D.22 Unique ground state wave func­tion

This de­riva­tion com­pletes {A.8}. In par­tic­u­lar, it proves that ground states are unique, given a real, non­in­fi­nite, po­ten­tial that only de­pends on po­si­tion. The de­riva­tion also proves that ground states can­not be­come zero. So they can be taken to be pos­i­tive.

The ba­sic idea is first to as­sume ten­ta­tively that there would be two in­de­pen­dent ground state wave func­tions. These could then be taken to be or­tho­nor­mal as usual. That means that the in­ner prod­uct of the two wave func­tions would be zero. How­ever, it can be shown, see be­low, that ground state wave func­tions can­not cross zero. That means that both wave func­tions can be taken to be every­where pos­i­tive. (The one ex­cep­tion is at im­pen­e­tra­ble bound­aries, where the wave func­tion is forced to be zero, rather than pos­i­tive. But that ex­cep­tion does not change the ar­gu­ment here.) Now if you check the de­f­i­n­i­tion of the in­ner prod­uct, you see that the in­ner prod­uct of two pos­i­tive wave func­tions is pos­i­tive, not zero. But the or­tho­nor­mal­ity says that it is zero. So there is a con­tra­dic­tion. That means that the made as­sump­tion, that there are two in­de­pen­dent ground states, must be wrong. So the ground state must be unique.

Fig­ure D.1: Right: the ab­solute value of the wave func­tion has a kink at a zero cross­ing. Mid­dle: the kink has been slightly blunted. Right: an al­ter­nate way of blunt­ing.

To fin­ish the proof then, it must still be shown that ground states that cross zero are not pos­si­ble. Ten­ta­tively sup­pose that you had a ground state whose wave func­tion did cross zero. Near a zero cross­ing, the wave func­tion $\psi$ will then look some­thing like the left part of fig­ure D.1. (For a multi-di­men­sion­al wave func­tion, you can take this fig­ure to be some ar­bi­trary one-di­men­sion­al cross sec­tion through the zero cross­ing.) Note from the fig­ure that the ab­solute value $\vert\psi\vert$ has a kink at the zero cross­ing.

Next re­call from {A.8} that $\vert\psi\vert$ has the ground state en­ergy just like $\psi$. So it should not be pos­si­ble to lower the en­ergy be­low that of $\vert\psi\vert$. But the prob­lem is now that the wave func­tion shown in the mid­dle in fig­ure D.1 does have less en­ergy. This wave func­tion looks gen­er­ally the same as $\vert\psi\vert$, ex­cept that the kink has been blunted just a lit­tle bit. More pre­cisely, this wave func­tion has been pre­vented from be­com­ing any smaller than some very small num­ber $\varepsilon$.

To see why this wave func­tion has less en­ergy, com­pare what hap­pens to the ki­netic and po­ten­tial en­er­gies. The en­ergy is the sum of a ki­netic en­ergy in­te­gral $I_T$ and a po­ten­tial en­ergy in­te­gral $I_V$. These are given by

$$I_T = \frac{\hbar^2}{2m} \int_{\rm all} (\nabla\psi)^2 { \... ...qquad I_V = \int_{\rm all} V \psi^2 { \rm d}^3{\skew0\vec r}$$

In the re­gion that is blunted, the in­te­grand of the ki­netic en­ergy in­te­gral is now zero, in­stead of what­ever pos­i­tive value it had in the left fig­ure. The con­stant $\varepsilon$ has zero de­riv­a­tives. So the ki­netic en­ergy has been de­creased no­tice­ably.

You might at first think that the po­ten­tial en­ergy can com­pen­sate by in­creas­ing more than the ki­netic en­ergy de­creases. But that does not work, be­cause the in­te­grand of the po­ten­tial en­ergy in­te­gral is pro­por­tional to $\vert\psi\vert^2$, and that is neg­li­gi­bly small in the blunted re­gion. In fact, $\vert\psi\vert^2$ is no larger than $\varepsilon^2$, and $\varepsilon$ was a very small num­ber. So, if the ki­netic en­ergy de­creases and the po­ten­tial en­ergy stays vir­tu­ally the same, the con­clu­sion is un­avoid­able. The en­ergy de­creases. The wave func­tion in the mid­dle in the fig­ure has less en­ergy than the ones on the left. So the ones on the left can­not be ground states. So ground state wave func­tions can­not cross zero.

