Wave function ψ — probability density - ψ - ²: Worked examples

The scenario matrix

Before working anything, let us map the whole territory. Every problem about $ψ$ and $∣ ψ ∣^{2}$ lands in one of these cells. The last column names the worked example that covers it.

#	Case class	What makes it tricky	Covered by
A	Real, everywhere-decaying $ψ$ (Gaussian)	infinite domain, Gaussian integral	Example 1
B	Real $ψ$ on a finite box	domain limits, $sin^{2}$ average	Example 2
C	Complex $ψ$ with a phase $e^{ik x}$	phase must cancel in $ψ^{*} ψ$	Example 3
D	Piecewise $ψ$ (two regions glued)	integrate region by region, continuity	Example 4
E	Degenerate input: constant $ψ$ on all of $R$	area is infinite → NOT normalizable	Example 5
F	Sub-interval probability + a symmetry shortcut	compute $P (a \leq x \leq b)$	Example 6
G	Superposition of two states	cross terms, orthogonality	Example 7
H	Limiting behaviour: what happens as a width $\to 0$ or $\to \infty$	localisation vs spreading	Example 8
I	Word problem / real detector	translate physics to an integral	Example 9
J	Exam twist: normalize then a trap about "probability at a point"	density vs probability	Example 10

We now fill every cell. Each example makes you Forecast first — guess the answer before reading the steps.

Example 1 — Cell A: real decaying Gaussian

Worked example Normalize

$ψ (x) = A e^{- x^{2} / (2 σ^{2})}$ $σ > 0$ is a fixed width. Find the positive constant $A$ .

Forecast: Does $A$ grow or shrink as the bell gets wider (bigger $σ$ )? Guess before reading.

Step 1. Write the density: since $ψ$ is real, $∣ ψ ∣^{2} = ψ^{*} ψ = ψ^{2} = A^{2} e^{- x^{2} / σ^{2}}$ . Why this step? $ψ^{*} = ψ$ for a real function, so squaring is safe here (see Complex Numbers).

Step 2. Demand total probability $1$ : $\int_{- \infty}^{\infty} A^{2} e^{- x^{2} / σ^{2}} d x = 1.$ Why this step? The particle must be somewhere — that is the normalization condition from the parent note.

Step 3. Use the Gaussian integral $\int_{- \infty}^{\infty} e^{- α x^{2}} d x = π / α$ with $α = 1/ σ^{2}$ : $\int_{- \infty}^{\infty} e^{- x^{2} / σ^{2}} d x = π σ^{2} = σ π .$ Why this tool? We need the area under a bell curve; the Gaussian integral is the one standard result that gives it in closed form — no elementary antiderivative exists, so this is the only route.

Step 4. So $A^{2} σ π = 1 \Rightarrow A = (σ π)^{- 1/2} = \frac{1}{( σ π ) ^{1/2}}$ .

Verify: As $σ$ grows, $A$ shrinks like $σ^{- 1/2}$ — a wider bell must be shorter to keep area $= 1$ . Units: $[σ] = m$ , so $[A] = m^{- 1/2}$ ✓, exactly the 1D unit of $ψ$ .

Figure — Wave function ψ — probability density - ψ - ²

Look at the figure: the blue $ψ$ and the orange $∣ ψ ∣^{2}$ . The orange curve is narrower (squaring pulls the tails down faster) and its total shaded area is exactly $1$ .

Example 2 — Cell B: real ψ on a finite box

Worked example Normalize the ground state in a box

$ψ (x) = A sin \frac{π x}{L}$ for $0 \leq x \leq L$ , and $ψ = 0$ outside. Find $A$ .

Forecast: Will $A$ depend on $L$ ? If the box doubles in width, does $A$ go up or down?

Step 1. Only the region $[0, L]$ contributes ( $ψ = 0$ elsewhere): $\int_{0}^{L} A^{2} sin^{2} \frac{π x}{L} d x = 1.$ Why this step? Outside the box $∣ ψ ∣^{2} = 0$ , so those parts add nothing — the integral collapses to $[0, L]$ .

Step 2. Use $sin^{2} θ = \frac{1}{2} (1 - cos 2 θ)$ : $\int_{0}^{L} sin^{2} \frac{π x}{L} d x = \int_{0}^{L} \frac{1}{2} d x - \frac{1}{2} \int_{0}^{L} cos \frac{2 π x}{L} d x = \frac{L}{2} - 0 = \frac{L}{2} .$ Why this tool? $sin^{2}$ has no simple antiderivative on its own; the double-angle identity turns it into a constant plus a cosine that integrates to zero over a full period. The cosine term vanishes because it completes exactly one cycle from $0$ to $L$ .

