Ehrenfest's Theorem and the Classical Limit
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You should be comfortable with the time-dependent Schrödinger equation, the definition of expectation values, and the commutation relation \([x, p] = i\hbar\).
Why this matters
Ehrenfest’s theorem is the bridge that proves quantum mechanics does not contradict classical mechanics but rather contains it as a limiting case.
At first glance, the probabilistic nature of quantum particles and the deterministic trajectories of classical objects seem irreconcilable. However, by examining the time evolution of the expectation values of position and momentum, we can demonstrate that the "average" behavior of a quantum system obeys equations strikingly similar to Newton’s second law. This insight is crucial for understanding why macroscopic objects appear to follow classical paths and for identifying the specific conditions under which quantum effects become negligible.
The Intuitive Picture
Imagine a ball rolling in a valley. In classical physics, if you know the ball's position and velocity, you can predict its exact path. In quantum mechanics, the "ball" is described by a wavefunction that spreads out, so we cannot pinpoint its exact location. Instead, we talk about the "center of mass" of the probability distribution, which is the expectation value \(\langle x \rangle\). Ehrenfest’s theorem tells us that this center of mass moves as if it were a classical particle, provided the potential energy landscape is not too "bumpy" or complex. It is the mathematical guarantee that the quantum world respects the classical world on large scales.
Deriving the Equation of Motion for \(\langle x \rangle\)
To find how the average position changes with time, we differentiate the expectation value \(\langle x \rangle = \int \psi^* x \psi \, dx\) with respect to time. Using the time-dependent Schrödinger equation, we can show that the rate of change of the average position is directly proportional to the average momentum. This result is independent of the specific form of the potential \(V(x)\).
This equation is the quantum analogue of the classical definition of velocity, \(v = p/m\). It suggests that if we track the center of the wave packet, it moves with the average momentum, just as a classical particle would.
The Momentum Equation and the Role of the Potential
The more subtle part of the theorem concerns the momentum. Differentiating \(\langle p \rangle\) with respect to time leads to an expression involving the gradient of the potential energy. Here, the quantum nature of the system becomes apparent. The equation of motion for the average momentum is not simply \(F = -\frac{dV}{dx}\) evaluated at the average position. Instead, it involves the expectation value of the force.
For a classical particle, the force is determined by the position at that instant. In quantum mechanics, the force is the average of the force over the entire probability distribution. If the potential \(V(x)\) is linear or quadratic, the average of the derivative is the derivative of the average, and the equations look exactly like Newton’s laws. However, for more complex potentials, the spread of the wavefunction matters.
When Does the Classical Limit Hold?
The classical limit is recovered when the wave packet is narrow enough that the potential energy does not change significantly across the width of the packet. Mathematically, this requires that the higher-order derivatives of the potential are small compared to the first derivative. If we expand \(\frac{dV}{dx}\) in a Taylor series around \(\langle x \rangle\), we see that the deviation from classical behavior depends on the variance of the position, \(\langle x^2 \rangle - \langle x \rangle^2\).
Thus, Ehrenfest’s theorem does not just state that quantum mechanics looks like classical mechanics; it quantifies the error. The "quantum correction" terms vanish as the wave packet becomes more localized, which typically happens for massive particles or in potentials that are nearly harmonic over the scale of the packet.
The derivation
Ehrenfest's Theorem and the Classical Limit
Derives the equations of motion for expectation values, showing they mirror Newton's laws and pin down the correspondence with classical mechanics.
A common misconception is that Ehrenfest’s theorem implies that the wave packet itself follows a classical trajectory. In reality, the theorem only governs the evolution of the first moments (mean position and mean momentum). The wave packet can spread, distort, or split, even if its center moves classically. The theorem does not prevent quantum interference or tunnelling, which are higher-order effects not captured by the first two moments. C
Worked example
Problem. Consider a particle of mass \(m = 1.0 \times 10^{-3} \ \mathrm{kg}\) moving in a potential \(V(x) = \frac{1}{2}kx^2\) with \(k = 100 \ \mathrm{N/m}\). At \(t=0\), the expectation values are \(\langle x \rangle_0 = 0.01 \ \mathrm{m}\) and \(\langle p \rangle_0 = 0.05 \ \mathrm{kg \cdot m/s}\). Calculate \(\langle x \rangle\) and \(\langle p \rangle\) at \(t = 2.0 \ \mathrm{ms}\).
