4.1.4.1 So­lu­tion harmd-a

Ques­tion:

Write out the ground state wave func­tion and show that it is in­deed spher­i­cally sym­met­ric.

An­swer:

Re­peat­ing an ear­lier ex­er­cise, tak­ing the generic ex­pres­sion

$$\psi_{n_xn_yn_z}=h_{n_x}(x) h_{n_y}(y) h_{n_z}(z)$$

and sub­sti­tut­ing $n_x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n_y$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n_z$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, you get the ground state eigen­func­tion

$$\psi_{000} = h_0(x) h_0(y) h_0(z).$$

Now sub­sti­tute for $h_0$ from ta­ble 4.1:

$$\psi_{000} = {\displaystyle\frac{1}{\left(\pi\ell^2\right)^{3/4}}} e^{-x^2/2\ell^2}e^{-y^2/2\ell^2}e^{-z^2/2\ell^2}$$

where the con­stant $\ell$ is as given in ta­ble 4.1. You can mul­ti­ply out the ex­po­nen­tials:

$$\psi_{000} = {\displaystyle\frac{1}{\left(\pi\ell^2\right)^{3/4}}} e^{-(x^2+y^2+z^2)/2\ell^2}.$$

Ac­cord­ing to the Pythagorean the­o­rem, $\sqrt{x^2+y^2+z^2}$ is the dis­tance from the ori­gin $r$, so

$$\psi_{000} = {\displaystyle\frac{1}{\left(\pi\ell^2\right)^{3/4}}} e^{-r^2/2\ell^2}.$$

It fol­lows that the wave func­tion only de­pends on the dis­tance $r$ from the ori­gin, not on the an­gu­lar ori­en­ta­tion com­pared to it. That is the de­f­i­n­i­tion of spher­i­cally sym­met­ric: it looks the same from any an­gle.