Physlets run in a Java-enabled browser on the latest Windows & Mac operating systems.

If Physlets do not run, click here for help on updating Java and setting Java security.

# Section 13.1: The Two-dimensional Infinite Square Well

Animation 1: Energy Eigenfunction | Animation 2: Energy Eigenfunction Contours | Animation 3: Probability Density | Animation 4: Probability Density Contours

Please wait for the animation to completely load.

Recall that in one-dimension, the infinite square well confines a particle to be between 0 to *L* in the *x* direction. In this case, the time-independent Schrödinger equation is

−(*ħ*^{2}/2*m*)(*d*^{2}/*dx*^{2}) ψ(*x*) = *E*ψ(*x*) ,

which for the application of the boundary conditions gives the solutions ψ_{n}(*x*) = (2/*L*)^{1/2} sin(*n*π*x*/*L*) and *E*_{n }= *n*^{2}π^{2}*ħ*^{2}/(2*mL*^{2}) where *n* = 1, 2, 3,….

We now seek to extend this to two dimensions for a symmetric two-dimensional infinite well, where *V*(x,*y*) = 0 for −*a* < *x* < *a* and −*b* < *y* < *b* and *V*(*x*,*y*) = ∞ elsewhere. The Hamiltonian has changed from the one-dimensional case with the addition of a term that involves the derivative in the* y* direction. The energy eigenfunction must therefore also be a function of both x and *y* and satisfy the two-dimensional time-independent Schrödinger equation:

[−(*ħ*^{2}/2*m*)(∂^{2}/∂*x*^{2}) − (*ħ*^{2}/2m)(∂^{2}/∂*y*^{2})] ψ(*x*,*y*) = *E*ψ(*x*,*y*) . (13.1)

How do we solve this time-independent Schrödinger equation? We begin by using the technique of *separation of variable*s to write the energy eigenfunction as ψ(*x*,*y*) = ψ(*x*)ψ(*y*). In doing so, we see that only the part of the energy eigenfunction that is of one variable or the other gets differentiated by the differential terms. Thus we expect

[*E*_{nx }+ *E*_{ny}]ψ(*x*)ψ(*y*) = *E*ψ(*x*)ψ(*y*) ,

which we can separate into

−(*ħ*^{2}/2*m*)(*d*^{2}/*dx*^{2}) ψ(x) = *E*_{x}ψ(*x*) and −(*ħ*^{2}/2*m*)(*d*^{2}/*dy*^{2}) ψ(*y*) = *E*_{y}ψ(*y*) .

For the individual boundary conditions yields the solutions ψ_{nx}(*x*) = (1/a)^{1/2} sin(*n*_{x}π(*x *+ 1)/2*a*) and *E*_{nx }= *n*_{x}^{2}π^{2}*ħ*^{2}/(8*ma*^{2}) and ψ_{ny}(*y*) = (1/*b*)^{1/2} sin(*n*_{x}π(*y* + 1)/2*b*) and *E*_{ny }= *n*_{y}^{2}π^{2}*ħ*^{2}/(8*mb*^{2}). We have for the entire solution, therefore,

*E*_{nx ny } = *n*_{x}^{2}π^{2}*ħ*^{2}/(8*ma*^{2}) + *n*_{y}^{2}π^{2}*ħ*^{2}/(8*mb*^{2}) = π^{2}*ħ*^{2}/(8*m*^{2}) [*n*_{x}^{2}/*a*^{2} + *n*_{y}^{2}/*b*^{2}] , (13.2)

and ψ_{nx ny}(*x*,*y*) = (1/*ab*)^{1/2} sin(*n*_{x}π(*x* + 1)/*a*)sin(*n*_{y}π(*y* + 1)/*b*).

Consider the case where *a* = *b* = *L* which is shown in the animation **( ħ = 2m = 1 and L = 1)**. You can view the energy eigenfunction in either a three-dimensional plot or as a contour plot or view the probability density as a three-dimensional plot or as a contour plot. For such a

*squar*e well, the energy eigenfunction and energy are simplification yields

ψ_{nx ny }(*x*,*y*) = (1/*L*) sin(*n*_{x}π(*x* + 1)/*L*)sin(*n*_{y}π(*y* + 1)/*L*) (13.3)

and

*E*_{nx ny } = = π^{2}*ħ*^{2}/(8mL^{2}) [*n*_{x}^{2} + *n*_{y}^{2}] , (13.4)

respectively. We note that this equation yields the energies

*E*_{11} = 2*E*_{1} *E*_{21 } = *E*_{12} = 5*E*_{1} *E*_{22 }= 8*E*_{1} *E*_{31} = *E*_{13 }= 10*E*_{1} ,

and so forth. We therefore find degeneracies of the energy eigenvalues which is due to the geometrical symmetry of the problem. The result that some energies are degenerate is called a symmetry degeneracy.

next »