4.6.6.1 So­lu­tion hione-a

Ques­tion:

The so­lu­tion for the hy­dro­gen mol­e­c­u­lar ion re­quires elab­o­rate eval­u­a­tions of in­ner prod­uct in­te­grals and a com­puter eval­u­a­tion of the state of low­est en­ergy. As a much sim­pler ex­am­ple, you can try out the vari­a­tional method on the one-di­men­sion­al case of a par­ti­cle stuck in­side a pipe, as dis­cussed in chap­ter 3.5. Take the ap­prox­i­mate wave func­tion to be:

$$\psi = a x(\ell -x)$$

Find $a$ from the nor­mal­iza­tion re­quire­ment that the to­tal prob­a­bil­ity of find­ing the par­ti­cle in­te­grated over all pos­si­ble $x$ po­si­tions is one. Then eval­u­ate the en­ergy $\langle{E}\rangle$ as $\langle\psi\vert H\vert\psi\rangle$, where ac­cord­ing to chap­ter 3.5.3, the Hamil­ton­ian is

$$H = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2}$$

Com­pare the ground state en­ergy with the ex­act value,

$$E_1=\hbar^2\pi^2/2m\ell^2$$

(Hints: $\int_0^{\ell}x(\ell -x){ \rm d}{x}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\ell^3$$\raisebox{.5pt}{$/$}$​6 and $\int_0^{\ell}x^2(\ell -x)^2{ \rm d}{x}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\ell^5$$\raisebox{.5pt}{$/$}$​30)

An­swer:

To sat­isfy the nor­mal­iza­tion re­quire­ment that the par­ti­cle must be some­where, you need $\langle\psi\vert\psi\rangle$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, or sub­sti­tut­ing for $\psi$,

$$1 = \langle a x(\ell -x) \vert a x(\ell -x)\rangle = \vert a\vert^2 \langle x(\ell -x) \vert x(\ell -x)\rangle$$

And by de­f­i­n­i­tion, chap­ter 2.3, the fi­nal in­ner prod­uct is just the in­te­gral $\int_0^{\ell}x^2(\ell -x)^2{ \rm d}{x}$ which is given to be $\ell^5$$\raisebox{.5pt}{$/$}$​30. So you must have

$$\vert a\vert^2 =\frac{30}{\ell^5}$$

Now eval­u­ate the ex­pec­ta­tion en­ergy:

$$\langle E \rangle = \langle a x(\ell -x) \vert H \vert a x(\... ...ac{\partial^2}{\partial x^2} \Bigg\vert x(\ell -x)\Bigg\rangle$$

You can sub­sti­tute in the value of $\vert a\vert^2$ from the nor­mal­iza­tion re­quire­ment above and ap­ply the Hamil­ton­ian on the func­tion to its right:

$$\langle E \rangle = \frac{30}{\ell^5} \frac{\hbar^2}{m} \langle x(\ell -x) \vert 1 \rangle$$

The in­ner prod­uct is by de­f­i­n­i­tion the in­te­gral $\int_0^{\ell}x(\ell -x){ \rm d}{x}$, which was given to be $\ell^3$$\raisebox{.5pt}{$/$}$​6. So the fi­nal ex­pec­ta­tion en­ergy is

$$\langle E \rangle = \frac{\hbar^2 10}{2m\ell^2} \mbox{ versus } \frac{\hbar^2\pi^2}{2m\ell^2} \mbox{ exact.}$$

The er­ror in the ap­prox­i­ma­tion is only 1.3%! That is a sur­pris­ingly good re­sult, since the parabola $ax(\ell -x)$ and the sine $a'\sin({\pi}x/\ell)$ are sim­ply dif­fer­ent func­tions. While they may have su­per­fi­cial re­sem­blance, if you scale each to unit height by tak­ing $a$ $\vphantom0\raisebox{1.5pt}{$=$}$ 4$\raisebox{.5pt}{$/$}$​$\ell^2$ and $a'$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, then the de­riv­a­tives at $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 and $\ell$ are 4$\raisebox{.5pt}{$/$}$​$\ell$ re­spec­tively $\pi$$\raisebox{.5pt}{$/$}$​$\ell$, off by as much as 27%.

If you go the next log­i­cal step, ap­prox­i­mat­ing the ground state with two func­tions as

$$\psi = a x(\ell -x) + b x^2 (\ell -x)^2$$

where $a$ and $b$ are re­lated by the nor­mal­iza­tion re­quire­ment $\langle\psi\vert\psi\rangle$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, you find a ground state en­ergy to a stun­ning, (for a two term ap­prox­i­ma­tion,) ac­cu­racy of 0.001 5%! How­ever, the al­ge­bra be­comes im­pos­si­bly messy, so it was left out of the ques­tions list. Sim­i­larly, a two point, lin­ear in­ter­po­la­tion, fi­nite el­e­ment ver­sion was left out, since there is so much bag­gage, it would dis­tract from the true pur­pose of this book, to bring across the ba­sic ideas of quan­tum me­chan­ics to en­gi­neers.

Now, if you read the next sub­sec­tion, you will see that in real-life, multi-​di­men­sion­al, prob­lems, get­ting re­sults this ac­cu­rate is dif­fi­cult. Still, if you are des­per­ate for a good so­lu­tion of these very com­plex prob­lems by a sim­ple and re­li­able means, vari­a­tional meth­ods are hard to beat.