Wavefunctions and Probability Density

A wavefunction is a coordinate-space probability amplitude. In one dimension, $\psi(x,t)$ assigns a complex amplitude to position $x$ at time $t$. The probability density for position is

$$\rho(x,t)=|\psi(x,t)|^2.$$

The density is not itself a probability. Probabilities are obtained by integrating the density over regions.

Required background. Coordinate Representation supplies $\psi(x)=\langle x|\psi\rangle$ and the coordinate measure.

Helpful background. Born Rule for Continuous Outcomes supplies the probability measure assigned to a spatial region.

Probability In One Dimension

For a particle on the real line, the probability of finding the particle in an interval $[a,b]$ is

$$P(a\le x\le b) =\int_a^b |\psi(x,t)|^2\,dx.$$

If the particle is certainly somewhere on the line, the wavefunction is normalized:

$$\int_{-\infty}^{\infty} |\psi(x,t)|^2\,dx=1.$$

This condition fixes the units of the wavefunction. Since $|\psi|^2 dx$ is dimensionless, a one-dimensional normalized wavefunction has units of length$^{-1/2}$.

On a finite interval $0<x<L$, the normalization condition becomes

$$\int_0^L |\psi(x,t)|^2\,dx=1.$$

The interval and boundary conditions are part of the physical problem; they are not merely integration limits.

Density, Probability, and Resolution

For an absolutely continuous position distribution, the cumulative probability is

$$F(x,t) = P(X\le x) = \int_{-\infty}^{x}\rho(x',t)\,dx'.$$

Where the derivative exists,

$$\rho(x,t)=\frac{\partial F(x,t)}{\partial x}.$$

The probability of an ideal point event is zero:

$$P(X=x_0) = \int_{\{x_0\}}\rho(x,t)\,dx =0.$$

This does not mean the density at $x_0$ is zero. It means that a density acquires probabilistic meaning only after integration against a region or detector response. A position-diagonal yes/no effect

$$E_w=\int dx\,w(x)|x\rangle\langle x|, \qquad 0\le w(x)\le1,$$

registers with probability

$$P_w = \int dx\,w(x)\rho(x,t).$$

An ideal interval detector corresponds to an indicator function. This is a restricted position-diagonal model, not every finite-resolution measurement. A classically blurred resolved readout $y$ can instead use a response kernel

$$p(y)=\int dx\,K(y|x)\rho(x), \qquad K(y|x)\ge0, \qquad \int dy\,K(y|x)=1.$$

General quantum detector effects need not be diagonal in position. Even this simple model shows why probabilities for bins or response profiles are more operationally direct than probability “at a point.”

Probability In Three Dimensions

For a particle in three-dimensional space, the wavefunction is $\psi(\mathbf r,t)$ and the probability density is

$$\rho(\mathbf r,t)=|\psi(\mathbf r,t)|^2.$$

The probability of finding the particle in a spatial region $\Omega$ is

$$P(\Omega) =\int_\Omega |\psi(\mathbf r,t)|^2\,d^3r.$$

In Cartesian coordinates, $d^3r=dx\,dy\,dz$. In spherical coordinates,

$$d^3r=r^2\sin\theta\,dr\,d\theta\,d\phi.$$

Thus a normalized three-dimensional wavefunction satisfies

$$\int_0^\infty dr \int_0^\pi d\theta \int_0^{2\pi} d\phi\, |\psi(r,\theta,\phi,t)|^2 r^2\sin\theta =1.$$

The factor $r^2\sin\theta$ is part of the volume element. Omitting it changes the probability distribution.

Densities Under Coordinate Changes

A probability is coordinate independent, but a density is defined relative to a measure. For a monotonic change $y=y(x)$,

$$\rho_y(y)\,dy=\rho_x(x)\,dx,$$

so

$$\rho_y(y) = \rho_x(x(y)) \left|\frac{dx}{dy}\right|.$$

The numerical value of a density can therefore change when coordinates change even though every region probability remains the same.

In three dimensions, the radial probability density for a general wavefunction is

$$p_r(r,t) = r^2 \int_{S^2} |\psi(r,\theta,\phi,t)|^2\,d\Omega,$$

and it is normalized with flat $dr$:

$$\int_0^\infty p_r(r,t)\,dr=1.$$

For a spherically symmetric state this reduces to

$$p_r(r,t)=4\pi r^2|\psi(r,t)|^2.$$

Thus $|\psi(r)|^2$ is a volume density, while $p_r(r)$ is a one-dimensional radial density. Their maxima need not occur at the same radius.

