In a previous article, I argued that the Ehrenfest theorem—despite its mathematical elegance—is fundamentally misleading about the quantum-to-classical transition. The theorem shows that quantum expectation values obey classical equations of motion, but this correspondence is both fragile (it breaks for nonlinear potentials like the Coulomb force) and superficial (it says nothing about decoherence, superconductivity, or why baseballs don’t diffract).
But if expectation values aren’t the answer, what is? How does the deterministic world of Newton emerge from the probabilistic world of Schrödinger?
There is another path, one that Newton himself might have found more satisfying. It doesn’t rely on averages or statistical ensembles. Instead, it asks: What if we demand that the wavefunction itself becomes localized, concentrated in a small region of space like a classical particle?
This seemingly simple requirement leads somewhere remarkable. When wavefunctions become sufficiently “punctual”—localized in both position and momentum, as much as Heisenberg’s uncertainty principle allows—they don’t merely have expectation values that obey classical laws. The wavefunction itself traces out a classical trajectory, carried along by the gradient of its quantum phase. The Schrödinger equation becomes the Hamilton-Jacobi equation, and quantum mechanics becomes classical mechanics, not through averaging, but through geometry.
This is the content of the WKB approximation, named after Wentzel, Kramers, and Brillouin, who developed it independently in 1926. It has been called “semiclassical” physics, but that undersells it. The WKB method reveals that classical mechanics is not a separate theory that emerges from quantum mechanics in some vague limit. Rather, it is already hidden inside the Schrödinger equation, encoded in the phase of the wavefunction, waiting to be liberated when quantum effects become small.
But this liberation comes with conditions. And when we trace those conditions to their logical conclusion, we discover something profound: the true boundary between quantum and classical is not size, or energy, or Planck’s constant. It is entanglement.
Part I: The Geometry of Quantum Phase
The Classical Action Hidden in the Wavefunction
Every wavefunction has two components: an amplitude and a phase. We typically write a wavefunction as $$\psi(x,t) = A(x,t) , e^{i\theta(x,t)}$$ where \(A\) is real and positive (the amplitude) and \(\theta\) is real (the phase). The probability density is \(|\psi|^2 = A^2\), which depends only on the amplitude. The phase seems like a mere mathematical artifact, an extra degree of freedom with no physical meaning.
But this is deceptive. The phase encodes the momentum of the particle. When we apply the momentum operator \(\hat{p} = -i\hbar \partial/\partial x\) to \(\psi\), we get $$\hat{p}\psi = -i\hbar \left(\frac{\partial A}{\partial x} + i A \frac{\partial \theta}{\partial x}\right) e^{i\theta} = \hbar \frac{\partial \theta}{\partial x} \psi - i\hbar \frac{\partial A}{\partial x} e^{i\theta}$$
For a wavefunction that is slowly varying in amplitude, the second term is negligible, and we have $$\hat{p}\psi \approx \hbar \frac{\partial \theta}{\partial x} \psi$$
The momentum is approximately \(p = \hbar \partial\theta/\partial x\). The phase gradient is the momentum density.
This suggests writing the phase in a specific form. Define a function \(S(x,t)\) by \(\theta = S/\hbar\), so that $$\psi(x,t) = A(x,t) , e^{iS(x,t)/\hbar}$$
Now the momentum becomes \(p = \partial S/\partial x\), which is precisely the relationship between momentum and action in classical mechanics. The function \(S\) is Hamilton’s principal function, the solution to the Hamilton-Jacobi equation.
Could it be that simple? Could the wavefunction be hiding a classical trajectory in its phase?
