# Tag Info

6

I have a heuristic justification, but no proof. One can consider individual quantum states, with uncertain position and momentum, as a distribution over phase space. Semiclassically, every quantum state occupies an area $h$ of this space, as justified here. This fact is frequently used in statistical mechanics and is basically equivalent to the WKB ...

6

If you go through the proof of the no-cloning theorem (see e.g. wikipedia) you will notice that only the following properties are used: Unitarity of the evolution operator Hilbert space axioms for a composite state $|\phi \rangle | 0 \rangle$ (which imply the Cauchy-Schwartz inequality) The no-cloning theorem than says that no unitary operator exists ...

4

Aaronson's claims are true, but your statement about what he means is not correct. If cloning were possible, the HUP would still exist but would pose a non-absolute limit on how much you could learn about a single copy of an unknown quantum state- that is, any limit on how much you learned about a state could be attributed to not doing a very good ...

4

Yes, light diffraction may be viewed both as a classical phenomenon and as a quantum mechanical consequence of Heisenberg's uncertainty principle. However, since both explanations work equally well, it doesn't provide any direct evidence for quantum mechanics. Let me explain why the two explanations are equivalent. I'll do the classical uncertainty bound ...

4

The slit is not of zero width, so there is still considerable uncertainty in the position of the particle. Moreover, the narrowing of the slit does lead to diffraction (i.e. a spread in the momentum).

4

You can work in the position representation. It is not difficult: $$\langle \psi \mid[x,p] \mid \psi \rangle = \langle \psi \mid (xp-px)\mid \psi \rangle =\\=\int dx \ \psi^*(x) \left(-i \hbar x \ \partial_x \psi(x) + i \hbar \ \partial_x (x \ \psi(x)) \right) = \\ =-i \hbar \int dx \ \psi^* \left(x \ \psi'-(\psi+x \ \psi')\right) = \\ =i \hbar \int dx \... 4 I think knzhou is right on the money in their answer regarding the core reason for this to be the case. Most of these systems are being probed in the WKB region, which means that, as a rule of thumb, going up by one energy eigenstate means that the state's accessible phase-space area goes up by \hbar. The second half of the argument is that, for the simple ... 3 The cone potential V(x) is exactly solvable, with eigenstates of the form$$ \psi(x)\propto\mathrm{Ai}(|x|-b)  in terms of the Airy $\mathrm{Ai}$ function, so this is rather easy to test. It is probably possible to produce explicit analytic expressions for the uncertainty product, but simple numerical evidence is plenty to see the behaviour here. ...

3

In a nutshell, no. Part of the problem seems to be that you misunderstand the fundamentals of string theory. The strings do vibrate. The frequency of these vibrations determines the type of particle and the energy of the string determines the energy of the particle. Second, your understanding of the uncertainty principle isn't quite right. Yes, we cannot ...

3

Wrote this to address the 2nd question in the OP: "Is it possible to derive a generalized HUP from a nonlinear modification of quantum theory (Invent some nonlinear observables that is neither position nor momentum)?", following the discussion on the topic in comments. Although it is not a clear-cut answer, perhaps it may help. But first a couple of ...

2

The Heisenberg Uncertainty Principle has two distinct aspects: One is the identification of matter as a wave and, in particular, the relationship between a particle's momentum $p$ and its wavelength $\lambda$ through de Broglie's relationship $p=h/\lambda$. This is the crucial bit of physical input. The second one is purely mathematical, and it's the ...

2

Of course you can and we do it all the time, for all practical purposes. Spin eigenstates exist everywhere, eg, the ground state of a hydrogen atom, for the electron state. That's the angular momentum that Gennaro was referencing (or one example). The angular position is then undetermined, the angular wave function is constant and the angular prob density is ...

2

There are two questions here, one of theory and one of practice. In theory, you can definitely imagine measuring a particle's position (or its momentum) exactly. Its wavefunction $\psi(x)$ is then represented by a Dirac delta function for the position you measured the particle at. That is, $\psi(x) \propto \delta(x - x_{m})$ in one dimension, where $x_m$ ...

2

Observables are operators, in particular they are of the self-adjoint type (with discrete spectrum spanning the entire Hilbert space). Given a normalised state $|\psi\rangle$, the expectation value of an operator $A$ thereupon is defined as $\langle A \rangle = \langle\psi |\, A\, |\psi\rangle$; equivalently, one can prove that the uncertainty on the ...

2

Firstly, as an aside, note that there is nothing special about bringing the electron to 'rest', as there is no such thing as absolute rest or motion. What you want to do is get the electron in a state of any precise velocity (and hence momentum). Now the way the accepted (thoroughly tested) quantum mechanical framework works is that if you tell me the state ...

1

The location of a particle cannot be given by the uncertainty principle. The uncertainty principle can be used to find out how accurately we can expect to be able to determine the position (given completely accurate uncertainty free experimental apparatus) on the basis of how accurately we know the momentum the particle. Experimentally a position sensitive ...

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