Difference between revisions 4452888 and 4472189 on simplewiki

[[File:Particle in a box.svg|thumb|300px| A classical particle, such as a ball, bouncing between two high walls. The particle moves forwards and backwards, and can not go out.]]

(contracted; show full)Now we'll focus on solving for <math>\phi(t)</math> the time dependent part of the equation:
:<math>\mathrm{i}\hbar\frac{\frac{\partial}{\partial t}\phi(t)}{\phi(t)}=C</math>

Some algebraic manipulation gives us:
:<math>\frac{\partial\phi(t)}{\phi(t)}=\frac{-\mathrm{i}C}{\hbar}\partial t</math> (because <math>(\frac{1}{i}=-i)</math>)

Solving this differential equation yields:
:<math>
\ln[\phi(t)]=\frac{-\mathrm{i}C}{\hbar}t</math>

We then raise e to the powers of both sides to isolate <math>\phi(t)</math>:
:<math>\phi(t)=e^{\frac{-\mathrm{i}C}{\hbar}t}</math>

We can then expand <math>\phi(t)</math> with Eulers equation, and we get:
:<math>\phi(t)=\cos\left(\frac{C}{\hbar}t)-i\sin\right) - i\sin\left(\frac{C}{\hbar}t\right)</math>

So we see that it's a wave with a frequency <math>\omega=\frac{C}{\hbar}</math> . Therefore, <math>C=\hbar\omega</math>

However, we can take it one step further, and say that in this case, <math>E=\hbar\omega</math>

Substituting back into <math>\phi(t)=e^{\frac{-\mathrm{i}C}{\hbar}t}</math> gives us our final solution for the time dependent part of the function:
:<math>\phi(t)=e^{-\mathrm{i(}\frac{E}{\hbar})t}</math>

==== Solving the Spatial Portion ====
Continuing from before, we have:
:<math>-\frac{\hbar^{2}}{2m}\frac{\frac{\partial^{2}}{\partial x^{2}}\psi(x)}{\psi(x)}+V(x)=E</math>

As mentioned earlier, the definition of the 'particle in a box' scenario is that there is an infinite potential outside of the box, and no potential within the box. Therefore:
:<math>V(x) =
\begin{cases}
0, & 0 < x < L,\\
\infty, & \text{otherwise,}
\end{cases},
</math>

Substituting <math>V(x)=0\;</math> within the box gives us:
:<math>-\frac{\hbar^{2}}{2m}\frac{\frac{\partial^{2}}{\partial x^{2}}\psi(x)}{\psi(x)}=E</math>

Doing some algebra gives us:
:<math>\frac{\partial^{2}}{\partial x^{2}}\psi(x)+\frac{2mE}{\hbar^{2}}\psi(x)=0</math>
The solutions for a differential equation of the form <math>ay''+by'+cy=0\;</math> where <math>b^{2}-4ac<0\;</math> are given to be
:<math>y(t)=c_{1}e^{\lambda t}\cos(\mu t)+c_{2}e^{\lambda t}\sin(\mu t)\;</math>, where <math>\lambda=\frac{-b}{2a}\;</math>, and <math>\mu=\frac{\sqrt{b^{2}-4ac}}{2a*i}\;</math>
In our case, <math>b=0\;</math>, and <math>c = \frac{2m}{\hbar^{2}}E\;</math>, so <math>\lambda=0\;</math> and <math>\mu=\sqrt{\frac{2m}{\hbar^{2}}E}\;</math>. For ease in writing, we'll set <math>k=\mu=\sqrt{\frac{2m}{\hbar^{2}}E}\;</math>, so our solution is <math>y(t)=c_{1}\cos(kt)+c_{2}\sin(kt)\;</math>

Therefore, we can say the form of <math>\psi(x)\;</math> is:
<math>\psi(x) = A\sin(kx) + B\cos(kx)\;</math>

The wave function needs to be continuous, and because it is 0 outside of the walls of the well, it needs to be 0 at the point <math>x = 0\;</math>. Therefore <math>B=0\;</math>, and <math>\psi(x) = A\sin(kx)\;</math>

On the same note, the wave function needs to be continuous, and as it is 0 outside of the walls of the well, it needs to be 0 at the point <math>x = L\;</math>. In order for that to be true, the sin must be 0. <math>\sin\;</math> is 0 at integer values of <math>\pi\;</math>. Therefore <math>kL=n\pi\;</math>, and <math>k=\frac{n\pi}{L}\;</math>where <math>n\;</math> is an integer.

Therefore, <math>\psi(x) = A\sin(\frac{n\pi x}{L})\;</math>

One of the conditions of the wave function is that the probability of finding it somewhere the enclosed well must be a total of 1 (100%).
The size (or [[amplitude]]) of the wavefunction at a given position is related to the probability of finding a particle there by <math>P(x) = |\psi(x)|^2\;</math>.
Therefore, <math>\int_{0}^{L}|\psi(x)|^{2}=1\;</math>, or
:<math>\int_{0}^{L}A^{2}\sin^{2}(kx)dx=1\;</math>. 
:<math>A^{2}\int_{0}^{L}\frac{1-\cos(2kx)}{2}dx=1\;</math>
:<math>\frac{A^{2}}{2}\int_{0}^{L}dx-\frac{A^{2}}{2}\int_{0}^{L}\cos(2kx)dx=1\;</math>

However, integrating the cos portion gives us:<math>\int_{0}^{L}\cos(2kx)=\frac{L\sin(2kL)}{2k}\;</math>. Substituting in <math>k=\frac{n\pi}{L}\;</math> gives us <math>\frac{L\sin(2\frac{n\pi}{L}L)}{2\frac{n\pi}{L}}=\frac{L\sin(2n\pi)}{2\frac{n\pi x}{L}}=0\;</math>, because <math>\sin(2n\pi)\;</math>, where n is an integer is always 0.

Therefore, the only part remaining is the <math>\frac{A^{2}}{2}\int_{0}^{L}dx\;</math>, which gives us <math>\frac{A^{2}L}{2}=1\;</math>, and after some algebraic manipulation, we reach our result of:
:<math>A=\sqrt{\frac{2}{L}}\;</math>

Now we have our solution for the spatial component:
<math>\psi(x) = \sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})\;</math>, where n is an integer.

[[File:particle in a box wavefunctions 2.svg|thumb|right|upright|Initial wavefunctions for the first four states in a one-dimensional particle in a box]]

==== Putting them together ====
The first thing we did was assume that <math>\Psi(x,t)\;</math> is the product of two functions, <math>\Psi(x,t)=\psi(x)\phi(t)\;</math>

:<math>\phi(t)=e^{-\mathrm{i(}\frac{E}{\hbar})t}</math>
:<math>\psi(x) = \sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})\;</math>, where n is an integer.

Therefore, our result for the time dependent Schrödinger equation is:
:<math>\Psi(x,t)=\psi(x)\phi(t)=\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})e^{-\mathrm{i(}\frac{E}{\hbar})t}\;</math>

=== Energy levels ===
[[File:Confined particle dispersion - positive.svg|thumb|upright|The energy of a particle in a box (black circles) and a free particle (grey line) both depend upon wavenumber in the same way. However, the particle in a box may only have certain, discrete energy levels.]]

(contracted; show full)[[it:Particella in una scatola]]
[[he:בור פוטנציאל אינסופי]]
[[ja:井戸型ポテンシャル]]
[[pl:Cząstka w studni potencjału]]
[[pt:Partícula em uma caixa]]
[[sv:Partikel i låda]]
[[uk:Квантовий рух в прямокутній потенційній ямі]]
[[zh:無限深方形阱]]