Schrödinger equation — time-dependent, time-independent: Worked examples

This page is a workout. The parent note built the two equations; here we throw the common exam situations at them so few cases surprise you.

Before we solve anything, three pieces of vocabulary that the whole page leans on. I define them now so no symbol is ever used before it is earned.

Now, what counts as a "case"?

The scenario matrix

Every problem below is one of these cells. This matrix is not exhaustive — three big topics (finite well & tunnelling, barrier scattering, the harmonic oscillator) have their own pages and are only signposted here — but it covers the cases most exams pack into a single Schrödinger question.

Cell	What makes it different	Example
A. Verify a given $Ψ$	plug into TDSE, check both sides match	Ex 1
B. $V = 0$ free particle	plane wave, continuous energy	Ex 2
C. Bound state, hard walls	quantized $E_{n}$ from $ψ = 0$ at edges	Ex 3
D. Normalization	fix the constant $A$ so total prob $= 1$	Ex 4
E. Excited / higher $n$	more wiggles, energy scales as $n^{2}$	Ex 5
F. Stationary vs moving fog	why $	\Psi
G. Degenerate / forbidden input	$n = 0$ , $E < V_{m i n}$ — what breaks	Ex 7
H. Real-world word problem	electron in a nanowire, numbers in eV	Ex 8
I. Exam twist	shift the box, or ask a photon energy	Ex 9

We reuse three constants throughout. Let me define them once, in plain words.

Cell A — Verify a given wavefunction

Before Ex 1 we must earn three symbols it uses.

Worked example Ex 1 — Does

$Ψ = A e^{i (k x - ω t)}$ satisfy the free TDSE? (Cell A, B) Forecast: guess before reading — will the two sides give the same number times $Ψ$ , or will an extra factor appear? Write down your bet.

Step 1. Compute the left side $i ℏ \partial_{t} Ψ$ . Why this step? The TDSE is left $=$ right; we must build each side separately. $\partial_{t} Ψ = - iω Ψ \Rightarrow i ℏ \partial_{t} Ψ = i ℏ (- iω) Ψ = ℏ ω Ψ.$ The $i \times (- i) = + 1$ , so the $i$ 's cancel and we get a real multiple $ℏ ω$ . Notice $A$ rides along untouched — that is why its value is irrelevant here.

Step 2. Compute the right side $- \frac{ℏ ^{2}}{2 m} \partial_{x}^{2} Ψ$ (with $V = 0$ , a free particle). Why this step? Each space derivative pulls down a factor $ik$ ; we need two of them for kinetic energy. $\partial_{x}^{2} Ψ = (ik)^{2} Ψ = - k^{2} Ψ \Rightarrow - \frac{ℏ ^{2}}{2 m} (- k^{2}) Ψ = \frac{ℏ ^{2} k ^{2}}{2 m} Ψ.$

Step 3. Set them equal. Why this step? The plane wave is a solution only if the two sides agree for all $x, t$ . $ℏ ω = \frac{ℏ ^{2} k ^{2}}{2 m} \Rightarrow E = \frac{p ^{2}}{2 m} (using E = ℏ ω, p = ℏ k) .$

Verify: This is exactly the free-particle energy–momentum relation. Units: $ℏ ω$ is (J·s)(1/s) = J; $\frac{ℏ ^{2} k ^{2}}{2 m}$ is $\frac{( J\cdots ) ^{2} ( 1/ m ) ^{2}}{kg} = J$ . ✓ Both sides are energies. The plane wave works, and it forces the correct energy.

Cell B — Free particle, real numbers

Worked example Ex 2 — Momentum and energy of a de Broglie electron

(Cell B) An electron has wavelength $λ = 1.0 \times 1 0^{- 10} m$ (1 Ångström, atomic size). Find $k$ , $p$ , and $E$ . Forecast: guess — will its energy be a few eV, or thousands? (Atomic-scale wavelengths tend to give tens of eV.)