Amplitude, Phase, And Interference

The probability density sees only the magnitude of the wavefunction at a point, but the phase of $\psi$ still matters. Write

$$\psi(x,t)=R(x,t)e^{iS(x,t)/\hbar},$$

where $R\ge 0$. Then

$$|\psi(x,t)|^2=R^2(x,t).$$

The local phase $S(x,t)$ does not change the density at that instant by itself, but phase gradients contribute to probability current and interference. A global phase $e^{i\alpha}\psi$ changes no physical probabilities. A relative phase between two components can move nodes, change interference fringes, or reverse current.

For two orthonormal components, let

$$\psi = \frac{1}{\sqrt2} \left(\psi_1+e^{i\varphi}\psi_2\right).$$

The position density is

$$\begin{aligned} |\psi|^2 ={}&\frac12 \left(|\psi_1|^2+|\psi_2|^2\right)\\ &+\operatorname{Re} \left(e^{i\varphi}\psi_1^*\psi_2\right). \end{aligned}$$

The cross term carries the relative phase. Global orthogonality makes its integral vanish, but it can remain nonzero locally and create interference fringes. A single position-density snapshot generally cannot reconstruct the phase of the state; measurements in other bases or at other times are needed.

Nodal Structure

A node is a point or surface where the wavefunction vanishes. At a node,

$$\psi=0, \qquad \rho=0.$$

Nodes are not merely graphical features. In bound-state problems they often encode excitation number, boundary conditions, parity, and orthogonality. For example, the $n$th eigenstate of the infinite square well has $n-1$ interior nodes. In central-potential problems, nodal surfaces can come from radial nodes or angular nodes.

For a complex wavefunction, a node requires both real and imaginary parts to vanish. A zero of $\operatorname{Re}\psi$ alone is not a node. Nodes can also move in time for nonstationary superpositions.

Internal Degrees of Freedom

If position is accompanied by a discrete internal label $s$, the components are

$$\psi_s(x,t)=\langle x,s|\psi(t)\rangle.$$

When position is measured without resolving the internal state, the density is

$$\rho(x,t)=\sum_s|\psi_s(x,t)|^2.$$

The sum is incoherent because the internal outcomes are orthogonal and unresolved. A unitary change of basis acting only on the internal factor leaves this summed density invariant. Interference can reappear in a conditional position distribution after the internal output is resolved or postselected, or after dynamics couples the internal and motional degrees of freedom.

For spin-$1/2$,

$$\psi(x,t) = \begin{pmatrix} \psi_\uparrow(x,t)\\ \psi_\downarrow(x,t) \end{pmatrix}, \qquad \rho=|\psi_\uparrow|^2+|\psi_\downarrow|^2.$$

Configuration-Space Densities

For two particles, the wavefunction

$$\Psi(\mathbf r_1,\mathbf r_2,t)$$

lives on six-dimensional configuration space. Its squared magnitude is a joint density:

$$P(\Omega_1,\Omega_2) = \int_{\Omega_1}d^3r_1 \int_{\Omega_2}d^3r_2\, |\Psi(\mathbf r_1,\mathbf r_2,t)|^2.$$

The one-particle marginal density is

$$\rho_1(\mathbf r_1,t) = \int d^3r_2\, |\Psi(\mathbf r_1,\mathbf r_2,t)|^2.$$

For a labeled coordinate this marginal integrates to one. For $N$ identical particles, the conventional one-body number density is instead normalized to $N$; for a symmetric or antisymmetric normalized wavefunction,

$$n(\mathbf r_1) = N\int d^3r_2\cdots d^3r_N\, |\Psi(\mathbf r_1,\ldots,\mathbf r_N)|^2.$$

Consequently,

$$\int d^3r\,n(\mathbf r)=N.$$

The unit-normalized marginal and the $N$-normalized number density answer different questions.

It is generally incorrect to picture a many-particle wavefunction as one classical field in ordinary three-dimensional space. Entanglement appears precisely because the joint amplitude need not factor into a product of one-particle amplitudes. Entangled States and Partial Trace develop the abstract state and marginal-state structures.