That is ba­si­cally it. Un­for­tu­nately, there are a few loose ends in the rea­son­ing above. That is even if you ig­nore that small is not a valid math­e­mat­i­cal con­cept. A num­ber is ei­ther zero or not zero; that is all in math­e­mat­ics. The cor­rect math­e­mat­i­cal state­ment is: “there is a num­ber $\varepsilon$ small enough that the ki­netic en­ergy de­crease ex­ceeds the po­ten­tial en­ergy in­crease.” (Note that small enough does not im­ply small. All pos­i­tive num­bers less than 1 000 are small enough to be less than 1 000.) But that is so picky.

More im­por­tantly, you might ob­ject that af­ter blunt­ing, the wave func­tion will no longer be nor­mal­ized. But you can cor­rect for that by di­vid­ing the given ex­pres­sion of the ex­pec­ta­tion en­ergy by the square norm of the wave func­tion. In par­tic­u­lar, us­ing a prime to in­di­cate a quan­tity af­ter blunt­ing the wave func­tion, the cor­rect en­ergy is

$$\left\langle{E}\right\rangle ' = \frac{I_T' + I_V'}{\left\l... ...ht\rangle = \int_{\rm all} (\psi')^2 { \rm d}^3{\skew0\vec r}$$

Now note that the square norm changes neg­li­gi­bly un­der the smooth­ing, be­cause its in­te­grand in­volves $\vert\psi\vert^2$ just like the po­ten­tial en­ergy. So di­vid­ing by the square norm should not make a dif­fer­ence.

In fact, there is a trick to avoid the nor­mal­iza­tion prob­lem com­pletely. Sim­ply re­de­fine the po­ten­tial en­ergy by a con­stant to make the ex­pec­ta­tion en­ergy zero. You can al­ways do that; chang­ing the de­f­i­n­i­tion of the po­ten­tial en­ergy by con­stant does not make a dif­fer­ence phys­i­cally. And if the ex­pec­ta­tion en­ergy $\left\langle{E}\right\rangle$ is zero, then so is $I_{\rm {T}}+I_{\rm {V}}$. There­fore the change in en­ergy due to blunt­ing be­comes

$$\left\langle{E}\right\rangle ' - \left\langle{E}\right\rang... ...{.03em}\right.\!\left\vert\vphantom{\psi'}\psi'\right\rangle }$$

Com­par­ing start and end, you see that the sign of the change in en­ergy is the same as the sign of the change in the ki­netic and po­ten­tial en­ergy in­te­grals. Re­gard­less of whether $\psi'$ is nor­mal­ized. And the sign of the change in en­ergy is all that counts. If it is neg­a­tive, you do not have a ground state. So if $I_T+I_V$ de­creases due to blunt­ing, you do not have a ground state. Be­cause of this trick, the nor­mal­iza­tion prob­lem can be ig­nored in the rest of the de­riva­tions.

You might fur­ther ob­ject that the given ar­gu­ments do not ac­count for the pos­si­bil­ity that the wave func­tion could cross zero with zero slope. In that case, the in­te­grand of the orig­i­nal ki­netic en­ergy would be van­ish­ingly small too. True.

But in one di­men­sion, you can use the Cauchy-Schwartz in­equal­ity of the no­ta­tions sec­tion on $\vert\psi\vert$ to show that the de­crease in ki­netic en­ergy will still be more than the in­crease in po­ten­tial en­ergy. For sim­plic­ity, the co­or­di­nate $x$ will be taken zero at the orig­i­nal zero cross­ing, as in the mid­dle graph of fig­ure D.1. Now con­sider the part of the blunted re­gion at neg­a­tive $x$. Here the orig­i­nal ki­netic en­ergy in­te­gral was:

$$\begin{eqnarray*} I_T = \frac{\hbar^2}{2m} \int_{-\delta}^0 \vert\psi_x\vert^2{... ...m d}x \right)^2 \\ & = & \frac{\hbar^2\varepsilon^2}{2m\delta} \end{eqnarray*}$$