Step 3. So $A^{2} \cdot \frac{L}{2} = 1 \Rightarrow A = \frac{2}{L}$ .

Verify: Bigger box $\Rightarrow$ smaller $A$ ( $A \propto L^{- 1/2}$ ): the same total probability is spread over more room, so the peak height drops. This matches Particle in a Box. Units: $[A] = m^{- 1/2}$ ✓.

Example 3 — Cell C: complex ψ, the phase cancels

Worked example Density of a travelling, decaying wave

$ψ (x) = A e^{ik x} e^{- ∣ x ∣/ L}$ with $A$ real $> 0$ . Find $∣ ψ ∣^{2}$ , then normalize.

Forecast: Does the momentum-carrying wiggle $e^{ik x}$ show up in the probability density?

Step 1. Conjugate: $ψ^{*} = A e^{- ik x} e^{- ∣ x ∣/ L}$ . Why this step? Conjugation flips $i \to - i$ , i.e. $e^{ik x} \to e^{- ik x}$ (see Complex Numbers).

Step 2. Multiply: $∣ ψ ∣^{2} = ψ^{*} ψ = A^{2} e^{- ik x} e^{ik x} e^{- 2∣ x ∣/ L} = A^{2} e^{- 2∣ x ∣/ L} .$ Why this step? $e^{- ik x} e^{ik x} = e^{0} = 1$ : the phase factor divides out. It carried the particle's momentum, but is invisible to $∣ ψ ∣^{2}$ .

Step 3. Normalize. By the even symmetry of $e^{- 2∣ x ∣/ L}$ : $\int_{- \infty}^{\infty} A^{2} e^{- 2∣ x ∣/ L} d x = 2 A^{2} \int_{0}^{\infty} e^{- 2 x / L} d x = 2 A^{2} \cdot \frac{L}{2} = A^{2} L = 1.$ Why the factor 2? The integrand is symmetric about $x = 0$ , so the whole line equals twice the right half — a shortcut that halves the work.

Step 4. $A = \frac{1}{L}$ .

Verify: $A^{2} L = (1/ L)^{2} \cdot L = 1$ ✓. The result is independent of $k$ — two waves with different momenta but same envelope share the same $∣ ψ ∣^{2}$ , exactly the "global-phase-invisible" lesson.

Example 4 — Cell D: piecewise ψ glued at a seam

Worked example Two-region wave function

$ψ (x) = ⎩ ⎨ ⎧ A x, A (2 - x), 0, 0 \leq x \leq 1 1 \leq x \leq 2 otherwise$ (lengths in metres). Find $A$ .

Forecast: This is a triangular "tent". Its peak is at $x = 1$ . Will $A$ be bigger or smaller than $1$ ?

Step 1. Split the integral at the seam $x = 1$ : $\int_{0}^{2} ∣ ψ ∣^{2} d x = A^{2} \int_{0}^{1} x^{2} d x + A^{2} \int_{1}^{2} (2 - x)^{2} d x .$ Why this step? $ψ$ has a different formula on each side, so you must integrate each region separately, then add.

Step 2. Left piece: $\int_{0}^{1} x^{2} d x = \frac{1}{3}$ . Right piece: substitute $u = 2 - x$ , $d u = - d x$ ; as $x : 1 \to 2$ , $u : 1 \to 0$ , giving $\int_{0}^{1} u^{2} d u = \frac{1}{3}$ . Why this step? The tent is symmetric, so each half contributes the same $\frac{1}{3}$ .

Step 3. Total: $A^{2} (\frac{1}{3} + \frac{1}{3}) = \frac{2}{3} A^{2} = 1 \Rightarrow A = \frac{3}{2} = 1.5 \approx 1.2247$ .

Verify: Continuity check — at $x = 1$ the left gives $A \cdot 1 = A$ and the right gives $A (2 - 1) = A$ : they match, so $ψ$ is continuous, one of the four physical conditions. $A > 1$ as forecast (a narrow tent must be tall). Units: area was in metres, $[A] = m^{- 1/2}$ ✓.

Example 5 — Cell E: the degenerate, NON-normalizable case

Worked example A constant everywhere — the trap

Can $ψ (x) = A$ (a nonzero constant, for all $x$ ) ever be normalized?

Forecast: Say yes or no before computing.

Step 1. Try the condition: $\int_{- \infty}^{\infty} ∣ A ∣^{2} d x = ∣ A ∣^{2} \cdot (\infty) = \infty$ . Why this step? A constant density over an infinite line encloses infinite area — no finite $A$ can rescale $\infty$ down to $1$ .

Step 2. Conclusion: not square-integrable ⇒ not a valid physical wave function. The only "constant" that normalizes is $A = 0$ , which describes no particle.

