5.3.1 So­lu­tion 2state-a

Ques­tion:

The ef­fec­tive­ness of mix­ing states was al­ready shown by the hy­dro­gen mol­e­cule and mol­e­c­u­lar ion ex­am­ples. But the gen­er­al­ized story above re­stricts the ba­sis states to be or­thog­o­nal, and the states used in the hy­dro­gen ex­am­ples were not.

Show that if $\psi_1$ and $\psi_2$ are not or­thog­o­nal states, but are nor­mal­ized and pro­duce a real and pos­i­tive value for $\langle\psi_1\vert\psi_2\rangle$, like in the hy­dro­gen ex­am­ples, then or­thog­o­nal states can be found in the form

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

For nor­mal­ized $\psi_1$ and $\psi_2$ the Cauchy-Schwartz in­equal­ity im­plies that $\langle\psi_1\vert\psi_2\rangle$ will be less than one. If the states do not over­lap much, it will be much less than one and $\varepsilon$ will be small.

(If $\psi_1$ and $\psi_2$ do not meet the stated re­quire­ments, you can al­ways re­de­fine them by fac­tors $ae^{{\rm i}{c}}$ and $be^{-{\rm i}{c}}$, with $a$, $b$, and $c$ real, to get states that do.)

An­swer:

The in­ner prod­uct of $\bar\psi_1$ and $\bar\psi_2$ must be zero for them to be or­thog­o­nal:

$$\alpha^2 \left\langle\psi_1-\varepsilon\psi_2\vert \psi_2-\varepsilon\psi_1 \right\rangle = 0$$

and this can be mul­ti­plied out, drop­ping the com­mon fac­tor $\alpha^2$ and not­ing that $\langle\psi_1\vert\psi_1\rangle$ and $\langle\psi_2\vert\psi_2\rangle$ are one, as

$$\langle\psi_1\vert\psi_2\rangle\varepsilon^2 - 2\varepsilon + \langle\psi_1\vert\psi_2\rangle = 0$$

for which the small­est root can be writ­ten as

$$\varepsilon = \frac{\langle\psi_1\vert\psi_2\rangle} {1+\sqrt{1-\langle\psi_1\vert\psi_2\rangle^2}}.$$

That is less than $\langle\psi_1\vert\psi_2\rangle$, hence $\varepsilon$ is small if the over­lap is small.

The con­stant $\alpha$ fol­lows from the re­quire­ment that the new states must still be nor­mal­ized, and is found to be

$$\alpha = \frac{1}{\sqrt{1-2\varepsilon\langle\psi_1\vert\psi_2\rangle +\varepsilon^2}}.$$

Note that the de­nom­i­na­tor is nonzero; the ar­gu­ment of the square root ex­ceeds $(1-\varepsilon)^2$.