The Schrödinger Equation Rearranged
To find out, we substitute \(\psi = A e^{iS/\hbar}\) into the time-dependent Schrödinger equation $$i\hbar \frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2} + V(x)\psi$$
Computing the derivatives: $$\frac{\partial \psi}{\partial t} = \left(\frac{\partial A}{\partial t} + \frac{iA}{\hbar}\frac{\partial S}{\partial t}\right) e^{iS/\hbar}$$ $$\frac{\partial \psi}{\partial x} = \left(\frac{\partial A}{\partial x} + \frac{iA}{\hbar}\frac{\partial S}{\partial x}\right) e^{iS/\hbar}$$ $$\frac{\partial^2 \psi}{\partial x^2} = \left(\frac{\partial^2 A}{\partial x^2} + \frac{2i}{\hbar}\frac{\partial A}{\partial x}\frac{\partial S}{\partial x} + \frac{iA}{\hbar}\frac{\partial^2 S}{\partial x^2} - \frac{A}{\hbar^2}\left(\frac{\partial S}{\partial x}\right)^2\right) e^{iS/\hbar}$$
Substituting into Schrödinger’s equation and multiplying through by \(e^{-iS/\hbar}\), we get $$i\hbar \frac{\partial A}{\partial t} - A\frac{\partial S}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2 A}{\partial x^2} - \frac{i\hbar}{m}\frac{\partial A}{\partial x}\frac{\partial S}{\partial x} - \frac{i\hbar A}{2m}\frac{\partial^2 S}{\partial x^2} + \frac{A}{2m}\left(\frac{\partial S}{\partial x}\right)^2 + V(x)A$$
Now separate into real and imaginary parts. The real part gives $$\frac{\partial S}{\partial t} + \frac{1}{2m}\left(\frac{\partial S}{\partial x}\right)^2 + V(x) = \frac{\hbar^2}{2mA}\frac{\partial^2 A}{\partial x^2}$$
The imaginary part gives $$\frac{\partial A}{\partial t} + \frac{1}{m}\frac{\partial A}{\partial x}\frac{\partial S}{\partial x} + \frac{A}{2m}\frac{\partial^2 S}{\partial x^2} = 0$$
which can be rewritten as $$\frac{\partial A^2}{\partial t} + \frac{1}{m}\frac{\partial}{\partial x}\left(A^2 \frac{\partial S}{\partial x}\right) = 0$$
This is a continuity equation for the probability density \(A^2\), with velocity \(v = \frac{1}{m}\frac{\partial S}{\partial x}\).
The Classical Limit Revealed
Look carefully at the real part: $$\frac{\partial S}{\partial t} + \frac{1}{2m}\left(\frac{\partial S}{\partial x}\right)^2 + V(x) = \frac{\hbar^2}{2mA}\frac{\partial^2 A}{\partial x^2}$$
On the left side, we have precisely the Hamilton-Jacobi equation: $$\frac{\partial S}{\partial t} + \frac{1}{2m}\left(\frac{\partial S}{\partial x}\right)^2 + V(x) = 0$$
On the right side, we have a quantum correction term proportional to \(\hbar^2\).
When is this correction negligible? When $$\left|\frac{\hbar^2}{2mA}\frac{\partial^2 A}{\partial x^2}\right| \ll \left|\frac{1}{2m}\left(\frac{\partial S}{\partial x}\right)^2\right|$$
If the amplitude \(A\) is slowly varying—meaning its second derivative is small—then the quantum correction vanishes, and we recover the classical Hamilton-Jacobi equation exactly.
The condition for slow variation can be expressed as $$\left|\frac{\lambda_{\text{deBroglie}}}{A} \frac{\partial A}{\partial x}\right| \ll 1$$
where \(\lambda_{\text{deBroglie}} = 2\pi\hbar/p = 2\pi\hbar/|\partial S/\partial x|\) is the de Broglie wavelength.
Interpretation: The wavefunction amplitude must vary slowly over distances comparable to the de Broglie wavelength. When this holds, the wavefunction is “punctual” enough that it follows classical trajectories defined by \(S(x,t)\).
What the Phase Describes
Once the Hamilton-Jacobi equation holds, the classical equations of motion follow immediately. The particle position evolves according to $$\frac{dx}{dt} = \frac{\partial S/\partial x}{m} = \frac{p}{m}$$
and the momentum evolves according to $$\frac{dp}{dt} = -\frac{\partial S}{\partial t}\bigg|_x = -\frac{\partial}{\partial x}\left(\frac{\partial S}{\partial t}\right)$$
Using the Hamilton-Jacobi equation, this becomes $$\frac{dp}{dt} = -\frac{\partial V}{\partial x}$$
These are Hamilton’s equations. The phase \(S\) generates classical trajectories.
The crucial difference from Ehrenfest is this: we are not talking about expectation values. We are saying that the wavefunction itself is localized and moves along a classical trajectory. The particle isn’t “on average” at position \(\langle x \rangle\); it is actually concentrated near a specific point \(x(t)\), which evolves classically.
This is a far more satisfying picture of the classical limit.
Part II: The Conditions for Punctuality
The WKB approximation works when the wavefunction is “punctual”—localized in both position and momentum. But how localized is enough? Let us make this precise.
