Observables, Expectation Values, and the Operator Dictionary
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You should be comfortable with the Born rule, normalisation of wavefunctions, and the definition of the inner product in one-dimensional Hilbert space. Familiarity with basic differentiation and integration by parts is required.
Why this matters
Quantum mechanics replaces definite trajectories with statistical predictions, requiring a precise mathematical link between physical quantities and mathematical operators.
In classical mechanics, we calculate the average value of a quantity by integrating its value over the probability distribution. In quantum mechanics, the wavefunction \(\psi(x)\) itself encodes the probability distribution, but the quantity being measured (like momentum) is not simply a function of position. We need a rule that tells us how to extract the average value of any physical observable from the state vector. This rule is the expectation value, and the specific mathematical objects used to represent observables are the operators. Establishing this "operator dictionary" is the bridge between the abstract Hilbert space formalism and concrete physical predictions.
The Core Idea: From Averages to Operators
Imagine you have a particle in a box. If you measure its position, you get a specific number. If you repeat the experiment many times with identically prepared particles, you get a spread of results. The "expectation value" is the average of all these results. It is not necessarily a value you will ever measure in a single trial, but it is the centre of the distribution. The key insight is that this average can be calculated without knowing the individual measurement outcomes, provided we know the wavefunction \(\psi(x)\) and the operator \(\hat{A}\) associated with the observable \(A\). The operator acts on the wavefunction, and the inner product of the original wavefunction with the result gives the average. This is a purely mathematical procedure that yields a physical prediction.
The General Formula for Expectation Values
For any observable \(A\) represented by a Hermitian operator \(\hat{A}\), the expectation value \(\langle A \rangle\) for a normalised state \(\psi(x)\) is given by the inner product of the state with the operator acting on the state. In the position representation, this becomes an integral over all space.
This formula is the central computational tool of quantum mechanics. It tells us that to find the average value of an observable, we apply the operator to the wavefunction, multiply by the complex conjugate of the wavefunction, and integrate. The Hermitian nature of \(\hat{A}\) guarantees that this result is a real number, which is essential because physical measurements yield real values.
Position and Momentum Operators
Two fundamental observables are position and momentum. The operator for position is the simplest: it is just multiplication by the coordinate \(x\). If you want the average position, you integrate \(x |\psi(x)|^2\).
The momentum operator is more subtle. It is derived from the de Broglie relation \(p = \hbar k\), where \(k\) is the wave number. Since the wavefunction of a momentum eigenstate is \(\psi(x) \propto e^{ikx}\), applying the derivative operator \(\frac{\partial}{\partial x}\) yields \(ik\). To recover \(p = \hbar k\), we must multiply by \(-i\hbar\). Thus, the momentum operator is a differential operator.
Consequently, the expectation value of momentum involves the derivative of the wavefunction:
This shows that momentum is related to the spatial variation (phase gradient) of the wavefunction, not its magnitude. A real wavefunction (constant phase) has zero average momentum.
Consistency and Boundary Conditions
When calculating expectation values involving differential operators, such as momentum or kinetic energy, integration by parts is often required to move the derivative from the wavefunction to its complex conjugate. This process relies on the boundary condition that \(\psi(x)\) vanishes at infinity (or at the boundaries of a finite domain). This ensures that the surface terms vanish, preserving the Hermiticity of the operator and the reality of the expectation value. If the wavefunction does not satisfy these boundary conditions, the operator may not be self-adjoint, and the physical interpretation breaks down.
The derivation
Expectation Values and the Operator Dictionary
For a normalised state ψ(x), the average value of an observable A over repeated identical measurements is the inner product ⟨A⟩ = ⟨ψ|Â|ψ⟩ = ∫ ψ*(x) Â ψ(x) dx. Imposing the Born rule for position fixes x̂ = x (multiplication), and imposing the de Broglie relation for momentum eigenstates fixes p̂ = −iℏ ∂/∂x. These two representations are the seed of the entire operator dictionary.
A common pitfall is assuming that \(\langle p^2 \rangle = \langle p \rangle^2\). This is generally false. The variance \(\sigma_p^2 = \langle p^2 \rangle - \langle p \rangle^2\) is non-negative, and it is zero only if the state is a momentum eigenstate. For a general wavepacket, \(\langle p^2 \rangle\) is strictly greater than \(\langle p \rangle^2\). Confusing the square of the average with the average of the square leads to incorrect uncertainty estimates. C
Worked example
Problem. Consider a particle in a one-dimensional infinite square well of width \(L\), where the potential is zero for \(0 < x < L\) and infinite elsewhere. The ground state wavefunction is \(\psi_1(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{\pi x}{L}\right)\). Calculate the expectation value of the momentum \(\langle p \rangle\) and the expectation value of the kinetic energy \(\langle T \rangle\), given \(\hbar = 1.05 \times 10^{-34} \ \mathrm{J \cdot s}\) and \(L = 1.0 \ \mathrm{nm}\).
