Wave function ψ — probability density - ψ - ²: Question bank

Before you start, three words earn their keep here:

Amplitude — the raw complex value $ψ$ at a point (can be negative or imaginary).
Density — probability per unit length, i.e. $∣ ψ ∣^{2}$ ; you must multiply by a length (integrate) to get an actual probability.
Global phase — multiplying all of $ψ$ by one fixed factor $e^{i θ}$ (same $θ$ everywhere).

True or false — justify

True or false:

∣ ψ (x) ∣^{2}

at a single point

x

is the probability of finding the particle exactly at

x

False.

∣ ψ ∣^{2}

is a density; the probability at an exact point is

∣ ψ ∣^{2} d x \to 0

. You must integrate over an interval to get a real probability.

True or false: If

\int_{- \infty}^{\infty} ∣ ψ ∣^{2} d x = 4

, then

ψ

is normalized by dividing it by

4

False. Normalization acts on

∣ ψ ∣^{2}

, which is quadratic in

ψ

. Divide

ψ

4 = 2

so that

∣ ψ ∣^{2}

divides by

4

True or false: Two wave functions

ψ

and

e^{i θ} ψ

(same fixed

θ

) predict identical measurement outcomes.

True.

∣ e^{i θ} ψ ∣^{2} = e^{- i θ} e^{i θ} ∣ ψ ∣^{2} = ∣ ψ ∣^{2}

, so every prediction built from the density is unchanged. A global phase is physically invisible.

True or false: A position-dependent phase

e^{ik x}

is also physically invisible.

False. It cancels in

∣ ψ ∣^{2}

of a single state, but it carries momentum information and matters in momentum and interference. Only a constant (global) phase is invisible.

True or false:

ψ

is the physical wave you could in principle measure directly, like a water ripple.

False.

ψ

is a complex probability amplitude, not a physical displacement. Only

∣ ψ ∣^{2}

and expectation values built from

ψ

are measurable.

True or false: If

ψ

is everywhere real, then

∣ ψ ∣^{2} = ψ^{2}

True. For real

ψ

ψ^{*} = ψ

, so

ψ^{*} ψ = ψ^{2}

. But for complex

ψ

ψ^{2} \neq = ∣ ψ ∣^{2}

(e.g.

ψ = i

gives

ψ^{2} = - 1

but

∣ ψ ∣^{2} = 1

True or false: A function that grows without bound as

x \to + \infty

can still be a valid physical wave function.

False. It fails square-integrability:

\int ∣ ψ ∣^{2} d x

would diverge, so it cannot be normalized to total probability 1.

True or false: A constant function

ψ (x) = c \neq = 0

over all of infinite space is a legal normalized wave function.

False.

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

, so it cannot be normalized. (It is a useful idealization for a plane wave, but not a genuine physical state.)

True or false: Multiplying

ψ

by a real constant

2

leaves the physics unchanged, just like a phase does.

False. A real factor

2

multiplies

∣ ψ ∣^{2}

4

, changing total probability from 1 to 4 — that breaks normalization. Only unit-modulus factors (

∣ e^{i θ} ∣ = 1

) leave physics intact.

Spot the error

Find the flaw: "Since

ψ

is a probability, it must satisfy

0 \leq ψ \leq 1

ψ

is not a probability; it is an amplitude that can be negative or complex. The constrained-to-

[0, 1]

object is a probability, obtained only after

\int ∣ ψ ∣^{2} d x

Find the flaw: "

P (a \leq x \leq b) = ∣ ψ (b) ∣^{2} - ∣ ψ (a) ∣^{2}

, like evaluating an antiderivative."

You must integrate the density:

P = \int_{a}^{b} ∣ ψ ∣^{2} d x

∣ ψ ∣^{2}

is the integrand, not the antiderivative, so subtracting its endpoint values is meaningless.

Find the flaw: "To normalize

ψ = A e^{- x^{2} / (2 σ^{2})}

I set

A e^{0} = 1

at the peak, giving

A = 1

Normalization is a condition on the total area

\int ∣ ψ ∣^{2} d x = 1

, not on the peak value. The peak height is whatever it must be so the area equals 1.

Find the flaw: "For

ψ = A e^{ik x}

, I compute

∣ ψ ∣^{2} = A^{2} e^{2 ik x}

You squared

ψ

instead of taking

ψ^{*} ψ

. Correctly

∣ ψ ∣^{2} = A^{*} e^{- ik x} \cdot A e^{ik x} = ∣ A ∣^{2}

, a constant — the phase cancels.

Find the flaw: "The units of

ψ

in 1D are the same as probability, dimensionless."

∣ ψ ∣^{2}

has units

m^{- 1}

(probability per length), so

ψ

has units

m^{- 1/2}

. It is not dimensionless.

Find the flaw: "

∣ ψ ∣^{2} = ψ \cdot ψ^{*}

can sometimes be negative if

ψ

is very negative."

