A.9 Wave function symmetries

Symmetries are very important in physics. For example, symmetries in wave functions are often quite helpful to understand the physics qualitatively.

As an example, the hydrogen molecular ion is mirror symmetric around its midplane. This midplane is the plane halfway in between the two nuclei, orthogonal to the line connecting them. To roughly understand what the mirror symmetry around this plane means, think of the midplane as an infinitely thin mirror. Take this mirror to be two-sided, so that you can look in it from either side. That allows you to see the mirror image of each side of the molecule. Simply put, the mirror symmetry of the ion means that the mirror image looks exactly the same as the original ion.

(If you would place the entire molecule at one side of the mirror, its entire mirror image would be at the other side of it. But except for this additional shift in location, everything would remain the same as in the case assumed here.)

Under the same terms, human beings are roughly mirror symmetric around the plane separating their left and right halves. But that symmetry is far from perfect. For example, if you part your hair at one side, your mirror image parts it at the other side. And your heart changes sides too.

To describe mirror symmetry more precisely, take the line through the nuclei to be the $z$-​axis. And take $z$ to be zero at the mirror. Then all that the mirror does mathematically is replace $z$ by $\vphantom0\raisebox{1.5pt}{$-$}$$z$. For example, the mirror image of the nucleus at positive $z$ is located at the corresponding negative $z$ value. And vice-versa.

The effect of mirroring on any molecular wave function $\Psi$ can be represented by a “mirror operator” ${\cal M}$. According to the above, all this operator does is replace $z$ by $\vphantom0\raisebox{1.5pt}{$-$}$$z$:

\begin{displaymath}
{\cal M}\Psi(x,y,z)=\Psi(x,y,-z)
\end{displaymath}

By definition a wave function is mirror symmetric if the mirror operator has no effect on it. Mathematically, if the mirror operator does not do anything, then ${\cal M}\Psi$ must be the same as $\Psi$. So mirror symmetry requires

\begin{displaymath}
{\cal M}\Psi(x,y,z)\equiv\Psi(x,y,-z)=\Psi(x,y,z)
\end{displaymath}

The final equality above shows that a mirror-symmetric wave function is the same at positive values of $z$ as at the corresponding negative values. Mathematicians might simply say that the wave function is symmetric around the $xy$-​plane, (i.e. the mirror). The ground state $\psi_{\rm {gs}}$ of the molecular ion is mirror symmetric in this sense. The big question to be addressed in this addendum is, why?

The fundamental reason why the ion is mirror symmetric is a mathematical one. The mirror operator ${\cal M}$ commutes with the Hamiltonian $H$. Recall from chapter 4.5.1 what this means:

\begin{displaymath}
{\cal M}H = H {\cal M}
\end{displaymath}

In words, it does not make a difference in which order you apply the two operators.

That can be seen from the physics. The Hamiltonian consists of potential energy $V$ and kinetic energy $T$. Now it does not make a difference whether you multiply a wave function value by the potential before or after you flip the value over to the opposite $z$-​position. The potential is the same at opposite $z$ values, because the nuclei at the two sides of the mirror are the same. As far as the kinetic energy is concerned, if it involved a first-order $z$-​derivative, there would be a change of sign when you flip over the sign of $z$. But the kinetic energy has only a second order $z$-​derivative. A second order derivative does not change. So all together it makes no difference whether you first mirror and then apply the Hamiltonian or vice-versa. The two operators commute.

Also according to chapter 4.5.1, that has a consequence. It implies that you can take energy eigenfunctions to be mirror eigenfunctions too. And the ground state is an energy eigenfunction. So it can be taken to be an eigenfunction of ${\cal M}$ too:

\begin{displaymath}
{\cal M}\psi_{\rm gs}(x,y,z) = \lambda \psi_{\rm gs}(x,y,z)
\end{displaymath}

Here $\lambda$ is a constant called the eigenvalue. But what would this eigenvalue be?

To answer that, apply ${\cal M}$ twice. That multiplies the wave function by the square eigenvalue. But if you apply ${\cal M}$ twice, you always get the original wave function back, because $-(-z)$ $\vphantom0\raisebox{1.5pt}{$=$}$ $z$. So the square eigenvalue $\lambda^2$ must be 1, in order that the wave function does not change when multiplied by it. That means that the eigenvalue $\lambda$ itself can be either 1 or $\vphantom0\raisebox{1.5pt}{$-$}$1. So for the ground state wave function $\psi_{\rm {gs}}$, either

\begin{displaymath}
{\cal M}\psi_{\rm gs}(x,y,z)\equiv\psi_{\rm gs}(x,y,-z)= 1\times\psi_{\rm gs}(x,y,z)
\end{displaymath}

or

\begin{displaymath}
{\cal M}\psi_{\rm gs}(x,y,z)\equiv\psi_{\rm gs}(x,y,-z)=-1\times\psi_{\rm gs}(x,y,z)
\end{displaymath}

If the first possibility applies, the wave function does not change under the mirroring. So by definition it is mirror symmetric. If the second possibility applies, the wave function changes sign under the mirroring. Such a wave function is called “mirror antisymmetric.” But the second possibility has wave function values of opposite sign at opposite values of $z$. That is not possible, because the previous addendum showed that the ground state wave function is everywhere positive. So it must be possibility one. That means that the ground state must indeed be mirror symmetric as claimed.

