Symmetry and parity

Chapter 3 ended with a promise: armed with the new language, one-dimensional problems can move faster. This section pays the first installment.

Start by revisiting something that appeared in chapter 2 but went unexplained at the time. Line up the eigenstates of the harmonic oscillator from lowest to highest: $\psi_0$ is a bell curve, mirror-symmetric about the origin; $\psi_1$ is negative on the left and positive on the right, changing sign under reflection; $\psi_2$ is symmetric again; $\psi_3$ antisymmetric again… Even, odd, even, odd, alternating without a single exception. The finite square well does the same — back then we even split the solutions outright into an “even class” and an “odd class” to solve them.

The pattern is too tidy to be a coincidence. But in the language of chapter 2 it was merely “that’s how the solutions happen to come out”: change the potential and you must solve again, observe again. What we were missing is an argument that delivers the conclusion in advance.

The commutators of chapter 3 exist for precisely this kind of job.

The mirror operator

Define the parity operator $\hat P$: its action is simply to hold the wavefunction up to a mirror,

$$\hat P\,\psi(x)=\psi(-x)\tag{4.1.1}$$

“Parity” sounds arcane, but it just means “does it change when left and right are swapped”. Two properties can be read off immediately:

First, reflecting twice is the same as not reflecting at all: $\hat P^2=\hat 1$. So if $\psi$ is an eigenstate of $\hat P$ with eigenvalue $\lambda$, then $\lambda^2=1$, leaving only

$$\lambda=+1\quad\text{or}\quad\lambda=-1\tag{4.1.2}$$

States with eigenvalue $+1$ satisfy $\psi(-x)=\psi(x)$: they are even functions (even-parity states). States with eigenvalue $-1$ satisfy $\psi(-x)=-\psi(x)$: odd functions (odd-parity states). Beyond $\pm1$ there is no third option.

Second, $\hat P$ is Hermitian (a change of variables $x\to-x$ verifies $\braket{\phi}{\hat P\psi}=\braket{\hat P\phi}{\psi}$), so by the standard of section 3.5 it is a legitimate observable — measure it, and the only possible answers are $+1$ and $-1$.

Any function splits into an even part plus an odd part: $\psi=\frac12(\psi(x)+\psi(-x))+\frac12(\psi(x)-\psi(-x))$. In chapter 3 language: the even and odd functions together form a complete basis, and the spectral decomposition of $\hat P$ is clean.

The key step: commuting with H

The central result of section 3.6 was: two operators that commute share a common eigenbasis; a commuting observable is conserved. So now we compute $[\hat H,\hat P]$.

Derivation: [H, P] = 0 when V(−x) = V(x)basic~6 minexpand

The Hamiltonian is $\hat H=-\frac{\hbar^2}{2m}\frac{\dd^2}{\dd x^2}+V(x)$. To compute $[\hat H,\hat P]$, apply $\hat H\hat P$ and $\hat P\hat H$ to an arbitrary $\psi(x)$ and check whether the results agree.

The kinetic term first. One order of operations is mirror first, then second derivative:

$$\frac{\dd^2}{\dd x^2}\big[\psi(-x)\big]\tag{4.1.3}$$

By the chain rule the first derivative gives $-\psi'(-x)$ (the inner function $-x$ contributes a minus sign), and differentiating again contributes another one: $(-1)^2\psi''(-x)=\psi''(-x)$.

The other order is second derivative first, then mirror: $\hat P\big[\psi''(x)\big]=\psi''(-x)$.

They agree. The minus sign appeared twice and cancelled — a second derivative is inherently blind to left versus right. So the kinetic term always commutes with $\hat P$, no conditions attached.

Now the potential term. One side is

$$V(x)\,\hat P\psi=V(x)\,\psi(-x)\tag{4.1.4}$$

The other is

$$\hat P\big[V(x)\psi(x)\big]=V(-x)\,\psi(-x)\tag{4.1.5}$$

Subtracting:

$$[V,\hat P]\,\psi=\big[V(x)-V(-x)\big]\,\psi(-x)\tag{4.1.6}$$

For this to vanish for every $\psi$, it is necessary and sufficient that

$$V(-x)=V(x)\tag{4.1.7}$$

Conclusion: the potential is left-right symmetric $\Longleftrightarrow$ $[\hat H,\hat P]=0$. This is the precise meaning of the word “symmetry” in operator language: the mirror transformation leaves the Hamiltonian unchanged.

What commuting buys you

Why even and odd alternate exactly

“Even or odd, nothing else” is settled, but we are still half a step from the opening observation: why, ordered by energy, does the sequence run even, odd, even, odd?

The answer hides in node counting. Chapter 2 showed the pattern repeatedly: ordering one-dimensional bound states by energy, the $n$-th state (counting from 0) has exactly $n$ nodes — the higher the energy, the more sharply the wavefunction bends and the more times it crosses zero. Now splice that together with parity:

• An odd function must pass through the origin: $\psi(-0)=-\psi(0)$ forces $\psi(0)=0$, so the origin is always a node. Whereas if an even function vanished at the origin, its derivative (the derivative of an even function is odd) would vanish there too, and by uniqueness of solutions the entire curve would be identically zero — so an even state can never vanish at the origin.
• In a symmetric potential nodes also come in mirror pairs: if $x_0$ is a node, so is $-x_0$. Hence an even state has an even number of nodes, an odd state an odd number.

