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

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 with the Nobel prize in Physics just 7 years later his breakthrough discovery). He discovered that a state of a quantum system composed by particles (such as electrons and nucleons) can 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 are solution of a wave equation: the so-called 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 molecular. 

Nowadays, computational quantum chemistry has surged to a level of a fundamental research discipline in chemistry and material science. In recognition of the 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 are going to study analytical and numerical solutions of SE for simple quantum systems in stationary conditions, e.g. whose state does not change with the time. For this purpose, we are going to 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 his 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 total energy of the system. I just recall you 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 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.  

Shall we look more in detail to the Hamiltonian operator, and, for the sake of simplicity, shall we limit ourselves to consider only a mono-dimensional physical system, namely a system defined only along the reference axis x. Therefore, the wave function depends on only 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 it is associated (in our case) with operator \bf \hat x. Therefore we can write the resulting SE as:


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) is equal 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 the Figure 1. 

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

Derivation of the wavefunction

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


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

we can write 


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


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


 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


 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 wavelength 

of the particle


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}


\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 its integral the probability to find an 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 of 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

\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 integral

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


  • 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.

About Danilo Roccatano

I have a Doctorate in chemistry at the University of Roma “La Sapienza”. I led educational and research activities at different universities in Italy, The Netherlands, Germany and now in the UK. I am fascinated by the study of nature with theoretical models and computational. For years, my scientific research is focused on the study of molecular systems of biological interest using the technique of Molecular Dynamics simulation. I have developed a server (the link is in one of my post) for statistical analysis at the amino acid level of the effect of random mutations induced by random mutagenesis methods. I am also very active in the didactic activity in physical chemistry, computational chemistry, and molecular modeling. I have several other interests and hobbies as video/photography, robotics, computer vision, electronics, programming, microscopy, entomology, recreational mathematics and computational linguistics.
This entry was posted in Science Topics, What is new. Bookmark the permalink.

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

  1. Marcello Colozzo says:

    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:

    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.


    • 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.


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

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