N.24 A prob­lem if the en­ergy is given

Ex­am­in­ing all shelf num­ber com­bi­na­tions with the given en­ergy and then pick­ing out the com­bi­na­tion that has the most en­ergy eigen­func­tions seems straight­for­ward enough, but it runs into a prob­lem. The prob­lem arises when it is re­quired that the set of shelf num­bers agrees with the given en­ergy to math­e­mat­i­cal pre­ci­sion. To see the prob­lem, re­call the sim­ple model sys­tem of chap­ter 11.3 that had only three en­ergy shelves. Now as­sume that the en­ergy of the sec­ond shelf is not $\sqrt{9}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 3 as as­sumed there, (still ar­bi­trary units), but slightly less at $\sqrt{8}$. The dif­fer­ence is small, and all fig­ures of chap­ter 11.3 are es­sen­tially un­changed. How­ever, if the av­er­age en­ergy per par­ti­cle is still as­sumed equal to 2.5, so that the to­tal sys­tem en­ergy equals the num­ber of par­ti­cles $I$ times that amount, then $I_2$ must be zero: it is im­pos­si­ble to take a nonzero mul­ti­ple of an ir­ra­tional num­ber like $\sqrt{8}$ and end up with a ra­tio­nal num­ber like $2.5I-I_1-4I_3$. What this means graph­i­cally is that the oblique en­ergy line in the equiv­a­lent of fig­ure 11.5 does not hit any of the cen­ters of the squares math­e­mat­i­cally ex­actly, ex­cept for the one at $I_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. So the con­clu­sion would be that the sys­tem must have zero par­ti­cles on the mid­dle shelf.

Of course, phys­i­cally this is ab­solute non­sense; the en­ergy of a large num­ber of per­turbed par­ti­cles is not go­ing to be cer­tain to be 2.5 $I$ to math­e­mat­i­cal pre­ci­sion. There will be some un­cer­tainty in en­ergy, and the cor­rect shelf num­bers are still those of the dark­est square, even if its en­ergy is 2.499 9...$I$ in­stead of 2.5 $I$ ex­actly. Here typ­i­cal text­books will pon­tif­i­cate about the ac­cu­racy of your sys­tem-en­ergy mea­sure­ment de­vice. How­ever, this book shud­ders to con­tem­plate what hap­pens phys­i­cally in your glass of ice wa­ter if you have three sys­tem-en­ergy mea­sure­ment de­vices, but your best one is in the shop, and you are un­cer­tain whether to be­lieve the unit you got for cheap at Wal-Mart or your backup unit with the stick­ing nee­dle.

To avoid these co­nun­drums, in this book it will sim­ply be as­sumed that the right com­bi­na­tion of shelf oc­cu­pa­tion num­bers is still the one at the max­i­mum in fig­ure 11.6, i.e. the max­i­mum when the num­ber of en­ergy eigen­func­tions is math­e­mat­i­cally in­ter­po­lated by a con­tin­u­ous func­tion. Sure, that may mean that the oc­cu­pa­tion num­bers are no longer ex­act in­te­gers. But who is go­ing to count 10$\POW9,{20}$ par­ti­cles to check that it is ex­actly right? (And note that those other books end up do­ing the same thing any­way in the end, since the math­e­mat­ics of an in­te­ger-val­ued func­tion de­fined on a strip is so much more im­pos­si­ble than that of a con­tin­u­ous func­tion de­fined on a line.)

If frac­tional par­ti­cles both­ers you, even among 10$\POW9,{20}$ of them, just fix things af­ter the fact. Af­ter find­ing the frac­tional shelf num­bers that have the biggest en­ergy, se­lect the whole shelf num­bers near­est to it and then change the given en­ergy to be 2.499 999 9...or what­ever it turns out to be at those whole shelf num­bers. Then you should have per­fectly cor­rect shelf num­bers with the high­est num­ber of eigen­func­tions for the new given en­ergy.