Particles in the box

The particle in the box is a model in quantum mechanics in which a free particle is in a box potential . It is a special case of the potential well , in which the potential is zero in a certain area and infinite outside of it. The model system makes the quantization of the energy understandable. As a one-dimensional model, it is comparatively easy to calculate.

The potential, here denoted by, is infinitely large outside the potential box and zero inside. ${\displaystyle V}$

Structure and requirements

The one-dimensional model system consists of a free particle , for example a gas molecule, which is located in the potential-free space between two infinitely large potentials. The boundaries called “walls” (one at and one at ) are orthogonal to the x-axis and thus parallel to each other. This simplistic model of a potential well is called a potential box . ${\displaystyle x=0}$ ${\displaystyle x=L}$

In the model, no forces act on the particle within the potential box of the length ( gravitation and electromagnetic fields are not taken into account). Since the potential outside the box is infinitely large, the particle cannot leave the box. It follows that the particle inside the box moves with constant speed and is reflected on the walls without any loss of energy. If one considers a vector quantity , then it applies that the amount of the speed remains constant. ${\displaystyle L}$ ${\displaystyle v}$ ${\displaystyle v}$

State function and probability of encounter

In the potential box there can only exist waves for which is a multiple of half its wavelength . ${\displaystyle L}$ ${\displaystyle \lambda }$

If one describes the particle, as is usual in quantum physics, with the help of a simple wave function , it follows that in the interior of the potential box only those energy eigenfunctions are permissible for which is an integral multiple of half its wavelength. ${\displaystyle L}$

Another quantum mechanical peculiarity in the model is the probability of encounter, i.e. the probability of encountering the particle at a certain location. The probability of finding the particle somewhere in the potential box is because it cannot leave the box. Everywhere outside the box, the probability of being hit is accordingly . The probability of encountering individual points within the box is different and depends on the state of the particle. ${\displaystyle 1}$ ${\displaystyle 0}$

Another peculiarity of quantum mechanics, the tunnel effect , does not occur with the potential described here, but only with a finitely high potential well.

energy

Because only certain individual eigenstates are allowed for a particle within a potential box , they can only have certain discrete , dependent energy values. This also applies to finitely high “walls” and has far-reaching effects on, for example, understanding the structure of atoms . With the assumptions made above, the energy of a particle can be derived depending on the following equation: ${\displaystyle n}$ ${\displaystyle n}$ ${\displaystyle n}$

${\displaystyle E_{n}={\frac {h^{2}}{8mL^{2}}}n^{2}\quad {\text{mit}}\quad n=1,2,3,\ldots }$

If a particle is excited , i.e. if energy is supplied to an atom through irradiation, it changes directly to a higher energy level without a “flowing” transition (“ quantum leap ”). If a particle changes to a lower energy level, it releases the energy released, for example in the form of a photon .

From the above equation, three simple conclusions can be drawn that describe the particle in the potential box qualitatively:

1. The energy of the particle is proportional to the square of the quantum number ( ) ${\displaystyle n}$ ${\displaystyle E\sim n^{2}}$
2. The longer the potential box, the smaller the energy of the particle ( ) ${\displaystyle E\sim L^{-2}}$
3. The longer the potential box, the smaller the difference between two energy levels and . ${\displaystyle E_{n}}$ ${\displaystyle E_{n+1}}$

These statements apply mutatis mutandis to other potential pots.

The solutions of the Schrödinger equation lead to the quantization of the energy

The Hamilton operator of the one-dimensional problem reads in position representation

${\displaystyle H=-{\frac {\hbar ^{2}}{2m}}{\frac {\mathrm {d} ^{2}}{\mathrm {d} x^{2}}}+V(x)\ ,\quad V(x)={\begin{cases}0&(0<x<L)\\\infty &(x\leq 0,\ L\leq x)\end{cases}}}$

The Schrödinger equation

${\displaystyle \mathrm {i} \hbar {\frac {\partial }{\partial t}}\Psi (x,t)=H\Psi (x,t)}$

goes with the approach

${\displaystyle \Psi (x,t)=\psi (x)\,\exp \left(-\mathrm {i} {\frac {E}{\hbar }}t\right)}$

into the time-independent (stationary) Schrödinger equation .