Verify: This is exactly why the parent's condition " $ψ \to 0$ as $x \to \pm \infty$ " exists. A plane wave $e^{ik x}$ (constant magnitude) is the same story — a useful idealisation, but strictly needs a box or a wave packet to be normalized. Cross-check with de Broglie Wavelength: an infinitely long pure wave has perfectly known momentum but completely unknown position — see Heisenberg Uncertainty Principle.

Example 6 — Cell F: sub-interval probability with a symmetry shortcut

Worked example Probability in the left half of the box

Using the normalized ground state $ψ = 2/ L sin \frac{π x}{L}$ (from Example 2), find $P (0 \leq x \leq L /2)$ .

Forecast: The density $sin^{2}$ is symmetric about the box centre $x = L /2$ . Guess $P$ before integrating.

Step 1. Symmetry argument: $sin^{2} \frac{π x}{L}$ has a mirror symmetry about $x = L /2$ , so the area on the left half equals the area on the right half. Why this step? If two halves are equal and add to $1$ , each is exactly $\frac{1}{2}$ .

Step 2. Confirm by integration: $P = \int_{0}^{L /2} \frac{2}{L} sin^{2} \frac{π x}{L} d x = \frac{2}{L} [\frac{x}{2} - \frac{L}{4 π} sin \frac{2 π x}{L}]_{0}^{L /2} .$ Why this step? Same double-angle trick as Example 2; the bracket is the antiderivative of $sin^{2}$ .

Step 3. At $x = L /2$ : $\frac{L /2}{2} - \frac{L}{4 π} sin π = \frac{L}{4} - 0 = \frac{L}{4}$ . At $x = 0$ : $0$ . So $P = \frac{2}{L} \cdot \frac{L}{4} = \frac{1}{2}$ .

Verify: $P = \frac{1}{2}$ matches the symmetry forecast ✓. A Forecast-then-Verify win: the picture told us the answer before the algebra.

Now a non-symmetric slice, to show the shortcut is not always available:

Example 7 — Cell G: superposition and cross terms

Worked example Equal mix of two box states

Let $ϕ_{1} = 2/ L sin \frac{π x}{L}$ and $ϕ_{2} = 2/ L sin \frac{2 π x}{L}$ be two normalized box states. Consider $ψ = c (ϕ_{1} + ϕ_{2})$ . Find the constant $c$ that normalizes $ψ$ .

Forecast: Naively you might guess the norm is $1 + 1 = 2$ so $c = 1/ 2$ . Is a cross term going to spoil that?

Step 1. Expand $∣ ψ ∣^{2} = c^{2} (ϕ_{1} + ϕ_{2})^{2} = c^{2} (ϕ_{1}^{2} + 2 ϕ_{1} ϕ_{2} + ϕ_{2}^{2})$ . Why this step? The middle cross term $2 ϕ_{1} ϕ_{2}$ is where interference lives — we must check it.

Step 2. Integrate term by term: $\int ϕ_{1}^{2} = 1, \int ϕ_{2}^{2} = 1, \int ϕ_{1} ϕ_{2} d x = \frac{2}{L} \int_{0}^{L} sin \frac{π x}{L} sin \frac{2 π x}{L} d x = 0.$ Why is the cross term zero? The two box states are orthogonal — $sin \frac{π x}{L}$ and $sin \frac{2 π x}{L}$ are different standing-wave modes; their product integrates to zero over the box. This is a key fact from Particle in a Box.

Step 3. So $\int ∣ ψ ∣^{2} = c^{2} (1 + 0 + 1) = 2 c^{2} = 1 \Rightarrow c = \frac{1}{2}$ .

Verify: The naive guess was right — but only because the states are orthogonal. If they weren't, the cross term would change $c$ . Always check.

Example 8 — Cell H: limiting behaviour of a spreading packet

Worked example What happens as a Gaussian width

$\to 0$ and $\to \infty$ ? Take the normalized Gaussian density from Example 1: $∣ ψ ∣^{2} = \frac{1}{σ π} e^{- x^{2} / σ^{2}}$ . Study the two limits.

Forecast: In which limit is the particle "sharply located"? In which is it "everywhere and nowhere"?

Step 1 — peak height at $x = 0$ : $∣ ψ (0) ∣^{2} = \frac{1}{σ π}$ . Why this step? The peak measures how "concentrated" the fog is.

Step 2 — limit $σ \to 0$ : peak $\to \infty$ , width $\to 0$ , but total area stays $1$ . This is a spike (a Dirac-delta-like localisation): the particle's position becomes almost certain. Why physically? Perfect position knowledge $\Rightarrow$ (via Heisenberg Uncertainty Principle) totally unknown momentum.

