# Can it be argued qualitatively that the ground state of a bound energy spectrum has zero nodes?

I am taking a 1-D system with a general potential $V(x)$ that results in a spectrum of bound energy eigenvalues. Without resorting to specific cases (such as the harmonic oscillator), how do you prove that the ground state is non-zero everywhere on the real axis? How do I prove this just from the postulates of Quantum Mechanics and the definition of a bound state?

• You can argue that a wavefunction with nodes can be modified to one without and aas a result, the second derivative (i.e. the energy) will decrease, since it is less wiggly. – NickD Jun 7 '17 at 1:49
• And how is a wavefunction with nodes modified into one without? – ferro11001 Jun 7 '17 at 2:17
• Duplicate 295857. – Cosmas Zachos Jun 7 '17 at 3:03
• @ferro11001: I'm talking about a mental transformation: you calculate the energy of a wiggly function, it's got a high second derivative, so the energy is high. Then consider a smooth function, it's got a small second derivative, hence the energy is smaller. The latter is a better approximation to the ground state than the former. – NickD Jun 7 '17 at 14:15
• – Qmechanic Jun 8 '17 at 14:44

"Qualitatively". So you want something less Courant-Hilbert and more seat-of-the-pants than the generic Sturm-Liouville analysis summarized in Messiah vI, Ch III, sec II.8,12.

I'll just outline a less specific/restrictive version of the Wikipedia proof, basically the Feynman-certified one of Appendix A, p 81-82, of M Cohen's 1956 Caltech PhD dissertation.

Without loss of generality, take the bound states to be real; and see that a state with a node at, say, x=0, so ψ(0)=0, cannot be the ground state, as its variational energy can always be lowered. That is, $$\langle E\rangle=\int dx \left ( \frac{\hbar^2}{2m} \left (\frac{d\psi}{dx}\right )^2 +\psi^2 V(x) \right)$$ is larger than the corresponding one for some continuous deformation φ of ψ which lacks a node, so ψ is not the ground state, after all.

You manufacture φ to equal |ψ|, except in a symmetric region [–ε,ε] around x=0, in which we take it, instead, to be, e.g., ψ(x)=ε, suitably normalized. (The WP choice. You could make the choice smoother at the edges of the interval by more elaborate choices, without significant differences.)

You may then readily convince yourself that, whatever the slope of ψ through 0, the normalization of φ will exceed that of ψ by $O(\epsilon^3)$.

Crucially, the kinetic energy is lowered by $O(\epsilon)$, $$\langle T\rangle_\psi - \langle T\rangle_\phi = \frac{\hbar^2}{2m} \int^\epsilon_{-\epsilon} dx \left ( \frac{d\psi (x)}{dx} \right ) ^2 \sim \epsilon \frac{\hbar^2}{m} \left(\frac{d\psi}{dx} (0)\right)^2,$$ times some finite fraction.

$\langle V\rangle$, however, changes by $O(\epsilon^3)$, so its contribution is subdominant, so, then, ignorable. Since $\langle E\rangle$ is lowered by $O(\epsilon)$, your nodeful ψ cannot be the ground state. In practice, the nodeless φ may be deformed further to a smoothest form avoiding energetically expensive undulations.

• Given this result, you may now further confirm the ground state's uniqueness. For, if it were degenerate, so, e.g., the lowest energy corresponded to two independent eigenstates, ψ and ρ, then ψ(0)ρ(x)–ψ(x)ρ(0) would have the same energy, but also a node at 0, which was just excluded.
• Why do you take square of $\psi'(x)$ instead of differentiating it twice? – Ruslan Nov 4 '18 at 18:09
• @Ruslan . What do you have in mind? The Sturm-Liouville surface term at infinity is ignorable. Are you asking about the detailed Messiah proof? – Cosmas Zachos Nov 4 '18 at 18:17
• I mean your expression for average energy, $\langle E\rangle$. By definition, it's $\left\langle \psi\right|\hat H\left|\psi\right\rangle$, where $\hat H$ contains a second-derivative operator. But your expression doesn't agree with this definition, instead having a first derivative, which is then squared. – Ruslan Nov 4 '18 at 18:23
• After you integrate by parts, you chuck the surface term at infinity : It is an S-L system. Again, what do you have in mind? – Cosmas Zachos Nov 4 '18 at 18:27
• OK, I see what you are doing here. Was confused by such a large step taken without notice. – Ruslan Nov 4 '18 at 18:46