In 3d with V(|r|)

m_i=m (reduced mass of nuclei measured in electron mass units)

                Equation 2.7 of Variance Minimization becomes

  (1.1)

Ground State

                Try         

                (1.2)

Equation (1.1) becomes

(1.3)

As in the one-d case a=Ö(m/2) cancels the variability of the variance, if V = r²,  leading to E=3/Ö(m2),  which is 3 times the zero point energy in 1 d.

                In the case that V is (r-c)², the form for Y is more complicated.  It is ÑY×ÑY that cancels with V for r² which is much more difficult to arrange in this case.

Excited State

In (1.1)try the form

 (2.1)

So that

(2.2)

Simplify

               (2.3)

If a-Ö(m/2) and V = r², then E = 5/Ö(2m) -- as it should.  If V(r) minimizes at other than r = 0, then the 1/x from the first derivative will interact with the x when the e_x coefficient is squared.

With Y_l^m

From Morse and Feshbach[1](2.5)

                    (2.6)

The operator[2]

(2.7)

So that

(2.8)

In Error! Reference source not found.try the form

 (2.9)

So that

(2.10)

In this case the components of the square are orthogonal so that

(2.11)

Then the process that led to the log terms can be reversed leading to

(2.12)

So that (2.11) becomes

(2.13)

Substituting -L²Y/r² for Ѳ from (2.7)

              (2.14)

Then using

      (2.15)

Try the form

(2.16)

So that

(2.17)

The del² of log(r) is given by

               (2.18)

Del of log(r) is

               (2.19)

Del of r² is

(2.20)

So that (2.17) becomes

     (2.21)

Collecting terms

                               (2.22)

This says that K=l, and for V=r², a = Ö(m/2).  Note that the term r^K is necessary.