I came across a problem on quantum mechanics that has stumped me.

Consider a single particle in a 1-D potential $U(x)$. The two lowest energy eigenstates are $\psi_S(x)$ and $\psi_A(x)$ with energies $\pm E$ $(E>0)$. Which eigenstate has lower energy and why?

I know the symmetric wavefunction is supposed to have lower energy, but I cannot come up with a good explanation to justify my answer.

I could say that $\psi_A$ has higher energy since it has a node unlike $\psi_S$, but then I would be stuck trying to explain why more nodes implies higher energy.

I feel like the solution is extremely simple, but the answer evades me. Any help with this problem will be appreciated.

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  • 1
    $\begingroup$ Hint: Think about the relative magnitudes of the average absolute values of the two wave function's first derivatives near the center (for your above diagrams). One wf has a minimum while the other has a point of inflection. What does a higher average derivative imply for the total energy? $\endgroup$ Commented Mar 16, 2017 at 15:36

2 Answers 2


Here's a hand-wavy argument that runs like this. In the Schrodinger equation: $$ -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi(x)+V(x)\psi(x)=E\psi(x) $$ the second derivative term $-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi(x)$ plays the role of the kinetic energy. This is basically the curvature of the wavefunction.

For, for fixed potential, smaller curvature, i.e. smaller values of $-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi(x)$, will produce a smaller energy $E$.

Next, note that odd solutions are necessarily $0$ at the origin, and that the solutions must all to go zero as $x\to \pm \infty$.

Given this, an even (or symmetric) solution can have very low curvature since it can "spread out" over the well without having to go through $0$ at the origin.

Indeed, a property of the solutions to the Schrodinger equation is that the energy of the solution must increase with the number of nodes. Following Albert Messiah's book on Quantum Mechanics, the oscillatory theorem states that, if \begin{equation} \hat H\psi _{1}(\xi )=E_{1}\psi _{1}(\xi )\, ,\qquad \hat H\psi_{2}(\xi)=E_{2}\psi _{2}(\xi)\, ,\qquad E_{1}<E_{2}, \end{equation} then $\psi _{2}(\xi )$ has more zeroes (nodes) than $\psi _{1}(\xi ).$ To show this, manipulate the Schrodinger equation so that \begin{equation} \frac{1}{\psi _{1}(\xi )}\frac{\hbar ^{2}}{2m}\frac{d^{2}}{d\xi ^{2}}\psi _{1}(\xi )+E_{1}=V(\xi )=\frac{1}{\psi _{2}(\xi )}\frac{\hbar ^{2}}{2m}\frac{% d^{2}}{d\xi ^{2}}\psi _{2}(\xi )+E_{2}, \end{equation} which one then reorganizes as \begin{eqnarray} \psi _{2}(\xi )\frac{d^{2}}{d\xi ^{2}}\psi _{1}(\xi )-\psi _{1}(\xi )\frac{ d^{2}}{d\xi ^{2}}\psi _{2}(\xi ) &=&\frac{d}{d\xi }\left( \psi _{2}(\xi ) \frac{d}{d\xi }\psi _{1}(\xi )-\psi _{1}(\xi )\frac{d}{d\xi }\psi _{2}(\xi )\right) \nonumber \\ &=&\frac{\hbar ^{2}}{2m}\left( E_{2}-E_{1}\right) \psi _{1}(\xi )\psi _{2}(\xi ). \end{eqnarray} Integrating from $\xi =a$ to $\xi =b,$ we find \begin{eqnarray} &&\frac{\hbar ^{2}}{2m}\left( E_{2}-E_{1}\right) \int_{a}^{b}d\xi \psi _{1}(\xi )\psi _{2}(\xi ) \nonumber \\ &&\qquad = \left( \psi _{2}(b)\frac{d}{d\xi }\psi _{1}(b)-\psi _{1}(b)\frac{d}{d\xi }\psi _{2}(b)\right)\nonumber \\ &&\qquad\qquad -\left( \psi _{2}(a)\frac{d}{d\xi }\psi _{1}(a)-\psi _{1}(a)\frac{d}{d\xi } \psi _{2}(a)\right) \end{eqnarray} If one chooses the points $a,$ $b$ to be consecutive zeroes of $\psi _{1},$ and if $\psi _{1}$ is continuous, then $\psi _{1}$ does not change sign over the interval $\left[ a,b\right] .$ We can assume, without loss of generality, that $\psi _{1}$ is positive on this interval, so that $\psi _{1}^{\prime }(a)\geq 0$ and $\psi _{1}^{\prime }(b)\leq 0.$ Then, using these with $\psi _{1}(a)=\psi _{1}(b)=0,$ we obtain \begin{equation} \frac{\hbar ^{2}}{2m}\left( E_{2}-E_{1}\right) \int_{a}^{b}d\xi \psi _{1}(\xi )\psi _{2}(\xi )=\psi _{2}(b)\frac{d}{d\xi }\psi _{1}(b)-\psi _{2}(a)\frac{d}{d\xi }\psi _{1}(a). \end{equation}

