Particle in a box — solving TISE, energy levels, wavefunctions: Question bank

Every reveal below is Statement ::: reasoning. Read the statement, decide, then uncover.

True or false — justify

The wavefunction

ψ_{n} (x) = 2/ L sin (nπ x / L)

is a probability.

False —

ψ

is an amplitude; the probability density is

∣ ψ_{n} ∣^{2}

and you must integrate it over a region to get an actual probability.

Doubling the box length

L

doubles every energy level.

False — since

E_{n} \propto 1/ L^{2}

, doubling

L

divides every level by

4

(not

2

); levels drop and crowd together.

The ground state

n = 1

has zero energy because the particle "isn't excited yet."

False —

E_{1} = h^{2} /8 m L^{2} > 0

; a confined particle can never sit perfectly still, this is the zero-point energy demanded by uncertainty.

The energy levels are equally spaced, like rungs on a ladder.

False — spacings grow:

E_{n + 1} - E_{n} = (2 n + 1) E_{1}

, so gaps are

3 E_{1}, 5 E_{1}, 7 E_{1}, \dots

Equal spacing belongs to the Quantum Harmonic Oscillator, not the box.

Between adjacent nodes of

ψ_{n}

the particle moves; at a node it is momentarily stopped.

False — this is classical thinking. A node is where the amplitude is zero, so the particle is simply never found there; there is no "moving between" —

∣ ψ ∣^{2}

describes a static probability landscape.

The wavefunction must be zero at

x = 0

and

x = L

but can be nonzero just outside the walls.

False — outside the infinite walls

V = \infty

, so

ψ = 0

everywhere there; the boundary values

ψ (0) = ψ (L) = 0

come from matching that outside-zero continuously.

For very large

n

, the probability density flattens out and looks classical.

True — with many wiggles packed in,

∣ ψ_{n} ∣^{2}

averaged over any small window approaches the uniform classical distribution (a particle bouncing at constant speed is equally likely everywhere). This is the correspondence principle.

Since

ψ_{n}

can be negative, there are places where "negative probability" occurs.

False — probability density is

∣ ψ_{n} ∣^{2} \geq 0

always; the sign of

ψ

affects interference when waves combine but never gives a negative probability.

A heavier particle in the same box has more widely spaced energy levels.

False —

E_{n} \propto 1/ m

, so a heavier particle has lower, more tightly packed levels; the classical limit is approached faster for heavy objects.

Spot the error

ψ (0) = A sin 0 + B cos 0 = 0

, so

A = 0

Wrong term killed.

sin 0 = 0

and

cos 0 = 1

, so

ψ (0) = B

; the boundary condition forces

B = 0

, not

A = 0

. Keeping

A

is what leaves us a live particle.

sin (k L) = 0

gives

k L = nπ

for

n = 0, 1, 2, \dots

, so the ground state is

n = 0

Including

n = 0

is the error —

k = 0

makes

ψ \equiv 0

(no particle anywhere). The lowest physical state is

n = 1

"Since both sine and cosine solve

ψ^{''} = - k^{2} ψ

, the final wavefunction keeps both terms."

The boundary condition

ψ (0) = 0

eliminates the cosine (

B = 0

). The general solution has both, but the physical solution in this box does not.

"Normalisation gives

A^{2} \cdot L = 1

, so

A = 1/ L

The integral

\int_{0}^{L} sin^{2} (nπ x / L) d x = L /2

, not

L

, because

sin^{2}

averages to

\frac{1}{2}

. Correct result is

A = 2/ L

E_{2} - E_{1} = 2 E_{1}

because we go from level 1 to level 2."

Energy scales as

n^{2}

, so

E_{2} - E_{1} = (4 - 1) E_{1} = 3 E_{1}

. The gap is

2^{2} - 1^{2} = 3

, not the difference in labels.

"Making the walls finite instead of infinite would only lower the energies slightly and change nothing else."

With finite walls the wavefunction leaks into the barrier (tunnelling), there are only finitely many bound states, and

ψ

is no longer exactly zero at the walls — a real qualitative change.