Never. Writing

ψ = a + ib

ψ^{*} ψ = a^{2} + b^{2} \geq 0

always. The whole point of squaring the modulus is to guarantee non-negativity.

Find the flaw: "In Particle in a Box with

ψ = A sin (π x / L)

, I normalize by integrating from

- \infty

\infty

ψ

is zero outside

[0, L]

, so the integral runs only over

[0, L]

\int_{0}^{L} A^{2} sin^{2} (π x / L) d x = A^{2} L /2 = 1

. Integrating the always-zero tails adds nothing but signals a misunderstanding of the domain.

Why questions

Why do we use

∣ ψ ∣^{2}

rather than

∣ ψ ∣

, which is also non-negative?

Interference experiments (e.g. electron diffraction) show amplitudes add first, then get squared. Predictions match

∣ ψ ∣^{2}

, not

∣ ψ ∣

; nature selects the squared modulus.

Why must

ψ \to 0

x \to \pm \infty

for a bound particle?

Otherwise

\int ∣ ψ ∣^{2} d x

diverges and cannot equal 1 (square-integrability fails). A localized particle must have vanishing probability infinitely far away.

Why does dividing

ψ

N

(not

N

) fix normalization when

\int ∣ ψ ∣^{2} d x = N

Because

∣ ψ ∣^{2}

is quadratic in

ψ

: scaling

ψ \to ψ / N

scales

∣ ψ ∣^{2} \to ∣ ψ ∣^{2} / N

, so the total area becomes

N / N = 1

Why does the phase factor

e^{ik x}

vanish from the probability density but still matter physically?

ψ^{*} ψ

it cancels (

e^{- ik x} e^{ik x} = 1

), so it never appears in position probabilities. But it encodes momentum (

p = ℏ k

via de Broglie Wavelength), which shows up in Expectation Values and interference.

Why must

ψ

be single-valued?

ψ

had two values at one point,

∣ ψ ∣^{2}

would give two conflicting probability densities there — physically nonsensical. One point, one probability.

Why do we require

ψ

(and often

ψ^{'}

) to be continuous?

A jump in

ψ

would make

ψ^{'}

infinite, and the Schrödinger Equation links

ψ^{''}

to finite energy — so discontinuities generally demand infinite energy, which is unphysical (except at infinite potential walls).

Edge cases

Edge case: a state made of two boxes,

ψ

nonzero in

[0, L]

and

[2 L, 3 L]

, zero between. Is this legal?

Yes, provided the total

\int ∣ ψ ∣^{2} d x = 1

. The particle can have probability spread over disconnected regions; nothing forbids gaps of zero probability.

Edge case:

ψ (x) = 0

at some interior point

x_{0}

(a node). Does the particle "avoid" that point?

The probability density is zero there, so finding it in an infinitesimal window around

x_{0}

has vanishing probability. This is exactly what happens at nodes of a Particle in a Box excited state — a real, observable feature.

Edge case: is

ψ \equiv 0

everywhere a valid wave function?

No. It gives

\int ∣ ψ ∣^{2} d x = 0 \neq = 1

and cannot be rescaled (dividing zero by anything stays zero). "No particle anywhere" is not a normalizable state.

Edge case: a

ψ

with a finite jump discontinuity but still square-integrable — is it acceptable?

The integral

\int ∣ ψ ∣^{2} d x

may exist, but the physical conditions demand continuity (so

ψ^{'}

stays finite for the Schrödinger equation). Such a

ψ

is rejected as unphysical despite being square-integrable.

Edge case: does

∣ ψ ∣^{2}

having a very tall, thin spike mean the particle "is" almost exactly there?

Only in the sense of high density. The actual probability is the spike's area (

∣ ψ ∣^{2} d x

); a tall spike over a tiny width can still carry little probability. Localizing tightly in

x

forces spread in momentum (Heisenberg Uncertainty Principle).

Recall wrap-up

Recall One-line takeaways

Density vs probability ::: $∣ ψ ∣^{2}$ is per-length; multiply by $d x$ (integrate) for a real probability.
Normalize by ::: dividing $ψ$ by $N$ , since $∣ ψ ∣^{2} \propto ψ^{2}$ .
Invisible thing ::: only a global phase $e^{i θ}$ ; position-dependent phase carries momentum.
Illegal states ::: $ψ \equiv 0$ , non-normalizable (constant / growing), multi-valued, discontinuous.

Connections

2.3.08 Wave function ψ — probability density - ψ - ² (Hinglish) — parent topic (Hinglish).
Schrödinger Equation — why continuity of $ψ$ and $ψ^{'}$ is enforced.
Heisenberg Uncertainty Principle — spike-in- $x$ ⇒ spread-in- $p$ edge case.
Particle in a Box — nodes, domain of integration, disconnected regions.
de Broglie Wavelength — the $k$ in $e^{ik x}$ that hides in the phase.
Complex Numbers — why $ψ^{*} ψ = a^{2} + b^{2} \geq 0$ .
Expectation Values — where the "invisible" phase reappears.