Expectation Values In Position Space

The expectation value of position in one dimension is

$$\langle x\rangle =\int_{-\infty}^{\infty} x|\psi(x,t)|^2\,dx.$$

The expectation value of a function of position is

$$\langle f(x)\rangle =\int_{-\infty}^{\infty} f(x)|\psi(x,t)|^2\,dx.$$

Momentum and energy usually require operators involving derivatives, not just multiplication by a function. In position representation,

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

so

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

provided the wavefunction lies in the domain where this expression is meaningful.

Normalization alone does not guarantee that every moment exists. A state can satisfy

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

while $\langle x^2\rangle$ or $\langle p^2\rangle$ diverges. Before quoting a mean, variance, kinetic energy, or uncertainty, check the convergence of the defining integral and the relevant operator domain.

Mixed States and Position Density

A general state is described by a density operator $\hat\rho$. Its position probability density is the diagonal of its coordinate kernel:

$$\rho_{\mathrm{pos}}(x) = \langle x|\hat\rho|x\rangle.$$

For a pure state $\hat\rho=|\psi\rangle\langle\psi|$, this reduces to

$$\rho_{\mathrm{pos}}(x)=|\psi(x)|^2.$$

For an ensemble $\hat\rho=\sum_k p_k|\psi_k\rangle\langle\psi_k|$,

$$\rho_{\mathrm{pos}}(x) = \sum_k p_k|\psi_k(x)|^2.$$

No interference terms appear between the displayed ensemble labels because probabilities, not amplitudes, are being mixed. Density-operator ensemble decompositions are nonunique, so these labels must not be interpreted as a privileged underlying decomposition. This differs from a coherent superposition, where amplitudes are added before taking the modulus squared. Density Operators owns the general distinction.

Worked Example: Normalized Exponential Tail

Consider an even one-dimensional bound-state shape

$$\psi(x)=A e^{-\kappa |x|}, \qquad \kappa>0.$$

Normalize it:

$$1=\int_{-\infty}^{\infty}|A|^2e^{-2\kappa|x|}\,dx =2|A|^2\int_0^\infty e^{-2\kappa x}\,dx =\frac{|A|^2}{\kappa}.$$

Thus $|A|=\sqrt{\kappa}$. Choosing $A=\sqrt{\kappa}$ gives

$$\psi(x)=\sqrt{\kappa}\,e^{-\kappa |x|}.$$

The probability density decays as $e^{-2\kappa |x|}$, twice as fast in the exponent as the amplitude. This distinction is common in tunneling and bound-state tail estimates.

Conservation of Normalization

Normalization at one time remains normalization at later times when the Hamiltonian generates unitary evolution and the boundary conditions prevent probability flux from escaping the modeled domain. In local nonrelativistic wave mechanics,

$$\frac{\partial\rho}{\partial t} +\nabla\cdot\mathbf j=0.$$

Integrating over a region $\Omega$ gives

$$\frac{d}{dt} \int_\Omega\rho\,d^3r = -\int_{\partial\Omega} \mathbf j\cdot d\mathbf S.$$

The total probability is conserved when the boundary flux vanishes or the full space is used with sufficiently decaying states. The detailed derivation belongs to Continuity Equation and Probability Current.

If numerical propagation causes the norm to drift, possible causes include a nonunitary time-stepper, insufficient resolution, absorbing boundaries, or a non-Hermitian effective Hamiltonian. Norm loss is not automatically a software defect; its interpretation depends on the model.

What Position Density Does Not Determine

Knowing $|\psi(x)|^2$ at one time does not generally determine:

• the local or relative phase;
• the momentum distribution;
• the probability current;
• off-diagonal coherence in the position basis;
• entanglement with an unobserved subsystem.

Position density is one measurement distribution, not a complete state description. State reconstruction requires an informationally complete set of measurements or additional dynamical assumptions.

Boundary of This Page

This page owns the practical spatial-density interpretation of coordinate wavefunctions. It does not own the derivation of the Born rule, the general measurement-update rule, probability-current dynamics, or density-operator theory. Born Rule for Continuous Outcomes, Probability Current, and Density Operators are their canonical homes.