Assume now that $\psi _{2}(\xi )$ is always positive on $\left[ a,b\right] .$ Then the integrand is positive, so the integral will be positive (the area under a positive curve is necessarily positive!) On the right hand side, however, the first term is negative by virtue of $\psi _{1}^{\prime }(b)\leq 0$ while the second is positive. Overall, the right hand side is therefore negative, while the left side is negative, a contradiction : $\psi _{2}(\xi )$ cannot be always positive on $\left[ a,b\right] ,$ meaning that between two consecutives zeroes of $\psi _{1}(\xi )$, the function $\psi _{2}(\xi )$ must itself go through zero. In other words, $\psi _{2}(\xi )$ has more zeroes than $\psi _{1}(\xi ).$ This proves the theorem.

This result, applied to your problem, shows that the most symmetric solution, which has no nodes, must have the lowest energy.

The theorem is also beautifully illustrated with the harmonic oscillator wave function. By inspection, the wave function for the lowest energy has no zeroes, the wave functions for the next lowest energies have one, two, three, etc. nodes, the number of nodes increasing with the energy. It is also easily verified that, between two nodes of a solution, another solution of higher energy will go through zero at least once.

BTW this also works for the Coulomb potential provided $V(r)$ is the effective potential, which includes the centrifugal term. For given fixed $\ell$ the number of nodes also increases with energy, as expected.


The answer is, in fact, simple. The ground state must have no nodes. (See wikipedia, or Mur-Petit, Poll & Mazzanti 2002, or, if you must, Courant-Hilbert v I Ch VI  §6, Nodes of Eigenfunctions -- "Knoten").

It is unique, as a second degenerate state thus also without nodes must be orthogonal to the first, but how could it be, if they are both of the same sign everywhere?

But all excited states must have nodes, as they must be orthogonal to this nodeless ground state, and they can only achieve that by possession of both positive and negative regions conspiring in their overlap with it.

The reason, reflected but not over-emphasized in the proofs, is, as @Lewis Miller points out, the dominant contribution of the curvature at nodes.

As @ZeroTheHero 's answer summarizes, the Messiah Quantum Mechanics vI Ch 3,  §8 and  §12 provides the nodal theorems based on orthogonality and the properties of the Wronskian of the Sturm-Liouville system:

If one arranges the eigenstates in the order of increasing energies, $\epsilon_1,\epsilon_2, \epsilon_3,...$, the eigenfunctions likewise fall in the order of increasing number of nodes; the nth eigenfunction has n-1 nodes, between each of which the following eigenfunctions have at least one node.


  • $\begingroup$ The link to the Mur-Petit paper seems to have rot. The full citation to the article is Mur-Petit J, Polls A, Mazzanti F. The variational principle and simple properties of the ground-state wave function. American Journal of Physics. 2002 Aug;70(8):808-10. There is also a reply to this paper which points to a minor technical bug that somewhat restricts the conclusions of Mur-Petit et al. $\endgroup$ Commented Dec 29, 2020 at 15:28
  • $\begingroup$ @ZeroTheHero Thanks so much... I supplanted the doi link, as I should have done in the first place.... $\endgroup$ Commented Dec 29, 2020 at 15:34
  • $\begingroup$ surely it is for me to thank you for this interesting reference. See also Foong SK, Kiang D, Nogami Y. Comment on “The variational principle and simple properties of the ground-state wave function,” by J. Mur-Petit, A. Polls, and F. Mazzanti [Am. J. Phys. 70 (8), 808–810 (2002)]. American Journal of Physics. 2003 Jul;71(7):731-2. for the technical bug. $\endgroup$ Commented Dec 29, 2020 at 15:37
  • $\begingroup$ Thanks for the debugging... $\endgroup$ Commented Dec 29, 2020 at 15:38

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