Step 1. Get the wavenumber $k = 2 π / λ$ . Why this step? $k$ (defined above) counts radians of wave per metre; it is the bridge from wavelength to momentum via $p = ℏ k$ . See de Broglie Hypothesis. $k = \frac{2 π}{1.0 \times 1 0 ^{- 10}} = 6.283 \times 1 0^{10} m^{- 1} .$

Step 2. Momentum $p = ℏ k$ . $p = (1.055 \times 1 0^{- 34}) (6.283 \times 1 0^{10}) = 6.63 \times 1 0^{- 24} kg\cdotm/s .$

Step 3. Energy $E = \frac{p ^{2}}{2 m}$ . Why this step? Free particle has only kinetic energy; TDSE Ex 1 proved this is the right formula. $E = \frac{( 6.63 \times 1 0 ^{- 24} ) ^{2}}{2 ( 9.11 \times 1 0 ^{- 31} )} = 2.41 \times 1 0^{- 17} J = 150 eV .$

Verify: Convert back: $2.41 \times 1 0^{- 17} /1.602 \times 1 0^{- 19} = 150.6 eV$ . This is the classic "electron microscope" energy — realistic. ✓

Cell C — Bound state, hard walls (quantization is born)

Worked example Ex 3 — Ground-state energy of an electron in a 1 nm box

(Cell C) Box width $L = 1.0 \times 1 0^{- 9} m$ , infinite walls (so $V = 0$ inside and $V = \infty$ outside — a valley with vertical cliffs). Find $E_{1}$ . Forecast: free electron at 1 Å was 150 eV. Trapped in 1 nm (ten times wider), guess whether $E_{1}$ is above or below 1 eV.

Step 1. Recall the trapped energies (parent note derived them): $E_{n} = \frac{n ^{2} π ^{2} ℏ ^{2}}{2 m L ^{2}} .$ Why this step? The walls force $ψ (0) = ψ (L) = 0$ , which only allows $k L = nπ$ — that is where $n^{2}$ comes from. See Energy Quantization and Particle in a Box.

Step 2. Set $n = 1$ and substitute. $E_{1} = \frac{π ^{2} ( 1.055 \times 1 0 ^{- 34} ) ^{2}}{2 ( 9.11 \times 1 0 ^{- 31} ) ( 1.0 \times 1 0 ^{- 9} ) ^{2}} .$

Step 3. Crunch it. $E_{1} = 6.03 \times 1 0^{- 20} J = 0.376 eV .$

Verify: Below 1 eV, as forecast — a wider box means longer allowed wavelength, smaller momentum, smaller energy. Units: $\frac{( J\cdots ) ^{2}}{kg\cdotm ^{2}} = J$ . ✓

The figure below draws this ground state. Look at the cyan curve: it is a single hump — exactly one half of a sine wave — pinned to zero at both amber walls. That "one half-wave fits the box" picture is the boundary condition $k L = π$ made visible, and it is the geometric reason $n = 1$ is the lowest allowed state.

Cell D — Normalization

Worked example Ex 4 — Normalize the ground state

$ψ_{1} = A sin (π x / L)$ (Cell D) Now we finally fix the amplitude $A$ that we left free back in Ex 1. Forecast: The particle is somewhere in the box, so $\int_{0}^{L} ∣ ψ_{1} ∣^{2} d x = 1$ . Guess whether $A$ grows or shrinks as the box $L$ widens.

Step 1. Write the normalization demand. Why this step? $∣ ψ ∣^{2}$ is a probability density (Wavefunction and Born Interpretation); total probability must equal 1 — that is what pins down $A$ . $\int_{0}^{L} A^{2} sin^{2} (\frac{π x}{L}) d x = 1.$

Step 2. Use $\int_{0}^{L} sin^{2} (π x / L) d x = L /2$ . Why this step? $sin^{2}$ averages to $\frac{1}{2}$ over a half-period, and the box holds exactly a half-period, so the integral is (average) $\times$ (width) $= \frac{1}{2} \cdot L$ . $A^{2} \cdot \frac{L}{2} = 1.$

Step 3. Solve for $A$ . $A = \frac{2}{L} .$

Verify: As $L$ grows, $A$ shrinks — a wider box spreads the same total probability thinner, so the peak height drops. Units: $1/ m$ , correct for a 1-D wavefunction so that $∣ ψ ∣^{2} d x$ is dimensionless. ✓ For $L = 1 nm$ : $A = 2/1 0^{- 9} = 4.47 \times 1 0^{4} m^{- 1/2}$ .