The Three Scales
For a particle in a potential \(V(x)\), there are three important length scales:
- The de Broglie wavelength: \(\lambda_{\text{dB}} = \frac{2\pi\hbar}{p}\), where \(p = \sqrt{2m(E-V)}\) is the classical momentum
- The wavefunction width: \(\sigma\), the spatial extent over which \(|\psi|^2\) is significant
- The potential scale: \(L\), the distance over which the potential \(V(x)\) varies appreciably
For classical behavior, we need $$\lambda_{\text{dB}} \ll \sigma \ll L$$
The first inequality, \(\lambda_{\text{dB}} \ll \sigma\), ensures that the wavefunction contains many wavelengths of oscillation. This makes the amplitude \(A\) slowly varying compared to the rapidly oscillating phase \(e^{iS/\hbar}\), which is the WKB assumption.
The second inequality, \(\sigma \ll L\), ensures that the wavefunction is localized within a region where the potential is approximately constant, so the particle “experiences” a well-defined force.
Together, these inequalities require $$\frac{2\pi\hbar}{p} \ll \sigma \ll L$$
Since \(p \sim \hbar/\sigma\) (from Heisenberg’s uncertainty principle), the first inequality gives $$\sigma \gg \hbar/p \sim \sigma$$
which is automatically satisfied. The real constraint is the second inequality, which can be written as $$pL \gg 2\pi\hbar$$
or equivalently, $$\frac{S}{\hbar} \gg 1$$
where \(S \sim pL\) is the classical action over the scale \(L\).
Conclusion: Classical behavior emerges when the action is large compared to \(\hbar\). This is the precise meaning of “quantum effects are small.”
Free Particle: The Fragility of Localization
Consider a free particle (\(V = 0\)). Suppose we prepare a Gaussian wave packet with initial width \(\sigma_0\) and momentum \(p_0\): $$\psi(x,0) \propto e^{ip_0 x/\hbar} e^{-x^2/4\sigma_0^2}$$
This evolves according to the Schrödinger equation. The center of the packet moves classically: \(x_{\text{center}}(t) = p_0 t/m\). But the width spreads: $$\sigma(t)^2 = \sigma_0^2 + \left(\frac{\hbar t}{2m\sigma_0}\right)^2$$
For short times \(t \ll t_{\text{spread}} = \frac{2m\sigma_0^2}{\hbar}\), the spreading is negligible, and the wave packet behaves like a classical particle. But eventually, \(\sigma(t)\) grows without bound. The wavefunction delocalizes, and the classical description fails.
Example: Macroscopic particle
- Baseball: \(m \sim 0.1 , \text{kg}\), \(v \sim 10 , \text{m/s}\), \(\sigma_0 \sim 1 , \text{mm}\)
- \(\lambda_{\text{dB}} = \frac{h}{mv} \sim 10^{-34} , \text{m}\) (incredibly small!)
- \(t_{\text{spread}} = \frac{2m\sigma_0^2}{\hbar} \sim 10^{26} , \text{years}\)
The baseball remains localized for longer than the age of the universe. It is effectively classical.
Example: Microscopic particle
- Electron: \(m \sim 10^{-30} , \text{kg}\), \(v \sim 10^6 , \text{m/s}\), \(\sigma_0 \sim 1 , \mu\text{m}\)
- \(\lambda_{\text{dB}} = \frac{h}{mv} \sim 10^{-9} , \text{m} = 1 , \text{nm}\)
- \(t_{\text{spread}} = \frac{2m\sigma_0^2}{\hbar} \sim 10^{-6} , \text{s}\)
The electron wave packet spreads appreciably within microseconds. Classical behavior is fleeting.
Bound States: The Correspondence Principle
For a particle bound in a potential (harmonic oscillator, hydrogen atom, etc.), the story is different. The energy eigenstates do not spread—they are stationary. But are they localized?
Harmonic Oscillator
The ground state has width \(\sigma_0 = \sqrt{\hbar/(2m\omega)}\), where \(\omega\) is the oscillator frequency. This is the zero-point motion, the minimum uncertainty allowed by quantum mechanics.
For a high-energy state with quantum number \(n \gg 1\), the energy is \(E_n \approx n\hbar\omega\), and the classical amplitude is \(A_n = \sqrt{2E_n/(m\omega^2)} = \sqrt{2n\hbar/(m\omega)}\). The wavefunction oscillates \(n\) times within the classical turning points, with an envelope width of \(\sigma_n \sim A_n/\sqrt{2} \sim \sqrt{n} , \sigma_0\).
The de Broglie wavelength is \(\lambda_{\text{dB}} \sim A_n/n \sim \sigma_0/\sqrt{n}\).
