# Tag Info

17

Heisenberg's Uncertainty Principle is, in essence, a consequence of two basic facts: de Broglie's relation between a particle's momentum and its associated wavelength, and a mathematical fact known as the bandwidth theorem, which states an equivalent uncertainty relation between a wave's position and its wavelength. The fundamental physical leap here is of ...

0

To begin with, entropy is a classical thermodynamics concept. Different statistical frameworks , assuming some postulates, can define an entropy. The basic formulation of entropy defined by statistical mechanics where kB is the Boltzmann constant, equal to 1.38065×10^−23 J K−1. The summation is over all the possible microstates of the system, and p_i ...

0

In quantum statistical mechanics entropy is not defined via the probability density of a single state but through the density matrix which talks about the "non-quantum uncertainty" over the states. This is the "probability density" over which all the entropy theorems are proven in the quantum world. That is, if we know the system to be in a sharp quantum ...

1

I try to measure the energy and the position of the system simultaneously The states with definite energy are not states with definite position so there is no particle state of both definite energy and definite position.

1

At the first look the question seemed very interesting, but later I found the mistake. You said you are measuring the position of the particle precisely. But how? You can tell that the particle is inside the well but you can not know the exact position of the particle. For more info read ...

2

If you are considering a system of a single particle in a potential well with infinitely high walls and with finite width, the energy operator is $H = \frac{p^2}{2m} + V(x)$ where $V(x)$ is the potential energy operator, vanishing inside the well and infinite outside it. Being that $\frac{p^2}{2m}$ does not commute with $x$, how are you saying that ...

1

Symplectic groups: Not directly. But symplectic manifolds (or even better: Poisson manifolds): Yes. The Poisson bracket $\{f,g\}_{PB}$ and the functions $f$, $g$ are the classical counterparts of the quantum commutator $[\hat{f},\hat{g}]$ and the operators $\hat{f}$, $\hat{g}$. The quantum commutator $[\hat{f},\hat{g}]$, in turn, gives rise to the Heisenberg ...

3

Yes, of course, symplectic groups describe generalized situations that reveal the uncertainty principle. The reason for the relationship is that the symplectic groups are defined by preserving an antisymmetric bilinear invariant, $$M A M^T = A$$ where $M$ is a matrix included into the symplectic group is the equation holds and $A$ is a non-singular ...

0

TL;DR: Feynman was talking about a gas, so he was correct: molecules of a gas are largely independent, they don't have any quantum zero point kinetic energy. Their mean (random) kinetic energy is temperature. If he had been talking about any other particles such as atoms in a solid he would not have been correct - they do have kinetic energy even at ...

6

Your question, I think, is best answered by separately addressing its many directly and indirectly implied questions. You are mixing up a lot of different notions together in your question as you've phrased it. So to clarify, I thought I would pull all of the pieces apart. To start: Do atoms have nonzero kinetic energy at absolute zero? Yes, they still ...

7

If carefully interpreted and converted to mathematics, the passages are not contradicting one another. The uncertainty principle guarantees that there is some zero-point energy that can't be eliminated (the second passage) – it is the energy of the ground state of the physical object (atom or a macroscopic piece of a material). On the other hand, the ...

0

I'm surprised that none of the other answers (from last year) mention interaction-free measurement. The way we know that the uncertainty principle is not a side effect of interactions during measurement is that it applies even when there is no interaction (and even when the measuring device is nowhere near any part of the system).

0

The time-energy uncertainty relation (and other time-"observable" uncertainty relations that can be constructed) is (considered) not to have same meaning as canonical uncertainty relations. Meaning uncertainty relations costructed from canonical dynamical variables/observables (in the Hamiltonian sense), like position and momentum, since time parameter is ...

1

A paper by Colosi and Rovelli that I read may have another interesting answer to the question. Consider space partitioned off (arbitrarily) into two regions $R_1$ and $R_2$. The global Fock vacuum is found by solving for the global Hamiltonian which includes the possible correlations between the regions. A local detector in a region $R_1$ is governed by a ...

0

The explanation is really very simple to understand intuitively, and very beautiful. Imagine that a particle an uncertainity in its velocity $v$ of $\delta v$. Suppose at $t=0$ we have $x=x_{0}$. After $t=T$, the location of the particle will be given by the range $(x_{0}+Tv-T\delta v,x_{0}+Tv+T\delta v)$, because we dont know the exact velocity the ...

Top 50 recent answers are included