Cell E — Excited state (higher $n$ )

Worked example Ex 5 — Energy jump from

$n = 1$ to $n = 2$ in the 1 nm box (Cell E) Forecast: since $E_{n} \propto n^{2}$ , going $1 \to 2$ quadruples $E$ . Guess the gap $Δ E = E_{2} - E_{1}$ .

Step 1. Use the ratio $E_{n} = n^{2} E_{1}$ . Why this step? Everything except $n^{2}$ is identical between levels, so we avoid recomputing constants. $E_{2} = 4 E_{1} = 4 (0.376) = 1.505 eV .$

Step 2. Subtract. $Δ E = E_{2} - E_{1} = (4 - 1) E_{1} = 3 E_{1} = 1.129 eV .$

Verify: $Δ E$ is three times the ground energy, not double — because energy grows as $n^{2}$ , not $n$ . ✓

The figure overlays $n = 1$ (cyan) and $n = 2$ (amber). Look at the white dot at the centre: the $n = 2$ curve passes through zero there — it has gained one extra node compared with the single-hump ground state. More nodes = tighter wiggles = larger $k$ = higher energy; the picture is the physical reason the level spacing grows.

Cell F — Stationary vs moving fog

Worked example Ex 6 — Show

$∣Ψ ∣^{2}$ is time-independent for a stationary state (Cell F) Take $Ψ (x, t) = ψ (x) e^{- i E t /ℏ}$ . Forecast: "stationary" — do you expect the probability cloud to drift, or sit still while $Ψ$ spins in phase?

Step 1. Write $∣Ψ ∣^{2} = Ψ^{*} Ψ$ . Why this step? Probability density is the modulus squared, and modulus squared always kills complex phases. $Ψ^{*} = ψ^{*} (x) e^{+ i E t /ℏ} .$

Step 2. Multiply. $∣Ψ ∣^{2} = ψ^{*} ψ e^{+ i E t /ℏ} e^{- i E t /ℏ} = ∣ ψ (x) ∣^{2} \cdot e^{0} = ∣ ψ (x) ∣^{2} .$ Why this step? The two opposite phases cancel: $e^{+ i θ} e^{- i θ} = 1$ . The $t$ vanishes.

Verify: No $t$ remains — the fog is frozen even though $Ψ$ rotates. This is exactly the meaning of "stationary state." At $E = 1.129 eV$ the phase spins at $ω = E /ℏ = 1.71 \times 1 0^{15} rad/s$ , yet $∣Ψ ∣^{2}$ never moves. ✓

The figure shows this directly: the amber curve $∣Ψ ∣^{2}$ sits perfectly still, while the faded cyan curves are $Re (Ψ)$ at three instants — clearly spinning up and down. The observable (amber) is frozen; the phase (cyan) is not.

Cell G — Degenerate / forbidden inputs

Worked example Ex 7 — What goes wrong at

$n = 0$ and at $E < V_{m i n}$ ? (Cell G — the case people skip) Here $V_{m i n} = 0$ (the flat floor of the box), so " $E < V_{m i n}$ " means " $E < 0$ ". Forecast: guess which of these is "allowed but boring," and which is "physically impossible."

Case $n = 0$ . Why check this? The quantization $k L = nπ$ seems to allow $n = 0$ . Then $k = 0$ , and $ψ = A sin (0 \cdot x) = 0$ everywhere. Total probability $\int ∣ ψ ∣^{2} = 0 \neq = 1$ . A particle that is nowhere is no particle. So $n = 0$ is rejected — the lowest real state is $n = 1$ , and it already has nonzero energy $E_{1} > 0$ .

Case negative $n$ . $sin (- θ) = - sin θ$ , so $ψ_{- n} = - ψ_{n}$ . Since an overall $- 1$ doesn't change $∣ ψ ∣^{2}$ , it is the same physical state. So $n = - 2$ gives nothing new; only $n = 1, 2, 3 \dots$ count.

Case $E < V_{m i n}$ (here $E < 0$ ). Why check this? Could a trapped particle have energy below the floor of its potential landscape? Then $k = 2 m E /ℏ$ is imaginary, so $ψ$ becomes a growing/decaying exponential, not a sine. It cannot be zero at both walls and normalizable at once — no valid state. So bound energies are always $E > V_{m i n}$ .