${\displaystyle H\,\psi (x)=E\,\psi (x)}$

In the following the time-independent Schrödinger equation has to be solved ( eigenvalue problem of the Hamilton operator)

Inside the box

The stationary Schrödinger equation corresponds within the box to that of a free particle ( ordinary differential equation of the 2nd order)

${\displaystyle -{\frac {\hbar ^{2}}{2m}}{\frac {\mathrm {d} ^{2}}{\mathrm {d} x^{2}}}\psi (x)=E\psi (x)\ ,\quad (0\leq x\leq L)}$

The following approach is chosen for the wave function inside the box ${\displaystyle \psi (x)}$

${\displaystyle \,\psi (x)=A\sin(kx)+B\cos(kx)}$

The approach with complex exponential functions would be equivalent . ${\displaystyle \psi (x)=A\exp(\mathrm {i} kx)+B\exp(-\mathrm {i} kx)}$

This approach is used in the Schrödinger equation, the second derivative being according to the location . ${\displaystyle {\tfrac {\mathrm {d} ^{2}}{\mathrm {d} x^{2}}}\psi (x)=-k^{2}\psi (x)}$

${\displaystyle -{\frac {\hbar ^{2}}{2m}}(-k^{2})\psi (x)=E\psi (x)}$

This gives the energy as a function of the wave number : ${\displaystyle E}$ ${\displaystyle k}$

${\displaystyle E={\frac {\hbar ^{2}k^{2}}{2m}}}$

Outside the box, continuity condition

Outside the box, the wave function must be identically zero due to the infinitely high potential.

${\displaystyle \,\psi (x)=0\ ,\quad (x<0,\ x>L)}$

However, since the wave function has to be continuous everywhere, boundary conditions are placed on the wave function in the box, namely that the wave function on the walls is equal to 0: ${\displaystyle \,\psi (x)}$

${\displaystyle \psi (x=0)=0\quad {\text{ und }}\quad \psi (x=L)=0}$.

Boundary condition 1

From the first boundary condition it follows for the wave function inside the box

${\displaystyle {\begin{matrix}\psi (x=0)&=&A\sin(k\cdot 0)+B\cos(k\cdot 0)\\&=&A\cdot 0+B\cdot 1\\&{\overset {!}{=}}&0\end{matrix}}}$.

For this equation to be satisfied, it must be. This simplifies the wave function to ${\displaystyle B=0}$

${\displaystyle \,\psi (x)=A\sin(kx)}$.

Boundary condition 2

With the help of the second boundary condition it then follows for the wave function inside the box

${\displaystyle \,\psi (x=L)=A\sin(kL)\,{\overset {!}{=}}\,0}$.

In order for this equation to be fulfilled, it has to be a whole multiple of (the trivial solution would mean that no wave exists), that is ${\displaystyle kL}$ ${\displaystyle \pi }$ ${\displaystyle A=0}$

${\displaystyle kL=n\pi \quad {\text{ wobei }}\quad n=1,2,3,\dots }$

The wavenumber can therefore only assume discrete values ${\displaystyle k}$

${\displaystyle k=k_{n}={\frac {\pi }{L}}n\ ,\quad n\in \mathbb {N} }$

Actually it only follows from the second boundary condition that is an integer. For , however, the wave function would be zero everywhere and thus the normalization condition would not have to be fulfilled, so it is not allowed. For negative , the wave function is the same except for the sign as for the positive , namely . Since wave functions that differ by a factor describe the same state, the negative integers do not produce new states. That's why you limit yourself to . ${\displaystyle n\in \mathbb {Z} }$ ${\displaystyle n=0}$ ${\displaystyle \psi (x)=A\sin(0\cdot x)=0}$ ${\displaystyle n=0}$ ${\displaystyle n^{\prime }=-n<0}$ ${\displaystyle n}$ ${\displaystyle \psi (x)=A\sin(k_{n^{\prime }}x)=A\sin(-k_{n}x)=-A\sin(k_{n}x)}$ ${\displaystyle n\in \mathbb {N} }$

