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PU-202 / L04 · lecture

Probability Current and the Continuity Equation

~7 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)

BEFORE YOU START

You should be comfortable with the time-dependent Schrödinger equation, complex conjugation of wavefunctions, and basic partial differentiation. Familiarity with the Born rule and the concept of probability density is assumed.

Why this matters

The continuity equation is the quantum mechanical guarantee that probability is never created or destroyed.

In classical fluid dynamics, the continuity equation expresses the conservation of mass: if fluid flows out of a region, the mass inside must decrease. In quantum mechanics, the "fluid" is probability. Since the total probability of finding a particle somewhere in space must always remain unity, the local probability density must obey a strict conservation law. This lecture establishes that law, defines the probability current, and demonstrates how it ensures that the normalization of the wavefunction is preserved over time.

The Idea in Plain Language

Imagine a region of space defined by a boundary. If the probability of finding a particle inside that region changes over time, it is not because the particle is being created or annihilated, but because probability is flowing across the boundary. The rate at which probability leaves the region is determined by the "flow" of the wavefunction at the boundary. This flow is described by a quantity called the probability current. The continuity equation simply states that the change in the amount of probability inside a volume is equal to the net flow of probability out of that volume. It is the local expression of the global conservation of probability.

Defining the Probability Current

To derive the continuity equation, we start with the time-dependent Schrödinger equation for a particle of mass \( m \) in a real potential \( V(x) \):

\[ i\hbar \frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2} + V(x)\psi \]

We also write the complex conjugate equation for \( \psi^* \):

\[ -i\hbar \frac{\partial \psi^*}{\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2 \psi^*}{\partial x^2} + V(x)\psi^* \]

The probability density is defined as \( \rho = |\psi|^2 = \psi^* \psi \). To find how this density changes in time, we calculate \( \frac{\partial \rho}{\partial t} \) using the product rule:

\[ \frac{\partial \rho}{\partial t} = \frac{\partial}{\partial t}(\psi^* \psi) = \psi^* \frac{\partial \psi}{\partial t} + \psi \frac{\partial \psi^*}{\partial t} \]

Substituting the expressions for \( \frac{\partial \psi}{\partial t} \) and \( \frac{\partial \psi^*}{\partial t} \) from the Schrödinger equations, the potential terms \( V(x) \) cancel out because \( V(x) \) is real. This leaves only the kinetic energy terms. After rearranging, we obtain:

\[ \frac{\partial \rho}{\partial t} = \frac{\hbar}{2mi} \left( \psi^* \frac{\partial^2 \psi}{\partial x^2} - \psi \frac{\partial^2 \psi^*}{\partial x^2} \right) \]

Notice that the right-hand side can be written as the spatial derivative of a quantity. By recognizing that \( \frac{\partial}{\partial x} \left( \psi^* \frac{\partial \psi}{\partial x} - \psi \frac{\partial \psi^*}{\partial x} \right) \) expands to the terms above, we identify the probability current \( j \):

\[ j = \frac{\hbar}{2mi} \left( \psi^* \frac{\partial \psi}{\partial x} - \psi \frac{\partial \psi^*}{\partial x} \right) \]

Using the identity \( \frac{1}{i} = -i \), this is often written more compactly as the imaginary part of the product of the conjugate wavefunction and its spatial derivative:

\[ j = \frac{\hbar}{m} \mathrm{Im} \left( \psi^* \frac{\partial \psi}{\partial x} \right) \]

With this definition, the time derivative of the density becomes \( \frac{\partial \rho}{\partial t} = -\frac{\partial j}{\partial x} \).

The Continuity Equation

Combining the results from the previous section, we arrive at the continuity equation:

\[ \frac{\partial \rho}{\partial t} + \frac{\partial j}{\partial x} = 0 \]

This equation has the same mathematical form as the conservation of charge in electromagnetism or the conservation of mass in fluid dynamics. It states that the local rate of change of probability density is balanced by the divergence of the probability current. If the current is flowing into a region (negative divergence), the probability density increases. If the current is flowing out (positive divergence), the probability density decreases.

To see the global conservation of probability, we integrate the continuity equation over all space from \( -\infty \) to \( \infty \):

\[ \int_{-\infty}^{\infty} \frac{\partial \rho}{\partial t} \, dx + \int_{-\infty}^{\infty} \frac{\partial j}{\partial x} \, dx = 0 \]

The first term is the time derivative of the total probability \( N(t) = \int_{-\infty}^{\infty} \rho \, dx \). The second term is the difference in the current at the boundaries: \( [j(\infty, t) - j(-\infty, t)] \). For any physically acceptable wavefunction, the wavefunction and its derivatives vanish at infinity, so \( j(\pm \infty, t) = 0 \). Therefore:

\[ \frac{dN}{dt} = 0 \]

This confirms that the total probability is conserved. If the wavefunction is normalized at \( t=0 \), it remains normalized for all time.

