New answers tagged

1

For a neutron of that speed, the uncertainty in the momentum is expected to be less than the momentum magnitude. Using the actual momentum will be an upper bound on the momentum uncertainty. That correlates to a lower bound on the position uncertainty. So, $\Delta x$ is lower bounded by $\hbar/(2p)$: $$\Delta x \ge \frac{\hbar}{2m_nv}.$$ $\Delta x$ could ...


-1

Gravity fluctuations will always cause vibrations in atoms and molecules limiting the lowest temperature obtainable. Closer to the mass source, the stronger the gravity field. As stated by Asaf earlier, evaporative cooling will lower the temperature only so far. Adding a magnetic field may temporarily increase temperature by increasing vibrations in the ...


0

I think you are probably misinterpreting the context here. If you read the previous line carefully it says "there is always an undetermined interaction between observer and observed; there is nothing we can do to avoid the interaction or to allow for it ahead of time. And later he just says due to the fact that photon can be scattered within the 2θ' angle ...


1

There is yet another solution (maybe more elementary)$^1$, with some components of the answers from Qmechanic and JoshPhysics (Currently I'm taking my first QM course and I don't quite understand the solution of Qmechanic, and this answer complement JoshPhysics's answer) the solution uses the Heisenberg Equation: The time evolution of an operator $\hat{A}$ ...


6

The temperature limit for laser cooling is not related to gravity but to the always-present momentum kick during absoprtion/emission of photons. Ultracold atom experiments typically use laser cooling at an initial stage and afterwards evaporative cooling is used to reach the lowest temperatures. In evaporative cooling the most energetic atoms are discarded ...


4

Summary Using the entropic uncertainty principle, one can show that $μ_qμ_p≥\frac{π}{4e}$, where $μ$ is the mean deviation. This corresponds to $F≥\frac{π^2}{4e}=0.9077$ using the notations of AccidentalFourierTransform’s answer. I don’t think this bound is optimal, but didn’t manage to find a better proof. To simplify the expressions, I’ll assume $ℏ=1$, ...


4

This is a great example of how hard it is to popularize quantum mechanics. Greene's example is not quite right, because classically, the butterfly does have a definite position and momentum, at all times. We can also measure these values simultaneously to arbitrary accuracy, as your friend says. (As for your concern about exposure time, we could decrease ...


0

The Heisenberg uncertainty principle is a basic foundation stone of quantum mechanics, and is derivable from the commutator relations of the quantum mechanical operators describing the pair of variables participating in the HUP. You are discussing the energy time uncertainty, . For an individual particle, it describes a locus in the time versus energy ...


3

It cannot be proven, because "wave-particle duality" is not a mathematical statement. It most definitely is not "logically true". Can you try to make it mathematical? A mathematical framework The "complementarity principle" was introduced in order to better understand some features of quantum mechanics in the early days. The problem is that if you consider ...


1

The uncertainty principle never said that nothing can be measured simultaneously with accuracy. Uncertainty principle states that it is not possible to measure two canonically conjugate quantities at the same time with accuracy. Like you cannot measure the x component of momentum $p_x$ and the x coordinate position simultaneously with accuracy. But the x ...


0

I am not satisfied of the published replies, so I will try my own, as a metrologist (expert in measurement units, but not in theoretical physics). The question clearly is referring to the experimental frame while all the answers are referring only to the theoretical frame, so they do not talk nor understand with each other. Here we are dealing with two ...


1

The point dipole is an approximation from classical physics - note that it also involves an infinite field strength in its center, where the field amplitude is not differentiable. I think such a source is not compatible with the common approach to quantum mechanics. If you take such a very small, subwavelength source, it is true that the evanescent near ...


11

We can assume WLOG that $\bar x=\bar p=0$ and $\hbar =1$. We don't assume that the wave-functions are normalised. Let $$ \sigma_x\equiv \frac{\int \mathrm dx\; |x|\;|\psi(x)|^2}{\int\mathrm dx\; |\psi(x)|^2} $$ and $$ \sigma_p\equiv \frac{\int \mathrm dp\; |p|\;|\tilde \psi(p)|^2}{\int\mathrm dx\; |\psi(x)|^2} $$ Using $$ \int\mathrm dp\ |p|\;\mathrm ...


4

I went back to the derivation of the Heisenberg uncertainty principle and tried to modify it. Not sure if what I've come up with is worth anything, but you'll be the judge: The original derivation Let $\hat{A} = \hat{x} - \bar{x}$ and $\hat{B} = \hat{p} - \bar{p}$. Then the inner product of the state $| \phi\rangle = \left(\hat{A} + i \lambda ...


0

As in the link you give, the functional form depends on the probability distribution used, and these differ widely, nothing as general as the Heisenberg form can appear. The quantum mechanical equivalent requires the solution for the specific boundary problem. In any case , the HUP is about deltas, i.e. uncertainties, and not only standard deviations as ...


2

I) In this answer we will consider the microscopic description of classical E&M only. The Lorentz force reads $$ \tag{1} {\bf F}~:=~q({\bf E}+{\bf v}\times {\bf B})~=~\frac{\mathrm d}{\mathrm dt}\frac{\partial U}{\partial {\bf v}}- \frac{\partial U}{\partial {\bf r}}~=~-q\frac{\mathrm d{\bf A}}{\mathrm dt} - \frac{\partial U}{\partial {\bf r}}, $$ ...



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