Chem 3240 · Lecture 3.2
The quantum world is fundamentally probabilistic.
Certainty is replaced by probabilities.
The wavefunction \(\psi\) is the central object.
Key question: what does \(\psi\) actually mean?
Classically, a wave amplitude is a physical displacement (a guitar string).
The quantum \(\psi\) is generally complex.
The absolute square is a probability density:
\[p(x) = \psi^{*}(x) \cdot \psi(x) = |\psi(x)|^2\]
Non-negative: \(\quad p(x) \geq 0\)
Normalized:
\[\int_{-\infty}^{\infty} p(x)\,dx = 1\]
A wavefunction must be normalizable to be a true probability distribution.
\[\int^{+\infty}_{-\infty} |\psi(x)|^2\, dx = 1\]
Normalize \(\psi(x) = C\,x\) on \([0,1]\):
\[C^2 \int_0^1 x^2\, dx = C^2 \cdot \frac{1}{3} = 1\]
\[C = \sqrt{3} \quad \Rightarrow \quad \psi(x) = \sqrt{3}\,x\]
Integrate the density over the region of interest:
\[p(a<x<b) = \int_a^b |\psi(x)|^2\, dx\]
For \(\psi = \sqrt{3}\,x\), so \(p(x) = 3x^2\):
\[P(0.3<x<0.6) = 0.6^3 - 0.3^3 = 0.189\]
Weight values by their probability:
\[\langle x \rangle = \int x \cdot p(x)\, dx\]
Any function of \(x\):
\[\langle f \rangle = \int f(x) \cdot p(x)\, dx\]
Momentum and energy are operators, not simple functions:
\[\langle A \rangle = \int \psi^{*}(x) \cdot \hat{A}\, \psi(x)\, dx\]
Particle in a box, \(\psi_n(x) = \sqrt{2}\sin(n\pi x)\):
\[\langle p \rangle = \int_0^1 \psi_n^*(-i\hbar)\psi_n'\, dx = 0\]
\[\langle p^2 \rangle = (n\pi\hbar)^2 \quad\Rightarrow\quad \langle E \rangle = \frac{(n\pi\hbar)^2}{2m}\]
The wavefunction is a probability amplitude: \(|\psi|^2\) is the probability density, normalization makes it certain the particle exists, and observables are expectation values of operators sandwiched in \(\psi^{*}\hat{A}\psi\).