The first in­equal­ity above is the Cauchy-Schwartz in­equal­ity. The fi­nal equal­ity ap­plies be­cause the change in $\psi$ is the in­te­gral of its de­riv­a­tive. Com­par­ing start and end above, the ki­netic en­ergy de­crease is at least $\hbar^2\varepsilon^2$$\raisebox{.5pt}{$/$}$​$2m\delta$. On the other hand for the in­crease in po­ten­tial en­ergy

$$I_V' - I_V = \int_{-\delta}^0 V (\varepsilon^2-\vert\psi\ve... ... \varepsilon^2{ \rm d}x = V_{\rm max} \varepsilon^2 \delta$$

where $V_{\rm {max}}$ is the max­i­mum (re­de­fined) po­ten­tial in the re­gion. (Note that if $\psi$ is not mo­not­o­n­u­ous, $\vphantom{0}\raisebox{1.5pt}{$-$}$$\delta$ and $\delta_2$ are de­fined as the points clos­est to the ori­gin where $\varepsilon$ is reached. So by de­f­i­n­i­tion $\vert\psi\vert$ does not ex­ceed $\varepsilon$.) It is seen that the max­i­mum po­ten­tial en­ergy de­crease is pro­por­tional to $\delta$. How­ever, the min­i­mum ki­netic en­ergy de­crease is pro­por­tional to 1/$\delta$. So for small enough $\delta$, po­ten­tial en­ergy in­crease can­not com­pete with ki­netic en­ergy de­crease. More specif­i­cally, tak­ing $\delta^2$ small com­pared to $\hbar^2$$\raisebox{.5pt}{$/$}$​$2mV_{\rm {max}}$ makes the po­ten­tial en­ergy in­crease small com­pared to the ki­netic en­ergy de­crease. So the net en­ergy will de­crease, show­ing that the orig­i­nal wave func­tion is in­deed not a ground state. (If $V_{\rm {max}}$ is neg­a­tive, the po­ten­tial en­ergy will de­crease, and net en­ergy de­crease is au­to­matic.)

The same ar­gu­ments nor­mally ap­ply for the blunted re­gion at pos­i­tive $x$. How­ever, there is a pos­si­ble ex­cep­tion. If af­ter $\psi$ reaches zero, it stays zero, there will be no po­si­tion $\delta_2$. At least not one van­ish­ingly close to zero. To deal with this pos­si­bil­ity, a slightly more so­phis­ti­cated blunt­ing can be used. That one is shown to the right in fig­ure D.1. Here the blunt­ing re­gion is taken to be sym­met­ric around the ori­gin. The value of $\delta$ is taken as the small­est dis­tance from the ori­gin where $\varepsilon$ is reached. There­fore once again $\vert\psi\vert$ does not ex­ceed $\varepsilon$. Note that the mod­i­fied wave func­tion now has some ki­netic en­ergy left. In par­tic­u­lar it has left

$$\frac{\hbar^2}{2m} \int_{-\delta}^\delta \left\vert\frac{\... ...ebox{-.7pt}{$\leqslant$}}\frac{\hbar^2\varepsilon^2}{4m\delta}$$

How­ever, as seen above, the neg­a­tive blunted part has ki­netic en­ergy of at least $\hbar^2\varepsilon^2$$\raisebox{.5pt}{$/$}$​$2m\delta$. So the ki­netic en­ergy de­crease is still at least half of what it was. That is enough not to change the ba­sic story.

Note that in nei­ther ap­proach, the zero cross­ing point can be at an im­pen­e­tra­ble bound­ary. Nei­ther blunted wave func­tion is zero at $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 at it should be at an im­pen­e­tra­ble bound­ary. That ex­plains why ground state wave func­tions can in­deed be­come zero at im­pen­e­tra­ble bound­aries. The ground state of the par­ti­cle in a pipe pro­vides an ex­am­ple, chap­ter 3.5.

Also note the need to as­sume that the po­ten­tial does not be­come pos­i­tive in­fin­ity. If the po­ten­tial is pos­i­tive in­fin­ity in a fi­nite re­gion, then the wave func­tion is in fact zero in­side that re­gion. The par­ti­cle can­not pen­e­trate into such a re­gion. Its sur­face acts just like an im­pen­e­tra­ble bound­ary.

