# When is it true that more nodes equals higher energy?

Consider all the MOs of some isolated molecule. (It could be a single atom too; I'll use MO to refer to AOs as well.) Number them in increasing order of the number of nodes (node = surface where the wave function has zero density). Orbitals with the same number of nodes can be numbered in any order. Now you have a sequence of orbitals $O_1, O_2, ...$. Let their respective energies be $E_1, E_2, ...$.

It seems to be "common knowledge" that $E_n \le E_{n+1}$ for any such system and any $n$. As Martin put it to me yesterday, "An orbital with 47 nodes can never be lower in energy than one with only 46." (Follow up on Counting Nodal Planes in cyclopropane.)

For various reasons, given below, I think this cannot be true in general, and I'd really like to know under what conditions it is known to be true. "Known" here can mean either a rigorous statement with a reference to a proof (a trivial example would be: It is true for a one-electron atom; we can calculate the energies exactly) or a precise statement with empirical justification (something like "no counterexamples are known for class X of molecules" - again with a reference).

Important: Please, I'm not looking for an explanation that re-states the rule in some equivalent or even looser fashion ("more nodes means the orbital is larger and less dense, therefore it must be higher in energy").

Why do I think the statement can't be always true? Well, a calcium atom has a filled 4s orbital and empty 3d orbitals. If this does not count as a counterexample, please explain what notion of orbital energy the statement applies to.

In general, I'm happy to believe that two MOs must satisfy the rule if they have "comparable" sets of nodes (say, constant $n_x$ and $n_y$ in the example I'm about to discuss), but I'd like to understand what is meant by "comparable" in general. In a molecule with no symmetry, are there comparable MOs at all? If so, how do we know if two given MOs are comparable?

The case of a much simpler system, the 3D rectangular box, may also be relevant. The energy levels for such a box are of course $$\frac{\hbar^2\pi^2}{2m}\biggl( \frac{n_x^2}{L_x^2}+ \frac{n_y^2}{L_y^2}+ \frac{n_z^2}{L_z^2} \biggr),$$ where $n_x$, $n_y$ and $n_z$ are one more (or one less, if you count the walls) than the number of nodal planes in the respective direction. If we take $L_x = L_y = 1$ and $L_z = 0.1$ (say), the wavefunction with $n_x=5$, $n_y=1$, $n_z =1$ has energy $5^2+ 1^2 + (1/0.1)^2 = 126$ and 4 nodes (or 10, if you count the walls), while the wavefunction for $n_x=1$, $n_y=1$, $n_z=2$ has energy $402$ and 1 node (or 7). So clearly the rule is not true here.

Granted that molecules are not boxes, but this does show that arguments based simply on the number of sign changes are not rigorous, so they don't answer my question.

• Martin - very generous of you to offer a bounty. I too am puzzled that nobody has replied or made comments so far! – Silvio Levy Jul 27 '14 at 17:16
• I really want a canonical answer myself. However, I believe you are correct and my previous assumption was too narrow minded. But I really would like to have a little more prove. And I know that there are people in this community that could give more insight. – Martin - マーチン Jul 29 '14 at 7:30
• Particle in a 3D box follows this rule where energy scales with nodes. So does particle in a 1D box. But that is because the waves you are comparing are comparable. With the 3D rectangle with different values of L (for x, y, and z), you can only compare the x-, y-, and z- components with each other. You can't sum over all the nodes and make this comparison. Just like you can compare 1s, 2s, 3,s, etc. with each other, p-orbitals with each other, etc. but not p- and d- orbitals. The correlation breaks down when you do that. – LordStryker Jul 29 '14 at 13:51
• @LordStryker - thanks. Reply: "Particle in a 3D box follows this rule" -- By "THIS rule" you don't mean the rule I wrote but your rule: "you can only compare the x-, y-, and z- components with each other." Fair enough. But if drive a nail into the box and destroy the symmetry of this toy example you no longer have x-, y-, and z- modes. So in this situation WHAT rule is followed? Regarding "You can't sum over all the nodes and make this comparison." -- Again: Can you state what makes two nodes comparable, i.e, how one can tell that two nodes are comparable? – Silvio Levy Jul 29 '14 at 17:23
• @SilvioLevy You bring up a good point. I've never worked PIAB examples in anything other than the simple 1D/2D examples, cubes, and rectangles. Particle in a 'funhouse'? I'd love to see someone work out an example like that. – LordStryker Jul 29 '14 at 19:09

