Assuming the time independent Schrodinger’s wave equation discuss the solution for a particle in one dimensional potential well of infinite height and hence obtain the normalized wave equation.

Consider a particle of mass m bouncing back and forth between the walls of one dimensional potential well as shown in figure. The particle is said to be under bound state. Let the motion of the particle be confined along the G-axis in between two infinitely hard walls at x = 0 and x = a.
Since the walls are infinitely hard, no energy is lost by the particle during the collision with walls and the total energy remains constant.

The description of the potential well is as follows. In between walls i.e. 0<x<a , the potential V=0. Beyond the walls i.e. x\le 0and x\ge a, the potential V=\infty.

Beyond the walls or outside the potential well

Since the principle is unable to penetrate the hard walls it exist only inside the potential well, Hence \psi=0 and the probability of finding the particle outside the potential well is zero.

one dimensional potential well of infinite height
figure : one dimensional potential well of infinite height

In between the walls or inside the potential well

the Schrodinger wave equation is given by

\frac{\partial ^{2}\psi}{\partial x^{2}}+\frac{8\Pi^{2}m(E-U)}{h^{2}}\psi=0

Since the potential inside the well V = 0, hence potential energy U = 0, the Schrodinger wave equation becomes

\frac{\partial ^{2}\psi}{\partial x^{2}}+\frac{8\Pi^{2}m(E-0)}{h^{2}}\psi=0

\frac{\partial ^{2}\psi}{\partial x^{2}}+\frac{8\Pi^{2}mE}{h^{2}}\psi=0

\frac{\partial ^{2}\psi}{\partial x^{2}}+k^{2}\psi=0


here k is a constant for a given value of energy E. The general solution for equation is given by

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

Here in the above equation A and B are arbitrary constants which can be evaluated by applying boundary conditions.

  1. The first boundary condition is at x=0,\psi(x)=0. applying this to general equation . we get 0=A\sin 0+B\cos 0 \Longrightarrow B=0 hence equation reduce to \psi\left( x \right)=A\sin kx
  2. The secondary boundary condition is at x=a,\psi\left( x \right)=0. applying this to equation .we get 0=A\sin ka, since A\neq 0 then \sin ka=0. this result is ka=n\Pi which further could be written as k=\frac{n\Pi}{a}, n can take integer value. hence equation could be written as
\psi(x)=A\sin(\frac{n\Pi x}{a})

also from equation k is compared



Substituting for n=1,2,3,… in the above equation Energy Eigen Values are obtained. The lowest energy state corresponding to lowest integral value of n=1 which is called as Zero point Energy is given by E_{1}=\frac{h^{2}}{8ma^{2}}. the energy values of a bound particle in one dimensional potential well are quantized and are represented by the equation E_{n}=n^{2}E_{1}

Normalization of wave equation

The wave function for a particle in one dimensional potential well of infinite height is given by the equation\psi(x)=A\sin(\frac{n\Pi x}{a}).

In this equation  is an arbitrary constant and it can take any value. The process of determination of value of the arbitrary constant is called Normalization of wave function. The particle has to exist some where inside the potential the probability of the finding the particle inside the potential well is given by

wave function and probability density for n=1,2,3

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