User:Rectified/Math display test

Particle in a box

https://infogalactic.com/info/Quantum_mechanics

1-dimensional potential energy box (or infinite potential well)

Main article: Particle in a box

The particle in a one-dimensional potential energy box in the $x$ direction, the time-independent Schrödinger equation may be written

Original:

$- \frac {\hbar ^2}{2m} \frac {d ^2 \psi}{dx^2} = E \psi.$

Copy:

$- \frac {\hbar ^2}{2m} \frac {d ^2 \psi}{dx^2} = E \psi.$

One change of exponent:

$- \frac {\hbar ^2}{2m} \frac {d ^2 \psi}{dx^n} = E \psi.$

Change \frac to \tfrac:

$\tfrac {\hbar ^2}{2m} \frac {d ^2 \psi}{dx^n} = E \psi.$

With the differential operator defined by

$\hat{p}_x = -i\hbar\frac{d}{dx}$

the previous equation is evocative of the classic kinetic energy analogue,

$\frac{1}{2m} \hat{p}_x^2 = E,$

with state $\psi$ in this case having energy $E$ coincident with the kinetic energy of the particle.

The general solutions of the Schrödinger equation for the particle in a box are

$\psi(x) = A e^{ikx} + B e ^{-ikx} \qquad\qquad E = \frac{\hbar^2 k^2}{2m}$

or, from Euler's formula,

$\psi(x) = C \sin kx + D \cos kx.\!$

The infinite potential walls of the box determine the values of C, D, and k at x = 0 and x = L where ψ must be zero. Thus, at x = 0,

$\psi(0) = 0 = C\sin 0 + D\cos 0 = D\!$

and D = 0. At x = L,

$\psi(L) = 0 = C\sin kL.\!$

in which C cannot be zero as this would conflict with the Born interpretation. Therefore, since sin(kL) = 0, kL must be an integer multiple of π,

$k = \frac{n\pi}{L}\qquad\qquad n=1,2,3,\ldots.$

The quantization of energy levels follows from this constraint on k, since

$E = \frac{\hbar^2 \pi^2 n^2}{2mL^2} = \frac{n^2h^2}{8mL^2}.$

Instantaneous acceleration

https://infogalactic.com/info/Acceleration

Instantaneous acceleration is a mathematical concept, the limit of the average acceleration over an infinitesimal interval of time. In the terms of calculus, instantaneous acceleration is the derivative of the velocity vector with respect to time:

Original:

$\mathbf{a} = \lim_{{\Delta t}\to 0} \frac{\Delta \mathbf{v}}{\Delta t} = \frac{d\mathbf{v}}{dt}$

Copy:

$\mathbf{a} = \lim_{{\Delta t}\to 0} \frac{\Delta \mathbf{v}}{\Delta t} = \frac{d\mathbf{v}}{dt}$

Concatenate:

$\mathbf{a} = \lim_{{\Delta t}\to 0} \frac{\Delta \mathbf{v}}{\Delta t} = \frac{d\mathbf{v}}{dt} \mathbf{a} = \lim_{{\Delta t}\to 0} \frac{\Delta \mathbf{v}}{\Delta t} = \frac{d\mathbf{v}}{dt}$

Pile on:

$\mathbf{a} = \lim_{{\Delta t}\to 0} \frac{\Delta \mathbf{v}}{\Delta t} = \frac{d\mathbf{v}}{dt} \mathbf{a} \lim_{{\Delta t}\to 0} \frac{\Delta \mathbf{v}}{\Delta t} = \frac{d\mathbf{v}}{dt}\frac{\hbar^2 \pi^2 n^2}{2mL^2} = \frac{n^2h^2}{8mL^2}\frac{1}{2m} \hat{p}_x^2 \psi(L)$

It can be seen that the integral of the acceleration function a(t) is the velocity function v(t); that is, the area under the curve of an acceleration vs. time (a vs. t) graph corresponds to velocity.

$\mathbf{v} = \int \mathbf{a} \ dt$

Given the fact that acceleration is defined as the derivative of velocity, v, with respect to time t and velocity is defined as the derivative of position, x, with respect to time, acceleration can be thought of as the second derivative of x with respect to t:

$\mathbf{a} = \frac{d\mathbf{v}}{dt} = \frac{d^2\mathbf{x}}{dt^2}$