## General case

There is indeed a mathematical theorem that deals with the number of nodes an eigenfunction corresponding to a certain eigenvalue can possess. It was laid down by Courant$^{[1, 2]}$ and it states the following:

Given the self-adjoint second order (partial) differential equation

\begin{equation} \left(\hat{L} + \lambda \rho(\mathbf{x}) \right) u(\mathbf{x}) = 0 \end{equation}

(where $\hat{L} = L(\mathbf{\Delta}, \mathbf{x})$ is a linear, hermitian differential operator, $\rho(\mathbf{x})$ is positive and bounded, and $\lambda$ is the eigenvalue) for a domain $G$ with homogeneous boundary conditions, that is $u(\mathbf{x}) = 0$ on the boundary of the region $G$; if its eigenfunctions are ordered according to increasing eigenvalues, then the nodes of the $n^{\text{th}}$ eigenfunction divide the domain into no more than $n$ subdomains. The nodal set of $u(\mathbf{x})$ is defined as the set of points $\mathbf{x}$ such that $u(\mathbf{x}) = 0$. No assumptions are made about the number of independent variables.

The proof is rather involved and so I won't show it here. But if you want you can look it up in  or here.

So, Courant's nodal line theorem tells us, that if we order the possible energy eigenvalues of the time-independent Schroedinger equation as $\lambda_1 \leq \lambda_2 \leq \lambda_3 \leq \dots$, then (depending on precisely how you set up the numbering) the $n^{\text{th}}$ eigenfunction, $\Psi_{n}$ (the one with energy eigenvalue $\lambda_n$) has at most $n$ nodes (including the trivial one at the boundary $\mathbf{x} \to \infty$). Unfortunately, this gives you only an upper bound for the number of nodes a wave function with a certain energy eigenvalue may possess. So, all we know is that the ground state wave function $\Psi_{1}$ cannot have any nodes within the region $G$ (in total it has one node, namely the one at $\mathbf{x} \to \infty$). Wave functions for higher $n$ may possess up to $n-1$ nodes within $G$ but may as well have less. Thus, we cannot in general say that if a wave function has more nodes than another one it will automatically correspond to a state with higher energy.

## Special case: Schroedinger equation in one dimension

There is however a special case: For the Sturm-Liouville eigenvalue problem (and thus for ordinary second order differential equations with homogeneous boundary conditions) we can strengthen Courant's nodal line theorem such that if we order the possible eigenvalues as $\lambda_1 \leq \lambda_2 \leq \lambda_3 \leq \dots$, then the $n^{\text{th}}$ eigenfunction (the one with energy eigenvalue $\lambda_n$) has exactly $n$ nodes (including the trivial one at the boundary $\mathbf{x} \to \infty$).

This is useful since the one-dimensional time-independent Schrödinger equation is a special case of a Sturm-Liouville equation. So, in the case of the inhomogeneous radial Schrödinger equation with a local potential and node-less inhomogeneity such as the radial Schrödinger equation for the hydrogenic atom

\begin{equation} \bigg( \frac{ - \hbar^{2} }{ 2 m_{\mathrm{e}} } \frac{ \mathrm{d}^{2} }{ \mathrm{d} r^{2} } + \frac{ \hbar^{2} }{ 2 m_{\mathrm{e}} } \frac{ \ell (\ell + 1) }{ r^{2} } - \frac{ Z e^{2} }{ 2 m_{\mathrm{e}} r } - E \bigg) r R(r) = 0 \end{equation}

it is generally true that a wavefunction with more (radial) nodes must always correspond to a state of higher energy than a wavefunction with less radial nodes. Also, it is clear that the wavefunctions of the one-dimensional particle-in-a-box must follow this rule. But for the three-dimensional particle-in-a-box this is not true anymore, since in that case the Schroedinger equation of the system is not an ordinary second-order differential equation but a partial-differential equation for which only the general version of Courant's nodal line theorem holds.

