Parity and Nodes

Parity and nodes are two of the most powerful organizing tools for one-dimensional bound states. Parity uses symmetry to split an even potential into even and odd sectors. Node counting orders the bound states by the number of times their wavefunctions cross zero.

This page is the focused one-dimensional solving guide. The broader qualitative overview is Qualitative Features of One-Dimensional Bound States, while the abstract symmetry principle is Parity.

Setting

Consider a real one-dimensional Hamiltonian

$$\hat H = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} +V(x),$$

with boundary conditions that define a self-adjoint bound-state problem. A bound state satisfies

$$\hat H\psi_n=E_n\psi_n, \qquad \int \lvert\psi_n(x)\rvert^2\,dx\lt\infty.$$

The statements below apply most directly to ordinary regular one-dimensional wells. Singular potentials, half-line problems, and radial reductions may require adjusted boundary conditions, but the same ideas remain useful diagnostics.

Parity Symmetry

Parity acts as

$$(\Pi\psi)(x)=\psi(-x).$$

If the potential is even,

$$V(-x)=V(x),$$

then

$$[\hat H,\Pi]=0.$$

Energy eigenstates can then be chosen as parity eigenstates:

$$\Pi\psi=\eta\psi, \qquad \eta=\pm1.$$

In position space,

$$\psi(-x)=\psi(x) \quad \text{for even states},$$

and

$$\psi(-x)=-\psi(x) \quad \text{for odd states}.$$

Parity is not available unless the Hamiltonian and its domain are symmetric. An asymmetric square well can still have ordered nodes, but its eigenstates are not generally even or odd.

Half-Line Boundary Conditions

For a smooth even potential, parity converts the full-line problem into two half-line problems on $x\ge0$.

An even state obeys

$$\psi'(0)=0.$$

An odd state obeys

$$\psi(0)=0.$$

Then one solves on $x\ge0$ and reflects the solution:

$$\psi_{\mathrm e}(-x)=\psi_{\mathrm e}(x), \qquad \psi_{\mathrm o}(-x)=-\psi_{\mathrm o}(x).$$

This is why symmetric wells often have separate even and odd transcendental equations. In the Finite Square Well, for example, even states use $q\tan(qa)=\kappa$, while odd states use $-q\cot(qa)=\kappa$.

If the origin contains a singular interaction, such as a delta potential, the origin condition must be derived from the singular matching rule rather than copied from the smooth case.

Nondegeneracy In One Dimension

Ordinary one-dimensional bound states are nondegenerate. A short Wronskian argument explains why.

Suppose $\psi$ and $\phi$ solve the same real Schrödinger equation with the same energy:

$$-\frac{\hbar^2}{2m}\psi''+V\psi=E\psi, \qquad -\frac{\hbar^2}{2m}\phi''+V\phi=E\phi.$$

Define the Wronskian

$$W(x)=\psi(x)\phi'(x)-\psi'(x)\phi(x).$$

Subtracting the two equations gives

$$W'(x)=0,$$

so $W$ is constant. For bound states on the line, both wavefunctions and their derivatives vanish at infinity fast enough that

$$W(\infty)=0.$$

Therefore $W(x)=0$ everywhere. The two solutions are linearly dependent, so they do not represent two independent bound states with the same energy.

This is why, in an even one-dimensional well, each bound-state energy has a definite parity. Degenerate subspaces do not appear for ordinary one-dimensional bound states, so there is no freedom to mix even and odd states at the same energy.

Node Theorem

For regular one-dimensional bound-state problems, the energy eigenstates can be ordered as

$$E_0\lt E_1\lt E_2\lt\cdots$$

and the corresponding eigenfunction $\psi_j$ has exactly $j$ interior nodes:

$$\psi_0:\;0\ \text{nodes}, \qquad \psi_1:\;1\ \text{node}, \qquad \psi_2:\;2\ \text{nodes}, \quad \ldots$$

The physical intuition is kinetic energy. Additional nodes require additional curvature, and curvature contributes to

$$\langle T\rangle = \frac{\hbar^2}{2m} \int \lvert\psi'(x)\rvert^2\,dx$$

when boundary terms vanish.

The mathematical statement is a Sturm–Liouville result: zeros of successive eigenfunctions interlace, and eigenvalues are ordered by oscillation count. The rigorous framework is Sturm–Liouville Theory.

Parity Alternation

In an even one-dimensional well, parity and node counting combine. The ground state is nodeless and can be chosen positive, so it is even. The first excited state has one node; by symmetry that node lies at the origin, so the state is odd. The next state has two nodes and is even, and so on:

$$\text{even ground state}, \quad \text{odd first excited state}, \quad \text{even second excited state}, \quad \text{odd third excited state}, \ldots$$

This alternation is a guide, not a replacement for boundary conditions. It assumes an ordinary symmetric one-dimensional bound problem. If the domain itself imposes a node or if the potential is singular at the symmetry point, check the domain before applying the smooth-well pattern.

Example: Centered Infinite Well

For an infinite square well centered at the origin, $-a\lt x\lt a$, the potential is even. The eigenfunctions can be chosen as alternating cosine and sine functions:

$$\psi_{\mathrm e}(x)\propto \cos(kx), \qquad \psi_{\mathrm o}(x)\propto \sin(kx),$$

with hard-wall conditions at $x=\pm a$.

The ground state is even and has no interior node. The first excited state is odd and has one node at $x=0$. The centered representation makes parity visible, while the interval $0\lt x\lt L$ representation used in Infinite Square Well makes boundary conditions at the walls direct.

