Solutions to practice problems on the quantum mechanics of atoms and molecules

1. The radial probability density is proportional to where R is the r-dependent part of the wavefunction. This passes through zero when R passes through zero. For the 2s wavefunction, we can find this point by inspection: must be zero for the wavefunction to be zero so there is a radial node at . For the 3s wavefunction, we need the polynomial in parentheses to be zero:

   > solve(27*a0^2-18*a0*r+2*r^2=0,r);

   

2. This is a straightforward exercise in integration in spherical polar coordinates:

   > psi2px := (r,theta,phi) ->
   > 1/sqrt(32*Pi*a0^5)*r*exp(-r/(2*a0))*sin(theta)*cos(phi);

   

   > V := r -> -e^2/(4*Pi*epsilon0*r);

   

   > assume(a0>0);
   > int(int(int(psi2px(r,theta,phi)*V(r)*psi2px(r,theta,phi)
   > *r^2*sin(theta),r=0..infinity),theta=0..Pi),phi=0..2*Pi);

   

3. First, define the wavefunction:

   > psi0 := x -> exp(-alpha*x^2/2);

   

   > assume(alpha>0);

   If we multiply by a normalization factor, then integrate the square of the wavefunction and force the result to be equal to 1, we see that the normalization constant must be

   > 1/sqrt(int(psi0(x)^2,x=-infinity..infinity));

   

4. Let us first define the trial wavefunction:

   > phi := x -> x*(L-x) + a*x^2*(L-x)^2;

   

   The particle-in-a-box Hamiltonian is all kinetic:

   > Evar := -hbar^2/(2*m)*int(phi(x)*diff(phi(x),x$2),x=0..L)
   > /int(phi(x)^2,x=0..L);

   

   > avalues := solve(diff(Evar,a)=0,a);

   

   The better solution is the one that gives the lowest energy:

   > E1 := simplify(subs(a=avalues[1],Evar));

   

   > E2 := simplify(subs(a=avalues[2],Evar));

   

   > evalf(E1);

   

   > evalf(E2);

   

   E1 is by far the better solution.

   The ground-state energy is given in terms of h, not , so we must make the conversion:

   > hbar := h/(2*Pi);

   

   > evalf(E1);

   

   This is extremely close to the ground-state energy:

   > percent_err := evalf((E1-h^2/(8*m*L^2))/(h^2/(8*m*L^2))*100);