Step 3 — limit $σ \to \infty$ : peak $\to 0$ , the bell flattens toward the constant of Example 5 — heading toward the non-normalizable case: the particle is smeared over all space, momentum sharply defined.

Verify: Compute area at any $σ$ : $\int \frac{1}{σ π} e^{- x^{2} / σ^{2}} d x = \frac{1}{σ π} \cdot σ π = 1$ ✓ — always exactly $1$ , no matter the width. The trade-off is height vs spread, never total probability.

Example 9 — Cell I: real-world detector word problem

Worked example A detector slit

An electron is in the box ground state $ψ = 2/ L sin \frac{π x}{L}$ with $L = 1.0 nm$ . A detector covers the central strip from $x = 0.4 nm$ to $x = 0.6 nm$ . What is the probability the electron is caught?

Forecast: The central strip is where the density peaks. Do you expect more or less than the $0.2/1.0 = 0.2$ "flat" guess?

Step 1. Translate to an integral (with $L = 1$ nm so $\frac{π x}{L} = π x$ , $x$ in nm): $P = \int_{0.4}^{0.6} 2 sin^{2} (π x) d x .$ Why this step? A physical "chance of detection over a region" is exactly $\int_{a}^{b} ∣ ψ ∣^{2} d x$ .

Step 2. Antiderivative: $2 \int sin^{2} (π x) d x = x - \frac{sin ( 2 π x )}{2 π}$ . Why this tool? Double-angle identity again — the workhorse for $sin^{2}$ .

Step 3. Evaluate: $P = [x - \frac{s i n 2 π x}{2 π}]_{0.4}^{0.6} = (0.6 - \frac{s i n 1.2 π}{2 π}) - (0.4 - \frac{s i n 0.8 π}{2 π}) .$ Numerically $sin 1.2 π \approx - 0.5878$ , $sin 0.8 π \approx 0.5878$ , so $P \approx (0.6 + 0.09355) - (0.4 - 0.09355) = 0.69355 - 0.30645 \approx 0.3871.$

Verify: $P \approx 0.39$ , which is bigger than the flat guess $0.20$ — correct, because the density peaks in the middle where the detector sits. Probability is dimensionless (the nm units in $x$ and in $ψ^{2} \sim nm^{- 1}$ cancel) ✓.

Example 10 — Cell J: the exam twist (density vs point-probability)

Worked example The classic trap

A student normalizes $ψ = 2/ L sin \frac{π x}{L}$ and is asked: "What is the probability of finding the particle exactly at $x = L /2$ ?" They answer $∣ ψ (L /2) ∣^{2} = \frac{2}{L} sin^{2} \frac{π}{2} = \frac{2}{L}$ . Mark it.

Forecast: Is $\frac{2}{L}$ a probability at all?

Step 1. Units check: $\frac{2}{L}$ has units $m^{- 1}$ . A probability is dimensionless. So $\frac{2}{L}$ cannot be a probability — it is a density. Why this step? Units are the fastest lie-detector in physics.

Step 2. The true answer: the probability at a single exact point is $P ({L /2}) = \int_{L /2}^{L /2} ∣ ψ ∣^{2} d x = 0.$ Why? An interval of zero width encloses zero area. You only get a nonzero number over a strip $[x_{0} - \frac{Δ}{2}, x_{0} + \frac{Δ}{2}]$ , where $P \approx ∣ ψ (x_{0}) ∣^{2} Δ = \frac{2}{L} Δ$ for small $Δ$ .

Verify: For the box, if $Δ = 0.01 L$ around the centre, $P \approx \frac{2}{L} \cdot 0.01 L = 0.02$ — small but nonzero, and dimensionless ✓. The student confused density ( $\frac{2}{L}$ ) with probability (needs the width factor).

Coverage check

Recall Did we fill every cell?

A Gaussian ::: Example 1 B finite box ::: Example 2 C complex phase ::: Example 3 D piecewise ::: Example 4 E degenerate/non-normalizable ::: Example 5 F sub-interval probability ::: Example 6 G superposition cross terms ::: Example 7 H limiting behaviour ::: Example 8 I word problem ::: Example 9 J exam density-vs-point trap ::: Example 10

Connections

Parent — Hinglish version
Particle in a Box — Examples 2, 6, 7, 9, 10 all live here.
Complex Numbers — conjugation and modulus in Example 3.
Heisenberg Uncertainty Principle — the width trade-off of Example 8.
de Broglie Wavelength — the plane wave behind the non-normalizable Example 5.
Schrödinger Equation — the equation these $ψ$ obey.
Expectation Values — the natural next step once $∣ ψ ∣^{2}$ is normalized.