Physical Chemistry. The Particle in a Box I: the Schrödinger Equation in One-dimension

/ Danilo Roccatano

In 1926, the Austrian physicist Erwin Schrödinger (1887-1961) made a fundamental mathematical discovery that had a profound impact on the study of the molecular world (in 1933, Schrödinger was awarded the Nobel Prize in Physics just seven years later, his breakthrough discovery). He discovered that a quantum system’s state composed of particles (such as electrons and nucleons) could be described by postulating the existence of a function of the particle coordinates and time, called state function or wave function (\Psi, psi function). This function is the solution of a wave equation: the Schrödinger equation (SE). Although the SE equation can be solved analytically only for relatively simple cases, the development of computer and numerical methods has made possible the application of SE to study complex molecules. 

Nowadays, computational quantum chemistry has surged to a level of a fundamental research discipline in chemistry and material science. In recognition of their contribution to the advancement of science, the pioneers of this discipline, Prof. Walter Kohn and Prof. John A. Pople, have been awarded the 2008 Nobel Prize in chemistry.

In this primer, we will study analytical and numerical solutions of SE for simple quantum systems in stationary conditions, e.g. whose state does not change with time. For this purpose, we will use the time-independent form of SE. This condition is an excellent approximation to describe the state of molecular systems. 

The time-independent SE is given in its compact form by the equation.

{\bf \hat H }\Psi(x,y,z)=E\Psi(x,yz),

where the \bf \hat H (H hat) indicates the time-independent Hamiltonian operator used to calculate the system’s total energy, I recall that an operator is a mathematical transformation that applies to what follows the operator symbol, in this case, the wave function (\Psi(x,y,z)). We will see in a moment how this looks. The application of the operator produces, as a result, the same wave function multiplied for a number $E.$ This number represented the total energy state of the system (or \em eigenstate\em). This equation is also called the eigenvalue equation, and, in this case, \Psi(x,y,z) is also said to be an eigenfunction of \bf \hat H with the associated \em eigenvalue,\em E.  

We now look more in detail at the Hamiltonian operator. For the sake of simplicity, we will limit ourselves to consider only a mono-dimensional physical system, namely a system defined only along a reference x-axis. Therefore, the wave function depends only on by x or \Psi(x). The Hamiltonian operator represents the total energy of the system and, therefore, is composed by the summation of a kinetic and potential energy operator: \bf \hat H = \hat K+ \hat V. The kinetic energy is expressed as  

E=p_x^2/2m where p_x  is the classic momentum that is associated with the partial differential operator  p_x \rightarrow \left(\hbar/i \right)\left( \partial /\partial x\right). The potential energy is usually a function of the coordinates and is associated (in our case) with operator \bf \hat x. Therefore, we can write the resulting SE as:

 -\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2}+V(x)\psi(x)=E\psi(x)

This differential equation can be solved analytically for different types of potentials. A simple classical example is a particle in a one-dimension box within infinitely repulsive barriers.

In this case, the value of V(x) equals zero for and infinite in the walls. The wave function beyond the walls of the box is null. The wave function and energies of the particle inside the box are obtained by the solution of SE with V(x)=0 reported in the central part of Figure 1. 

Figure 1: Schematic representation of the one-dimensional box system.

Derivation of the wavefunction

 The one-dimension SE is a linear second-order differential equation with solutions 

 \psi(x)=Ae^{ikx}+Be^{-ikx}

using the Euler relation e^{\pm ikx}=\cos(kx)\pm i\sin(kx)

we can write 

\psi(x)=(A+B)\cos(kx)+(A-B)\sin(kx)

or called C=A+B and D=A-B we can write

\psi_k(x)=C\sin(kx)+D\cos(kx)

by substituting in SE with V(x)=0 (e.g. particle inside the box)

 -\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2}=E\psi(x)

 we obtain 

-\frac{\hbar^2}{2m}\left( -C\sin(kx)-D\cos(kx)\right) = E_k\left(C\sin(kx)+D\cos(kx)\right)

that gives

E_k=\frac{k^2\hbar^2}{2m}

 the quantised total energy of the system.