Physical Interpretation of the Current

The probability current \( j \) is, in one dimension, a probability per unit time (units \( \mathrm{s^{-1}} \)); in three dimensions it is a probability flux, per unit area per unit time. It tells us the rate at which probability flows past a point. For a plane wave \( \psi = A e^{i(kx - \omega t)} \), the current is:

\[ j = \frac{\hbar k}{m} |A|^2 \]

Since \( \hbar k = p \) is the momentum, this can be written as \( j = v |A|^2 \), where \( v = p/m \) is the classical velocity. This shows that for a free particle, the probability current is simply the probability density multiplied by the particle's velocity. This is a direct link to classical mechanics.

However, for a standing wave, such as \( \psi = A \sin(kx) \), the current is zero everywhere. This makes sense: a real standing wave is an equal superposition of waves moving left and right, whose currents cancel at every point. Nothing flows anywhere, which is why its probability density does not change in time.

The derivations

◆ DERIVATION EMBED · D-007 · not re-derived here

Probability Current and the Continuity Equation

Derives the probability current density and the local continuity equation guaranteeing conservation of probability under Schrödinger evolution.

∂ρ/∂t + ∂j/∂x = 0,    j = (ħ/m) Im( ψ* ∂ψ/∂x )
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◆ DERIVATION EMBED · D-138 · not re-derived here

Born Rule and Conservation of Normalisation

For a single particle described by a wavefunction ψ(x, t) obeying the time-dependent Schrödinger equation with a real potential, the quantity ρ = |ψ|2 is a probability density, and its space integral N(t) = ∫|ψ|2 dx is independent of time: dN/dt = 0. Consequently a state normalised so that N(0) = 1 remains normalised, N(t) = 1, for all t.

d/dt ∫−∞∞ |ψ(x, t)|2 dx = 0  ⇒  ∫−∞∞ |ψ|2 dx = 1  (∀ t)
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Tier 3 · watch out

The continuity equation relies on the potential \( V(x) \) being real. If the potential has an imaginary part, representing absorption or emission (e.g., in scattering with inelastic channels), the continuity equation acquires a source term, and probability is no longer conserved locally. C

Worked example

Problem. A particle of mass \( m = 9.11 \times 10^{-31} \, \mathrm{kg} \) is described by the wavefunction \( \psi(x,t) = A e^{i(kx - \omega t)} \) with \( k = 2.0 \times 10^{10} \, \mathrm{m^{-1}} \). Calculate the probability current \( j \) if the normalization constant is \( A = 1.0 \times 10^{-10} \, \mathrm{m^{-1/2}} \).

  1. Identify the formula for the probability current for a plane wave: \( j = \frac{\hbar k}{m} |A|^2 \).
  2. Substitute the given values: \( \hbar = 1.055 \times 10^{-34} \, \mathrm{J \cdot s} \), \( k = 2.0 \times 10^{10} \, \mathrm{m^{-1}} \), \( m = 9.11 \times 10^{-31} \, \mathrm{kg} \), and \( |A|^2 = (1.0 \times 10^{-10})^2 = 1.0 \times 10^{-20} \, \mathrm{m^{-1}} \).
  3. Calculate the numerator: \( \hbar k = (1.055 \times 10^{-34})(2.0 \times 10^{10}) = 2.11 \times 10^{-24} \, \mathrm{kg \cdot m/s} \) (a momentum).
  4. Divide by mass to get velocity-like term: \( \frac{2.11 \times 10^{-24}}{9.11 \times 10^{-31}} \approx 2.32 \times 10^6 \, \mathrm{m/s} \), the classical velocity \( v = \hbar k/m \).
  5. Multiply by \( |A|^2 \): \( j = (2.32 \times 10^6\ \mathrm{m/s})(1.0 \times 10^{-20}\ \mathrm{m^{-1}}) = 2.32 \times 10^{-14} \, \mathrm{s^{-1}} \).
\[ j = 2.32 \times 10^{-14} \, \mathrm{s^{-1}} \]

Units check. In one dimension \( \psi \) has units \( \mathrm{m^{-1/2}} \), so \( \rho = |\psi|^2 \) is a probability per metre. Then \( j = v\rho \) is \( (\mathrm{m/s})(\mathrm{m^{-1}}) = \mathrm{s^{-1}} \): the probability per second that passes a given point. This matches the continuity equation, where \( \partial\rho/\partial t \) and \( \partial j/\partial x \) are both in \( \mathrm{m^{-1}\,s^{-1}} \).

CHECK YOURSELF

Three quick tests

  1. What is the probability current for a real-valued wavefunction \( \psi(x) \)?
    AnswerZero, because \( \psi^* = \psi \), so \( \psi^* \psi' - \psi \psi'^* = 0 \).
  2. If \( \frac{\partial \rho}{\partial t} > 0 \) at a point, what does this imply about \( \frac{\partial j}{\partial x} \)?
    AnswerIt implies \( \frac{\partial j}{\partial x} < 0 \), meaning the current is flowing into the point (net inflow).
  3. Why does the total probability \( N(t) \) remain constant for a normalizable wavefunction?
    AnswerBecause the boundary terms \( j(\pm \infty) \) vanish, so the integral of the continuity equation over all space yields \( dN/dt = 0 \).
PROBLEMS

Problem set for PU-202 →

Each derivation in this unit ends with five graded problems with full solutions.

NOTES

Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.

Derivations cited: D-007 · D-138 · Threads: matter, waves, energy