As calculated above, the energy depends on the wave number , inserting gives: ${\displaystyle E}$ ${\displaystyle k}$

${\displaystyle E=E_{n}={\frac {\hbar ^{2}k_{n}^{2}}{2m}}={\frac {\hbar ^{2}\pi ^{2}}{2mL^{2}}}n^{2}={\frac {h^{2}}{8mL^{2}}}n^{2}\ ,\quad n\in \mathbb {N} }$

Since only integer values ​​can be assumed, the energy can also only assume certain values. The energy of the particle is thus quantized, the energy levels are "discrete". ${\displaystyle n}$

Normalization

The amplitude can still be determined using the normalization condition: ${\displaystyle A}$

${\displaystyle {\begin{aligned}1&{\overset {!}{=}}\int _{\mathbb {R} }\psi _{n}^{*}(x)\psi _{n}(x)\mathrm {d} x=|A|^{2}\int _{0}^{L}\sin ^{2}\left(n{\frac {\pi }{L}}x\right)\mathrm {d} x=|A|^{2}\left[{\frac {x}{2}}-{\frac {L}{4n\pi }}\sin \left(2n{\frac {\pi }{L}}x\right)\right]_{0}^{L}\\&=|A|^{2}\left({\frac {L}{2}}-{\frac {L}{4n\pi }}\underbrace {\sin \left(2n\pi \right)} _{=0}\right)=|A|^{2}{\frac {L}{2}}\end{aligned}}}$

Since is a complex number, only its amount is fixed, the phase is arbitrary: ${\displaystyle A}$ ${\displaystyle \phi \in \mathbb {R} }$

${\displaystyle A={\sqrt {\frac {2}{L}}}e^{i\phi }}$

Wave functions that differ only by a constant phase factor describe the same state. Therefore one can bet and thus choose real. ${\displaystyle \phi =0}$ ${\displaystyle A={\sqrt {\tfrac {2}{L}}}}$

Summary

The eigenvalues ​​(= possible energy values) and eigenfunctions (= wave functions) of the Hamilton operator for a particle in the box with infinitely high potential walls are:

${\displaystyle E_{n}={\frac {\hbar ^{2}\pi ^{2}}{2mL^{2}}}n^{2}\ ,\quad \psi _{n}(x)={\begin{cases}{\sqrt {\tfrac {2}{L}}}\sin(n{\frac {\pi }{L}}x)&,\ (0\leq x\leq L)\\0&,\ (x<0,\ x>L)\end{cases}}\ ,\quad n\in \mathbb {N} \setminus \{0\}}$

Basic state

The ground state energy (lowest possible energy) is not zero ( is not allowed because of the Heisenberg uncertainty principle), but ${\displaystyle n=0}$

${\displaystyle E_{1}={\frac {\hbar ^{2}\pi ^{2}}{2mL^{2}}}}$

This can also be obtained from considering Heisenberg's uncertainty principle : The particle is restricted to the spatial area . Then the minimum impulse results from . Inside the box, the potential is zero, so the total energy is equal to the kinetic energy . ${\displaystyle \Delta x\Delta p\geq h/2}$ ${\displaystyle x_{\text{max}}=L}$ ${\displaystyle x_{\text{max}}p_{\text{min}}=h/2}$ ${\displaystyle E=p^{2}/2m}$

${\displaystyle p_{\text{min}}={\frac {h}{2x_{\text{max}}}}={\frac {h}{2L}}\quad \Rightarrow \quad E_{\text{min}}={\frac {p_{\text{min}}^{2}}{2m}}={\frac {h^{2}}{8mL^{2}}}={\frac {\hbar ^{2}\pi ^{2}}{2mL^{2}}}}$

