A.8 Positive ground state wave function

This addendum discusses why in at least the simplest cases a ground state wave function can be assumed to be real, positive, and unique (i.e. nondegenerate). It is assumed that the potential is a real function of position. That is true for the hydrogen molecular ion. It is also true for a single hydrogen atom and most other simple systems, at least in the nonrelativistic approximations normally used in this book.

It should first be noted that if potentials are allowed that are positive infinity in a finite region, nonunique ground states that cannot be taken positive may in fact possible. Such a potential can provide an impenetrable boundary, completely separating one region of space from another. In that case the ground state wave functions at the two sides of the boundary become decoupled, allowing for indeterminacy in the combined ground state. Such artificial cases are not covered here. But you can readily find examples in lots of books on quantum mechanics, especially in one dimension. Here it will be assumed that the potentials stay away from positive infinity. For practical purposes, it may also be noted that if the potential becomes positive infinity at just a few points, it is usually not a problem unless the approach to singularity is very steep.

There is however a much more important restriction to the conclusions in this note: ground states may not be positive if you go to many-particle systems. That is discussed further in the final paragraphs of this addendum.

First consider why the ground state, and any other energy eigenstate, can be assumed to be real without loss of generality. Suppose that you had a complex eigenfunction $\psi$ for the eigenvalue problem $H\psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E\psi$. Write the eigenfunction as

\begin{displaymath}
\psi = \psi_{\rm {r}} + {\rm i}\psi_{\rm {i}}
\end{displaymath}

where the real part $\psi_{\rm {r}}$ and the imaginary part $\psi_{\rm {i}}$ are real functions. Plugging this into the eigenvalue problem, you see that the real and imaginary parts each separately must satisfy the eigenvalue problem. (For the complex number $H\psi-E\psi$ to be zero, both its real and imaginary parts have to be zero.) So each of $\psi_{\rm {r}}$ and $\psi_{\rm {i}}$ is just as good an eigenfunction as $\psi$ and each is real. Since the original complex $\psi$ is a linear combination of the two, you do not need it separately. (If either $\psi_{\rm {r}}$ or $\psi_{\rm {i}}$ is zero, it is not an eigenfunction, but then it is not needed to describe $\psi$ either. And if it is nonzero, it can be normalized.)

Next consider why the ground state can be taken to be positive, assuming, for now, that it is unique. What characterizes the ground state $\psi_{\rm {gs}}$ is that it has the lowest possible expectation value of the energy. The expectation energy can be written for arbitrary, but normalized, wave functions $\psi$ as

\begin{displaymath}
\big\langle E\big\rangle = \frac{\hbar^2}{2m} \int_{\rm al...
...\vec r}
+ \int_{\rm all} V \psi^2 {\,\rm d}^3{\skew0\vec r}
\end{displaymath}

In the first term, the kinetic energy, integrations by part have been used to get rid of the second order derivatives. Note that $(\nabla\psi)^2$ stands for the sum of the square partial derivatives of $\psi$. Now by definition $\psi_{\rm {gs}}$ has the lowest possible value of the above expectation energy among all possible normalized functions $\psi$. But only terms square in $\psi$ appear in the expectation energy. So $\vert\psi_{\rm {gs}}\vert$ has the same expectation energy. That means that $\vert\psi_{\rm {gs}}\vert$ is a ground state wave function too. Under the given assumption that the ground state is unique, $\vert\psi_{\rm {gs}}\vert$ can be taken to be the ground state. That makes the ground state positive. (Note that a constant of magnitude one does not make a difference in an eigenfunction. So the original $\psi_{\rm {gs}}$ might well have been equal to $-\vert\psi_{\rm {gs}}\vert$. But $\vert\psi_{\rm {gs}}\vert$ is equivalent to that.)

Finally, it needs to be shown that the ground state is indeed unique as assumed above. That is really messy, so it has been banned to derivation {D.22}. It is based on the same idea that the absolute value of a ground state is a ground state too.

Regrettably the arguments above stop working for more than two electrons. To really understand the reason, you will first need to read chapter 5.6 on multiple-particle systems. But in a nutshell, the wave function for systems with multiple electrons must satisfy additional requirements, called the antisymmetrization requirements. These requirements normally turn $\vert\psi_{\rm {gs}}\vert$ into an unacceptable wave function. Then obviously the above arguments fall apart. Fortunately, for just two electrons, there is a loophole in the requirement called spin. That allows the hydrogen molecule, with two electrons, still to be covered.

The same problem occurs for atomic nuclei that contain multiple protons and/or neutrons. (For atomic nuclei, the potentials also tend to be far more complicated than assumed here. But that is another matter.) In general, particles for which antisymmetrization requirements apply are called fermions.

There is however a different class of particles called ”bosons.” For those, the wave function has to satisfy symmetrization requirements. Symmetrization requirements are still OK if you replace $\psi$ by $\vert\psi\vert$. So the ideas above are helpful for understanding the ground state of large numbers of bosons. For example, they are helpful in understanding the superfluidity of liquid helium near its ground state, [18, pp. 321-323]. Complete helium atoms are bosons.