Tag Info

Hot answers tagged

33

The first paragraph is basically right, but I wouldn't ascribe the uncertainty principle to particles, just to the universe/physics in general. You can no more get arbitrarily good, simultaneous measurements of position and momentum (of anything) than you can construct a function with an arbitrarily narrow peak whose Fourier transform is also arbitrarily ...


31

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 ...


19

The uncertainty principle should be understood as follows: The position and momentum of a particle are not well-defined at the same time. Quantum mechanically, this is expressed through the fact that the position and momentum operators don't commute: $[x,p]=i\hbar$. The most intuitive explanation, for me, is to think about it in terms of wave-particle ...


18

This is really a footnote to Chris' answer but it got a bit long for a comment. It sounds odd to claim that a particle has no position, but it's easier to understand if you appreciate that a particle is just an excitation in a quantum field. When Heisenberg was developing his ideas physicists still thought of particles as little billiard balls. With the ...


18

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 ...


18

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) $$ ...


16

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 ...


15

You are misunderstanding the Uncertainty Principle. The Uncertainty Principle says that a particle cannot simultaneously have a definite momentum and a definite position. This is not due to our incomplete knowledge of parameters. This is a fundamental law of the universe and arises from the fact that the momentum and position operators do not commute in ...


14

The reason for many contradictory statements regarding the nature of virtual particles is that they are often invoked for heuristical explanations of phenomena that arise within the framework of quantum field theory. One then tries to justify those explanations by attributing certain properties to virtual particles they do not actually possess. What ...


13

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 ...


12

The Schrodinger equation is time-symmetric. The answer is therefore No. From all of the comments, I feel like I must be oversimplifying or missing something, but I can't see what.


12

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 ...


12

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 ...


11

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 ...


11

If you know its instantaneous velocity at every moment of time in a given interval AND you knew the exact position of a particle at the start of the interval, then you could use integrals to find its exact position at any moment in the interval. That is a logically true statement. There is no contradiction here with the Uncertainty principle because you ...


11

This is an estimation tool not uncommon in theoretical physics. Namely, one knows the value of some quantity for a given problem and therefore assumes that the scale of the problem with regards to that quantity is of the same order of magnitude as the known value. In other words, we assume that the error in our known value must not be too much greater than ...


10

The uncertainty principle can be seen as a result of space $x$ and momentum $p$ being a Fourier transform pair. The free-particle wave function has, similarly to the exponential $e^{-\frac{i}{\hbar} px}$ an exponential $e^{-\frac{i}{\hbar} E t}$. Thus one could expect a similar uncertainty relation for the variable pair $(E, t)$. An immediate result is that ...


10

So, why can't the uncertainty relations be violated in such a case, if I could, say, measure the position of the object with this wave function That's the catch. You can't. Or rather, you can measure the position, but the result you get will vary from one measurement to the next, because the wavefunction $\exp(x^2/2i - cx)$ is not an eigenstate of ...


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 ...


9

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 ...


9

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. ...


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

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 ...


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 ...


8

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, ...



Only top voted, non community-wiki answers of a minimum length are eligible