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.
\begin{figure}
\centering
\setlength{\unitlength}{1pt}
\begin{picture}(...
...(166.8,43.7){\makebox(0,0)[l]{$\varepsilon_2$}}
\end{picture}
\end{figure}

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

\begin{displaymath}
I_T = \frac{\hbar^2}{2m} \int_{\rm all} (\nabla\psi)^2 {\,...
...uad
I_V = \int_{\rm all} V \psi^2 {\,\rm d}^3{\skew0\vec r}
\end{displaymath}

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

\begin{displaymath}
\big\langle E\big\rangle ' = \frac{I_T' + I_V'}{\big\langl...
...g\rangle = \int_{\rm all} (\psi')^2 {\,\rm d}^3{\skew0\vec r}
\end{displaymath}

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 $\big\langle E\big\rangle $ is zero, then so is $I_{\rm {T}}+I_{\rm {V}}$. Therefore the change in energy due to blunting becomes

\begin{displaymath}
\big\langle E\big\rangle ' - \big\langle E\big\rangle = \b...
...V' - (I_T + I_V)}{\big\langle\psi'\big\vert\psi'\big\rangle }
\end{displaymath}

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...
...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

\begin{displaymath}
I_V' - I_V = \int_{-\delta}^0 V (\varepsilon^2-\vert\psi\v...
...\varepsilon^2{\,\rm d}x
= V_{\rm max}\,\varepsilon^2 \delta
\end{displaymath}

where $V_{\rm {max}}$ is the maximum (redefined) potential in the region. (Note that if $\psi$ is not monotonuous, $\vphantom0\raisebox{1.5pt}{$-$}$$\delta$ and $\delta_2$ are defined as the points closest to the origin where $\varepsilon$ is reached. So by definition $\vert\psi\vert$ does not exceed $\varepsilon$.) It is seen that the maximum potential energy decrease is proportional to $\delta$. However, the minimum kinetic energy decrease is proportional to 1/$\delta$. So for small enough $\delta$, potential energy increase cannot compete with kinetic energy decrease. More specifically, taking $\delta^2$ small compared to $\hbar^2$$\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

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

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 $\vphantom0\raisebox{1.5pt}{$-$}$$\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:

\begin{displaymath}
{\cal V}_n = C_n \delta^n \qquad A_n = n C_n \delta^{n-1}
\end{displaymath}

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

\begin{displaymath}
{\rm d}^n {\skew0\vec r}= {\rm d}A_n {\rm d}r
\end{displaymath}

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 $r$:

\begin{displaymath}
\varphi(r) \equiv \int \vert\psi\vert \frac{{\rm d}A_n}{A_n}
\end{displaymath}

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:

\begin{displaymath}
\psi' \equiv \vert\psi\vert - \varphi(r) + \varepsilon
\end{displaymath}

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

\begin{displaymath}
I_V' - I_V =
\int\left[ V(\vert\psi\vert-\varphi(r)+\var...
...on)^2 - V (\vert\psi\vert)^2\right] {\,\rm d}^3{\skew0\vec r}
\end{displaymath}

Multiplying out the square, that becomes:

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

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

\begin{displaymath}
I_V' - I_V \mathrel{\raisebox{-.7pt}{$\leqslant$}}2 V_{\rm...
...ew0\vec r}\qquad \int {\,\rm d}^3{\skew0\vec r}= C_n \delta^n
\end{displaymath}

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

\begin{displaymath}
\int \vert\psi\vert {\,\rm d}^3{\skew0\vec r}= \mathop{\in...
...athrel{\raisebox{-.7pt}{$\leqslant$}}\varepsilon C_n \delta^n
\end{displaymath}

So finally

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

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

\begin{displaymath}
I_T' = \frac{\hbar^2}{2m} \int
\Bigg[
\bigg(\frac{\par...
...}}{\partial\phi}\bigg)^2
\Bigg]
{\,\rm d}^3{\skew0\vec r}
\end{displaymath}

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:

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

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

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

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

\begin{displaymath}
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
\end{displaymath}

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

\begin{displaymath}
I_T - I_T' \mathrel{\raisebox{-1pt}{$\geqslant$}}\frac{\hb...
...a \bigg(\frac{\partial{\varphi}}{\partial r}\bigg)^2 {\rm d}r
\end{displaymath}

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

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

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

\begin{displaymath}
\delta_1 \le \frac14 \sqrt{\frac{n}{3\,2^{n-1}} \frac{\hbar^2}{m V_{\rm max}}}
\end{displaymath}

In that case,

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

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

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

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

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

(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:

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

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

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

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

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

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:

\begin{displaymath}
\ln\varepsilon_i =
\ln(1{-}{\textstyle\frac{1}{4}}) + \l...
...\ln(1{-}{\textstyle\frac{1}{32}}) + \ldots + \ln\varepsilon_1
\end{displaymath}

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$}$ $\vphantom0\raisebox{1.5pt}{$-$}$$x$:

\begin{displaymath}
\ln(1{-}{\textstyle\frac{1}{4}}) + \ln(1{-}{\textstyle\fra...
...{\textstyle\frac{1}{16}} + {\textstyle\frac{1}{32}} + \ldots]
\end{displaymath}

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

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

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

Woof.