Temporal development of a wave packet in the box potential

The temporal development of the wave function is given by

${\displaystyle \Psi (x,t)=\sum _{n=1}^{\infty }a_{n}e^{-{\frac {\mathrm {i} }{\hbar }}E_{n}t}\psi _{n}(x)}$

where the coefficients result from the initial condition: ${\displaystyle a_{n}}$

${\displaystyle a_{n}=\int _{-\infty }^{\infty }\psi _{n}^{\dagger }(x)\Psi (x,0)\mathrm {d} x=\int _{0}^{L}\psi _{n}^{\dagger }(x)\Psi (x,0)\mathrm {d} x}$

Using a variable transformation , the problem can be posed in such a way that it is symmetrical around zero. It is then ${\displaystyle x\to y=x-{\bar {x}}}$

${\displaystyle \psi (y)={\sqrt {\frac {2}{L}}}{\begin{cases}\sin(n\pi y/L)&n{\text{ ungerade}}\\\cos(n\pi y/L)&n{\text{ gerade}}\end{cases}}}$

This leads to the total wave function

${\displaystyle \Psi (y,t)={\frac {2}{L}}\sum _{n\ {\text{gerade}}}^{\infty }\left[\int _{-L/2}^{L/2}\mathrm {d} y'\Psi (y',0)\cos({\tfrac {n\pi y'}{L}})\right]e^{-{\frac {\mathrm {i} \hbar \pi ^{2}n^{2}t}{2mL^{2}}}}\cos({\tfrac {n\pi y}{L}})+{\frac {2}{L}}\sum _{n\ {\text{ungerade}}}^{\infty }\left[\int _{-L/2}^{L/2}\mathrm {d} y'\Psi (y',0)\sin({\tfrac {n\pi y'}{L}})\right]e^{-{\frac {\mathrm {i} \hbar \pi ^{2}n^{2}t}{2mL^{2}}}}\sin({\tfrac {n\pi y}{L}})}$.

The total wave function is periodic in time with a period

${\displaystyle T_{R}={\frac {4mL^{2}}{\pi \hbar }}}$,

called the revival time . That is, it applies . This is a characteristic property of the box potential, since here all energy eigenvalues ​​are integral multiples of the ground state energy . ${\displaystyle \Psi (y,t+T_{R})=\Psi (y,t)}$ ${\displaystyle E_{1}}$

Interesting structures can also develop for rational multiples of . If natural numbers are with , then applies ${\displaystyle T_{R}}$ ${\displaystyle k,l}$ ${\displaystyle k<l}$

${\displaystyle e^{-{\frac {\mathrm {i} }{\hbar }}E_{n}\left(t+{\frac {k}{l}}T_{R}\right)}=e^{-{\frac {\mathrm {i} }{\hbar }}E_{n}t}e^{-2\pi \mathrm {i} n^{2}{\frac {k}{l}}}}$

For and even the phase factor gives the value , for odd the value . So it is ${\displaystyle l=2}$ ${\displaystyle n}$ ${\displaystyle +1}$ ${\displaystyle n}$ ${\displaystyle -1}$

${\displaystyle \Psi (y,t+T_{R}/2)=\Psi ^{\text{gerade}}(y,t)-\Psi ^{\text{ungerade}}(y,t)=\Psi (-y,t)}$

and the wave function is mirrored around the center of the box. That is, a wave packet that was initially located in the left half of the box appears on the right after half the revival time. This is called a mirror revival . Trivially, the following applies to the probability density:

${\displaystyle \rho (y,t+T_{R}/2)=\rho (-y,t)}$

For and even the phase factor gives the value and for odd the value . If real, then: ${\displaystyle l=4}$ ${\displaystyle n}$ ${\displaystyle +1}$ ${\displaystyle n}$ ${\displaystyle -\mathrm {i} }$ ${\displaystyle \Psi (x,t)}$

${\displaystyle \rho (y,t+T_{R}/4)=\rho ^{\text{gerade}}(y,t)+\rho ^{\text{ungerade}}(y,t)={\frac {1}{2}}\rho (y,t)+{\frac {1}{2}}\rho (-y,t)}$

In this case, the wave packet is divided into two parts, each with half the probability density on both sides. This case is called fractional revival .