Common Mistakes

• Calling $|\psi(x)|^2$ the probability of being at exactly $x$ in a continuous problem.
• Forgetting that probability in a region is an integral of the density.
• Dropping the coordinate measure in three-dimensional or spherical-coordinate problems.
• Confusing zeros of the real part of $\psi$ with nodes of $\psi$ itself.
• Assuming that a real-looking probability density determines the whole state; phases matter.
• Forgetting the units of $\psi$ when changing dimensions or normalization conventions.
• Using bound-state normalization for plane waves on the full line.
• Comparing density values in different coordinates without transforming the measure.
• Confusing the radial density $p_r(r)$ with the volume density $|\psi(\mathbf r)|^2$.
• Interpreting a many-particle configuration-space wavefunction as a scalar field on ordinary space.
• Assuming normalization guarantees finite means, variances, or kinetic energy.
• Adding probabilities where coherent amplitudes should be added, or vice versa.
• Treating one position-density snapshot as complete information about the state.

Where This Is Used

• Coordinate Representation defines $\psi(x)=\langle x|\psi\rangle$.
• Normalization Conventions compares square, box, flux, and delta normalization.
• Expectation Values explains the abstract expectation-value formula.
• Time-Dependent Schrödinger Equation in Coordinate Space develops wave-packet evolution and verification.

Exercises

1. Let $|\Psi\rangle$ have position-dependent spin components $f(x)$ and $g(x)$ in the $|\uparrow\rangle,|\downarrow\rangle$ basis. Show that changing to $|\pm\rangle=(|\uparrow\rangle\pm|\downarrow\rangle)/\sqrt2$ leaves the unresolved position density invariant, while postselection on $+$ exposes an interference term.

Solution

The amplitudes in the new basis are

$$\psi_+(x)=\frac{f(x)+g(x)}{\sqrt2}, \qquad \psi_-(x)=\frac{f(x)-g(x)}{\sqrt2}.$$

Their unresolved sum is

$$|\psi_+|^2+|\psi_-|^2=|f|^2+|g|^2.$$

The joint density for position and the $+$ result is

$$|\psi_+|^2 = \frac12\left(|f|^2+|g|^2+2\operatorname{Re}(f^*g)\right).$$

After division by the total probability of the $+$ outcome, this gives the conditional position density. The interference appears only because the internal output is resolved and postselected.

2. In three dimensions, suppose $\psi$ is spherically symmetric: $\psi(\mathbf r)=f(r)$. Write the normalization condition.

Solution

Use $d^3r=r^2\sin\theta\,dr\,d\theta\,d\phi$. Since $f$ is independent of the angles,

$$1=\int |\psi(\mathbf r)|^2\,d^3r =4\pi\int_0^\infty |f(r)|^2r^2\,dr.$$

3. Let $\psi_1$ and $\psi_2$ be orthonormal. Find the density of

$$\psi = \frac{\psi_1+e^{i\varphi}\psi_2}{\sqrt2}$$

and explain why normalization is independent of $\varphi$ even though the local density may not be.

Solution

Expanding the modulus squared gives

$$\begin{aligned} |\psi|^2 ={}&\frac12 \left(|\psi_1|^2+|\psi_2|^2\right)\\ &+\operatorname{Re} \left(e^{i\varphi}\psi_1^*\psi_2\right). \end{aligned}$$

The last term depends on the relative phase and can alter the local density. Its integral is

$$\operatorname{Re} \left( e^{i\varphi}\langle\psi_1|\psi_2\rangle \right)=0,$$

so the total norm remains one for every $\varphi$.

4. A one-dimensional density is uniform on $0<x<L$. Transform it to the coordinate $y=x^2$.

Solution

The original density is $\rho_x(x)=1/L$. Since $x=\sqrt y$,

$$\left|\frac{dx}{dy}\right| = \frac{1}{2\sqrt y}.$$

Therefore, on $0<y<L^2$,

$$\rho_y(y) = \frac{1}{2L\sqrt y}.$$

It is normalized because

$$\int_0^{L^2} \frac{dy}{2L\sqrt y}=1.$$

The divergence near $y=0$ is integrable and reflects coordinate compression, not an infinite probability.

5. For a normalized two-particle wavefunction $\Psi(\mathbf r_1,\mathbf r_2)$, show that the first-particle marginal is normalized.

Solution

Define

$$\rho_1(\mathbf r_1) = \int d^3r_2\, |\Psi(\mathbf r_1,\mathbf r_2)|^2.$$

Then

$$\begin{aligned} \int d^3r_1\,\rho_1(\mathbf r_1) &=\int d^3r_1\int d^3r_2\, |\Psi(\mathbf r_1,\mathbf r_2)|^2\\ &=1. \end{aligned}$$

The marginal is therefore a valid one-particle position density even when $\Psi$ is entangled and cannot be factorized.

References

• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• 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.