5
$\begingroup$

Usually, the energy-time analogue of the position-momentum uncertainty relation is quoted as $\Delta E \Delta t \geq \frac{h}{4 \pi}$. This has interpretational issues and such. But, with a suitable definition of $\Delta t$, this can be derived. See The energy-time uncertainty relation for instance. The interpretation for $\Delta t$ is then the amount of time needed for an arbitrary observable to change by one standard deviation.

When popular science writers try to heuristically explain Hawking radiation, they mention the same uncertainty relation and say that for time scales $\Delta t \sim \frac{1}{\Delta E}$, particles with energy $\Delta E$ can be created. This happens even in vacuum. Can someone please explain how the second relation follows (heuristically, at least) from the first?

Improve this question
$\endgroup$

3 Answers 3

Reset to default
6
$\begingroup$

The Heisenberg uncertainty principle applies only to operators satisfying canonical commutation rules. This is the case for corresponding components of position and momentum operators, but not for the energy operator (Hamiltonian), which has no associated conjugate partner. (Conjugate pairs of selfadjoint operators necessarily have unbounded spectrum, while a good Hamiltonian must be bounded below. This argument is due to
W. Pauli, Die allgemeinen Prinzipien der Wellenmechanik, Handbuch der Physik (S. Flügge, ed.), Vol V/1, p. 60, Springer, Berlin 1958.
English translation: The general principles of quantum mechanics, p. 63, Springer, Berlin 1980.

Time measurements do not need a time operator, but are captured well by a positive operator-valued measure (POVM) for the time observable modeling properties of the measuring clock.

The problem of extending Hamiltonian mechanics to include a time operator, and to interpret a time-energy uncertainty relation, first posited (without clear formal discussion) in the early days of quantum mechanics, has a large associated literature; a survey article by Busch
http://lanl.arxiv.org/abs/quant-ph/0105049
carefully reviews the literature up to the year 2000. (The book in which Busch's survey appeared discusses related topics.) There is no natural operator solution in a Hilbert space setting, as shown by Pauli's argument mentioned above.

However, a well-defined time-energy uncertainty relation resembling that of Heisenberg has been rigorously established in the context of statistical mechanics by Gilmore, see
http://einstein.physics.drexel.edu/~bob/Thermodynamics/p1985_3237_1.pdf

All this has nothing at all to do with any short-time creation of energy from the vacuum. The latter is a popular misinterpretation of this relation. See the discussion in Creation of particle anti-particle pairs.

Improve this answer
$\endgroup$
1
$\begingroup$

The HUP has a rigorous position in the mathematical formulation of quantum mechanics and quantum field theory. It appears where the commutator of two observables is non zero, which is a mathematical relationship imposed on the eigenstates of the system. The matrix mechanics paragraph in the link describes it. The wave mechanics paragraph in this entry . is maybe simpler to understand . The discussion is about the momentum position uncertainty, but it also holds for time and energy. It describes how a standard deviation can be defined from the quantum mechanical wavefunctions.

The values of E and t are expectation values of the eigenstates whether of a vacuum state or hole horizon. i.e. there exists a wavefunction on which the expectation value can be calculated and therefor a standard deviation can be defined for E and t .

Improve this answer
$\endgroup$
10
  • 1
    $\begingroup$ If I understand correctly, the similarity between momentum-position uncertainty and energy-time uncertainty isn't that straightforward. This is because time isn't an operator like momentum or position. It is an external parameter. So, a standard deviation for t doesn't make sense. Which is why one has to interpret what one means by $\Delta t$ in the uncertainty relation. Interpretations similar to the one I have referred to are the only ones I have seen that sound reasonable to me. But, I don't understand how that is related to what I mention in the second paragraph. Thanks. $\endgroup$
    – contrariwise
    Commented Oct 27, 2012 at 5:09
  • 1
    $\begingroup$ You're quite correct that $t$ isn't usually treated as an operator so there isn't a direct analogy between the two forms of the HUP. Over the years I've seen many papers on this subject but I don't think there's any explanation that triggers your "oh yes, of course" detectors. Most of us just shrug and accept that it works. $\endgroup$
    – John Rennie
    Commented Oct 27, 2012 at 6:37
  • 1
    $\begingroup$ you are not clear with what you mean by "the second relation": as you have written it is an algebraic manipulation keeping the inequalities. If you mean how could particles be presumed within the delta(E) it is another story and has to do with creation and annihilation operators in quantum field theory. The delta(E) just says that there exists an uncertain energy, QFT creates particle pairs within that uncertainty. $\endgroup$
    – anna v
    Commented Oct 27, 2012 at 6:46
  • $\begingroup$ continued: that these virtual particles are "real" comes from experimental verification of quantities computed by QFT. $\endgroup$
    – anna v
    Commented Oct 27, 2012 at 6:47
  • $\begingroup$ @annv, I understand how particles are created. My question is about $\Delta E$ itself. I have similar-looking relations in the first and the second paragraph. But, the interpretation for $\Delta t$ in the first is very different from that in the second - at least at first glance. I am basically asking how they can be reconciled. Thanks. $\endgroup$
    – contrariwise
    Commented Oct 27, 2012 at 7:05
0
$\begingroup$

There are not interpretational issues regarding $\Delta E \Delta \tau_Q \geq {h}/{4 \pi}$. Here

$$\Delta \tau_Q \equiv \Delta_Q \left( \frac{\mathrm{d} \langle Q \rangle}{\mathrm{d} t} \right)^{-1}$$

is a characteristic time for variability of phenomena in this state. Note that $\Delta \tau_Q$ depends on both the particular dynamical variable $Q$ and the state, and that it may vary with time.

Consider that $Q$ is the projection operator for a given initial state $\Psi(0)$. Then $\langle Q \rangle$ gives the survival probability of the initial state. If you want know the shortest time at which the survival probability drops to half-life, $\tau_{1/2}$, just substitute the survival probability and integrate the above expression. The result is

$$ \Delta \tau_{1/2} \geq \frac{h}{8 \Delta_E}$$

This $\Delta \tau_{1/2}$ gives the half-life of an unstable state $\Psi(0)$ (e.g. one with a virtual pair particle-antiparticle) with energy spread $\Delta_E $. By symmetry considerations this is the same scale of time for the spontaneous creation of the same pair. Notice that the above inequality can be approximated by $\Delta \tau_{1/2} \sim 1/ \Delta_E$.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.