Hot answers tagged uncertainty-principle
23
The question asks about the time dependence of the function
$$f(t) := \langle\psi(t)|(\Delta \hat{x})^2|\psi(t)\rangle
\langle\psi(t)|(\Delta \hat{p})^2|\psi(t)\rangle,$$
where
$$\Delta \hat{x} := \hat{x} - \langle\psi(t)|\hat{x}|\psi(t)\rangle, \qquad
\Delta \hat{p} := \hat{p} - \langle\psi(t)|\hat{p}|\psi(t)\rangle, \qquad
...
14
There is a fair amount of background mathematics to this question, so it will be a while before the punch line.
In quantum mechanics, we aren't working with numbers to represent the state of a system. Instead we use vectors. For the purpose of a simple introduction, you can think of a vector as a list of several numbers. Therefore, a number itself is a ...
10
There is a definine velocity and momentum, we just don't know it.
Nope. There is no definite velocity--this was the older interpretation. The particle has all (possible) velocities at once;it is in a wavefunction, a superposition of all of these states. This can actually be verified by stuff like the double-slit experiment with one photon--we cannot ...
10
No, the uncertainty principle isn't wrong. The PRL paper doesn't suggest that the original uncertainty principle relating uncertainties of position and momentum fails. It "only" questions a modified interpretation of the principle that says that the momentum is disturbed at least by $\hbar / 2 \Delta x$ for a given precision of the position measurement ...
10
it is the error created by photons striking on elementary particles
It's not. Heisenberg's uncertainty principle actually has nothing to do with any particular experiment, or any particular interaction. It's a purely mathematical statement about waves.
Its true meaning is explained in detail on the Wikipedia page, but the gist is that if you have a ...
9
Laplace's determinism is not physically correct over long periods of time. That is, it neglects chaos/"sensitive dependence on initial conditions"/exponential growth of microscopic perturbations already in Newtonian dynamics, which was seriously thought about only in the 20th century. Being true, this also will not be overcome. Stochasticity enters ...
8
Good question! For the properties related by the uncertainty principle, there are two reasons why they come in pairs:
Intuitively, the uncertainty principle relates the variance of a function to the variance of its Fourier transform. And, up to a couple of numerical factors, the Fourier transform of a Fourier transform is the original function. ...
8
The uncertainty principle, in the variance formulation, states that in any quantum state $|\rangle$, the quantity
$$\langle (p-<p>)^2 \rangle \langle (x-\langle x\rangle)^2\rangle \ge {\hbar^2 \over 4} $$
To understand why shifting p and x by their expected value and squaring gives the squared uncertainty, see this answer.
The proof is by noting the ...
8
In Physics "nothing" is generally taken to be the lowest energy state of a theory. We wouldn't normally use the word "nothing" but instead describe the lowest energy state as the "vacuum". I can't think of an intuitive way to describe the QM vacuum because all the obvious analogies have "something" instead of nothing "nothing", so I'll do my best but you may ...
8
In quantum mechanics, two observables that cannot be simultaneously determined are said to be non-commuting. This means that if you write down the commutation relation for them, it turns out to be non-zero. A commutation relation for any two operators $A$ and $B$ is just the following $$[A, B] = AB - BA$$ If they commute, it's equal to zero. For ...
7
A simple example of non-commutativity is rotations in 3D, cf. figure.
Physically, the rotations around the $x$- and the $y$-axis are generated by angular momentum operators $\hat{L}_x$ and $\hat{L}_y$, respectively, which do not commute.
From the mathematical expressions for $\hat{L}_x$ and $\hat{L}_y$, you may proceed with the mathematical derivation, ...
7
A short answer (which has just become somewhat longer) is that there is more than just the Uncertainty Principle thought experiment involved in the non-determinism deduction. If it were only that then physicists might just conclude that it was some classical wave phenomenon (which it is in a way) that gave rise to the HUP. The key other factor at work is the ...
7
It is well established by now that the speed of light barrier does not apply to quantum particles, and this property makes the construction of relativistic quantum field theories and other relativistic quantum systems tightly constrained. The argument that you can't transmit signals faster than light is fine, but particles are not necessarily signals, ...
7
Well, if the mode is speed-questioning, I'll attempt speed-answering:
If photon will travel for million years without collisions, what subtle effects can be accumulated?
The wave packet keeps expanding (or your uncertainty about location, take your pick), and the frequency drops due to space expansion. Nothing else that anyone knows about.