- Calculate \(\langle p \rangle\). The momentum operator is \(\hat{p} = -i\hbar \frac{d}{dx}\). \[ \langle p \rangle = \int_0^L \psi_1^*(x) \left( -i\hbar \frac{d}{dx} \right) \psi_1(x) \, dx \] Since \(\psi_1(x)\) is real, \(\psi_1^* = \psi_1\). The derivative is \(\frac{d}{dx} \sin\left(\frac{\pi x}{L}\right) = \frac{\pi}{L} \cos\left(\frac{\pi x}{L}\right)\). \[ \langle p \rangle = -i\hbar \frac{2}{L} \frac{\pi}{L} \int_0^L \sin\left(\frac{\pi x}{L}\right) \cos\left(\frac{\pi x}{L}\right) \, dx \] Using \(\sin(2\theta) = 2\sin\theta\cos\theta\), the integrand is proportional to \(\sin\left(\frac{2\pi x}{L}\right)\). The integral of \(\sin\left(\frac{2\pi x}{L}\right)\) from \(0\) to \(L\) is zero. Thus, \(\langle p \rangle = 0\). This makes sense due to symmetry; the particle is equally likely to move left or right.
- Calculate \(\langle T \rangle\). The kinetic energy operator is \(\hat{T} = \frac{\hat{p}^2}{2m} = -\frac{\hbar^2}{2m} \frac{d^2}{dx^2}\). First, find the second derivative: \[ \frac{d^2}{dx^2} \sin\left(\frac{\pi x}{L}\right) = -\left(\frac{\pi}{L}\right)^2 \sin\left(\frac{\pi x}{L}\right) \] So, \[ \hat{T} \psi_1(x) = -\frac{\hbar^2}{2m} \left[ -\left(\frac{\pi}{L}\right)^2 \psi_1(x) \right] = \frac{\hbar^2 \pi^2}{2m L^2} \psi_1(x) \] The expectation value is: \[ \langle T \rangle = \int_0^L \psi_1^*(x) \left( \frac{\hbar^2 \pi^2}{2m L^2} \psi_1(x) \right) \, dx = \frac{\hbar^2 \pi^2}{2m L^2} \int_0^L |\psi_1(x)|^2 \, dx \] Since the wavefunction is normalised, the integral is 1. \[ \langle T \rangle = \frac{\hbar^2 \pi^2}{2m L^2} \]
- Substitute numerical values. Assume the particle is an electron, \(m = 9.11 \times 10^{-31} \ \mathrm{kg}\). \[ L = 1.0 \ \mathrm{nm} = 1.0 \times 10^{-9} \ \mathrm{m} \] \[ \langle T \rangle = \frac{(1.05 \times 10^{-34})^2 \pi^2}{2 (9.11 \times 10^{-31}) (1.0 \times 10^{-9})^2} \] \[ \langle T \rangle = \frac{1.1025 \times 10^{-68} \times 9.87}{2 \times 9.11 \times 10^{-31} \times 10^{-18}} \] \[ \langle T \rangle = \frac{1.088 \times 10^{-67}}{1.822 \times 10^{-48}} \approx 5.97 \times 10^{-20} \ \mathrm{J} \] Convert to electron-volts: \(1 \ \mathrm{eV} = 1.60 \times 10^{-19} \ \mathrm{J}\). \[ \langle T \rangle \approx \frac{5.97 \times 10^{-20}}{1.60 \times 10^{-19}} \approx 0.373 \ \mathrm{eV} \]
Units check. For \(\langle T \rangle\), \(\frac{(\mathrm{J \cdot s})^2}{\mathrm{kg \cdot m^2}} = \frac{\mathrm{J^2 \cdot s^2}}{\mathrm{kg \cdot m^2}}\). Since \(\mathrm{J} = \mathrm{kg \cdot m^2/s^2}\), \(\mathrm{J^2 \cdot s^2} = \mathrm{kg^2 \cdot m^4}\). Thus, \(\frac{\mathrm{kg^2 \cdot m^4}}{\mathrm{kg \cdot m^2}} = \mathrm{kg \cdot m^2/s^2} = \mathrm{J}\). The units are correct.
Three quick tests
- Is the expectation value of an observable always equal to one of the possible measurement outcomes?
Answer
No. It is the statistical average over many trials. For a superposition of eigenstates, it lies between the eigenvalues but is not necessarily an eigenvalue itself. - Why must the operator for a physical observable be Hermitian?
Answer
Hermitian operators guarantee that their eigenvalues are real, which is necessary because measurement results are real numbers. They also ensure that the expectation value is real. - If \(\psi(x)\) is a real-valued function, what is \(\langle p \rangle\)?
Answer
Zero. The integrand \(\psi^* (-i\hbar \psi')\) becomes \(-i\hbar \psi \psi'\), and \( \psi\psi' = \tfrac{1}{2}(\psi^2)' \), so \( \langle p \rangle = -\tfrac{i\hbar}{2}\big[\psi^2\big]_{-\infty}^{\infty} = 0 \) for any normalisable real wavefunction. (It had to be: \( \langle p \rangle \) is real, but the integral is \( -i\hbar \) times a real number.)
Problem set for PU-202 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.