The node counts $0,1,2,3,\dots$ climb one at a time; even counts belong to even states, odd counts to odd states — the alternation is locked in. The ground state has 0 nodes and is always even; the first excited state has 1 node and is always odd. The harmonic oscillator, the finite well, and every symmetric one-dimensional potential you will ever meet all obey this, with no further checking required.

Step through $n$ in the simulation below and count the nodes, watch the parity, with your own eyes:

◈The one-dimensional harmonic oscillator

Equally spaced levels, zero-point energy, tails reaching into the classically forbidden region, and the one quantum state that really does behave like a classical particle. Units: ħ = m = 1.

Eₙ = (n+½)ħω

0.500

Level spacing ħω

1.00

Classical turning point ±A

±1.00

Zero-point energy E₀

0.500

Angular frequency ω1.00

A steeper well → larger spacing ħω and a narrower wavefunction

Quantum number n0

nodes = n = 0

View

Level ladderSingle eigenstateCoherent state

Show ψ

Three benefits, free of charge

The second benefit is half the work. Since the eigenstates of a symmetric potential are either even or odd, it suffices to solve on the half-axis $x\ge0$:

• For even states: impose $\psi'(0)=0$ at the origin (an even function is flat at its centre of symmetry);
• For odd states: impose $\psi(0)=0$.

Section 2.8 split the finite well into a $\tan z$ family and a $-\cot z$ family, which looked like an algebraic trick at the time — now it is transparent: that was parity doing the sorting. The $\tan$ family are the even states, the $\cot$ family the odd ones, and the two families take the stage alternately along the energy axis. Each time the dimensionless parameter $z_0=a\sqrt{2mV_0}/\hbar$ crosses another $\pi/2$, one more state appears, even and odd taking turns.

The third benefit is cashed in during the numerical work of section 4.4: solving on the half-axis halves the matrix, and blocking by parity also keeps nearly degenerate even and odd states from getting numerically mixed.

∎Key formulas

• Parity operator

  $$\hat P\,\psi(x)=\psi(-x),\qquad \hat P^2=\hat 1\ \Rightarrow\ \lambda=\pm1$$

  Hermitian and unitary; its eigenstates are the even and odd functions

• Symmetry condition

  $$V(-x)=V(x)\ \Longleftrightarrow\ [\hat H,\hat P]=0$$

  The kinetic term always commutes; the condition falls entirely on the potential

• Bound-state parity

  $$\hat H\psi_n=E_n\psi_n\ \Rightarrow\ \psi_n(-x)=\pm\psi_n(x)$$

  1D bound states are non-degenerate, so parity is forced, not chosen; even and odd alternate with n

• Parity selection rule

  $$\int\psi_m\,x\,\psi_n\,\dd x\ne0\ \text{requires}\ \psi_m,\psi_n\ \text{of opposite parity}$$

  An odd integrand integrates to zero; the embryo of spectroscopic selection rules

?Self-check4 questions

1. 1.

   Why can the eigenvalues of the parity operator only be +1 or −1?

   • Because P² = 1, so eigenvalues must satisfy λ² = 1
   • Because the wavefunction must be normalised
   • Because energy eigenvalues are real
   • Because one-dimensional bound states are non-degenerate

   Submit
2. 2.

   For one-dimensional bound states in a symmetric potential V(−x) = V(x), which statements are correct? (Select all that apply.)

   Select all that apply

   • The ground state is always even parity
   • ⟨x⟩ = 0 in every eigenstate
   • Ordered by energy, the eigenstates alternate even and odd
   • Any superposition of two eigenstates also has definite parity

   Submit
3. 3.

   When solving for the even-parity bound states of a symmetric potential on the half-axis x ≥ 0, the boundary condition to impose at the origin is:

   • ψ'(0) = 0
   • ψ(0) = 0
   • ψ(0) = ψ'(0) = 0
   • No condition is needed

   Submit
4. 4.

   A one-dimensional finite symmetric square well has dimensionless depth parameter z₀ = a√(2mV₀)/ħ = 4.0. How many bound states does it hold in total? (Hint: each time z₀ crosses another π/2 ≈ 1.571, the even and odd families take turns gaining one state.)

   40% relative tolerance

   Submit

What comes next

Parity has put the bound states in perfect order: even or odd, alternating level by level, with a pile of integrals now self-evidently zero.

But the other half of chapter 2’s world — the continuum states of tunneling and scattering — is still lying about loose. There we had only two numbers, the transmission $T$ and reflection $R$; they said “how many particles got through” but not what the potential did to the wave. The next section introduces a sharper quantity — the phase shift: everything a potential does to an incident wave ends up condensed into a single angle, “how far did it pull the waveform in, or push it out”. And you will see that parity has a part in the continuum too — it splits scattering into two channels that never interfere with each other’s business.