Editing Particle in a box (section)

=== Position wave function ===

In quantum mechanics, the [[wave function]] gives the most fundamental description of the behavior of a particle; the measurable properties of the particle (such as its position, momentum and energy) may all be derived from the wave function.<ref name="Davies1">Davies, p. 1</ref>
The wave function <math>\psi(x,t)</math> can be found by solving the [[Schrödinger equation]] for the system
<math display="block">i\hbar\frac{\partial}{\partial t}\psi(x,t) = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x,t) +V(x)\psi(x,t),</math>
where <math>\hbar</math> is the [[reduced Planck constant]], <math>m</math> is the [[mass]] of the particle, <math>i</math> is the [[imaginary unit]] and <math>t</math> is time.

Inside the box, no forces act upon the particle, which means that the part of the wave function inside the box oscillates through space and time with the same form as a [[free particle]]:<ref name="Davies4" /><ref name = "Bransden157">Bransden and Joachain, p. 157</ref>
{{NumBlk2||<math display="block"> \psi(x,t) = \left[A \sin(kx) + B \cos(kx)\right]e^{-i\omega t},</math>|1}} where <math>A</math> and <math>B</math> are arbitrary [[complex number]]s.  The frequency of the oscillations through space and time is given by the [[wavenumber|wave number]] <math>k</math> and the [[angular frequency]] <math>\omega</math> respectively.  These are both related to the total energy of the particle by the expression
<math display="block">E = \hbar\omega = \frac{\hbar^2 k^2}{2m},</math>
which is known as the [[dispersion relation]] for a free particle.<ref name = "Davies4" /> However, since the particle is not entirely free but under the influence of a potential, the energy of the particle is 
<math display="block"> E = T + V,</math> 
where ''T'' is the kinetic and ''V'' the potential energy. Therefore, the energy of the particle given above is ''not'' the same thing as <math> E =p^2 / 2m </math> (i.e. the momentum of the particle is not given by <math> p = \hbar k </math>). Thus the wave number ''k'' above actually describes the energy states of the particle and is not related to momentum like the "wave number" usually is. The rationale for calling ''k'' the wave number is that it enumerates the number of crests that the wave function has inside the box, and in this sense it is a wave number. This discrepancy can be seen more clearly below, when we find out that the energy spectrum of the particle is discrete (only discrete values of energy are allowed) but the momentum spectrum is continuous (momentum can vary continuously), i.e., <math>E \neq p^2 / 2m </math>. 

[[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]]
The [[amplitude]] of the wave function at a given position is related to the probability of finding a particle there by <math>P(x,t) = |\psi(x,t)|^2</math>.  The wave function must therefore vanish everywhere beyond the edges of the box.<ref name="Davies4" /><ref name="Bransden157" />  Also, the amplitude of the wave function may not "jump" abruptly from one point to the next.<ref name="Davies4" />  These two conditions are only satisfied by wave functions with the form{{sfn|Cohen-Tannoudji|Diu|Laloë|2019|p=271}}
<math display="block">\psi_n(x,t) =
\begin{cases}
A \sin\left(k_n \left(x-x_c+\tfrac{L}{2}\right)\right) e^{-i\omega_n t}\quad & x_c-\tfrac{L}{2} < x < x_c+\tfrac{L}{2}\\
0 & \text{otherwise}
\end{cases},</math>
where
<math display="block">k_n=\frac{n \pi}{ L},</math>
and
<math display="block">E_n=\hbar \omega_n=\frac{k_{n}^{2} \hbar^2}{2 m}=\frac{n^2 \pi^2 \hbar^2}{2 m L^2},</math>
for positive [[integer]]s <math>n \in \mathbb{Z}_{>0}</math>. The simplest solutions, <math>k_n = 0</math> or <math>A = 0</math> both yield the trivial wave function <math>\psi(x) = 0</math>, which describes a particle that does not exist anywhere in the system.<ref name="Bransden158">Bransden and Joachain, p.158</ref>   Here one sees that only a discrete set of energy values and wave numbers ''k'' are allowed for the particle. Usually in quantum mechanics it is also demanded that the derivative of the wave function in addition to the wave function itself be continuous; here this demand would lead to the only solution being the constant zero function, which is not what we desire, so we give up this demand (as this system with infinite potential can be regarded as a nonphysical abstract limiting case, we can treat it as such and "bend the rules"). Note that giving up this demand means that the wave function is not a differentiable function at the boundary of the box, and thus it can be said that the wave function does not solve the Schrödinger equation at the boundary points <math> x = 0 </math> and <math> x = L </math> (but does solve it everywhere else).

Finally, the unknown constant <math>A</math> may be found by [[Wavefunction#Normalization condition|normalizing the wave function]]. That is, it follows from
<math display="block">\int_0^L \left\vert \psi(x) \right\vert^2 dx = 1,</math>
that any complex number <math>A</math> whose [[absolute value]] is
<math display="block">\left| A \right| = \sqrt{\frac{2 }{L}},</math>
yields the same normalized state.

It is expected that the ''eigenvalues'', i.e., the energy <math>E_n</math> of the box should be the same regardless of its position in space, but <math>\psi_n(x,t)</math> changes. Notice that <math>x_c - \tfrac{L}{2}</math> represents a phase shift in the wave function. This phase shift has no effect when solving the Schrödinger equation, and therefore does not affect the ''eigenvalue''.

If we set the origin of coordinates to the center of the box, we can rewrite the spatial part of the wave function succinctly as:
<math display="block">\psi_n (x) = \begin{cases}
\sqrt{\frac{2}{L}} \sin(k_nx) \quad{} \text{for } n \text{ even} \\
\sqrt{\frac{2}{L}} \cos(k_nx) \quad{} \text{for } n \text{ odd}.
\end{cases}</math>