How about wave func­tions in more than one di­men­sion? That is easy, if you will al­low a very mi­nor as­sump­tion. The mi­nor as­sump­tion is that there is at least a sin­gle cross­ing point where the gra­di­ent of $\psi$ is con­tin­u­ous and nonzero. It does not have to be true at all the zero cross­ing points, just at one of them. And in fact it does not even have to be true for ei­ther one of the two sup­posed ground states. It is enough if it is true for a sin­gle point in some lin­ear com­bi­na­tion of them. So it is very hard to imag­ine ground states for which the as­sump­tion would not ap­ply.

Ac­cept­ing that as­sump­tion, things are straight­for­ward. Take the blunted wave func­tion es­sen­tially like the mid­dle graph in fig­ure D.1. The $x$-​di­rec­tion is now the di­rec­tion of the gra­di­ent at the point. How­ever, rather than lim­it­ing the wave func­tion to stay above $\varepsilon$, de­mand that it stays above $\varepsilon(\ell^2-r^2)$$\raisebox{.5pt}{$/$}$​$\ell^2$. Here $r$ is the dis­tance from the con­sid­ered zero cross­ing point, and $\ell$ is a num­ber small enough that the vari­a­tion in the gra­di­ent is no more than say 50% within a dis­tance $\ell$ from the zero cross­ing point. There is then again some ki­netic en­ergy left, but it is neg­li­gi­bly small. The es­ti­mates in each cross sec­tion of the blunted re­gion are es­sen­tially the same as in the one-di­men­sion­al case.

How­ever, all that does re­quire that one mi­nor as­sump­tion. You might won­der about patho­log­i­cal cases. For ex­am­ple, what if one wave func­tion is only nonzero where the other is zero and vice-versa? With zero gra­di­ent at every sin­gle point of the zero cross­ings to boot? Of course, you and I know that ground states are not just stu­pidly go­ing to be zero in siz­able parts of the re­gion. Why would the elec­tron stay out of some re­gion com­pletely? Would not its un­cer­tainty in po­si­tion at least pro­duce a very tiny prob­a­bil­ity for the ele­cron to be in­side them? But prov­ing it rig­or­ously is an­other mat­ter. Then there are some­what more rea­son­able con­jec­tures, like that a wave func­tion would be­come zero at a sin­gle point not at the bound­ary. (That would still give a unique ground state. But would you not want to know whether it re­ally could hap­pen?) How about frac­tal wave func­tions? Or just dis­con­tin­u­ous ones? In one di­men­sion the wave func­tion must be con­tin­u­ous, pe­riod. A dis­con­ti­nu­ity would pro­duce a delta func­tion in the de­riv­a­tive, which would pro­duce in­fi­nite ki­netic en­ergy. But in mul­ti­ple di­men­sions, things be­come much less ob­vi­ous. (Note how­ever that in real life, a no­tice­ably sin­gu­lar­ity in $\psi$ at a point would re­quire quite a sin­gu­lar po­ten­tial at that point.)

You might guess that you could use the ap­proach of the right graph in fig­ure D.1 in mul­ti­ple di­men­sions, tak­ing the $x$ co­or­di­nate in the di­rec­tion nor­mal to the zero cross­ing sur­face. But first of all that re­quires that the zero cross­ing sur­face is rec­ti­fi­able. That ex­cludes lone zero cross­ing points, or frac­tal cross­ing sur­faces. And in ad­di­tion there is a ma­jor prob­lem with try­ing to show that the de­riv­a­tives in di­rec­tions other than $x$ re­main small.

There is how­ever a method some­what sim­i­lar to the one of the right graph that con­tin­ues to work in more than one di­men­sions. In par­tic­u­lar, in three di­men­sions this method uses a small sphere of ra­dius $\delta$ around the sup­posed point of zero wave func­tion. The method can show in, any num­ber of di­men­sions, that $\vert\psi\vert$ can­not be­come zero. (Ex­cept at im­pen­e­tra­ble bound­aries as al­ways.) The method does not make any un­jus­ti­fied a pri­ory as­sump­tions like a nonzero gra­di­ent. How­ever, be warned: it is go­ing to be messy. Only math­e­mat­i­cally in­clined stu­dents should read the rest of this de­riva­tion.