Also for the other times with (small) whole numbers and typically several approximate reproductions of the wave packet arise. If one plots the development over time (ordinate) against the spatial distribution of the wave packet in a diagram, a strong structure in place and time can be seen as pronounced trenches in which the density of the probability of presence is very small. The shape of the diagram is reminiscent of the shape of an oriental rug. One therefore also speaks of the quantum carpet . ${\displaystyle \textstyle {\frac {k}{l}}T_{R}}$ ${\displaystyle k}$ ${\displaystyle l}$

Three-dimensional case (cuboid)

In the three-dimensional box (cuboid) the Hamilton operator looks like this:

${\displaystyle H=-{\frac {\hbar ^{2}}{2m}}\left({\frac {\partial ^{2}}{\partial x_{1}^{2}}}+{\frac {\partial ^{2}}{\partial x_{2}^{2}}}+{\frac {\partial ^{2}}{\partial x_{3}^{2}}}\right)+V_{1}(x_{1})+V_{2}(x_{2})+V_{3}(x_{3})}$

Here is the potential

${\displaystyle V_{i}(x_{i})={\begin{cases}0&(0\leq x_{i}\leq L_{i})\\\infty &(x_{i}<0,\ x_{i}>L_{i})\end{cases}}}$

The complete Hamilton operator can be found using

${\displaystyle H_{i}=-{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial x_{i}^{2}}}+V_{i}(x_{i})}$

write as the sum of three one-dimensional Hamilton operators:

${\displaystyle H=H_{1}+H_{2}+H_{3}\,}$

Separation approach

The stationary Schrödinger equation (three-dimensional)

${\displaystyle H\,\psi ({\vec {r}}\,)=E\,\psi ({\vec {r}}\,)}$

can be achieved with the following product approach

${\displaystyle \psi ({\vec {r}}\,)=\psi _{1}(x_{1})\psi _{2}(x_{2})\psi _{3}(x_{3})}$

separate into three one-dimensional problems.

To do this, insert the product approach into the stationary Schrödinger equation and take advantage of the fact that only acts on, i.e. H. the others can be passed by the Hamilton operator. ${\displaystyle H_{i}}$ ${\displaystyle \psi _{i}}$ ${\displaystyle \psi _{j}}$

${\displaystyle {\psi _{2}(x_{2})\psi _{3}(x_{3})\,H_{1}\psi _{1}(x_{1})+\psi _{1}(x_{1})\psi _{3}(x_{3})\,H_{2}\psi _{2}(x_{2})+\psi _{1}(x_{1})\psi _{2}(x_{2})\,H_{3}\psi _{3}(x_{3})=E\psi _{1}(x_{1})\psi _{2}(x_{2})\psi _{3}(x_{3})}}$

Share by supplies: ${\displaystyle \psi _{1}(x_{1})\psi _{2}(x_{2})\psi _{3}(x_{3})}$

${\displaystyle \underbrace {\frac {H_{1}\psi _{1}(x_{1})}{\psi _{1}(x_{1})}} _{E_{1}}+\underbrace {\frac {H_{2}\psi _{2}(x_{2})}{\psi _{2}(x_{2})}} _{E_{2}}+\underbrace {\frac {H_{3}\psi _{3}(x_{3})}{\psi _{3}(x_{3})}} _{E_{3}}=E}$

In this separation, the three constants , , defines the sum total energy yields: ${\displaystyle E_{1}}$ ${\displaystyle E_{2}}$ ${\displaystyle E_{3}}$ ${\displaystyle E}$

${\displaystyle E=E_{1}+E_{2}+E_{3}\,}$

One-dimensional problems

Now the one-dimensional problem has to be solved separately for each spatial direction, as already done above:

${\displaystyle H_{i}\,\psi _{i}(x_{i})=E_{i}\,\psi _{i}(x_{i})}$

Their solution is:

${\displaystyle E_{i,l}={\frac {\hbar ^{2}\pi ^{2}}{2mL_{i}^{2}}}n_{i,l}^{2}\ ,\quad \psi _{i,l}(x_{i})={\begin{cases}{\sqrt {\tfrac {2}{L_{i}}}}\sin(n_{i,l}{\frac {\pi }{L_{i}}}x_{i})&,\ (0\leq x_{i}\leq L_{i})\\0&,\ (x_{i}<0,\ x_{i}>L_{i})\end{cases}}\ ,\quad n_{i}\in \mathbb {N} }$

Stationary overall solution

The solution of the three-dimensional box is the product of the one-dimensional wave functions for the total wave function

${\displaystyle {\begin{aligned}\psi _{l_{1},l_{2},l_{3}}({\vec {r}}\,)&=\psi _{1}(x_{1})\psi _{2}(x_{2})\psi _{3}(x_{3})\\&={\begin{cases}{\sqrt {\frac {8}{L_{1}L_{2}L_{3}}}}\sin(n_{1,l_{1}}{\frac {\pi }{L_{1}}}x_{1})\sin(n_{2,l_{2}}{\frac {\pi }{L_{2}}}x_{2})\sin(n_{3,l_{3}}{\frac {\pi }{L_{3}}}x_{3})&,\quad 0\leq x_{1}\leq L_{1},\ 0\leq x_{2}\leq L_{2},\ 0\leq x_{3}\leq L_{3}\\0&,\quad {\text{sonst}}\end{cases}}\end{aligned}}}$

and for the total energy the sum of the one-dimensional energy eigenvalues:

${\displaystyle E_{l_{1},l_{2},l_{3}}=E_{1,l_{1}}+E_{2,l_{2}}+E_{3,l_{3}}={\frac {\hbar ^{2}\pi ^{2}}{2m}}\left({\frac {n_{1,l_{1}}^{2}}{L_{1}^{2}}}+{\frac {n_{2,l_{2}}^{2}}{L_{2}^{2}}}+{\frac {n_{3,l_{3}}^{2}}{L_{3}^{2}}}\right)\ ,\quad n_{1},n_{2},n_{3}\in \mathbb {N} }$

degeneration

The energy eigenvalues ​​can be degenerate , i.e. H. different wave functions have the same energy. For the three-dimensional box, this means that different quantum numbers lead to the same sum . ${\displaystyle n_{1},n_{2},n_{3}}$ ${\displaystyle {\tfrac {n_{1}^{2}}{L_{1}^{2}}}+{\tfrac {n_{2}^{2}}{L_{2}^{2}}}+{\tfrac {n_{3}^{2}}{L_{3}^{2}}}}$

For example, in the special case of the cube, that is , degeneracy occurs. The energy is given by: ${\displaystyle L_{1}=L_{2}=L_{3}}$

${\displaystyle E={\frac {\hbar ^{2}\pi ^{2}}{2mL^{2}}}\left(n_{1}^{2}+n_{2}^{2}+n_{3}^{2}\right)\ ,\quad n_{1},n_{2},n_{3}\in \mathbb {N} }$

For degeneracy, different quantum numbers must lead to the same sum . ${\displaystyle n_{1},n_{2},n_{3}}$ ${\displaystyle n_{1}^{2}+n_{2}^{2}+n_{3}^{2}}$

The lowest energy value is not degenerate (= simply degenerate) thus and . ${\displaystyle n_{1}=n_{2}=n_{3}=1}$ ${\displaystyle n_{1}^{2}+n_{2}^{2}+n_{3}^{2}=3}$ ${\displaystyle E=3{\tfrac {\hbar ^{2}\pi ^{2}}{2mL^{2}}}}$

The next higher energy value has already degenerated three times: thus and . ${\displaystyle (n_{1},n_{2},n_{3})=(2,1,1),\,(1,2,1),\,(1,1,2)}$ ${\displaystyle n_{1}^{2}+n_{2}^{2}+n_{3}^{2}=6}$ ${\displaystyle E=6{\tfrac {\hbar ^{2}\pi ^{2}}{2mL^{2}}}}$

