> restart:

In one dimension, the Hamiltonian operator H acts on an eigenfunction in the following way:

> H:=(-h^2/(2*m))*diff(phi(x),x$2)+V(x);

Here, is the spatial component of the time-dependant solution of the Schrodinger Equation, namely . For a free particle, there are no external interaction potentials, so making the Hamiltonian

> V(x):=0:

> H;

For to be an eigenfunction of the Hamiltonian operator (and thus an eigenfunction of any operators related to the Hamiltonian, such as momentum, kinetic energy, etc.) it must satisfy the eigenvalue equation where is ths eigenvalue ascociated with the Hamiltonian. In other words, if we find a function such that when the Hamiltonian acts on it, it yields a scalar multiple of itself( ), we have found an eigenfunction. Let's try where k is the wave number and A is an arbitrary constant. Letting the Hamiltonian act on yields:

> phi(x):=A*exp(-I*k*x):

> k:=p/h:

> H;

>

We see that the result is a solution of the form . Thus, since the Hamiltonian yields a solution of the form ( ), is indeed an eigenfunction. Furthermore, is the eigenvalue ascociated with the energy of the particle. It should be noted that there is a continuous (i.e. non-quantized) spectra of allowed eigen-energies. Let us now test the claim that if is an eigenfunction of the Hamiltonian, it is also an eigenfunction of the operators related to the Hamiltonian. In terms of the momentum operator , the Hamiltonian may be written as . Thus, should be an eigenfunction of the momentum operator, i.e. . When acts on , we get:

> Px:=-I*h*diff(phi(x),x):

> Px;

Since the momentum operator provides a result of ( ,i.e. the momentum), then is also an eigenfunction of . Let us now determine the probability density of a free particle. The probability density is defined as (where denotes the complex conjugate of , namely in this case ). Using the wavefunction (which has been shown to be an eigenfunction of this system), we find:

> P:=simplify(conjugate(phi(x))*phi(x));

> P:= simplify(abs(A)*exp(-I*k*x)*abs(A)*exp(I*k*x));

> plot(subs(A=1,P),x=-1..1, title = "Graph of P(x) vs. x in units of A^2");

In other words, the particle is equally likely to be anywhere in the 1-D universe (i.e. it's free!); there is no "preferred" location for the particle. To normalize , we require ; i.e. we must find the particle somewhere. Let us denote the entire span of the 1-D universe as L (L must be sufficiently large). Thus, the normalization condition becomes , giving .

> normalize:=int(P,x=-L/2..L/2)=1;

> solve(normalize, abs(A));

The presence of the negative solution (i.e. ) is merely a syntatical error due to Maple. In reality, the only acceptable solution is , since . Thus, the normalized wavefunction is .

We will now investigate the familiar case of a particle in a 1 dimensional box of length 2L. The particle is free in the region , but an infinte potential barrier prevents the particle from existing in the regions . In these regions, the wavefunction must be zero (to account for the absense of the particle in these regions, and thus giving zero probability of finding it here), giving the boundary conditions . In the region , the particle remains free, so the wavefunction defined in this region is the familiar form ; . It is beneficial for imposing boundary conditions to express in terms of periodic functions: . Imposing the boundary conditions yeilds the following: , .

This results in two conditions: to avoid trivial solutions, we can only stipulate that ONE of the constants be set to zero, implying the argument of the remaining term must yeild a value of zero when evaluated, i.e. the two cases are: , ; or , where n is any positive integer. This yields two solutions for : a symmetric solution when n=1, 3, 5,..., 2K-1 , and an anti-symmetric solution when n= 2, 4, 6,..., 2K (where K is a positive integer). To normalize these wavefunctions, we require the normalization conditions: , . Note that we need only to integrate within the boundaries of the box, since our boundary conditions already prevent us from having a probabilty of finding the particle outside, i.e. for .Doing so yields:

> restart:

> assume(n,integer):

> phi_s:=B*cos(n*Pi*x/(2*L)): phi_a:=A*sin(n*Pi*x/(2*L)):

> P_s:=(abs(B))^2*(cos(n*Pi*x/(2*L)))^2:

> P_a:=(abs(A))^2*(sin(n*Pi*x/(2*L)))^2:

> norm_s:=int(P_s,x=-L..L)=1:

> norm_a:=int(P_a,x=-L..L)=1:

> norm_s;

> norm_a;

Recognizing that for any value of n, one of the two periodic terms will become zero, the normalization equations reduce to: , . Thus, . Finally, we obtain the set of normalized wavefunctions for even and odd states: for odd states n=1, 3, 5,..., 2K-1 and for even states n=2, 4, 6,..., 2K.

> L:=1:B:=1/sqrt(L):A:=1/sqrt(L):

> plot([subs(n=1,phi_s),subs(n=2,phi_a),subs(n=3,phi_s),subs(n=4,phi_a),subs(n=5,phi_s)],x=-L..L, title="First 5 eigenstates of a particle in a 1-D box");

>