ψ_{3}

has 3 nodes inside the box."

ψ_{n}

has

n - 1

interior nodes, so

ψ_{3}

has 2 interior nodes (the two ends at

x = 0, L

are boundary zeros, not counted as interior).

Why questions

Why does confinement force the energy to be quantised at all?

The walls demand

ψ = 0

at both ends, and only sine waves whose half-wavelengths fit a whole number of times satisfy this — like standing waves on a string, each fit is one allowed

k

, hence one allowed energy.

Why is the ground-state energy nonzero even though the box has

V = 0

inside?

A zero-energy state would need

ψ =

constant (no curvature), but that cannot vanish at both walls; the smallest wave that fits already has curvature, and curvature is kinetic energy in the

ψ^{''}

term.

Why does higher

n

mean higher energy, geometrically?

n

= more wiggles = sharper curvature

ψ^{''}

, and the TISE ties larger

∣ ψ^{''} ∣

to larger kinetic energy — so more bends cost more energy.

Why do we throw away negative

n

sin (- nπ x / L) = - sin (nπ x / L)

, just an overall sign flip; since only

∣ ψ ∣^{2}

is physical, the negative label describes the same state.

Why is the probability of finding the

n = 1

particle in the left third less than

1/3

The density

∣ ψ_{1} ∣^{2}

humps in the middle and thins near the edges, so the outer third holds less than its "fair share" a flat distribution would give.

Why does making the box bigger make the particle behave more classically?

Larger

L

shrinks and crowds the levels (

E_{n} \propto 1/ L^{2}

), so the discrete spacing becomes negligible compared to typical energies — the ladder blurs into a continuum.

Why do we call

k

the "wave number" rather than leaving it as an unnamed constant?

Because

k = nπ / L

turns out to equal

2 π / λ_{n}

, directly counting how many radians of wave fit per metre — it is the spatial frequency of the standing wave.

Why must

ψ

be continuous at the walls but its slope is allowed to jump there?

Continuity of

ψ

is required so probability is well-defined; the slope can jump only because the potential is infinite — a finite well would demand continuous slope too.

Edge cases

What is

ψ (x)

exactly at

x = 0

and

x = L

for every

n

It is zero for all

n

— these are the enforced boundary nodes; every allowed standing wave shares these two zeros.

If we tried

n = 0

, what does the "wavefunction" describe?

Nothing —

ψ \equiv 0

everywhere means the particle has no probability of being anywhere, which is not a physical state, so

n = 0

is discarded.

For a macroscopic ball (

m \sim

1 kg) in a 1 m box, why don't we see quantised energies?

The spacing

\sim h^{2} / (8 m L^{2})

becomes absurdly tiny (

\sim 1 0^{- 67}

J), far below any measurable energy, so the levels appear perfectly continuous.

In the limit

n \to \infty

, what happens to the number of nodes and the classical resemblance?

Nodes (

n - 1

) grow without bound and the fine-scale density averages to uniform, matching a classical particle bouncing at fixed speed — the correspondence principle in action.

At the exact centre

x = L /2

, is

ψ_{2}

a node or an antinode?

A node —

sin (2 π \cdot \frac{1}{2}) = sin π = 0

, so the

n = 2

state has zero probability at the middle, unlike

n = 1

which peaks there.

What happens to the ground state energy if

L \to \infty

(walls removed to infinity)?

E_{1} = h^{2} /8 m L^{2} \to 0

and the levels merge into a continuum — the particle becomes free, recovering the unbound plane-wave picture with no quantisation.

Connections

Particle in a box — solving TISE, energy levels, wavefunctions — the parent derivation these traps test.
Wavefunction and Born Rule — why $∣ ψ ∣^{2}$ , not $ψ$ , is the probability.
Heisenberg Uncertainty Principle — the reason the ground state can't be zero energy.
Quantum Harmonic Oscillator — the equal-spacing contrast trap.
Quantum Tunnelling and Finite Well — what breaks when walls go finite.
Standing Waves on a String — the classical analogue for quantisation.
Quantum Dots — where box-like confinement shows up in the real world.