Degenerations higher than three can also occur, e.g. B. 4-fold thus and . ${\displaystyle (n_{1},n_{2},n_{3})=(3,3,3),\,(5,1,1),\,(1,5,1),\,(1,1,5)}$ ${\displaystyle n_{1}^{2}+n_{2}^{2}+n_{3}^{2}=27}$ ${\displaystyle E=27{\tfrac {\hbar ^{2}\pi ^{2}}{2mL^{2}}}}$

Three-dimensional case (sphere)

For the three-dimensional spherical box with a radius , it makes sense to represent the Hamilton operator in spherical coordinates : ${\displaystyle L}$

${\displaystyle H=-{\frac {\hbar ^{2}}{2m}}\left({\frac {1}{r^{2}}}{\frac {\partial }{\partial r}}\left(r^{2}{\frac {\partial }{\partial r}}\right)+{\frac {1}{r^{2}\sin \vartheta }}{\frac {\partial }{\partial \vartheta }}\left(\sin \vartheta {\frac {\partial }{\partial \vartheta }}\right)+{\frac {1}{r^{2}\sin ^{2}\vartheta }}{\frac {\partial ^{2}}{\partial \varphi ^{2}}}\right)+V}$

Here is the potential

${\displaystyle V(r)={\begin{cases}0&(0\leq r\leq L)\\\infty &(r>L)\end{cases}}}$

Separation approach

As with the hydrogen atom , the Schrödinger equation can be separated into two independent equations, where the wave function results from the product of a radius-dependent function and the spherical surface functions: ${\displaystyle R_{nl}(r)}$ ${\displaystyle Y_{lm}(\vartheta ,\varphi )}$

${\displaystyle \Psi _{nlm}(r,\vartheta ,\varphi )=R_{nl}(r)Y_{lm}(\vartheta ,\varphi )}$

Here, too, is the main or energy quantum number , the angular momentum quantum number and the magnetic quantum number. ${\displaystyle \,n}$ ${\displaystyle \,l}$ ${\displaystyle \,m}$

The following radial Schrödinger equation remains for the radius-dependent function (whereby V = 0 within the box was taken into account):

${\displaystyle -{\frac {\hbar ^{2}}{2m}}\left({\frac {1}{r^{2}}}{\frac {\partial }{\partial r}}\left(r^{2}{\frac {\partial }{\partial r}}\right)\right)R(r)-AR(r)=ER(r)}$

A results from solving the angle-dependent Schrödinger equation:

${\displaystyle A=-{\frac {\hbar ^{2}}{2m}}{\frac {l(l+1)}{r^{2}}}}$

Spherically symmetric solutions

At first only the simple case is considered (s-like wave functions). The term disappears from the radial Schrödinger equation. ${\displaystyle l=0}$ ${\displaystyle AR(r)}$

Additionally be set. It follows: ${\displaystyle u(r)=rR(r)}$

${\displaystyle R(r)={\frac {u(r)}{r}};\;{\frac {\partial }{\partial r}}R(r)={\frac {1}{r}}{\frac {\partial }{\partial r}}u(r)-{\frac {u(r)}{r^{2}}};\;}$

${\displaystyle r^{2}{\frac {\partial }{\partial r}}R(r)=r{\frac {\partial }{\partial r}}u(r)-u(r);\;}$

${\displaystyle {{\frac {\partial }{\partial r}}\left(r^{2}{\frac {\partial }{\partial r}}\right)R(r)={\frac {\partial }{\partial r}}\left(r{\frac {\partial }{\partial r}}u(r)-u(r)\right)={\frac {\partial }{\partial r}}u(r)+r{\frac {\partial ^{2}}{\partial r^{2}}}u(r)-{\frac {\partial }{\partial r}}u(r)=r{\frac {\partial ^{2}}{\partial r^{2}}}u(r);}}$