- Identify the force: \(F = -\frac{dV}{dx} = -kx\). The equations of motion are \(\frac{d\langle x \rangle}{dt} = \frac{\langle p \rangle}{m}\) and \(\frac{d\langle p \rangle}{dt} = -k\langle x \rangle\).
- Combine these to form a second-order differential equation for \(\langle x \rangle\): \(\frac{d^2\langle x \rangle}{dt^2} = -\frac{k}{m}\langle x \rangle\). This is the simple harmonic oscillator equation with \(\omega^2 = k/m\).
- Calculate \(\omega\): \(\omega = \sqrt{\frac{100}{1.0 \times 10^{-3}}} = \sqrt{10^5} \approx 316.2 \ \mathrm{rad/s}\).
- The general solution is \(\langle x \rangle(t) = A \cos(\omega t) + B \sin(\omega t)\). Using initial conditions: \(A = \langle x \rangle_0 = 0.01 \ \mathrm{m}\). Since \(\frac{d\langle x \rangle}{dt}\big|_{t=0} = \frac{\langle p \rangle_0}{m} = \frac{0.05}{10^{-3}} = 50 \ \mathrm{m/s}\), we have \(B\omega = 50\), so \(B = \frac{50}{316.2} \approx 0.158 \ \mathrm{m}\).
- At \( t = 2.0\ \mathrm{ms} \): \( \omega t = 316.23 \times 0.0020 = 0.6325\ \mathrm{rad} \), so \( \cos(\omega t) = 0.8066 \) and \( \sin(\omega t) = 0.5911 \). Then \(\langle x \rangle = 0.01(0.8066) + 0.1581(0.5911) = 0.0081 + 0.0935 = 0.1015\ \mathrm{m}\).
- Momentum: \(\langle p \rangle = m \frac{d\langle x \rangle}{dt} = m\omega(-A \sin\omega t + B \cos\omega t) = (10^{-3})(316.23)(-0.01 \times 0.5911 + 0.1581 \times 0.8066) = 0.0385\ \mathrm{kg \cdot m/s}\).
- Check with energy: \( \tfrac{1}{2}k\langle x \rangle^2 + \langle p \rangle^2/(2m) \) is \( 0.515 + 0.741 = 1.255\ \mathrm{J} \) at \( t = 2.0\ \mathrm{ms} \), and \( 0.005 + 1.250 = 1.255\ \mathrm{J} \) at \( t = 0 \): for a quadratic potential the centre of the packet moves exactly like a classical oscillator.
Units check. \( \sqrt{k/m} \) is \( \sqrt{(\mathrm{N/m})/\mathrm{kg}} = \mathrm{s^{-1}} \), so \( \omega t \) is dimensionless; \( B = v_0/\omega \) is in metres, and \( m\omega \times \) length is in \( \mathrm{kg\cdot m/s} \).
Three quick tests
- Does Ehrenfest’s theorem hold for a potential \(V(x) = x^4\)?
Answer
Yes, the theorem always holds, but the classical limit is less accurate because the force is highly non-linear. - If \(\langle x \rangle\) is constant, what must \(\langle p \rangle\) be?
Answer
\(\langle p \rangle\) must be zero, since \(\frac{d\langle x \rangle}{dt} = \frac{\langle p \rangle}{m}\). - Why does the classical limit work better for heavier particles?
Answer
Heavier particles have smaller de Broglie wavelengths, leading to narrower wave packets where the potential variation across the packet is smaller.
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.