Expectation Values in Wave Mechanics

Expectation values translate the abstract formula $\langle\psi\vert \hat A\vert\psi\rangle$ into integrals over wavefunctions. In wave mechanics this translation is practical: it tells how to compute average position, momentum, energy, widths, and consistency checks for model solutions.

For a normalized one-dimensional wavefunction and an operator $\hat A$ acting in position representation,

$$\langle A\rangle = \int \psi^*(x)(\hat A\psi)(x)\,dx,$$

provided $\psi$ is in the domain needed by $\hat A$ and the integral exists.

The abstract probability interpretation belongs to Expectation Values. This page is the coordinate-space working version.

Position Moments

The position operator acts by multiplication:

$$(\hat x\psi)(x)=x\psi(x).$$

Therefore

$$\langle x\rangle = \int_{-\infty}^{\infty} \psi^*(x)x\psi(x)\,dx = \int_{-\infty}^{\infty} x\lvert\psi(x)\rvert^2\,dx.$$

Higher position moments are similarly direct:

$$\langle x^n\rangle = \int x^n\lvert\psi(x)\rvert^2\,dx.$$

On a finite interval, replace the integration limits by the interval. In three dimensions,

$$\langle \mathbf r\rangle = \int_{\mathbb R^3} \mathbf r\,\lvert\psi(\mathbf r)\rvert^2\,d^3r.$$

In curvilinear coordinates the measure must be included. In spherical coordinates, $d^3r=r^2\sin\theta\,dr\,d\theta\,d\phi$.

Momentum Expectation

In one-dimensional position representation,

$$\hat p=-i\hbar\frac{d}{dx}.$$

Thus

$$\langle p\rangle = \int \psi^*(x) \left( -i\hbar\frac{d}{dx} \right) \psi(x)\,dx.$$

The formula is compact, but it has domain assumptions. The derivative must exist in the appropriate weak or ordinary sense, and the boundary behavior must make the operator symmetric on the chosen domain.

For a real bound-state wavefunction that vanishes or decays at the endpoints,

$$\langle p\rangle = -i\hbar\int \psi(x)\psi'(x)\,dx = -\frac{i\hbar}{2} \int \frac{d}{dx}\psi^2(x)\,dx =0.$$

This is a common reason one-dimensional stationary bound states in real potentials often have zero average momentum even though their kinetic energy is not zero.

Energy Expectation

For a one-dimensional Hamiltonian

$$\hat H = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} +V(x),$$

the energy expectation is

$$\langle H\rangle = \int \psi^*(x) \left[ -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} +V(x) \right] \psi(x)\,dx.$$

It is often useful to separate kinetic and potential contributions:

$$\langle H\rangle=\langle T\rangle+\langle V\rangle,$$

where

$$\langle V\rangle = \int V(x)\lvert\psi(x)\rvert^2\,dx.$$

If boundary terms vanish, integration by parts gives

$$\langle T\rangle = \frac{\hbar^2}{2m} \int \lvert\psi'(x)\rvert^2\,dx.$$

This form makes the kinetic-energy cost of rapid spatial variation visible. It is one of the simplest ways to understand why confinement raises energy.

Boundary Terms And Hermiticity Checks

Derivative operators require boundary checks. For the momentum operator on an interval $[a,b]$,

$$\int_a^b \phi^*(-i\hbar\psi')\,dx = \int_a^b (-i\hbar\phi')^*\psi\,dx -i\hbar\left[\phi^*(x)\psi(x)\right]_a^b.$$

The boundary term must vanish for $\hat p$ to be symmetric on the chosen domain. On a periodic domain, it can vanish because endpoint values match. On an infinite line, it can vanish because normalizable wavefunctions decay sufficiently fast. On an interval with hard walls, $\psi$ and $\phi$ vanish at the endpoints.

For the kinetic-energy operator, integration by parts produces a different boundary term:

$$\int_a^b \phi^*(-\psi'')\,dx = \int_a^b \phi'^*\psi'\,dx -\left[\phi^*(x)\psi'(x)\right]_a^b.$$

A more symmetric self-adjointness check compares both $\langle\phi\vert \hat T\psi\rangle$ and $\langle \hat T\phi\vert\psi\rangle$. The point for wave-mechanics calculations is practical: do not drop boundary terms unless the domain justifies doing so.

Variance And Standard Deviation

The variance of an observable is

$$(\Delta A)^2 = \langle(\hat A-\langle A\rangle)^2\rangle = \langle A^2\rangle-\langle A\rangle^2.$$

For position,

$$(\Delta x)^2 = \int (x-\langle x\rangle)^2 \lvert\psi(x)\rvert^2\,dx.$$

For momentum,

$$(\Delta p)^2 = \langle p^2\rangle-\langle p\rangle^2,$$

with

$$\langle p^2\rangle = \int \psi^*(x) \left( -\hbar^2\frac{d^2}{dx^2} \right) \psi(x)\,dx.$$

These quantities measure the spread of ideal measurement outcomes in the state, not instrument error bars. The abstract statistical definition is Variance and Standard Deviation.

