Editing Particle in a box (section)

== One-dimensional solution ==
[[File:Infinite potential well-en.svg|thumb|right|The barriers outside a one-dimensional box have infinitely large potential, while the interior of the box has a constant, zero potential. Shown is the shifted well, with <math  display="inline">x_c = L/2</math>]]

The simplest form of the particle in a box model considers a one-dimensional system.  Here, the particle may only move backwards and forwards along a straight line with impenetrable barriers at either end.<ref name="Davies4">Davies, p.4</ref>
The walls of a one-dimensional box may be seen as regions of space with an infinitely large [[potential energy]].  Conversely, the interior of the box has a constant, zero potential energy.<ref>Actually, any constant, finite potential <math>V_0</math> can be specified within the box.  This merely shifts the energies of the states by <math>V_0</math>.</ref>  This means that no forces act upon the particle inside the box and it can move freely in that region.  However, infinitely large [[force]]s repel the particle if it touches the walls of the box, preventing it from escaping.  The potential energy in this model is given as
<math display="block">V(x) = \begin{cases}
0, & x_c-\tfrac{L}{2} < x <x_c+\tfrac{L}{2},\\
\infty, & \text{otherwise,}
\end{cases},</math>
where ''L'' is the length of the box, ''x<sub>c</sub>'' is the location of the center of the box and ''x'' is the position of the particle within the box. Simple cases include the centered box (''x<sub>c</sub>'' = 0) and the shifted box (''x<sub>c</sub>'' = ''L''/2) (pictured).

=== 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>

=== Momentum wave function ===

The momentum wave function is proportional to the [[Fourier transform]] of the position wave function. With <math>k = p / \hbar</math> (note that the parameter {{mvar|k}} describing the momentum wave function below is not exactly the special {{math|''k<sub>n</sub>''}} above, linked to the energy eigenvalues), the momentum wave function is given by
<math display="block">\phi_n(p,t)
= \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^\infty \psi_n(x,t)e^{-ikx}\,dx 
= \sqrt{\frac{L}{\pi \hbar}} \left(\frac{n\pi}{n\pi+k L}\right)\,\operatorname{sinc}\left(\tfrac{1}{2}(n\pi-k L)\right)e^{-i k x_c}e^{i (n-1) \tfrac{\pi}{2}}e^{-i\omega_n t} ,
</math>
where sinc is the cardinal sine [[sinc function]], {{math|1=sinc(''x'') = sin(''x'')/''x''}}. For the centered box ({{math|1=''x<sub>c</sub>'' = 0}}), the solution is real and particularly simple, since the phase factor on the right reduces to unity. (With care, it can be written as an even function of {{mvar|p}}.)

It can be seen that the momentum spectrum in this wave packet is continuous, and one may conclude that for the energy state described by the wave number {{math|''k<sub>n</sub>''}}, the momentum can, when measured, also attain ''other values'' beyond <math> p = \pm \hbar k_n </math>.

Hence, it also appears that, since the energy is <math display="inline"> E_n = \frac{\hbar^2 k_n^2}{2m} </math> for the ''n''th eigenstate, the relation <math display="inline"> E = \frac{p^2}{2m} </math> does not strictly hold for the measured momentum {{mvar|p}}; the energy eigenstate <math> \psi_n </math> is not a momentum eigenstate, and, in fact, not even a superposition of two momentum eigenstates, as one might be tempted to imagine from equation ({{EquationNote|1}}) above: peculiarly, it has no well-defined momentum before measurement!

=== Position and momentum probability distributions ===
{{More citations needed section|date=November 2024|talk=Talk:Particle_in_a_box#Essential_Self_Adjointness_and_Boundary_Conditions}}
In classic physics, the particle can be detected anywhere in the box with equal probability.  In quantum mechanics, however, the probability density for finding a particle at a given position is derived from the wave function as <math>P(x) = |\psi(x)|^2.</math>  For the particle in a box, the probability density for finding the particle at a given position depends upon its state, and is given by
<math display="block">P_n(x,t) = \begin{cases}
\frac{2}{L} \sin^2\left(k_n \left(x-x_c+\tfrac{L}{2}\right)\right), & x_c-\frac{L}{2} < x < x_c+\frac{L}{2},\\
0, & \text{otherwise,}
\end{cases}</math>

Thus, for any value of ''n'' greater than one, there are regions within the box for which <math>P(x)=0</math>, indicating that ''spatial nodes'' exist at which the particle cannot be found. If [[relativistic wave equations]] are considered, however, the probability density not go to zero at the nodes (apart from the trivial case <math>n=0</math>).<ref>{{cite journal|doi=10.1088/0143-0807/17/1/004 |title=Relativistic particle in a box |year=1996 |last1=Alberto |first1=P | last2=Fiolhais |first2=C |last3=Gil |first3=V M S |url=https://estudogeral.sib.uc.pt/bitstream/10316/12349/1/Relativistic%20particle%20in%20a%20box.pdf |journal=European Journal of Physics |volume=17 |issue=1 |pages=19–24 | bibcode=1996EJPh...17...19A | hdl=10316/12349 |s2cid=250895519 | hdl-access=free }}</ref>

In quantum mechanics, the average, or [[expectation value]] of the position of a particle is given by
<math display="block">\langle x \rangle = \int_{-\infty}^{\infty} x P_n(x)\,\mathrm{d}x.</math>

For the steady state particle in a box, it can be shown that the average position is always <math>\langle x \rangle =x_c</math>, regardless of the state of the particle.  For a superposition of states, the expectation value of the position will change based on the cross term, which is proportional to <math>\cos(\omega t)</math>.