The dis­cus­sion will use three di­men­sions as an ex­am­ple. That cor­re­sponds, for ex­am­ple, to the elec­tron of the hy­dro­gen mol­e­c­u­lar ion. But the same ar­gu­ments can be made in any num­ber of di­men­sions. For ex­am­ple, you might have a par­ti­cle con­fined in a two-di­men­sion­al quan­tum well. In that case, the sphere around the point of zero wave func­tion be­comes a cir­cle. Sim­i­larly, in a one-di­men­sion­al quan­tum wire, the sphere be­comes the line seg­ment $\vphantom{0}\raisebox{1.5pt}{$-$}$$\delta$ $\raisebox{-.3pt}{$\leqslant$}$ $x$ $\raisebox{-.3pt}{$\leqslant$}$ $\delta$. If you have two non­con­fined elec­trons in­stead of just one, you are in six di­men­sions. All these cases can be cov­ered math­e­mat­i­cally by gen­er­al­iz­ing the three-di­men­sional sphere to a hy­per­sphere. A two-di­men­sion­al hy­per­sphere is phys­i­cally a cir­cle, and a one-di­men­sion­al hy­per­sphere is a line seg­ment. As dis­cussed in the no­ta­tions sec­tion, a gen­eral $n$-​di­men­sion­al hy­per­sphere has an $n$-​di­men­sion­al vol­ume and sur­face area given by:

$${\cal V}_n = C_n \delta^n \qquad A_n = n C_n \delta^{n-1}$$

For ex­am­ple, $C_3$ $\vphantom0\raisebox{1.5pt}{$=$}$ $4\pi$/3, so the above ex­pres­sions give the cor­rect vol­ume and sur­face of a sphere in three di­men­sions. In two di­men­sions, the vol­ume is phys­i­cally the area of the cir­cle, and the area is its perime­ter. The de­riva­tions will need that the $n$-​di­men­sion­al in­fin­i­tes­i­mal in­te­gra­tion el­e­ment is

$${\rm d}^n {\skew0\vec r}= {\rm d}A_n {\rm d}r$$

Here ${\rm d}{A}_n$ is an in­fin­i­tes­i­mal seg­ment of the spher­i­cal sur­face of ra­dius $r$. You can re­late this to the way that you do in­te­gra­tion in po­lar or spher­i­cal co­or­di­nates. How­ever, the above ex­pres­sion does not de­pend on ex­actly how the an­gu­lar co­or­di­nates on hy­per­sphere ar­eas are de­fined.

To show that points of zero wave func­tion are not pos­si­ble, once again first the op­po­site will be as­sumed. So it will be as­sumed that there is some point where $\vert\psi\vert$ be­comes zero. Then a con­tra­dic­tion will be es­tab­lished. That means that the as­sump­tion must be in­cor­rect; there are no points of zero wave func­tion.

To find the con­tra­dic­tion, de­fine a ra­dial co­or­di­nate $r$ as the dis­tance away from the sup­posed point of zero $\vert\psi\vert$. Next at every dis­tance $r$, de­fine $\varphi(r)$ as the av­er­age value $\vert\psi\vert$ on the spher­i­cal sur­face of ra­dius $r$:

$$\varphi(r) \equiv \int \vert\psi\vert \frac{{\rm d}A_n}{A_n}$$

Func­tion $\varphi(r)$ will need to be con­tin­u­ous for $r$ $\raisebox{.2pt}{$\ne$}$ 0, oth­er­wise the im­plied jump in wave func­tion val­ues would pro­duce in­fi­nite ki­netic en­ergy. For $\vert\psi\vert$ to be­come zero at $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, as as­sumed, re­quires that $\varphi(r)$ is also con­tin­u­ous at $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 and that $\varphi(0)$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. (Note that $\vert\psi\vert$ must be con­tin­u­ous and zero at $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. Oth­er­wise it would have val­ues that stay a fi­nite amount above zero how­ever close you get to $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. Then $\vert\psi\vert$ would not be zero in a mean­ing­ful sense. And here we want to ex­clude points of zero wave func­tion. Ex­clud­ing points of in­de­ter­mi­nate wave func­tion will be left as an ex­er­cise for the reader. But as al­ready men­tioned, that sort of sin­gu­lar be­hav­ior would re­quire quite a sin­gu­lar po­ten­tial.)

