uncertainty relationsYou're correct of course, but am I to conclude then that this is just a coincidence?

- #7

- 91

- 14

I understand that, the coincidence I was referring to was that each component of the 4 momentum has an uncertainty relation (of sorts) with its corresponding component of 4 position. Notwithstanding their differences, notwithstanding that for three components we have quantum operators and for the fourth we don't. Notwithstanding all that, we can still write them all asThe two uncertainty relationsare noton the equal footing. So there is nothing for you to consider as a coincidence or not!

Δχ^μ * Δp_μ >= h/2

Might be meaningless coincidence , might be worth thinking about.

- #8

- 12,580

- 4,945

This is not meaningful in standard formulation of QM, but makes sense in a non-standard one:

https://arxiv.org/abs/0811.1905

- #9

radium

Science Advisor

Education Advisor

- 765

- 243

The uncertainty principle can be thought of as a property of Fourier transforms and the bandwidth theorem, which is a kind of uncertainty relation. So while t is not an operator, it is the conjugate observable to energy in the Fourier transform. This also can be used to explained momentum position uncertainty principle as they are also conjugate variables.

- #10

Physics Footnotes

Gold Member

- 37

- 81

Although Schrodinger's equation can be made relativistic by using e.g. the Dirac Hamiltonian, it is not *manifestly *relativistic. An important consequence of this is that although you know that deep down space and time are behaving symmetrically, the formalism does not reveal that symmetry explicitly and can therefore be very misleading.

Heisenberg's Uncertainty Principle (HUP) is a perfect example of this nefarious state of affairs because it is proved by exploiting the asymmetric character of the formalism (namely by the use of operators on a Hilbert Space of functions defined over the*spatial *configuration of the system).

More specifically, although for each time ##t##, ##\psi_t(x)## belongs to a Hilbert Space, the space-time functions ##\Psi(x,t)## do*not*. This means that the operator theory (i.e. Hilbert Space functional analysis) available to us in the traditional form of the HUP is no longer meaningful when we introduce operators (like ##T## and ##E##) acting on the ##\Psi(x,t)## instead of the ##\psi_t(x)##.

There may well be a Lorentz covariant uncertainty relation defined over space-time in QM, but we will not be able to establish it (at least not rigorously) in the language of orthodox Hilbert Space quantum mechanics.

Heisenberg's Uncertainty Principle (HUP) is a perfect example of this nefarious state of affairs because it is proved by exploiting the asymmetric character of the formalism (namely by the use of operators on a Hilbert Space of functions defined over the

More specifically, although for each time ##t##, ##\psi_t(x)## belongs to a Hilbert Space, the space-time functions ##\Psi(x,t)## do

There may well be a Lorentz covariant uncertainty relation defined over space-time in QM, but we will not be able to establish it (at least not rigorously) in the language of orthodox Hilbert Space quantum mechanics.

Last edited:

- #11

bhobba

Mentor

- 10,061

- 3,162

Although Schrodinger's equation can be made relativistic by using e.g. the Dirac Hamiltonian, it is notmanifestlyrelativistic

How true that is.

Its based on the Galilean transformations as carefully explained in Ballentine. The Galilean transformation are explicitly not relativistic.

In classical mechanics this is explained, admittedly in Landau's usual terse style, in his very beautiful book on Mechanics:

https://www.amazon.com/dp/0750628960/?tag=pfamazon01-20

However for some reason it's not usually found in textbooks - don't know why because its very important.

To merge QM with relativity you need QFT where locality is a bit more subtle than in SR being part of the so called cluster decomposition property:

https://www.physicsforums.com/threads/cluster-decomposition-in-qft.547574/.

Its an interesting discussion relating that to Bells theorem but needs a new thread.

Thanks

Bill

Last edited by a moderator:

- #12

- 80

- 5

There is too much beauty in these three relations, relating rotational invariance, time reversal and translational symmetry. Should they be interpreted in terms of "fluctuations" maybe? since x and p are standard operators in QM, and E (the hamiltonian) too, but what about time¿ Can time at some scale fluctuate? (Like a quantum theory of gravity would suggest)

- #13

- 20,123

- 10,860

http://arxiv.org/abs/quant-ph/0510234

http://arxiv.org/abs/quant-ph/0307069

The "natural" uncertainty relation for angular momenta are

$$\delta l_j \delta l_k \geq \frac{\hbar}{2} |\epsilon_{jkl} l_l|.$$

The last uncertainty relation should read

$$\Delta x \Delta p \geq \frac{\hbar}{2},$$

and the 2nd one has to be derived with great care in the context of the proper meaning of ##\Delta E## and ##\Delta t## since time is not an observable in quantum theory, and the general Heisenberg-Robertson uncertainty relation

$$\Delta A \Delta B \geq \frac{1}{2} |\langle [\hat{A},\hat{B}]\rangle|$$

is not applicable.

For a thorough discussion of this issue, see

http://arxiv.org/abs/quant-ph/0105049v3

Share: