# Particle in a one dimensional box

Schrodinger Equation for a particle-in-a-box:

• Confine a particle to a finite region in which it has no potential energy. Its energy is only due to motion—-kinetic energy. It experiences no potentials due to any external force, i.e., its potential energy is zero inside the region.
• Ensure that the particle stays in the region, i.e., it does not jump out no matter what the particle’s energy is. How? (Assume that the potential energy along the boundary and outside of the region, namely the walls of the box and elsewhere, is infinite, repulsive). Thus the model is that of a particle trapped inside a box with no possibility of being located anywhere outside the box.
• Assume that the particle has only one degree of freedom and call it the x coordinate!

Examine the figure given earlier, again.

The particle has only kinetic energy and no potential energy inside the box. Therefore, inside the box ${\color {Black}{E = \frac{{p^2 }}{{2m}} = \frac{{\vec p.\vec p}}{{2m}} = \frac{{p_x ^2 }}{{2m}}}}$ .

(The particle wave function is dependent on only one coordinate, therefore, only x coordinate and likewise x component only for momentum p). So the one dimensional Schrödinger equation for this particle is obtained by using the operator form for the momentum in the kinetic energy expression for the particle: ${\color {Black}H = \frac{{p^2 }}{{2m}}\, \, {\rm{ (in \, \, operator \, \,form);}}}$ ${\color {Black} \hat{H}\Psi (x) = E\Psi (x) \Rightarrow {\rm{ }} - \frac{{\hbar ^2 }}{{2m}}\frac{{d^2 }}{{dx^2 }}\Psi (x) = E\Psi (x)}$ .

Solution: Rearrange the equation: ${\color {Black}\frac{{d^2 }}{{dx^2 }}\Psi (x) + k^2 \Psi (x) = 0\, \,{\rm{ where}}}$ ${\color {Black}k^2 = \frac{{2mE}}{{\hbar ^2 }}\, \, }$ .

General Solution: ${\color {Black} \Psi (x) = A{\rm{ cos}}(kx) + B{\rm{ sin}}(kx)}$

where $A \, {\rm{and}} \, B \, {\rm{are}} \, {\rm{arbitrary}} \, {\rm{constants}}$ .

Specific solution of differential equation has to be obtained with the requirement that the particle cannot penetrate the box. (The particle is confined to the box, or bounded to within the box.) The wave has zero amplitude at the ends of the box.The answer is ${\color {Black}\Psi (x = 0) = 0{\rm{ }} \Rightarrow {\rm{ }}A\cos (0) + B\sin (0) = 0;{\rm{ }}\, \therefore A = 0}$ ${\color {Black}\Psi (x = L) = 0{\rm{ }} \Rightarrow {\rm{ }}B\sin (kL) = 0;{\rm{ }} \, \, \therefore kL = n\pi ;{\rm{ }}\, \,n = 1,2,3, \ldots}$ ${\color {Black}{kL = n\pi }}$ .

What does the above condition mean? ${\color {Black}{k^2 = \frac{{2mE}}{{\hbar ^2 }} = \frac{{n^2 \pi ^2 }}{{L^2 }}}}$ .

if you thought you could put any particle in any such box and give it any energy and expect it to satisfy the Schrödinger equation, well, you have problems. Only certain discrete values of energies are solutions consistent with our model. Also the likelihood of finding the particle in different regions of the box are going to be different, contrary to what one would expect if classical equations were to be obeyed

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