# Tag Info

0

As Anna points out, the physical position of the particle is described by a probability distribution, say $$\rho= \psi^* \psi$$ and sticking to your example, we can define $$dP=\rho dx = \frac{2}{L} sin^2 (\frac{n\pi x}{L}) dx$$ The probability of finding the particle at an exact location is not defined due to the fact that the particle will be localised (...

0

Are you aware that the quantum mechanical wave function is a mathematical function that cannot be measured? It is the probability distribution, $Ψ^*Ψ$ that is measurable , which means the accumulation of many measurements of particles with the exact same boundary conditions. In quantum mechanics individual particles do not travel in space, do not form ...

1

Very Interesting. I cant think of a solid answer now, but I do have a proposal, although I find my own hypothesis very strange. What if the particle prior to measurement exists only in between two nodes and doesn't traverse the nodes at all. Now to explain why repeated observation result in the particle appearing in between any two nodes, I call upon the ...

1

Now, in order to calculate the probability current, I want to use $J(x)= \frac{\hbar}{m}|\psi(x)|^2\vec{k}$. I don't know why you think this formula is applicable here. As @Jakob already stated in his comments, this formula is not correct in general. The correct general formula can also be found e.g. in Wikipedia - Probability current: $$J(x) = \frac{\hbar}{... 1 The Fermi-Dirac distribution gives the probability of an energy level being occupied. You have to combine it with the density of states g(\epsilon) at that energy to get the physical properties you could be interested in, for example, the internal energy U can be computed as U = \int \epsilon\, f_{FD}(\epsilon)\, g(\epsilon)\,\mathrm{d}\... 0 Your approach is right while calculating a(p). Just be careful about the limits of integration. Don't assume wave function is spread everywhere but it's between 0 to a(the dimensions of the infinite square well). Detailed solution of your problem can be found on the link https://www.sarthaks.com/445606/a-particle-is-trapped-in-an-infinitely-deep-square-well-... 1 I wrote a quick python script (see below), which finds that there are \Omega_{\text{dist.}}(10) = 1001 inequivalent ways of writing E_A = 10 as a sum of 5 non-negative integers. Similarly, for E_B = 5 the script gives \Omega_{\text{dist.}}(5) = 126. For indistinguishable particles, we do the same but insist that n_1 \le n_2 \le \ldots \le n_5 to ... 0 So that picture is a way to understand quantum mechanics, and in fact it's a way that Feynman used to explain it to non-technical people, which you can see in his New Zealand lectures that were videotaped, and they became the book QED: The Strange Theory of Light and Matter. However it does not shed much light on the origins of amplitudes since that is just ... 0 In order to calculate the absorption one has to start with the atom/system in its ground state, c_g(0)=0 and observe how fast the probability to be found in this state decreases or how fast the probability to be found in the excited state increases:$$ w_{g\rightarrow e} = -\frac{d|c_g(t)|^2}{dt}|_{t=0} \left(= \frac{d|c_e(t)|^2}{dt}|_{t=0}\right) $$To ... 0 Eq. 2 of Ref [1] (which is by the same first author as Ref [2], which the OP mentioned in the comments) defines the Gibbs measure for the canonical ensemble $$p(H,\beta) = q(x) \exp\left[-\beta H(x)-W(\beta)\right]$$ where x ranges over configuration space, H(x) is the energy, \beta=1/kT is the inverse temperature, and W(\... 0 TL;DR The energy eigenstates for the infinite square well between 0 and a are$$E(x)=\sqrt{\frac{2}{a}}\sin\left(\frac{n\pi x}{a}\right)$$with energies$$E_n=\frac{n^2\pi^2\hbar^2}{2ma}$$The general solution for any initial wavefuntion \psi(x,0) is$$\psi(x,t)=\frac{2}{a}\sum_{n=1}^\infty \left[\exp\left(-i\frac{n^2\pi^2}{2ma}t\right)\sin\left(\frac{...

4

(1) Nothing wrong there. Plane waves are states of infinitely precise momentum and cannot be properly normalized in position space due to having infinite spread from Heisenberg uncertainty. In practice they still help e.g. in the scattering matrix formalism to get an amplitude for reflection and an amplitude for transmission, and to settle e.g. the basic ...

Top 50 recent answers are included