Consider a particle constrained in a spherical box of radius a. The Hamiltonian

Question

Consider a particle constrained in a spherical box of radius a. The Hamiltonian operator for this system is H= -(hbar^/2mr^2)d/dr(r^2)d/dr+(hbar^2)(l)(l+1)/2mr^2, 0 r a. The second term is the centrifugal potential term. In the ground state, l = 0, so only the first term matters. The boundary condition requires that (r=a)=0. Use (r)=a-r to calculate an upper bound to the groundstate energy of this system. There is no variational parameter in this case, but the calculated energy is still an upper bound to the groundstate energy, (pi^2)(hbar^2)/(2ma^2) Compare the values.

A student uses the trial function (r)=a-cr with c being a variational parameter. Upon minimizing the energy with respect to c, s/he finds that c1 and the trial energy can be lower than the exact energy and even become negative for certain a values. Certainly this cannot be correct, but what was wrong?

Explanation / Answer

A trial function that meets this requirement is f[x]:=N1 x (L-x)

where where N1 is a normalization constant.

Here we see that the trial function is not according to the lines of the proper trial function, hence this is incorrect.

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