Take now some small sphere, of some small ra­dius $\delta$, around the sup­posed point of zero wave func­tion. The value of $\varphi$ on the outer sur­face of this sphere will be called $\varepsilon$. It will be as­sumed that there are no val­ues of $\varphi(r)$ greater than $\epsilon$ in­side the sphere. (If there are, you can al­ways re­duce the value of $\delta$ to get rid of them.)

The blunt­ing in­side this sphere will now be achieved by re­plac­ing the $\varphi(r)$ part of $\vert\psi\vert$ by $\varepsilon$. So the blunted wave func­tion is:

$$\psi' \equiv \vert\psi\vert - \varphi(r) + \varepsilon$$

Con­sider now first what the cor­re­spond­ing in­crease in the po­ten­tial en­ergy in­te­gral in­side the sphere is:

$$I_V' - I_V = \int\left[ V(\vert\psi\vert-\varphi(r)+\varepsilon)^2 - V (\vert\psi\vert)^2\right] { \rm d}^3{\skew0\vec r}$$

Mul­ti­ply­ing out the square, that be­comes:

$$I_V' - I_V = \int 2V\vert\psi\vert(\varepsilon-\varphi(r))... ... \int V(\varepsilon - \varphi(r))^2 { \rm d}^3{\skew0\vec r}$$

Since $\varphi(r)$ is non­neg­a­tive, it fol­lows that the in­crease in po­ten­tial en­ergy is lim­ited as

$$I_V' - I_V \mathrel{\raisebox{-.7pt}{$\leqslant$}}2 V_{\rm ... ...kew0\vec r}\qquad \int { \rm d}^3{\skew0\vec r}= C_n \delta^n$$

Note that the hy­per­sphere for­mula for the vol­ume of the sphere has been used. The pur­pose is to make the fi­nal re­sult valid in any num­ber $n$ of di­men­sions, not just three di­men­sions. The re­main­ing in­te­gral in the above ex­pres­sion can be rewrit­ten as

$$\int \vert\psi\vert { \rm d}^3{\skew0\vec r}= \mathop{\int... ...mathrel{\raisebox{-.7pt}{$\leqslant$}}\varepsilon C_n \delta^n$$

So fi­nally

$$I_V' - I_V \mathrel{\raisebox{-.7pt}{$\leqslant$}}3 V_{\rm max} \varepsilon^2 C_n \delta^n$$

The next ques­tion is what hap­pens to the ki­netic en­ergy. In three-di­men­sion­al spher­i­cal co­or­di­nates, the ki­netic en­ergy af­ter blunt­ing is

$$I_T' = \frac{\hbar^2}{2m} \int \Bigg[ \bigg(\frac{\partia... ...si'}}{\partial\phi}\bigg)^2 \Bigg] { \rm d}^3{\skew0\vec r}$$

The ini­tial ki­netic en­ergy is given by a sim­i­lar ex­pres­sion, with $\vert\psi\vert$ re­plac­ing $\psi'$. Now the ex­pres­sion for the an­gu­lar de­riv­a­tives in the in­te­grand will be dif­fer­ent in a dif­fer­ent num­ber of di­men­sions. For ex­am­ple, in two-di­men­sion­al po­lar co­or­di­nates, there will be no $\phi$-​de­riv­a­tive. But these an­gu­lar de­riv­a­tives are un­changed by the blunt­ing and drop out in the dif­fer­ence in ki­netic en­ergy. So the de­crease in ki­netic en­ergy be­comes, af­ter sub­sti­tut­ing for $\psi'$ and sim­pli­fy­ing:

$$I_T - I_T' = \frac{\hbar^2}{2m} \mathop{\int\kern-7pt\int}\... ...c{\partial{\varphi}}{\partial r}\bigg)^2 { \rm d}A_n {\rm d}r$$

