4.1.5.1 So­lu­tion harme-a

Ques­tion:

Just to check that this book is not ly­ing, (you can­not be too care­ful), write down the an­a­lyt­i­cal ex­pres­sion for $\psi_{100}$ and $\psi_{010}$ us­ing ta­ble 4.1. Next write down $\left(\psi_{100}+\psi_{010}\right)$$\raisebox{.5pt}{$/$}$​$\sqrt 2$ and $\left(\psi_{010}-\psi_{100}\right)$$\raisebox{.5pt}{$/$}$​$\sqrt 2$. Ver­ify that the lat­ter two are the func­tions $\psi_{100}$ and $\psi_{010}$ in a co­or­di­nate sys­tem $(\bar{x},\bar{y},z)$ that is ro­tated 45 de­grees counter-clock­wise around the $z$-​axis com­pared to the orig­i­nal $(x,y,z)$ co­or­di­nate sys­tem.

An­swer:

Take the ro­tated co­or­di­nates to be $\bar{x}$ and $\bar{y}$ as shown:

$$\begin{picture}(200,200)(-100,-100)\thinlines \put(-100,0... ...ar x$}} \put(-64,60){\makebox(0,0)[rt]{$\bar y$}} \end{picture}$$

A vec­tor dis­place­ment of mag­ni­tude $x$ in the $x$-​di­rec­tion has a com­po­nent along the $\bar{x}$-​axis of mag­ni­tude $x\cos 45^\circ$, equiv­a­lent to $x$$\raisebox{.5pt}{$/$}$​$\sqrt 2$. Sim­i­larly, a vec­tor dis­place­ment of mag­ni­tude $y$ in the $y$-​di­rec­tion has a com­po­nent along the $\bar{x}$-​axis of mag­ni­tude $y\cos 45^\circ$, equiv­a­lent to $y$$\raisebox{.5pt}{$/$}$​$\sqrt 2$. So in gen­eral, for any point $(x,y)$,

$$\bar x = \frac{x+y}{\sqrt 2}.$$

Sim­i­larly you get

$$\bar y = \frac{y-x}{\sqrt 2}.$$

Turn­ing now to the eigen­func­tions, tak­ing the generic ex­pres­sion

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

and sub­sti­tut­ing $n_x$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, $n_y$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n_z$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, you get

$$\psi_{100} = h_1(x) h_0(y) h_0(z).$$

Now sub­sti­tute for $h_0$ and $h_1$ from ta­ble 4.1:

$$\psi_{100} = {\displaystyle\frac{\sqrt{2}x/\ell}{\left(\pi\e... ...ht)^{3/4}}} e^{-x^2/2\ell^2}e^{-y^2/2\ell^2}e^{-z^2/2\ell^2}$$

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

$$\psi_{100} = {\displaystyle\frac{\sqrt{2}x/\ell}{\left(\pi\ell^2\right)^{3/4}}} e^{-(x^2+y^2+z^2)/2\ell^2}.$$

The same way, you get

$$\psi_{010} = {\displaystyle\frac{\sqrt{2}y/\ell}{\left(\pi\ell^2\right)^{3/4}}} e^{-(x^2+y^2+z^2)/2\ell^2}.$$

So, the com­bi­na­tion $\left(\psi_{100}+\psi_{010}\right)$$\raisebox{.5pt}{$/$}$​$\sqrt 2$ is

$$\frac{\psi_{100}+\psi_{010}}{\sqrt 2} = {\displaystyle\frac{... ...}{\left(\pi\ell^2\right)^{3/4}}} e^{-(x^2+y^2+z^2)/2\ell^2}.$$

Now $x^2+y^2+z^2$ is ac­cord­ing to the Pythagorean the­o­rem the square dis­tance from the ori­gin, which is the same as $\bar{x}^2+\bar{y}^2+z^2$. And since $\bar{x}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(x+y)$$\raisebox{.5pt}{$/$}$​$\sqrt 2$, the sum $x+y$ in the com­bi­na­tion eigen­func­tion above is $\sqrt 2\bar{x}$. So the com­bi­na­tion eigen­func­tion is

$$\frac{\psi_{100}+\psi_{010}}{\sqrt 2} = {\displaystyle\frac{... ...i\ell^2\right)^{3/4}}} e^{-(\bar x^2+\bar y^2+z^2)/2\ell^2}.$$

which is ex­actly the same as $\psi_{100}$ above, ex­cept in terms of $\bar{x}$ and $\bar{y}$. So it is $\psi_{100}$ in the ro­tated frame.

The other com­bi­na­tion goes the same way.