Verify: The forbidden inputs ( $n = 0$ , $E < V_{m i n}$ ) yield either the zero function or a non-normalizable one — mathematics itself refuses them. The ground energy $E_{1} > 0$ is the zero-point energy, forced by Heisenberg Uncertainty Principle: a confined particle can never be perfectly at rest. ✓

Cell H — Real-world word problem

Worked example Ex 8 — Photon emitted when the nanowire electron drops

$n = 2 \to n = 1$ (Cell H) Same 1 nm box. The electron falls from the excited state to the ground state, emitting one photon. Find the photon wavelength. Forecast: visible light is ~400–700 nm. Guess whether this photon is visible, infrared, or ultraviolet.

Step 1. Photon energy $=$ the level gap. Why this step? Energy conservation: the electron loses $Δ E$ , the photon carries it away. $E_{γ} = Δ E = 1.129 eV = 1.808 \times 1 0^{- 19} J .$

Step 2. Photon wavelength from $E_{γ} = h c / λ$ . Why this formula? A photon's energy is set by its frequency $E = h f = h c / λ$ ; we invert to get $λ$ . $λ = \frac{h c}{E _{γ}} = \frac{( 6.626 \times 1 0 ^{- 34} ) ( 3.0 \times 1 0 ^{8} )}{1.808 \times 1 0 ^{- 19}} .$

Step 3. Compute. $λ = 1.099 \times 1 0^{- 6} m \approx 1100 nm .$

Verify: ~1100 nm is near-infrared — invisible to the eye, as forecast. Quick sanity shortcut: $λ (nm) \approx 1240/ E (eV) = 1240/1.129 = 1098 nm$ . ✓ Matches.

Cell I — Exam twist

Worked example Ex 9 — Shift the box: does

$- \frac{L}{2} < x < \frac{L}{2}$ change the energies? (Cell I) The same infinite well, but centred on the origin. Forecast: guess yes or no before solving — do the energies depend on where you place the box?

Step 1. Same TISE inside, new boundaries $ψ (\pm L /2) = 0$ . Why this step? Physics is unchanged; only the coordinate labels moved. But the form of $ψ$ may look different.

Step 2. Now both $sin$ and $cos$ survive. Even states $ψ = B cos (k x)$ need $cos (k L /2) = 0 \Rightarrow k L /2 = \frac{π}{2}, \frac{3 π}{2}, \dots$ ; odd states $ψ = A sin (k x)$ need $sin (k L /2) = 0 \Rightarrow k L /2 = π, 2 π, \dots$ . Why this step? A symmetric box admits symmetric (cos) and antisymmetric (sin) solutions separately.

Step 3. Combine both families: $k L /2 = nπ /2$ , i.e. $k L = nπ$ , $n = 1, 2, 3 \dots$ — identical to Ex 3. $E_{n} = \frac{n ^{2} π ^{2} ℏ ^{2}}{2 m L ^{2}} .$

Verify: The energies are exactly the same. Energy depends only on the box width $L$ , never on its position — shifting the origin is just relabelling coordinates, which cannot change physics. The wavefunctions merely change from all-sines to alternating cos/sin. ✓

Active Recall

Recall Which cell breaks, and why?

Why is $n = 0$ forbidden? ::: it gives $ψ \equiv 0$ , a particle that exists nowhere — not normalizable.
Why can't a bound particle have $E < V_{m i n}$ ? ::: $k$ becomes imaginary, $ψ$ blows up, cannot satisfy both walls and normalization.
Does shifting the box change $E_{n}$ ? ::: no — energy depends only on width $L$ .
Why does $∣Ψ ∣^{2}$ not move for a stationary state? ::: the $e^{\pm i E t /ℏ}$ phases cancel in $Ψ^{*} Ψ$ .
What does TISE need that TDSE does not? ::: a time-independent potential $V (x)$ .

Connections

Parent (Hinglish)
Particle in a Box — Cells C, E, G, I; also finite wells & tunnelling.
Quantum Harmonic Oscillator — the valley landscape $V = \frac{1}{2} m ω^{2} x^{2}$ .
Energy Quantization — why the $n^{2}$ ladder appears.
Wavefunction and Born Interpretation — Cells D, F.
de Broglie Hypothesis — Cell B, the $p = ℏ k$ bridge.
Heisenberg Uncertainty Principle — why $E_{1} > 0$ (zero-point energy).
Hamiltonian Operator — the $\hat{H}$ every case solves.

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