Since $\varphi$ does not de­pend on the an­gu­lar co­or­di­nates, that can be writ­ten

$$I_T - I_T' = \frac{\hbar^2}{2m} \int 2 \left[ \int\bigg(\... ...gg(\frac{\partial{\varphi}}{\partial r}\bigg)^2 A_n {\rm d}r$$

The ex­pres­sion be­tween square brack­ets is just the $r$-​de­riv­a­tive of $\varphi$. So the de­crease in ki­netic en­ergy be­comes, sub­sti­tut­ing in for $A_n$,

$$I_T - I_T' = \frac{\hbar^2}{2m} \int \bigg(\frac{\partial{\varphi}}{\partial r}\bigg)^2 n C_n r^{n-1} {\rm d}r$$

Note that the ki­netic en­ergy does de­crease. The right hand side is pos­i­tive. And if the max­i­mum po­ten­tial $V_{\rm {max}}$ in the vicin­ity of the point is neg­a­tive, the po­ten­tial en­ergy de­creases too. So that can­not be a ground state. It fol­lows that the ground state wave func­tion can­not be­come zero when $V_{\rm {max}}$ is neg­a­tive or zero. (Do re­call that the po­ten­tial $V$ was re­de­fined. In terms of the orig­i­nal po­ten­tial, there can­not be a zero if the po­ten­tial is less than the ex­pec­ta­tion value of en­ergy.)

But how about pos­i­tive $V_{\rm {max}}$? Here the fac­tor $r^{n-1}$ in the ki­netic en­ergy in­te­gral is a prob­lem in more than one di­men­sion. In par­tic­u­lar, if al­most all the changes in $\varphi$ oc­cur at small $r$, the fac­tor $r^{n-1}$ will make the ki­netic en­ergy change small. There­fore there is no as­sur­ance that the ki­netic en­ergy de­crease ex­ceeds the po­ten­tial en­ergy in­crease. So a ground state can­not im­me­di­ately be dis­missed like in one di­men­sion.

The so­lu­tion is a trick. You might say that only a math­e­mati­cian would think up a trick like that. How­ever, the au­thor in­sists that he is an aero­space en­gi­neer, not a math­e­mati­cian. The first thing to note that there is a con­straint on how much $\varphi$ can change in the outer half of the sphere, for $r$ $\raisebox{-.5pt}{$\geqslant$}$ $\delta$/2. There the fac­tor $r^{n-1}$ is at least $\delta^{n-1}$$\raisebox{.5pt}{$/$}$​2$\POW9,{n-1}$. So the ki­netic en­ergy de­crease is at least

$$I_T - I_T' \mathrel{\raisebox{-1pt}{$\geqslant$}}\frac{\hba... ...ta \bigg(\frac{\partial{\varphi}}{\partial r}\bigg)^2 {\rm d}r$$

Now the re­main­ing in­te­gral can be es­ti­mated by the Cauchy-Schwartz in­equal­ity as be­fore. Com­par­ing this with the max­i­mum pos­si­ble in­crease in po­ten­tial en­ergy will give a limit on the max­i­mum change in $\varphi$ in the outer half of the sphere. In par­tic­u­lar

$$\varepsilon - \varepsilon_{\rm mid} \mathrel{\raisebox{-.7p... ...n-1}}{n} \frac{m V_{\rm max}}{\hbar^2}}\: \delta\: \varepsilon$$

where $\varepsilon_{\rm {mid}}$ de­notes the value of $\varphi$ at the mid­point $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\delta$$\raisebox{.5pt}{$/$}$​2.

If the above in­equal­ity is not sat­is­fied, the ki­netic en­ergy de­crease would ex­ceed the po­ten­tial en­ergy in­crease and it can­not be a ground state. Note that if the sphere is cho­sen small, the rel­a­tive de­crease in $\varphi$ is small too. For ex­am­ple, sup­pose you choose a sphere, call it sphere 1, with a ra­dius

$$\delta_1 \le \frac14 \sqrt{\frac{n}{3 2^{n-1}} \frac{\hbar^2}{m V_{\rm max}}}$$

In that case,

$$\varepsilon_1 - \varepsilon_{\rm mid,1} \mathrel{\raisebox{-.7pt}{$\leqslant$}}{\textstyle\frac{1}{4}} \varepsilon_1$$