## Some concluding remarks

For real-world system like molecules or crystals the Schroedinger equation is a partial differential equation for which the special case outlined above doesn't apply so that only Courant's nodal line theorem in its general form holds which doesn't give a strict justification for the statement that more nodes mean higher energy. Yet it is very often observed that the number of nodes indeed increases with increasing energy. The reason for this can be motivated in the following way: The kinetic energy $E_{\mathrm{kin}}$ of a state is proportional to $\int \Psi \Delta \Psi \, d^{3} r$. Via Gauss's theorem it can be shown that $\int \Psi \Delta \Psi \, d^{3} r \propto \int |\nabla \Psi |^{2} \, d^{3} r$ and so $E_{\mathrm{kin}} \propto \int | \nabla \Psi |^{2} d^{3} r$. Now, nodes force a wavefunction to change it's sign. This often means that the value of $\Psi$ has to increase/decrease rather rapidly thus leading to areas with high absolute values of the gradient and thus to high kinetic energy. Since the potential energies shouldn't differ too much between the different states the higher kinetic energy usually also entails a higher total energy. As an example consider the bonding and antibonding wavefunctions of a homonuclear diatomic molecule whose atoms are placed at the positions $r_{\mathrm{A}}$ and $r_{\mathrm{B}}$. The bonding wavefunction has no nodes. Its value between the atoms doesn't have to undergo a rapid change and thus the slope is rather low. The antibonding wavefunction has one node between the atoms. Its value between the atoms must change rapidly from its positive to its negative maximum thus entailing a very high slope. The slopes of the tail regions are comparable for the bonding and antibonding wavefunctions since it can smoothly fall of to zero at infinity and is not required to go from a maximum value to zero within a very confined region of space - thus even if one wavefunction has to start of at a higher maximum value the gradient will not be much higher. It follows that the antibonding wavefunction has a higher kinetic energy than the bonding wavefunction.

## References

 R. Courant, D. Hilbert, Methods of Mathematical Physics, Vol. 1, Interscience, New York, 1953, p. 451-455.

 R. Courant, "Ein allgemeiner Satz zur Theorie der Eigenfunktionen Selbstadjungierter Differentialausdrücke", Nachr. v. d. Ges. d. Wiss. zu Göttingen 1923, p. 81.

• Thanks for the response, Philipp. Courant's theorem is beautiful but it's far weaker than the general claim, since usually in dimension d > 1 the number of eigenwaves with n nodes seems to grow roughly with $n^d$. I was hoping there might be some rigorous results that have practical use. Your concluding remarks are quite interesting but again the intuitive justification seems to be rigorous only in 1D. Also your first comment to the question provides a counterexample when the symmetries differ, even for homonuclear 2-atom systems. – Silvio Levy Aug 2 '14 at 20:45
• Another question for @Philipp: IIRC the hypotheses of Courant's theorem are satisfied by the time-independent, nonrelativistic Schrödinger equation, but not by relativistic versions such as the Klein-Gordon equation. Is this right, or am I confused? – Silvio Levy Aug 2 '14 at 21:08
• @SilvioLevy I fear you can't do better than Courant's theorem. It strongly suggests that it is simply not generally true that more nodes mean higher energy. But the connection between kinetic energy and gradient of the wavefunction gives some justification why it is true most of the time. How much more practical use would you want? – Philipp Aug 2 '14 at 21:08
• @SilvioLevy Also, why do you think that my intuitive justification is only rigorous in 1D? It involves the gradient, not the first derivative. I only chose a 1D example because it was easier to visualize. – Philipp Aug 2 '14 at 21:10
• To be clear, the issue is the $E_{\mathrm{kin}} \propto |\nabla\Psi |^{2}$, which should have an integral on the right side. You could have high $|\nabla \Psi|$ in a small region only, and still get a lower overall energy than on another wavefunction with slightly lower $|\nabla\Psi|$ over a larger region. – Silvio Levy Aug 2 '14 at 21:30