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Quantum Foundations

Start

Use the photoelectric effect and single-particle interference to motivate quantum ideas.

Wavefunctions

Here

Interpret amplitudes, probabilities, and measurement in a one-dimensional quantum state.

Quantization

Finish

Use bound-state potentials to build intuition for discrete energies in quantum systems.

Wavefunctions

The experimental clues from quantum foundations need a mathematical language. That language is the wavefunction $\psi$ . You can first think of it as a mysterious function of position $\psi(x)$ , in order to help make the connection to our classical descriptions of nature.

However, the new quantum ideas are much more general. The same quantum state can be expressed in whatever basis matches the question: position, momentum, energy, spin, or even the orbitals used in chemistry. In every case, $\psi$ is not a classical material wave. Instead, it expresses a probability amplitude whose squared magnitude tells us how measurement outcomes are distributed.

From Amplitude to Probability

Loosely speaking, $\psi(x)$ tells you where you can expect the corresponding particle to be located in space. In quantum mechanics, it is no longer correct to assume every particle has an exact location or every system has a precise state. Instead, we describe things using the wavefunction and then when a measurement is made, we convert the wavefunction to a probability distribution to predict the measurement outcomes.

For a normalized wavefunction, the corresponding probability density is

P(x) = |\psi(x)|^2.

That single rule matters more than almost anything else in introductory quantum mechanics. It says quantum predictions come from amplitudes first and probabilities second. Since amplitudes can interfere, the probability pattern can contain peaks, dips, and cancellations that a classical probability mixture would miss. One useful example is a localized wave packet:

\psi(x) = A e^{-(x-x_0)^2/(2\sigma^2)} \sin(kx).

Example wave packet

Amplitude View

Probability Density

Measurement hits land according to |psi|^2

Expected Position

0.500

This probability-weighted average gives a sense of where position measurements cluster.

Middle Third

0.885

Probability of landing in the central third of the plot.

Read the plots

This is a localized wave packet: the sine factor provides the wiggles, while the Gaussian envelope keeps the state concentrated in one region.

Controls

Packet Center0.50

Envelope Width0.16

Oscillations5

Position Measurements

Repeated measurements build up a set of hits that follows the probability plot.

Normalization and Probablity

If a particle must be found somewhere, then the total probability must add to $1$ . In continuous language, that becomes

\int |\psi(x)|^2\,dx = 1.

This condition is called normalization. It does not tell you the shape of the state by itself, but it fixes the overall scale.

Once you have a normalized wavefunction, the probability of measuring the location to be between $x$ and $x + \Delta x$ is

P(x,x + \Delta x) \approx |\psi(x)|^2 \Delta x

The Same Rule In Other Contexts

The amplitude-to-probability rule shows up in a variety of settings:

Position space: $\psi(x,t)$ describes amplitudes across space, and $|\psi(x,t)|^2 dx$ gives the probability of finding the particle in a small interval near $x$ at time $t$ .
Chemistry: a molecular orbital can be written as $\psi(\mathbf{r})$ . Then $|\psi(\mathbf{r})|^2 d^3r$ gives the probability of finding an electron in a small region of space near the point $\mathbf{r}$ . Electron-cloud pictures in chemistry come from this idea.
Two-state example: sometimes the state is not written as a function at all. A “quantum coin” can be written as

|\psi\rangle = \sqrt{0.8}\,|H\rangle + e^{i\phi}\sqrt{0.2}\,|T\rangle.

In that basis,

P(\text{heads}) = 0.8, \qquad P(\text{tails}) = 0.2.

Problem Solving

Normalize A Simple Box State

Problem

A particle in a one-dimensional box has wavefunction

\psi(x) = A \sin\!\left(\frac{\pi x}{L}\right)

for $0 \le x \le L$ , and $\psi(x)=0$ elsewhere. Find the normalization constant $A$ .

Given

$\psi(x) = A\sin(\pi x/L)$ in the box
the state must satisfy $\int_0^L |\psi(x)|^2\,dx = 1$

Use

\int_0^L A^2 \sin^2\!\left(\frac{\pi x}{L}\right) dx = 1.

Also,

\int_0^L \sin^2\!\left(\frac{\pi x}{L}\right) dx = \frac{L}{2}.

Solve

Substitute the integral result:

A^2 \frac{L}{2} = 1.

A^2 = \frac{2}{L} \qquad \Rightarrow \qquad A = \sqrt{\frac{2}{L}}.

Check

The amplitude scale depends on box width. A wider box spreads the same total probability over a larger region, so the normalized amplitude becomes smaller.

Energy Probabilities

Problem

A state is written as

\psi = \sqrt{0.75}\,\phi_1 + e^{i\pi/3}\sqrt{0.25}\,\phi_2,

where $\phi_1$ and $\phi_2$ are energy eigenstates. What are the probabilities of measuring the energies associated with $\phi_1$ and $\phi_2$ ?

Given

coefficient of $\phi_1$ : $\sqrt{0.75}$
coefficient of $\phi_2$ : $e^{i\pi/3}\sqrt{0.25}$

Use

For an energy measurement in this basis, the probabilities are the squared magnitudes of the coefficients.

Solve

P(E_1) = \left|\sqrt{0.75}\right|^2 = 0.75, \qquad P(E_2) = \left|e^{i\pi/3}\sqrt{0.25}\right|^2 = 0.25.

So the outcomes are

$75\%$ for the energy of $\phi_1$
$25\%$ for the energy of $\phi_2$

Check

The phase factor changes the amplitude but not the coefficient magnitude, so it does not change these basis-measurement probabilities.

Wavefunctions Checkpoint

Question 1 of 3

What does $|\psi(x)|^2$ represent in introductory quantum mechanics?

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