Now we  estimate the values of  constant k, C, and  D of the solution.

By applying the boundary conditions

\psi(0)=0; ~\psi(L)=0

we obtain for x=0 that 

\psi(0) =C\sin(0)+D\cos(0)=D=0

It is more convenient to represent the parameter k as a circular frequency with \lambda the De Broglie’s wavelength of the particle

k=\frac{2\pi}{\lambda}

For x=L the solution gives

\psi(L)=C\sin \left( \frac{2\pi L}{\lambda} \right) =0

This is accomplished for \frac{2\pi L}{\lambda} = n\pi or

\lambda =\frac{2L}{n}

and

\psi(x)=C\sin \left( \frac{n\pi x}{L} \right) \text{with} ~n=1,2, \dots

We still need to calculate the value of the constant C.  As the square of the wave function \psi(x) gives the probability density and the integral the probability of finding a particle in a region of the box, then if we integrate over all the x-axis, the probability should be equal to one:

\int _{0}^{L}\psi^2(x) dx = C^2 \int_{0}^{L} \sin^2 \left( \frac{n \pi x}{L} \right) dx =1 (1)

Using this integral we can find the value of C that normalize the function. The integral  \int_{0}^{L} \sin^2 (Kx) dx = \int_{0}^{L} \sin (x) \sin(Kx)dx with K=\frac{n\pi }{L} can be calculated by part 

  as \int uw' dx = uw- \int u'w with in this case u=sin(Kx); w'=K sin(Kx) therefore 

\int_{0}^L \sin^2 (x) dx = -\frac{\sin (Kx) \cos(Kx)}{K} \bigg|_{0}^L + K\int_0^L \cos^2(x) dx =  -\frac{\sin (Kx) \cos(Kx)}{K} \bigg|_{0}^L  + \int_0^L (1-\sin^2(Kx)) dx

  so we have  that 

   \int_0^L \sin^2 (Kx) dx =-\frac{\sin (Kx) \cos(Kx)}{K} \bigg|_{0}^L + \int_0^L dx - \int_0^L sin^2(Kx) dx

   taking the last integral in the second member of the equation to the first member we have 

2\int_0^L \sin^2 (Kx) dx = -\frac{\sin (Kx) \cos(Kx)}{K} \bigg|_{0}^L + x \bigg|_{0}^L

    but the first term in the second member is equal to zero  and the second equal to L, hence

\int_0^L \sin^2 (Kx) dx = \frac{L}{2}.

    Therefore from Equ. (1), we have

C^2 \frac{L}{2}=1

and C=\sqrt{\frac{2}{L}}.

Therefore the wavefunctions of a particle in a box are given by

\psi(x)=\sqrt{\frac{2}{L}} \sin \left( \frac{n\pi x}{L} \right) \text{with} ~n=1,2, \dots,~ \text {and} ~0\le x \le L.

and the associated energies are 

E_n=\frac{n^2 h^2}{8mL^2}.

 The quantum number n is an integer associated with the different energy levels of the particle. Larger n values correspond to higher energy levels of the system. Also, note in the graphs the increase in the number of nodes in the wave function with the increase of the value of n.

In Figure 2, the wavefunctions for the first three energy level of the particle in a box are shown.

Figure 2: Wavefunctions of the first 3 energy levels of a particle in a box.

The separation between the adjacent energy level with quantum number $n$ and $n+1$ is

\Delta E=E_{n+1}-E_n =(2n+1)\frac{h^2}{8mL^2}

The probability density for a particle in a box is given by the square of the wavefunction as follows

\psi^2(x)=\frac{2}{L}\sin^2 \left(\frac{n\pi x}{L} \right)

and the probability to find the particle in a region [x_1,x_2] of the box can be calculated using the following integral

P=\int_{x_1}^{x_2} \psi^2(x) dx.

