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.

Consider first the mirror symmetry of the hydrogen molecular ion around its mid-plane. Suppose that you look at the hydrogen molecular ion in a mirror. Assume that the plane of the mirror is normal to the $z$-axis, i.e. to the line through the two nuclei. To simplify the mathematics assume that the mirror is two-sided and infinitely thin. Then you can put it right in between the two nuclei and look in the mirror from either side. That allows you to see the mirror image of each side. Taking the coordinate $z$ to be zero in the center, what such a 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.

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

Now the hydrogen molecular ion is “mirror” symmetric with respect to the mirror. Roughly speaking, that means that the mirror image is exactly the same as the original ion.

To understand more precisely what it means requires consideration of the operator associated with the mirroring. The effect of the mirroring on any molecular wave function $\Psi$ can be represented by a “mirror operator,” call it ${\cal M}$. According to the above, all this operator does is replace $z$ by $-z$:

$${\cal M}\Psi(x,y,z)=\Psi(x,y,-z)$$

Now 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

$${\cal M}\Psi(x,y,z)\equiv\Psi(x,y,-z)=\Psi(x,y,z)$$

In other words, a mirror-symmetric wave function is the same at positive values of $z$ as at the corresponding negative values. Mathematians might simply say that the wave function is symmetric around the $xy$-plane, (i.e. the mirror).

Now you may have noted that we humans are not mirror symmetric. Just try to shake hands with your mirror image. You will be shaking hands with a left-handed person. So it is not immediately obvious why the ground state wave function of the molecular ion would be mirror symmetric.

The fundamental reason that the ground state of the molecular ion is mirror symmetric is a mathematical one. The mirror operator ${\cal M}$ commutes with the Hamiltonian $H$. Recall from chapter 4.5.1 that this means that it does not make a difference in which order you apply the two operators.

More importantly, 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:

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

where $\lambda$ 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 itself can be either 1 or $\vphantom0\raisebox{1.5pt}{$-$}$1. So for the ground state wave function $\psi_{gs}$, either

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

or

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

As already noted, if the first possibility applies, the wave function is mirror symmetric. If the second possibility applies, the 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, the ground state is mirror symmetric.

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 and 3.12, is antisymmetric.

That leaves one question unanswered: why does it not make a difference in which order you apply the Hamiltonian and the mirror operator? The answer follows 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 indeed it makes no difference whether you first mirror and then apply the Hamiltonian or vice-versa. The two operators commute.

Next consider the rotational symmetry of the hydrogen molecular ion around the axis through the nuclei. The ground state of the molecular ion is rotationally symmetric. That means that the wave function does not change if you rotate the ion around the $z$-axis through tne nuclei. The question is again, why not?

In this case, let ${\cal R}_\varphi$ be the operator that rotates a wave function $\Psi$ over a small angle $\varphi$ around the $z$-axis. This operator commutes again 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.

So once again the ground state must be an eigenfunction of the symmetry operator:

$${\cal R}_\varphi \psi_{\rm gs} = \lambda \psi_{\rm gs}$$

The question is again, what is that eigenvalue $\lambda$? First note that the magnitude of all eigen­values of ${\cal R}_\varphi$ must be 1, since wave functions must stay normalized to 1 after the rotation. Complex numbers of magnitude 1 can always be written as $e^{{{\rm i}}a}$ where $a$ is some real number. Number $a$ must be propor­tional to $\phi$, 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 $e^{{{\rm i}}m\phi}$, where $m$ is a real constant independent of $\phi$.

So the rotated ground state is some multiple $e^{{{\rm i}}m\phi}$ 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 $e^{{{\rm i}}m\phi}$ is real and positive too. And if you check the Euler formula (2.5), you see that $e^{{{\rm i}}m\phi}$ is only real and positive if it is 1. Then the ground state wave function does not change when rotated; it is rotationally symmetric around the $z$-axis through the nuclei.

You might of course wonder about excited energy states. For those there is another relevant observation: rotating over a full turn of $2\pi$ puts each point back where it came from. That should produce the original wave function. So the eigenvalues $e^{{{\rm i}}m2\pi}$ for a full turn must be 1 in order that the wave function stays the same. 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 $e^{{{\rm i}}m\phi}$ equal to 1 for all angles $\varphi$. But for excited states, $m$ can be nonzero. 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.

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 one, with positions ${\skew0\vec r}_1$ and ${\skew0\vec r}_2$. 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, because the two electrons are identical. So it 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 were they were originally. So the ground state is symmetric under particle exchange, just like it was symmetric under mirroring. In other words, swapping the particle positions does nothing to the wave function.

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 requirement, chapter 5.6. In fact they must be antisymmetric under an expanded definition of the exchange operator. Similar problems would apply for systems involving three or more protons or neutrons. However, for some particle types, like three or more helium atoms, the symmetry continues to apply, [17, p. 321].