For WKB validity, we need \(\lambda_{\text{dB}} \ll \sigma_n\), which gives $$\frac{\sigma_0}{\sqrt{n}} \ll \sqrt{n} , \sigma_0$$ $$n \gg 1$$
At high quantum numbers, the harmonic oscillator behaves classically. This is Bohr’s correspondence principle, now rigorously justified.
Hydrogen Atom
The Bohr radius is \(a_0 = \hbar^2/(me^2)\). For a state with principal quantum number \(n\), the orbital radius is \(r_n \sim n^2 a_0\), and the uncertainty in radius is \(\Delta r \sim n a_0\).
The ratio \(\Delta r / r_n \sim 1/n\) becomes small for large \(n\). At \(n = 100\) (a Rydberg atom), the fractional uncertainty is only 1%, and the electron behaves almost like a classical particle in a definite orbit.
Summary: The Quantum Number as the Classical Limit
A universal pattern emerges: classical behavior requires large quantum numbers. For free particles, this means large momentum and large spatial extent (\(p\sigma \gg \hbar\)). For bound states, this means high excitation (\(n \gg 1\)).
In all cases, the criterion is \(S/\hbar \gg 1\). When the action is large, quantum mechanics becomes classical mechanics, not through statistical averaging, but through localization and the dominance of the phase over the amplitude.
Part III: Entanglement as the Absolute Boundary
We have seen that a single particle can become “punctual” and behave classically under the right conditions. But what about two particles? Can they both be punctual simultaneously?
The answer depends on whether they are entangled.
Separable States: Independent Punctuality
Consider two particles described by a joint wavefunction \(\Psi(x_1, x_2)\). If the wavefunction factorizes, $$\Psi(x_1, x_2) = \psi_1(x_1) , \psi_2(x_2)$$
then the particles are independent. Each can be in a localized state \(\psi_i(x_i)\) satisfying the WKB conditions independently. Particle 1 follows a classical trajectory in its own phase space, and particle 2 does likewise. They behave like two independent classical particles, each with its own position and momentum.
This is the classical world: objects with definite states, evolving independently according to their own equations of motion.
Entangled States: The Impossibility of Punctuality
Now suppose the wavefunction does not factorize. A simple example is $$\Psi(x_1, x_2) = \frac{1}{\sqrt{2}} \left[\psi_L(x_1) \psi_R(x_2) + \psi_R(x_1) \psi_L(x_2)\right]$$
where \(\psi_L\) and \(\psi_R\) are localized wave packets centered at positions \(x_L\) and \(x_R\) far apart.
This is an entangled state. The particles are correlated: if particle 1 is found at \(x_L\), then particle 2 is at \(x_R\), and vice versa. But before measurement, neither particle has a definite position.
To see this, compute the reduced density matrix for particle 1 by tracing out particle 2: $$\rho_1(x, x’) = \int dx_2 , \Psi(x, x_2) \Psi^*(x’, x_2)$$
For the entangled state above, $$\rho_1(x, x’) = \frac{1}{2}\left[\psi_L(x)\psi_L^(x’) + \psi_R(x)\psi_R^(x’)\right]$$
This is a mixed state, not a pure state. Particle 1 is in a statistical mixture: 50% probability of being in \(\psi_L\), 50% in \(\psi_R\). It does not have a wavefunction of its own.
Can we apply the WKB approximation to this? No. The WKB method assumes a wavefunction of the form \(\psi = A e^{iS/\hbar}\), where \(S\) defines a classical trajectory. But a mixed state cannot be written this way. There is no single phase \(S\) that generates a trajectory. Particle 1 is fundamentally delocalized, split between two possible regions.
Even if the individual wave packets \(\psi_L\) and \(\psi_R\) are highly localized and satisfy all the WKB conditions, the entangled particle cannot be described classically.
Bell’s Theorem: No Classical Model of Entanglement
The failure of punctuality for entangled particles is not just a technical issue with the WKB approximation. It reflects something deeper: entanglement has no classical analog.
In 1964, John Bell proved that no local hidden variable theory—no classical model in which particles have definite but unknown properties—can reproduce the correlations predicted by quantum mechanics for entangled states. Experiments testing Bell’s inequalities have confirmed quantum mechanics and ruled out local classicality.
What does this mean for our picture of punctual particles? It means that entanglement is the absolute boundary of classical physics. No matter how localized the joint wavefunction \(\Psi(x_1, x_2)\) might be in the six-dimensional configuration space, if it is entangled, the individual particles cannot be assigned classical trajectories.