That means that $\varphi$ can at most de­crease by 25% go­ing from the out­side sur­face to the mid­point:

$$\varepsilon_{\rm mid,1} \mathrel{\raisebox{-1pt}{$\geqslant$}}{\textstyle\frac{3}{4}} \varepsilon_1$$

You might say, Why not? And in­deed, there would be noth­ing wrong with the idea that al­most all the change would oc­cur in the in­ner half of the sphere. But the idea is now to drive the math­e­mat­ics into a cor­ner from which even­tu­ally there is no es­cape. Sup­pose that you now de­fine the in­ner half of sphere 1 to be a sphere 2. So the ra­dius $\delta_2$ of this sphere is half that of sphere 1, and its value of $\varepsilon$ is

$$\varepsilon_2 = \varepsilon_{\rm mid,1} \mathrel{\raisebox{-1pt}{$\geqslant$}}{\textstyle\frac{3}{4}} \varepsilon_1$$

(If there are $\varphi$ val­ues in the sec­ond sphere that ex­ceed $\varepsilon_2$, you need to fur­ther re­duce $\delta_2$ to get rid of them. But all that does is re­duce the pos­si­ble changes in $\varphi$ even more.) In this sec­ond sphere, the al­lowed rel­a­tive de­crease in its outer half is a fac­tor 2 smaller than in sphere 1, be­cause $\delta$ is a fac­tor two smaller:

$$\varepsilon_{\rm mid,2} \mathrel{\raisebox{-1pt}{$\geqslant$}}{\textstyle\frac{7}{8}} \varepsilon_2$$

Now take the mid­point as the ra­dius of a sphere 3. Then

$$\varepsilon_3 = \varepsilon_{\rm mid,2} \mathrel{\raisebox{... ...}}{\textstyle\frac{3}{4}} {\textstyle\frac{7}{8}}\varepsilon_1$$

Keep do­ing this and for sphere num­ber $i$ you get

$$\varepsilon_i = {\textstyle\frac{3}{4}}{\textstyle\frac{7}{... ...c{31}{32}} \ldots {\textstyle\frac{2^i-1}{2^i}} \varepsilon_1$$

This must be­come zero for in­fi­nite $i$, be­cause the sphere radii con­tract to zero and $\varphi$ is zero at $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. But it does not! The al­lowed changes are sim­ply too small to reach zero. Just take the log­a­rithm:

$$\ln\varepsilon_i = \ln(1{-}{\textstyle\frac{1}{4}}) + \ln(... ... \ln(1{-}{\textstyle\frac{1}{32}}) + \ldots + \ln\varepsilon_1$$

If $\varepsilon_i$ be­comes zero, its log­a­rithm must be­come mi­nus in­fin­ity. But the in­fi­nite sum does not be­come in­fi­nite. Just use the Tay­lor se­ries ap­prox­i­ma­tion $\ln(1-x)$ $\vphantom0\raisebox{1.1pt}{$\approx$}$ $\vphantom{0}\raisebox{1.5pt}{$-$}$$x$:

$$\ln(1{-}{\textstyle\frac{1}{4}}) + \ln(1{-}{\textstyle\frac... ... {\textstyle\frac{1}{16}} + {\textstyle\frac{1}{32}} + \ldots]$$

The sum within square brack­ets is a geo­met­ric se­ries that has a fi­nite limit, not an in­fi­nite one.

So there is a con­tra­dic­tion. At some stage the de­crease in ki­netic en­ergy must ex­ceed the in­crease in po­ten­tial en­ergy. At that stage, the en­ergy can be re­duced by ap­ply­ing the blunt­ing. So the as­sumed wave func­tion can­not be a ground state.

You might still ob­ject that a Tay­lor se­ries ap­prox­i­ma­tion is not ex­act. But in the re­gion of in­ter­est

$$\ln(1-x) \mathrel{\raisebox{-1pt}{$\geqslant$}}- \frac{\ln(1-{\textstyle\frac{1}{4}})}{-{\textstyle\frac{1}{4}}} x$$

and the ad­di­tional ra­tio is just a con­stant, about 1.15, that does not make a dif­fer­ence.

Woof.