Schrödinger equation — time-dependent, time-independent: Question bank

True or false — justify

The Schrödinger equation is second order in time, like the wave equation for light

False. It is first order in time; the single

\partial_{t}

(with the

i

) is exactly what makes

Ψ

now fully determine all future

Ψ

The wavefunction

Ψ

is itself the probability of finding the particle

False.

Ψ

is complex; the probability density is

∣Ψ ∣^{2} = Ψ^{*} Ψ

, which is real and non-negative —

Ψ

can be negative or imaginary.

A stationary state has a wavefunction that does not change with time

False.

Ψ = ψ (x) e^{- i E t /ℏ}

keeps rotating in phase; only the observable

∣Ψ ∣^{2} = ∣ ψ ∣^{2}

is time-independent.

The Time-Independent Schrödinger Equation applies to every quantum problem

False. The TISE only exists when

V

is time-independent, because it comes from separating variables — a time-varying

V (x, t)

needs the full TDSE.

Energies in a box are quantized because we assumed they must be discrete

False. Quantization emerges from the boundary conditions

ψ (0) = ψ (L) = 0

; we assumed continuity of

ψ

, not discreteness of

E

The constant

E

from separation of variables could be any complex number

False.

E

must be real — it equals

\frac{1}{ψ} \hat{H} ψ

, the total energy, and complex

E

would make

∣Ψ ∣^{2}

grow or decay, breaking probability conservation.

A plane wave

A e^{i (k x - ω t)}

is normalizable over all space

False.

∣Ψ ∣^{2} = ∣ A ∣^{2}

is constant everywhere, so

\int_{- \infty}^{\infty} ∣ A ∣^{2} d x = \infty

— it is an idealized state, not a physical square-integrable one (see Wavefunction and Born Interpretation).

Two different quantum states can share the same energy

True. Distinct eigenfunctions can share one eigenvalue — this is called degeneracy and is common in 2D/3D wells and the Quantum Harmonic Oscillator in higher dimensions.

Multiplying a normalized

ψ

e^{i θ}

(constant phase) gives a physically different state

False.

∣ e^{i θ} ψ ∣^{2} = ∣ ψ ∣^{2}

, so all predictions are identical — an overall constant phase carries no physical information.

Spot the error

i ℏ \partial_{t}^{2} Ψ = \hat{H} Ψ

is the TDSE" — what's wrong?

The time derivative is first order, not squared:

i ℏ \partial_{t} Ψ = \hat{H} Ψ

. A second-order-in-time PDE needs two initial conditions (

Ψ

and

\partial_{t} Ψ

) to fix the future, so a single given

Ψ (t_{0})

would no longer determine the evolution — determinism from one state is lost.

"For the infinite well I keep the

cos

term because both sine and cosine solve

ψ^{''} = - k^{2} ψ

" — what's wrong?

The boundary

ψ (0) = 0

kills the cosine, since

cos 0 = 1 \neq = 0

; only

ψ = A sin (k x)

survives (see Particle in a Box).

n = 0

is a valid ground state of the box, energy zero" — what's wrong?

n = 0

gives

ψ = A sin 0 = 0

everywhere, meaning no particle at all; the lowest allowed state is

n = 1

(see Energy Quantization).

Ψ = ψ (x) + ϕ (t)

separates the TDSE" — what's wrong?

Separation of variables uses a product

Ψ = ψ (x) ϕ (t)

, not a sum; only the product lets you divide through and split

t

-only from

x

-only terms.

"Since

E = \frac{p ^{2}}{2 m}

always, the potential term is unnecessary" — what's wrong?

That relation holds only for a free particle; a real particle has

E = \frac{p ^{2}}{2 m} + V (x)

, which is why

\hat{H}

carries the

+ V

term.

"The Hamiltonian operator is just the number

E

" — what's wrong?

\hat{H} = - \frac{ℏ ^{2}}{2 m} \partial_{x}^{2} + V

is a differential operator;

E

is the eigenvalue it produces when it acts on an eigenfunction

ψ

(see Hamiltonian Operator).

"Probability is

\int Ψ d x = 1

" — what's wrong?

Normalization uses the squared magnitude:

\int ∣Ψ ∣^{2} d x = 1

. Integrating

Ψ

itself is meaningless for a complex field and need not even be real.