Uncertainty Check

For normalized one-dimensional states with finite variances and suitable domains,

$$\Delta x\,\Delta p\ge\frac{\hbar}{2}.$$

In wave mechanics this inequality is a useful diagnostic. If a proposed localized packet has both extremely small $\Delta x$ and extremely small $\Delta p$, either the calculation is wrong or the state is not in the assumed domain. The detailed derivation is Position-Momentum Uncertainty.

The free-particle Gaussian packet is the model example where the bound can be saturated at an initial time. See Gaussian Wave Packets for the dynamical version.

Example: Infinite Square Well Symmetry

For the infinite square well on $0<x<L$,

$$\psi_n(x)=\sqrt{\frac{2}{L}}\sin\frac{n\pi x}{L}.$$

The probability density is symmetric about $L/2$, so

$$\langle x\rangle=\frac{L}{2}.$$

The wavefunction can be chosen real and vanishes at both endpoints, so

$$\langle p\rangle=0.$$

But the kinetic energy is nonzero:

$$\langle H\rangle = \frac{n^2\pi^2\hbar^2}{2mL^2}.$$

This example is a useful warning: zero average momentum does not mean zero kinetic energy. Momentum spread, not average momentum, controls kinetic energy in a standing wave.

Three-Dimensional And Radial Expectations

In three dimensions, the expectation value of an operator is

$$\langle A\rangle = \int \psi^*(\mathbf r)(\hat A\psi)(\mathbf r)\,d^3r.$$

For a central-potential state

$$\psi(r,\theta,\phi)=R(r)Y_\ell^m(\theta,\phi),$$

radial expectations of functions $f(r)$ use

$$\langle f(r)\rangle = \int_0^\infty \lvert R(r)\rvert^2 f(r) r^2\,dr$$

when the angular part is normalized. If $u(r)=rR(r)$, the same expectation is

$$\langle f(r)\rangle = \int_0^\infty \lvert u(r)\rvert^2 f(r)\,dr.$$

Confusing $R(r)$ and $u(r)$ changes expectation values as well as normalization.

Numerical Expectations

On a uniform grid with spacing $\Delta x$, the continuum integral is approximated by

$$\langle A\rangle \approx \sum_i \psi_i^*(A\psi)_i\,\Delta x.$$

For multiplication by $x$,

$$\langle x\rangle \approx \sum_i x_i\lvert\psi_i\rvert^2\Delta x.$$

If a numerical eigenvector is normalized as $\sum_i\lvert v_i\rvert^2=1$, then it is not necessarily the sampled continuum wavefunction $\psi_i$. The grid weight must be tracked before comparing numerical and analytic expectation values.

Common Mistakes

• Treating $\langle A\rangle$ as a single-measurement outcome rather than an ensemble average.
• Forgetting to normalize the wavefunction before computing expectations.
• Dropping boundary terms from derivative operators without checking the domain.
• Using $p=-i\hbar d/dx$ on a wavefunction that is not differentiable enough.
• Confusing $\langle p\rangle=0$ with zero kinetic energy.
• Forgetting the radial measure $r^2dr$ when using $R(r)$.
• Forgetting the grid spacing $\Delta x$ in numerical expectation values.

Where This Is Used

• Wavefunctions and Probability Density supplies the density interpretation behind position expectations.
• Hamiltonians in Coordinate Space supplies the differential operators whose expectations are computed here.
• Boundary Conditions explains when boundary terms vanish.
• Infinite Square Well and Quantum Harmonic Oscillator provide first exact-model applications.
• Numerical Notebooks Index points to grid-based checks where normalization weights matter.
• Expectation Value gives the compact reference-card version.

References

• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.

Exercises

1. Show that a normalized real wavefunction on $[a,b]$ with $\psi(a)=\psi(b)=0$ has $\langle p\rangle=0$.

Solution

For real $\psi$,

$$\langle p\rangle = -i\hbar\int_a^b \psi(x)\psi'(x)\,dx = -\frac{i\hbar}{2} \int_a^b \frac{d}{dx}\psi^2(x)\,dx.$$

Thus

$$\langle p\rangle = -\frac{i\hbar}{2} \left[\psi^2(x)\right]_a^b =0$$

because $\psi(a)=\psi(b)=0$.

2. Derive the positive kinetic-energy form $\langle T\rangle=(\hbar^2/2m)\int\lvert\psi'\rvert^2dx$ for a hard-wall interval.

Solution

Start from

$$\langle T\rangle = -\frac{\hbar^2}{2m} \int_a^b \psi^*(x)\psi''(x)\,dx.$$

Integrating by parts gives

$$\int_a^b \psi^*\psi''\,dx = \left[\psi^*\psi'\right]_a^b - \int_a^b \lvert\psi'\rvert^2\,dx.$$

For hard-wall wavefunctions $\psi(a)=\psi(b)=0$, the boundary term vanishes. Therefore

$$\langle T\rangle = \frac{\hbar^2}{2m} \int_a^b \lvert\psi'\rvert^2\,dx.$$

3. If a radial state is written with $u(r)=rR(r)$, express $\langle r\rangle$ in terms of $u(r)$.

Solution

For normalized $u(r)$,

$$\int_0^\infty \lvert u(r)\rvert^2\,dr=1.$$

Since $\lvert R(r)\rvert^2r^2=\lvert u(r)\rvert^2$, the radial expectation is

$$\langle r\rangle = \int_0^\infty r\,\lvert u(r)\rvert^2\,dr.$$