${\displaystyle {\frac {1}{r^{2}}}{\frac {\partial }{\partial r}}\left(r^{2}{\frac {\partial }{\partial r}}\right)R(r)={\frac {1}{r}}{\frac {\partial ^{2}u(r)}{\partial r^{2}}}.}$

This simplifies the radial Schrödinger equation to:

${\displaystyle -{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial r^{2}}}u(r)=Eu(r)}$

As can be seen directly, the solution is the same as for the particle in the linear box: resp. ${\displaystyle u(r)}$ ${\displaystyle u(r)=a\sin(kr)+b\cos(kr)}$

${\displaystyle R(r)={\frac {a\sin(kr)+b\cos(kr)}{r}}}$

Since the potential is continuous at the origin, the wave function must not become singular there, so that the term is omitted. In addition, the boundary condition applies because of the continuity of the wave function. It follows for : ${\displaystyle \cos }$ ${\displaystyle R(L)=0}$ ${\displaystyle k}$

${\displaystyle k={\frac {n\pi }{L}}}$

Substituting into the radial Schrödinger equation yields: ${\displaystyle u(r)}$

${\displaystyle -{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial r^{2}}}\left(a\cdot \sin \left({\frac {n\pi }{L}}r\right)\right)=E_{n0}u(r)}$,

hence the energy eigenvalues with can be determined. ${\displaystyle E_{nl}}$ ${\displaystyle l=0}$

In summary: For (spherically symmetric solutions) the wave functions with the normalization constant and the energy eigenvalues result in: ${\displaystyle l=0}$ ${\displaystyle R_{n0}}$ ${\displaystyle a}$ ${\displaystyle E_{n0}}$

${\displaystyle R_{n0}(r)=a\cdot {\frac {\sin({\frac {n\pi }{L}}r)}{r}}}$

${\displaystyle E_{n0}={\frac {n^{2}h^{2}}{8mL^{2}}}}$

Non-spherically symmetric solution

For the solution of the Schrödinger equation is considerably more complicated. For spherical Bessel functions , which are related to the normal Bessel functions as follows: ${\displaystyle l>0}$ ${\displaystyle R_{nl}(r)}$ ${\displaystyle j_{l}}$ ${\displaystyle J_{l}}$

${\displaystyle j_{l}(r)={\sqrt {\frac {\pi }{2r}}}J_{l+1/2}(r).}$

${\displaystyle E_{nl}}$depends on the square of the -th root of these functions because of the boundary condition : ${\displaystyle R_{nl}(L)=0}$ ${\displaystyle n}$ ${\displaystyle x_{nl}}$

${\displaystyle E_{nl}={\frac {x_{nl}^{2}\hbar ^{2}}{2mL^{2}}},}$

which can not be determined analytically. ${\displaystyle x_{nl}}$

Model for conjugated systems

The particle in the box can be used as a simple model for a conjugated molecule, e.g. B. hexatriene , can be used to estimate its energy. It is assumed that the electrons in a conjugated molecule can move freely in it, but cannot leave it. You formally add half an atom at each end of the molecule. The length of this particle then corresponds to the box in which the electron is located.

Examples

An example from crystallography is the color center , where an electron is locked in an anion void and which can be described as a particle in a box to a good approximation. The color of dyes with linear conjugated pi systems can also be determined by considering the pi system as a one-dimensional particle in the box problem.

See also

• Particles on the ring
• Harmonic oscillator
• Anharmonic oscillator
• Born von Kármán model

literature

• BH Bransden, CJ Joachain: Quantum mechanics , 2nd. Edition, Pearson Education, Essex 2000, ISBN 0-582-35691-1 .
• John H. Davies: The Physics of Low-Dimensional Semiconductors: An Introduction , 6th reprint. Edition, Cambridge University Press, 2006, ISBN 0-521-48491-X .
• David J. Griffiths: Introduction to Quantum Mechanics , 2nd. Edition, Prentice Hall, 2004, ISBN 0-13-111892-7 .

Web links

Commons : 1D infinite square wells  - collection of images, videos and audio files

Individual evidence

1. ^ Abramowitz and Stegun: Page 437 .