READINGS

  • Atkins, P. and Paula, J. (2010) Physical Chemistry. 9th Edition, W. H. Freeman Co., New York.
  • I. Levine. Quantum Chemistry. VI edition. Pearson International edition.

5 thoughts on “Physical Chemistry. The Particle in a Box I: the Schrödinger Equation in One-dimension”

  1. Esercizio interessante (problemi unidimensionali in meccanica quantistica nonrelativistica). Tempo addietro ne avevo svolto uno simile. Precisamente, un oscillatore armonico unidimensionale, però considerando uno stato quantistico descritto da una funzione d’onda che pur essendo normalizzabile, non si annulla all’infinito: http://www.extrabyte.info/2019/03/01/gli-stati-ghost-delloscillatore-armonico-unidimensionale/

    Una tale soluzione ha un significato fisico? A me sembra di no, per la semplice ragione che si dovrebbe congetturare l’esistenza di un processo di tunnelling in presenza di un potenziale divergente all’infinito.

    Like

    Reply

    • Grazie Marcello per il tuo commento. I tuoi posts sono molto interessanti e decisamente piu’ avanzati del mio. Aggiungero’ dei link per chi e’ interessato ad approfondire l’argomento. Lo scopo di questi miei articoli e’ quello di dare una introduzione a concetti di base della chimica fisica e, anche, una occasione per presentare qualche vecchio programma in BASIC sviluppato in passato per scopi didattici.

      Riguardo la tua domanda, in realta soluzioni semplici come questa della equazioni di SE non dipendente dal tempo sono state molto importanti nello sviluppo della chimica teorica prima dell’avvento dei calcolatori elettronici. In una seconda parte del mio articolo (e grazie al tuo commento mi hai dato lo stimolo per scriverlo!) mostrerò un esempio classico di applicazione con il solito programmino aggiunto per fare qualche esperimento computazionale.

      Like

      Reply

  2. Ciao Danilo.
    Im paio si precisazioni e errata corrige:

    nella frase: “This function are solution of a wave equation: the so-called the Schrödinger equation (SE)”
    occorre sostituire “this function” con la forma plurale “these functions”.

    nella frase: “The application of the operator produce as results the same wave function multiplied for a number $E.$ This number represented the total energy state of the system (or \em eigenstate\em). This equation is also called eigenvalue equation, and, in this case, \Psi(x,y,z) is also said to be an eigenfunction of \bf \hat H with the associated \em eigenvalue,\em E. ”
    ci sono alcuni tag html che non vengono correttamente letti dal browser web – vedi i segni di dollaro e \em, \b \Psi(x,y,z)

    nella frase: “In this case, the value of V(x) is equal zero for and infinite in the walls.”
    manca una parte della descrizione della funzione potenziale: V(x) = 0 per 0 <= x <= L

    le formule :

    \int_{0}^L \sin^2 (x) dx = -\frac{\sin (Kx) \cos(Kx)}{K} \bigg|_{0}^L + K\int_0^L \cos^2(x) dx = -\frac{\sin (Kx) \cos(Kx)}{K} \bigg|_{0}^L + \int_0^L (1-\sin^2(Kx)) dx

    e

    \int_0^L \sin^2 (Kx) dx = \frac{L}{2}.

    non vengono visualizzate. Compare in entrambi i casi un orribile riquadro giallo con la scritta in rosso "Formula does not parse".

    Quando esegui la differenziazione della funzione d'onda in forma trigonometrica:

    -\frac{\hbar^2}{2m}\left( -C\sin(kx)-D\cos(kx)\right) = E_k\left(C\sin(kx)+D\cos(kx)\right)

    hai dimenticato a moltiplicare per k al quadrato il primo membro.

    Spero che le mie indicazioni aiutino a migliorare la qualità della tua pagina.

    grazie.

    Like

    Reply

  3. Pingback: Examples of the Particle in a Box model applications |

Leave a comment

This site uses Akismet to reduce spam. Learn how your comment data is processed.