Gravity ...
7
Let a quantum system with Hamiltonian $H$ be given. Suppose the system occupies a pure state $|\psi(t)\rangle$ determined by the Hamiltonian evolution. For any observable $\Omega$ we use the shorthand
$$
\langle \Omega \rangle = \langle \psi(t)|\Omega|\psi(t)\rangle.
$$
One can show that (see eq. 3.72 in Griffiths QM)
$$
...
7
We can satisfy your requirement "the photon was emitted at a correct angle" by "the photon was prepared in a momentum eigenstate". If the photon has definite momentum $\bf{k}$, then its direction of travel is well defined, as you have specified. A photon is a discrete excitation of a "mode", i.e. a solution of Maxwell's equations. For a photon in a ...
6
Noether theorem is as valid in CM(*) as in QM(**). It deals with conservation laws and symmetries. In CM the variables are certain, in QM they may be uncertain.
HUP belongs to QM and gives a limitation on canonically conjugated variable uncertainties in a given state.
If some variable in QM is uncertain, it does not mean its expectation value is not ...
6
It seems that problem here is with mishandling vector quantities. We want to compute things such as $\left<p^2\right>$ but these are in fact $\sum_i \left<p_i^2\right>$ and so the problem decomposes into components where the standard HUP and minimality conditions can be applied. But what you've done is that you applied one-dimensional HUP to ...
6
Manishearth's answer is correct, and this is just a minor extension of it. Manishearth correctly points out that the problem is your statement:
There is a definine velocity and momentum, we just don't know it.
Your statement is the hidden variables idea, and courtesy of Bell's theorem we currently believe that hidden variables are impossible.
Take the ...
6
The relation $p={h\over \lambda}$ applies to photons, it has nothing to do with the uncertainty principle. The issue is localizing the photons, finding out where the are at any given time.
The position operator for a photon is not well defined in any usual sense, because the photon position does not evolve causally, the photon can go back in time. The same ...
6
The statement is that if $K$ and $L$ are Hermitian operators – which means
$$ K = K^\dagger, \quad L = L^\dagger$$
and it implies that the eigenvalues of $K,L$ are real and the eigenvectors with different eigenvalues are orthogonal to each other, then $i(KL-LK)$ (the same as yours) is also Hermitian. This is easily proved by computing the Hermitian ...
6
Lubos' answer is correct: information is not an observable so does not have fluctuations in the sense that could enter an uncertainty relation. However, there does exist a relationship between 'information' and the uncertainty principle, although not of the type that it seems the OP expects.
First of all, note that 'information conservation' could never be ...
6
An observation is an act by which one finds some information – the value of a physical observable (quantity). Observables are associated with linear Hermitian operators.
The previous sentences tautologically imply that an observation is what "collapses" the wave function. The "collapse" of the wave function isn't a material process in any classical sense ...
6
The route to the uncertainty principle went something like this:
In Heisenberg's brilliant 1925 paper [1], he addresses the problem of line spectra caused by atomic transitions. Starting with the known $$\omega(n, n-\alpha) = \frac{1}{\hbar}\{W(n)-W(n-\alpha) \} $$ where $\omega$ are the angular frequencies, $W$ are the energies and $n, \alpha$ are integer ...
6
Yes, your interpretation heuristically makes sense. As you may already know, as a consequence of Heisenberg's uncertainty principle, that an electron has a wave and particle nature. When you think of the wave nature of single particle states you are talking about Bloch states. When you're thinking about the particle nature you are talking about Wannier ...
6
Let us be clear about the problem.
A photon is a quantum mechanical entity and follows the laws of quantum mechanics. There is always a probability attached to any possible path it can take so the strict answer is "no, the path of the photon is not deterministic".
BUT the problem changes when speaking of a large ensemble of photons, which is any light we ...
6
Your interpretation is not quite right. One sharp interpretation one can give to this "cutting" of phase space into cubes of size $h^{2N}$ (here $N$ is the dimension of the system's configuration space), is that it allows one to use classical phase space to count the number of energy eigenstates of the corresponding quantum hamiltonian. Instead of trying ...
5
The uncertainty principle arises from non-commuting observables. More information can be seen in a discussion and derivation at wikipedia: here The result is the general uncertainty principle for two observables A,B
$$\sigma_A\sigma_B \ge \frac{1}{2} \left|\left\langle\left[{A},{B}\right]\right\rangle\right|$$
Since $p_y$ and $x$ commute, in principle ...
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