Re: Amazing Photograph.
he particle in a one-dimensional potential energy box is the most mathematically simple example where restraints lead to the quantization of energy levels. The box is defined as having zero potential energy everywhere inside a certain region, and therefore infinite potential energy everywhere outside that region. For the one-dimensional case in the {\displaystyle x} x direction, the time-independent Schrödinger equation may be written[90]
{\displaystyle -{\frac {\hbar ^{2}}{2m}}{\frac {d^{2}\psi }{dx^{2}}}=E\psi .} -{\frac {\hbar ^{2}}{2m}}{\frac {d^{2}\psi }{dx^{2}}}=E\psi .
With the differential operator defined by
{\displaystyle {\hat {p}}_{x}=-i\hbar {\frac {d}{dx}}} {\hat {p}}_{x}=-i\hbar {\frac {d}{dx}}
the previous equation is evocative of the classic kinetic energy analogue,
{\displaystyle {\frac {1}{2m}}{\hat {p}}_{x}^{2}=E,} {\frac {1}{2m}}{\hat {p}}_{x}^{2}=E,
with state {\displaystyle \psi } \psi in this case having energy {\displaystyle E} E coincident with the kinetic energy of the particle.
The general solutions of the Schrödinger equation for the particle in a box are
{\displaystyle \psi (x)=Ae^{ikx}+Be^{-ikx}\qquad \qquad E={\frac {\hbar ^{2}k^{2}}{2m}}} \psi (x)=Ae^{ikx}+Be^{-ikx}\qquad \qquad E={\frac {\hbar ^{2}k^{2}}{2m}}
or, from Euler's formula,
{\displaystyle \psi (x)=C\sin kx+D\cos kx.\!} \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,
{\displaystyle \psi (0)=0=C\sin 0+D\cos 0=D\!} \psi (0)=0=C\sin 0+D\cos 0=D\!
and D = 0. At x = L,
{\displaystyle \psi (L)=0=C\sin kL.\!} \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,
{\displaystyle k={\frac {n\pi }{L}}\qquad \qquad n=1,2,3,\ldots .} k={\frac {n\pi }{L}}\qquad \qquad n=1,2,3,\ldots .
The quantization of energy levels follows from this constraint on k, since
{\displaystyle E={\frac {\hbar ^{2}\pi ^{2}n^{2}}{2mL^{2}}}={\frac {n^{2}h^{2}}{8mL^{2}}}.} E={\frac {\hbar ^{2}\pi ^{2}n^{2}}{2mL^{2}}}={\frac {n^{2}h^{2}}{8mL^{2}}}.
The ground state energy of the particles is E1 for n=1.
Energy of particle in the nth state is En =n2E1, n=2,3,4,.....
Particle in a box with boundary condition V(x)=0 -a/2<x<+a/2
A particle in a box with a little change in the boundary condition.
In this condition the general solution will be same, there will a little change to the final result, since the boundary conditions are changed
{\displaystyle \psi (x)=C\sin kx+D\cos kx.\!} \psi (x)=C\sin kx+D\cos kx.\!
At x=0, the wave function is not actually zero at all value of n.
Clearly, from the wave function variation graph we have,
At n=1,3,4,...... the wave function follows a cosine curve with x=0 as origin
At n=2,4,6,...... the wave function follows a sine curve with x=0 as origin
Variation of wave function with x and n.
Wave Function Variation with x and n.
From this observation we can conclude that the wave function is alternatively sine and cosine.
So in this case the resultant wave equation is
ψn(x) = Acos(knx) n=1,3,5,.............
= Bsin(knx) n=2,4,6,.............
NoI don't understand it either