Example: Finite Square Well

The finite square well is the first place parity saves serious algebra. The full-line matching problem separates into two equations:

$$q\tan(qa)=\kappa \quad \text{even},$$

and

$$-q\cot(qa)=\kappa \quad \text{odd}.$$

The roots alternate by parity as the energy increases. Newly appearing bound states enter from threshold with the next node count, as explained in Bound-State Counting.

Example: Double Well

In a symmetric double well, the exact eigenstates have definite parity. The lowest pair is approximately

$$\psi_+ \approx \frac{1}{\sqrt2} \left( \phi_L+\phi_R \right),$$

and

$$\psi_- \approx \frac{1}{\sqrt2} \left( \phi_L-\phi_R \right),$$

where $\phi_L$ and $\phi_R$ are approximate localized states in the left and right wells.

The even state is lower because it has no node between the wells. The odd state has a central node and lies higher. The localized states $\phi_L$ and $\phi_R$ are useful approximations, but they are not exact stationary states of a symmetric finite double well.

The exactly solvable Double Delta Potential gives explicit even and odd equations; the smoother Double-Well Potential gives the qualitative tunneling picture.

When Parity Is Broken

If $V(x)\ne V(-x)$, parity no longer commutes with the Hamiltonian. The eigenstates generally have no definite parity, and the half-line reduction is unavailable.

The node theorem still orders ordinary one-dimensional bound states. The ground state is still nodeless, the first excited state still has one node, and so on. The nodes need not sit at the origin, and the wavefunction need not have any simple left-right symmetry.

This distinction is important in asymmetric wells: losing parity does not destroy one-dimensional ordering, but it does remove the even-odd sector decomposition.

Practical Checklist

When solving a one-dimensional bound problem, ask:

• Is the potential and domain invariant under $x\mapsto -x$?
• If yes, should the state be even or odd?
• For a smooth origin, should the half-line condition be $\psi'(0)=0$ or $\psi(0)=0$?
• How many nodes should this energy level have?
• Does a claimed degeneracy contradict ordinary one-dimensional nondegeneracy?
• Are localized left-right states being mistaken for exact parity eigenstates?

These checks often catch errors before numerical values are computed.

Common Mistakes

• Assuming every one-dimensional potential has parity symmetry.
• Using even-state boundary conditions for an odd state, or the reverse.
• Forgetting that ordinary one-dimensional bound states are nondegenerate.
• Drawing a ground-state wavefunction with an interior node.
• Treating localized double-well states as exact energy eigenstates in a symmetric well.
• Applying smooth-origin conditions at a singular point without deriving the matching rule.
• Confusing the number of nodes with the principal quantum-number convention used by a particular page.

Where This Is Used

• Parity gives the general symmetry operator and selection-rule language.
• Qualitative Features of One-Dimensional Bound States gives the broader intuition for tails, localization, and deformations.
• Finite Square Well uses separate even and odd matching equations.
• Bound-State Counting uses nodes and thresholds to count states.
• Double-Well Potential uses parity eigenstates to explain tunneling splitting.
• Sturm–Liouville Theory gives the mathematical basis for node ordering.

References

• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• L. D. Landau and E. M. Lifshitz, Quantum Mechanics: Non-Relativistic Theory, 3rd ed., Pergamon, 1977.
• G. Teschl, Mathematical Methods in Quantum Mechanics, 2nd ed., American Mathematical Society, 2014.

Exercises

1. Show that a smooth even wavefunction obeys $\psi'(0)=0$ and a smooth odd wavefunction obeys $\psi(0)=0$.

Solution

If $\psi$ is even, then

$$\psi(-x)=\psi(x).$$

Differentiate with respect to $x$:

$$-\psi'(-x)=\psi'(x).$$

Set $x=0$ to get

$$-\psi'(0)=\psi'(0),$$

so $\psi'(0)=0$.

If $\psi$ is odd, then

$$\psi(-x)=-\psi(x).$$

Setting $x=0$ gives

$$\psi(0)=-\psi(0),$$

so $\psi(0)=0$.

2. Use the Wronskian argument to explain why two independent one-dimensional bound states cannot have the same energy.

Solution

Let $\psi$ and $\phi$ solve the same real Schrödinger equation at energy $E$. Their Wronskian

$$W=\psi\phi'-\psi'\phi$$

has derivative $W'=0$, so $W$ is constant. For bound states on the line, the wavefunctions and derivatives vanish at infinity, so $W=0$ there. Hence $W=0$ everywhere. A zero Wronskian means the two solutions are linearly dependent, not independent states.

3. What parities should the first four bound states of a smooth even one-dimensional well have?

Solution

The ground state is nodeless and even. The first excited state has one node and is odd. The second excited state has two nodes and is even. The third excited state has three nodes and is odd. The sequence is therefore

$$\text{even},\quad \text{odd},\quad \text{even},\quad \text{odd}.$$

4. In a symmetric double well, why are $\phi_L$ and $\phi_R$ not exact energy eigenstates?

Solution

The Hamiltonian has parity symmetry, and ordinary one-dimensional bound states are nondegenerate. Therefore exact energy eigenstates can be chosen with definite parity. A state localized only in the left well is not even or odd; parity maps it to a right-localized state. The exact low-energy states are approximately the even and odd combinations

$$\frac{1}{\sqrt2} \left( \phi_L+\phi_R \right), \qquad \frac{1}{\sqrt2} \left( \phi_L-\phi_R \right).$$

5. An asymmetric well has three bound states. What can you say about their node counts?

Solution

Parity is unavailable because the potential is asymmetric, but ordinary one-dimensional node ordering still applies. Ordered by increasing energy, the three states have

$$0,\quad 1,\quad 2$$

interior nodes respectively. The nodes need not be placed symmetrically.