The variance in the position is a measure of the uncertainty in position of the particle:
<math display="block">\mathrm{Var}(x) = \int_{-\infty}^\infty (x-\langle x\rangle)^2 P_n(x)\,dx = \frac{L^2}{12}\left(1-\frac{6}{n^2\pi^2}\right)</math>

The probability density for finding a particle with a given momentum is derived from the wave function as <math>P(x) = |\phi(x)|^2</math>. As with position, the probability density for finding the particle at a given momentum depends upon its state, and is given by
<math display="block">P_n(p)=\frac{L}{\pi \hbar} \left(\frac{n\pi}{n\pi+k L}\right)^2\,\textrm{sinc}^2\left(\tfrac{1}{2}(n\pi-k L)\right)</math>
where, again, <math>k = p / \hbar</math>. The expectation value for the momentum is then calculated to be zero, and the variance in the momentum is calculated to be:
<math display="block">\mathrm{Var}(p)=\left(\frac{\hbar n\pi}{L}\right)^2</math>

The uncertainties in position and momentum (<math>\Delta x</math> and <math>\Delta p</math>) are defined as being equal to the square root of their respective variances, so that:
<math display="block">\Delta x \Delta p = \frac{\hbar}{2} \sqrt{\frac{n^2\pi^2}{3}-2}</math>

This product increases with increasing ''n'', having a minimum for ''n'' = 1. The value of this product for ''n'' = 1 is about equal to 0.568 <math>\hbar</math>, which obeys the [[Heisenberg uncertainty principle]], which states that the product will be greater than or equal to <math>\hbar/2</math>.

Another measure of uncertainty in position is the [[Entropy (information)|information entropy]] of the probability distribution ''H''<sub>x</sub>:<ref name="Majernik1997">{{cite journal |last1=Majernik |first1=Vladimir |last2=Richterek |first2=Lukas | date=1997-12-01 |title=Entropic uncertainty relations for the infinite well |url=https://www.researchgate.net/publication/231121036 |journal=J. Phys. A |volume=30 |issue=4 |pages= L49|doi=10.1088/0305-4470/30/4/002 |access-date=11 February 2016| bibcode = 1997JPhA...30L..49M }}</ref>
<math display="block">H_x=\int_{-\infty}^\infty P_n(x) \log(P_n(x) x_0)\,dx =\log\left(\frac{2 L}{e \,x_0}\right)</math>
where ''x''<sub>0</sub> is an arbitrary reference length.

Another measure of uncertainty in momentum is the [[Entropy (information)|information entropy]] of the probability distribution ''H<sub>p</sub>'':
<math display="block">H_p(n)=\int_{-\infty}^\infty P_n(p) \log(P_n(p) p_0)\,dp</math>
<math display="block">\lim_{n\to\infty} H_p(n) = \log\left(\frac{4 \pi \hbar\, e^{2(1-\gamma)}}{ L\, p_0}\right)</math>
where ''γ'' is [[Euler-Mascheroni constant|Euler's constant]]. The quantum mechanical [[Uncertainty principle#Quantum entropic uncertainty principle|entropic uncertainty principle]] states that for <math>x_0\,p_0 = \hbar</math>
<math display="block">H_x+H_p(n) \ge \log(e\,\pi) \approx 2.14473...</math> ([[nat (unit)|nats]])

For <math>x_0\,p_0=\hbar</math>, the sum of the position and momentum entropies yields:
<math display="block">H_x+H_p(\infty) = \log\left(8\pi\, e^{1-2\gamma}\right) \approx 3.06974...</math>
where the unit is [[nat (unit)|nat]], and which satisfies the quantum entropic uncertainty principle.

=== 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.]]
The energies that correspond with each of the permitted wave numbers may be written as{{sfn|Hall|2013|p=81}}
<math display="block">E_n = \frac{n^2\hbar^2 \pi ^2}{2mL^2} = \frac{n^2 h^2}{8mL^2}.</math>
The energy levels increase with <math>n^2</math>, meaning that high energy levels are separated from each other by a greater amount than low energy levels are.  The lowest possible energy for the particle (its ''[[zero-point energy]]'') is found in state 1, which is given by<ref name="Bransden159">Bransden and Joachain, p. 159</ref>
<math display="block">E_1 = \frac{\hbar^2\pi^2}{2mL^2} = \frac{h^2}{8mL^2}.</math>
The particle, therefore, always has a positive energy.  This contrasts with classical systems, where the particle can have zero energy by resting motionlessly.  This can be explained in terms of the [[uncertainty principle]], which states that the product of the uncertainties in the position and momentum of a particle is limited by
<math display="block">\Delta x\Delta p \geq \frac{\hbar}{2}</math>
It can be shown that the uncertainty in the position of the particle is proportional to the width of the box.<ref name="Davies15">Davies, p. 15</ref>  Thus, the uncertainty in momentum is roughly inversely proportional to the width of the box.<ref name="Bransden159" />  The [[kinetic energy]] of a particle is given by <math>E = p^2/(2m)</math>, and hence the minimum kinetic energy of the particle in a box is inversely proportional to the mass and the square of the well width, in qualitative agreement with the calculation above.<ref name="Bransden159" />