Entanglement is irreducibly quantum. It cannot be “smoothed out” by localization or large quantum numbers. It represents correlations that have no classical counterpart.
Decoherence: How the Classical World Emerges
If entanglement prevents classicality, and if everything in the universe is ultimately quantum and potentially entangled, how does the classical world exist at all?
The answer, as we saw in the Ehrenfest article, is decoherence. But now we can understand it more precisely through the lens of punctuality and entanglement.
Consider a quantum system (say, an electron) interacting with an environment (say, air molecules). Initially, the system might be in a superposition: $$|\psi_{\text{sys}}\rangle = \alpha |A\rangle + \beta |B\rangle$$
where \(|A\rangle\) and \(|B\rangle\) are two localized states at different positions.
After interaction with the environment, the joint state becomes $$|\Psi_{\text{total}}\rangle = \alpha |A\rangle |\mathcal{E}_A\rangle + \beta |B\rangle |\mathcal{E}_B\rangle$$
where \(|\mathcal{E}_A\rangle\) and \(|\mathcal{E}_B\rangle\) are orthogonal environment states (the air molecules have scattered differently depending on where the electron was).
The system is now entangled with the environment. The reduced density matrix for the system is $$\rho_{\text{sys}} = |\alpha|^2 |A\rangle\langle A| + |\beta|^2 |B\rangle\langle B|$$
The off-diagonal terms (\(|A\rangle\langle B|\) and \(|B\rangle\langle A|\)), which represent quantum coherence and interference, have vanished. The system behaves as if it is in a classical statistical mixture: either in state \(|A\rangle\) with probability \(|\alpha|^2\), or in state \(|B\rangle\) with probability \(|\beta|^2\).
Crucially, if \(|A\rangle\) and \(|B\rangle\) are themselves punctual states satisfying the WKB conditions, then after decoherence, the system looks classical. It is in one of two localized states, each following a classical trajectory, with no quantum interference between them.
Decoherence converts quantum superpositions into classical mixtures by entangling the system with the environment. The environment “measures” the system continuously, destroying coherence and selecting pointer states—which are typically the localized, punctual states.
Why Don’t We See Entangled Baseballs?
A baseball is made of roughly \(10^{25}\) atoms. For it to behave quantum mechanically, all these atoms would need to be in a coherent quantum state—a macroscopic superposition or entanglement.
But each atom is constantly interacting with the environment: photons, air molecules, thermal vibrations. These interactions entangle the baseball with its surroundings on timescales of \(10^{-40}\) seconds or faster. Any quantum coherence is obliterated almost instantaneously.
After decoherence, the baseball is in a classical mixture of localized states. Each possible state is punctual (satisfying \(S/\hbar \gg 1\)), and WKB applies. The baseball follows a classical trajectory.
Contrast this with a pair of entangled photons in a carefully controlled laboratory experiment. The photons are isolated from the environment (no decoherence), and they remain entangled over arbitrarily long distances. They exhibit quantum correlations that violate Bell’s inequalities. No classical description is possible.
Size is not the issue; isolation is. Macroscopic quantum phenomena like superconductivity and superfluidity exist precisely because the system is cold (thermal noise is minimal) and coherent (decoherence is suppressed). The entanglement persists, and classicality fails.
Conclusion: A More Honest Picture
The path from quantum to classical is not smooth or universal. It depends on:
- Localization: The wavefunction must be concentrated in phase space, with \(\sigma\) satisfying \(\lambda_{\text{dB}} \ll \sigma \ll L\).
- Large action: The dimensionless ratio \(S/\hbar\) must be large, which typically requires large quantum numbers or macroscopic scales.
- Separability: The system must not be entangled with other systems (or the entanglement must be destroyed by decoherence).
When all three conditions hold, the WKB approximation applies, and the wavefunction follows a classical trajectory defined by its phase. This is a far more satisfying picture than the Ehrenfest theorem’s claim that “averages behave classically.” Here, the particle itself is localized and moves classically.
But the conditions are stringent. Entanglement—the defining feature of quantum mechanics, the source of its strangeness and power—cannot be tamed by localization or large quantum numbers. It is the boundary where classical intuition ends.
In the end, the classical world is an approximation, sustained by decoherence and the selection of punctual pointer states. Quantum mechanics is the fundamental theory. The deterministic trajectories of Newton are shadows cast by the phases of wavefunctions, visible only when entanglement is absent and action is large.
We live on the border between two worlds: one quantum, one classical. Understanding where that border lies—and why it appears where it does—remains one of the deepest questions in physics.