5.3.2 So­lu­tion 2state-b

Ques­tion:

Show that it does not have an ef­fect on the so­lu­tion whether or not the ba­sic states $\psi_1$ and $\psi_2$ are nor­mal­ized, like in the pre­vi­ous ques­tion, be­fore the state of low­est en­ergy is found.

This re­quires no de­tailed analy­sis; just check that the same so­lu­tion can be de­scribed us­ing the nonorthog­o­nal and or­thog­o­nal ba­sis states. It is how­ever an im­por­tant ob­ser­va­tion for var­i­ous nu­mer­i­cal so­lu­tion pro­ce­dures: your set of ba­sis func­tions can be cleaned up and sim­pli­fied with­out af­fect­ing the so­lu­tion you get.

An­swer:

Us­ing the orig­i­nal ba­sis states, the so­lu­tion, say the ground state of low­est en­ergy, can be writ­ten in the form

$$c_1 \psi_1 + c_2 \psi_2$$

for some val­ues of the con­stants $c_1$ and $c_2$. Now the ex­pres­sion for the or­thog­o­nal­ized func­tions,

$$\bar\psi_1 = \alpha\left(\psi_1 - \varepsilon\psi_2\right) \qquad\bar\psi_2 = \alpha\left(\psi_2 - \varepsilon\psi_1\right),$$

can for given $\bar\psi_1$ and $\bar\psi_2$ be thought of as two equa­tions for $\psi_1$ and $\psi_2$ that can be solved. In par­tic­u­lar, adding $\varepsilon$ times the sec­ond equa­tion to the first gives

$$\psi_1 = \frac{\bar\psi_1+\varepsilon\bar\psi_2}{\alpha(1-\varepsilon^2)}.$$

Sim­i­larly, adding $\varepsilon$ times the first equa­tion to the sec­ond gives

$$\psi_2 = \frac{\bar\psi_2+\varepsilon\bar\psi_1}{\alpha(1-\varepsilon^2)}.$$

If this is plugged into the ex­pres­sion for the so­lu­tion, $c_1\psi_1+c_2\psi_2$, it takes the form

$$\bar c_1 \bar\psi_1 + \bar c_2 \bar\psi_2$$

where

$$\bar c_1 = \frac{c_1+\varepsilon c_2}{\alpha(1-\varepsilon^2... ...bar c_2 = \frac{c_2+\varepsilon c_1}{\alpha(1-\varepsilon^2)}.$$

So, while the con­stants $\bar{c}_1$ and $\bar{c}_2$ are dif­fer­ent from $c_1$ and $c_2$, the same so­lu­tion can be found equally well in terms of $\bar\psi_1$ and $\bar\psi_2$ as in terms of $\psi_1$ and $\psi_2$.

In the terms of lin­ear al­ge­bra, $\bar\psi_1$ and $\bar\psi_2$ span the same func­tion space as $\psi_1$ and $\psi_2$: any wave func­tion that can be de­scribed as a com­bi­na­tion of $\psi_1$ and $\psi_2$ can also be de­scribed in terms of $\bar\psi_1$ and $\bar\psi_2$, al­though with dif­fer­ent con­stants. This is true as long as the de­f­i­n­i­tions of the new func­tions can be solved for the old func­tions as above. The ma­trix of co­ef­fi­cients, here

$$\left( \begin{array}{cc} \alpha & -\alpha\varepsilon \ -\alpha\varepsilon & \alpha \end{array}\right)$$

must have a nonzero de­ter­mi­nant.