# Classical formulation of mechanics applied to Quantum Mechanics

According to Ehrenfest's theorem, the expectation values of observables such as position ($$x$$), momentum ($$p$$), etc. behave not only in a deterministic way but in fact in a classical way. Therefore, would it be of any interest to study Quantum Mechanics by applying the classical formulation to the expectation values? Initially, I'd say no, except for maybe some special cases, since we would essentially be supressing the probabilistic nature of QM and therefore excluding phenomena such as quantum tunelling from our formulation. Nonetheless, some cases, such as a microssopic particle subject to a potencial $$V$$ could still be studied as I propose, simply writing Newton's second law, in this case not for the observables themselves but rather for their expectation values:

$$\frac{d}{dt}\langle \vec{p}\rangle = -\langle \vec{\nabla} V \rangle$$

And then we could find the position by the usual:

$$\frac{d}{dt}\langle \vec x\rangle = \frac{1}{m}\langle \vec p\rangle$$

Again, is it of any interest to know a particle's expectation value for position for every moment in time?

• Careful: Ehrenfest's theorem does not say that the quantum expectation values of every system obey the classical equations of motion because $\langle V(x)\rangle \neq V(\langle x\rangle)$ in general. Commented Feb 6 at 17:37
• You might, or might not, be interested in the phase-space formulation of QM. Commented Feb 6 at 17:42
• Just as a note: Quantum mechanics is entirely deterministic as anything in classical theory provided the system is left undisturbed, i.e. once we know the initial wave state, the state at any later term develops deterministically according to the unitary time development operator. Commented Feb 6 at 18:01
• Additionally, if an observable's expectation value follows its classical trajectory, the indeterminism is not in any way hampered. The fact that you are dealing with an expectation value underscores the probabilistic nature of the system. Commented Feb 6 at 18:09

Actually your premise is false. Even in terms of expected values, a quantum system is usually different from a classical system. As ACuriousMind commented, setting $$F(x)=-\nabla V(x)$$, you should expect that $$\langle F(x)\rangle \neq F(\langle x\rangle)$$. Therefore, your system of equations is not closed. You’ll need to calculate the time derivative of $$\langle F(x)\rangle$$. You can again use Ehrenfest’s theorem to get: \begin{align} \frac{d}{dt}\langle F(x)\rangle &= -i\langle [F(x),H]\rangle \end{align} The new observable is in turn independent from the rest so you’d need to repeat the process ad infinitum. This is the closure problem, where to get the equations of motion for the moments, you need to solve a hierarchical, infinite dimensional system. This is mathematically similar to turbulence and BBGKY.
There is one special case where the system is closed which is the harmonic oscillator. As soon as $$V$$ is non quadratic, you should expect to have a closure problem.
Take for example an anharmonic oscillator: $$V=\frac14x^4$$ You get (I’ve set $$m=1$$): \begin{align} \frac d{dt} \langle x \rangle &= \langle p\rangle\\ \frac d{dt} \langle p \rangle &= -\langle x^3\rangle\\ \frac d{dt} \langle x^3\rangle &= \frac32 \langle x^2p+px^2\rangle\\ \frac d{dt} \frac12\langle x^2p+px^2\rangle &= \frac12\langle xp^2+p^2x\rangle +\langle pxp\rangle -\langle x^5\rangle\\ &… \end{align} The equations will not close because the degree increases by 2 after each iteration (except for the first ones).
Therefore, even in terms of expected value, a quantum system is typically very different from a classical one. Your reasoning would work in the semiclassical limit where $$\hbar\to0$$, so that $$\langle f(x,p)\rangle = f(\langle x\rangle,\langle p\rangle)$$. You can recover for example asymptotics of the density of states. By studying the leading order in $$\hbar$$, you can even obtain tunneling and energy spectra. Check out the WKB method for more.