Why questions

Why must the Schrödinger equation be first order in time?

So that knowing

Ψ (x, t_{0})

alone determines all future

Ψ

— matching the deterministic predictive power of Newton's law, with randomness living only in

∣Ψ ∣^{2}

Why does the imaginary

i

have to appear in the TDSE?

It makes

\hat{H}

's action generate phase rotation

e^{- i E t /ℏ}

rather than growth/decay, which keeps

\int ∣Ψ ∣^{2} d x

constant in time — probability is conserved.

Why does a time-independent

V

let us separate variables?

V

doesn't depend on

t

, the spatial operator

\hat{H}

has no time in it, so the

t

-dependence factors out cleanly as

ϕ (t) = e^{- i E t /ℏ}

Why is the separation constant labeled

E

(energy)?

The space side equals

\frac{1}{ψ} \hat{H} ψ

; since

\hat{H}

is the energy operator, this constant is the total energy of that stationary state.

Why do stationary states have a frozen probability cloud despite

Ψ

evolving?

Their time part is a pure phase

e^{- i E t /ℏ}

of magnitude 1, so

∣Ψ ∣^{2} = ∣ ψ ∣^{2}

loses all time dependence even though

Ψ

keeps rotating.

Why does trapping a particle force discrete energies?

Boundary conditions require the wave to fit an exact number of half-wavelengths, and only special

k

(hence special

E

) satisfy them — like standing waves on a fixed string (see de Broglie Hypothesis).

Why start the derivation from a free-particle plane wave?

Because we know its

E

and

p

exactly (

p = ℏ k

E = ℏ ω

), so we can substitute

Ψ = A e^{i (k x - ω t)}

into the free TDSE and check it works:

i ℏ \partial_{t} Ψ = ℏ ω Ψ

on the left equals

- \frac{ℏ ^{2}}{2 m} \partial_{x}^{2} Ψ = \frac{ℏ ^{2} k ^{2}}{2 m} Ψ

on the right, forcing the correct dispersion

ℏ ω = \frac{ℏ ^{2} k ^{2}}{2 m}

, i.e.

E = \frac{p ^{2}}{2 m}

Edge cases

What happens to the well's energies as

L \to \infty

(walls infinitely far apart)?

E_{n} = \frac{n ^{2} π ^{2} ℏ ^{2}}{2 m L ^{2}} \to 0

and the levels crowd together — the spectrum becomes effectively continuous, recovering the free particle.

What is the ground-state energy of the infinite well, and can it be zero?

E_{1} = \frac{π ^{2} ℏ ^{2}}{2 m L ^{2}} > 0

; it cannot be zero because a confined particle must have nonzero momentum spread — a zero-energy state would violate the Heisenberg Uncertainty Principle.

If the potential is constant

V = V_{0}

everywhere, how do the energies change?

A constant just shifts every eigenvalue:

E \to E + V_{0}

, while the eigenfunctions

ψ (x)

stay identical — only differences in energy are physical.

What does the TISE give if

E

is chosen not to match any eigenvalue in a bound problem?

The solution fails to satisfy the boundary conditions (blows up or won't vanish at infinity), so it isn't normalizable — such

E

are simply not allowed.

ℏ \to 0

(classical limit), what happens to the spacing between box levels?

E_{n} \propto ℏ^{2}

so all gaps shrink toward zero, energies blur into a continuum and quantum discreteness disappears — the classical picture re-emerges.

For a normalized state, what must

∣Ψ ∣^{2}

do as

x \to \pm \infty

It must fall to zero fast enough for

\int ∣Ψ ∣^{2} d x

to be finite (=1); a probability density that stays nonzero at infinity cannot be normalized.

A visual map of the traps

The diagram below groups the misconceptions on this page by the concept they attack — use it as a checklist of "where you can slip".

Figure — Schrödinger equation — time-dependent, time-independent

Connections

Wavefunction and Born Interpretation — the $∣Ψ ∣^{2}$ traps.
Particle in a Box — source of the boundary-condition and $n = 0$ traps.
Energy Quantization — why discreteness emerges, not assumed.
Hamiltonian Operator — operator vs eigenvalue confusion.
Heisenberg Uncertainty Principle — why zero ground-state energy is forbidden.
de Broglie Hypothesis — standing-wave picture behind quantization.

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