It may be noted that the state of second lowest energy will be antisymmetric. You can see the same thing happening for the eigenfunctions of the particle in a pipe. The ground state figure 3.8, (or 3.11 in three dimensions), is symmetric around the center cross-section of the pipe. The first excited state, at the top of figures 3.9, (or 3.12), is antisymmetric. (Note that the grey tones show the square wave function. If the wave function is antisymmetric, the square wave function is symmetric. But it will be zero at the symmetry plane.)

Next consider the rotational symmetry of the hydrogen molecular ion around the axis through the nuclei. The ground state of the molecular ion does not change if you rotate the ion around the $z$-​axis through the nuclei. That makes it rotationally symmetric. The big question is again, why?

In this case, let ${\cal R}_\varphi$ be the operator that rotates a wave function $\Psi$ over an angle $\varphi$ around the $z$-​axis. This operator too commutes with the Hamiltonian. After all, the only physically meaningful direction is the $z$-​axis through the nuclei. The angular orientation of the $xy$ axes system normal to it is a completely arbitrary choice. So it should not make a difference at what angle around the $z$ axis you apply the Hamiltonian.

Therefore the ground state must be an eigenfunction of the rotation operator just like it is one of the mirror operator:

\begin{displaymath}
{\cal R}_\varphi \psi_{\rm gs} = \lambda \psi_{\rm gs}
\end{displaymath}

But now what is that eigenvalue $\lambda$? First note that the magnitude of all eigenvalues of ${\cal R}_\varphi$ must be 1. Otherwise the magnitude of the wave function would change correspondingly during the rotation. However, the magnitude of a wave function does not change if you simply rotate it. And if the eigenvalue is a complex number of magnitude 1, then it can always be written as $e^{{\rm i}\alpha}$ where $\alpha$ is some real number. So the rotated ground state is some multiple $e^{{\rm i}\alpha}$ of the original ground state. But the values of the rotated ground state are real and positive just like that of the original ground state. That can only be true if the multiplying factor $e^{{\rm i}\alpha}$ is real and positive too. And if you check the Euler formula (2.5), you see that $e^{{\rm i}\alpha}$ is only real and positive if it is 1. Since multiplying by 1 does not change the wave function, the ground state does not change when rotated. That then makes it rotationally symmetric around the $z$-​axis through the nuclei as claimed.

You might of course wonder about the rotational changes of excited energy states. For those a couple of additional observations apply. First, the number $\alpha$ must be proportional to the rotation angle $\varphi$, since rotating $\Psi$ twice is equivalent to rotating it once over twice the angle. That means that, more precisely, the eigenvalues are of the form $e^{{{\rm i}}m\varphi}$, where $m$ is a real constant independent of $\varphi$. Second, rotating the ion over a $2\pi$ full turn puts each point back to where it came from. That should reproduce the original wave function. So an eigenvalue $e^{{{\rm i}}m2\pi}$ for a full turn must be 1. According to the Euler formula, that requires $m$ to be an integer, one of ..., $\vphantom0\raisebox{1.5pt}{$-$}$2, $\vphantom0\raisebox{1.5pt}{$-$}$1, 0, 1, 2, .... For the ground state, $m$ will have to be zero; that is the only way to get $e^{{{\rm i}}m\varphi}$ equal to 1 for all angles $\varphi$. But for excited states, $m$ can be a nonzero integer. In that case, these states do not have rotational symmetry.

Recalling the discussion of angular momentum in chapter 4.2.2, you can see that $m$ is really the magnetic quantum number of the state. Apparently, there is a connection between rotations around the $z$-​axis and the angular momentum in the $z$-​direction. That will be explored in more detail in chapter 7.3.

For the neutral hydrogen molecule discussed in chapter 5.2, there is still another symmetry of relevance. The neutral molecule has two electrons, instead of just one. This allows another operation: you can swap the two electrons. That is called “particle exchange.” Mathematically, what the particle exchange operator ${\cal P}$ does with the wave function is swap the position coordinates of electron 1 with those of electron 2. Obviously, physically this does not do anything at all; the two electrons are exactly the same. It does not make a difference which of the two is where. So particle exchange commutes again with the Hamiltonian.

The mathematics of the particle exchange is similar to that of the mirroring discussed above. In particular, if you exchange the particles twice, they are back to where they were originally. From that, just like for the mirroring, it can be seen that swapping the particle positions does nothing to the ground state. So the ground state is symmetric under particle exchange.

It should be noted that the ground state of systems involving three or more electrons is not symmetric under exchanging the positions of the electrons. Wave functions for multiple electrons must satisfy special particle-exchange requirements, chapter 5.6. In fact they must be antisymmetric under an expanded definition of the exchange operator. This is also true for systems involving three or more protons or neutrons. However, for some particle types, like three or more helium atoms, the symmetry under particle exchange continues to apply. This is very helpful for understanding the properties of superfluid helium, [18, p. 321].