Expectation values (what you "expect" to measure)
How to compute average measurement outcomes from a wave function.
In quantum mechanics, each measurable quantity (position, momentum, energy, etc.) is represented by an operator \( \hat{A} \). If the particle is in the normalized state \( \psi(x) \), expectation value (average over many identical measurements) is
\[ \langle A \rangle = \int \psi^*(x)\,\hat{A}\,\psi(x)\,dx. \]
For a particle in a 1D infinite box from x=0 to x=L, the integral runs over the box:
\[ \langle A \rangle = \int_{0}^{L} \psi^*(x)\,\hat{A}\,\psi(x)\,dx. \]
Two very common operators are:
- Position: \( \hat{A}=x \) so \( \langle x\rangle=\int_{0}^{L} x\,|\psi(x)|^2\,dx \).
- Momentum (1D): \( \hat{p}_x=-i\hbar\,\frac{d}{dx} \) so \( \langle p\rangle=\int_{0}^{L}\psi^*(x)\left(-i\hbar\frac{d}{dx}\right)\psi(x)\,dx \).
For the particle-in-a-box stationary states, \( \psi_n(x)=\sqrt{\frac{2}{L}}\sin\!\left(\frac{n\pi x}{L}\right) \) (with n=1,2,3,\dots), symmetry of the standing wave about x=L/2 gives a simple result:
\[ \langle x\rangle=\frac{L}{2}, \qquad \langle p\rangle=0. \]
Variance
A useful "spread" measure is the variance \( (\Delta x)^2=\langle x^2\rangle-\langle x\rangle^2 \). For box states one finds
\[ \langle x^2\rangle = L^2\left(\frac{1}{3}-\frac{1}{2n^2\pi^2}\right), \qquad (\Delta x)^2 = L^2\left(\frac{1}{12}-\frac{1}{2n^2\pi^2}\right). \]
Big idea: expectation values come from \( \psi^*\,\hat{A}\,\psi \). The "probability density" is \( |\psi(x)|^2 \). The variance can be calculated from \( (\Delta a)^2=\langle a^2\rangle-\langle a\rangle^2 \).