I want to show an equality:

We know from Ehrenfest's theorem that $$ \frac{d \langle x \rangle(t)}{dt}= \left\langle \frac{\partial H}{\partial p} \right\rangle $$ I'd like to derive the following statement: $$\frac{ d\langle x \rangle(t)}{dt} = \frac{\partial \langle H \rangle}{\partial\langle p \rangle} $$

Because this would be in my Opinion the "total" classical limit: We observe a quantity energy and a quantity momentum, and that the derivation of energy with respect to momentum gives as the change of position with respect to time.

Is it possible to derive the statement that I want to derive? If not exactly, what are the approximations that have to be made? I could show it under the condition that the hamilton operator and the momentum operator share eigenstates, but I'd like to know how mutch it is possible in general.

Edit: By $$ \frac{\partial \langle H \rangle}{\partial\langle p \rangle} $$ I mean that I look at the change of $\langle H \rangle$, when I change the state $\Psi$ in a way that changes the mean value of momentum. I could do that by constructing a function $f(p)$ that maps the number $p$ to a state $\Psi_p$ in a way that $\langle \Psi_p | \hat{p} | \Psi_p \rangle = p$ suffices. One example could be a gaussian wave package with a mean momentum p. Then I can take the derivative $ \frac{d}{dp} \langle \Psi_p | \hat{H} | \Psi_p \rangle $.

There is definitely not a unique way to define such a function $\Psi_p$, but if one can show the the equality I am looking for for one such function $\Psi_p$, without making use of the exact form that $\Psi_p$ takes, wouldn't the quantity $ \frac{\partial \langle H \rangle}{\partial\langle p \rangle} = \frac{d}{dp} \langle \Psi_p | \hat{H} | \Psi_p \rangle $ still be well defined?

Second edit: The origin of my question is the semiclassical model of electron dynamics, as it is explained in Chapter 12 of solid state physics by Ashcroft. Here the electron energy $\epsilon_k$ with $\hat{H} \Psi_k = \epsilon _k \Psi_k$ is effectively interpreted as hamiltonian, and wavepackets of electrons with mean crystal momentum $k$ are interpreted as classical particles.


closed as unclear what you're asking by ACuriousMind, user36790, John Rennie, honeste_vivere, Gert Jul 5 '16 at 2:42

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    $\begingroup$ Your statement is non-sensical - $\langle H \rangle$ is not a function of $\langle p \rangle$ in general, so how are you taking that derivative? $\endgroup$ – ACuriousMind Jul 4 '16 at 14:41
  • $\begingroup$ I can define a function that maps $p$ to a state $\Psi_{p}$, whose mean value then is p (I construct the function in a way that $\langle H \rangle = p $ holds. There are many functions that can do that. Then I can derive < H >, which is a function of p, with respect to p. $\endgroup$ – Quantumwhisp Jul 4 '16 at 14:45
  • $\begingroup$ Please do define this operation explicitly, because currently I don't understand at all how you want to do it or why this should be a physically meaningful operation. $\endgroup$ – ACuriousMind Jul 4 '16 at 14:47
  • $\begingroup$ I don't understand why it is unclear what I'm asking for. It wrote down an equality and I want to know if this equality holds, or what approximations have to made that it holds. I edited the post to explain every term that might be unclear. $\endgroup$ – Quantumwhisp Jul 5 '16 at 9:57
  • $\begingroup$ If there are multiple ways to deform $|\Psi⟩$ into $|\Psi_p⟩$ that do the same increase in $⟨\Psi_p|\hat p|\Psi_p⟩$ but different increases in $⟨\Psi_p|\hat H|\Psi_p⟩$ (which is pretty much correct), then that would mean that $\partial ⟨H⟩/\partial ⟨p⟩$ cannot actually be defined, no? $\endgroup$ – Emilio Pisanty Jul 5 '16 at 18:17

First of all, you can't expect to recover general classical mechanics by simply making averages in quantum mechanics. Apart from very special cases, you can recover it only in the limit $\hslash\to 0$.

In such limit, something similar of what you expect can be proved. In particular, it holds when considering (squeezed) coherent states $C_{\hslash}(q,\xi)$ centred on $(q,\xi)\in\mathbb{R}^{2d}$ (with a suitable $\hslash$-dependence). First of all we suppose that $C_{\hslash}(q,\xi)$ is in the form domain of the Hamiltonian. Then $$\lim_{\hslash\to 0}\langle H_{\hslash}\rangle_{C_{\hslash}(q,\xi)}=: h(q,\xi)$$ defines a function of the phase space. Such function has many nice properties of the classical energy, e.g. $$h(t,q,\xi):=\lim_{\hslash\to 0}\langle H_{\hslash}\rangle_{e^{-i\frac{t}{\hslash}H_{\hslash}}C_{\hslash}(q,\xi)}= h(q,\xi)\; ;$$ i.e. it is an integral of motion.

On the other hand: $$\lim_{\hslash\to 0}\langle x_{\hslash}\rangle_{e^{-i\frac{t}{\hslash}H_{\hslash}}C_{\hslash}(q,\xi)}=: q(t)\; ,\\ \lim_{\hslash\to 0}\langle p_{\hslash}\rangle_{e^{-i\frac{t}{\hslash}H_{\hslash}}C_{\hslash}(q,\xi)}=: \xi(t)\; ;$$ with $q(0)=q$ and $\xi(0)=\xi$.

Finally, it is possible to prove that $q(t)$ and $\xi(t)$ obey the expected equations of motion: $$\left\{\begin{align}\dot{q}(t)&=\partial_{\xi}h(q(t),\xi(t))\\\dot{p}(t)&=-\partial_{q}h(q(t),\xi(t))\end{align}\right .\; .$$

  • $\begingroup$ I just wanted to underline that an average means many-many measurements (divided by the number of measurements), i.e., the CM is an inclusive picture. When you look at a moving classical body, you actually measure its positions with help of photons and add the separate (or one-photon) points together. Doing so, we associate a finite-sized body with only one point. When the body becomes small, the spread of different observed points becomes essential (Heisenberg uncertainty). $\endgroup$ – Vladimir Kalitvianski Jul 4 '16 at 18:38
  • $\begingroup$ Where can I find out more about what you describe? Has this some name? What do you mean by $C_h$ centered on $(q, \xi)$? $\endgroup$ – Quantumwhisp Jul 4 '16 at 21:20
  • $\begingroup$ What I describe is usually called classical limit ;-) In the simple formulation with coherent states, it can be found in this paper by Klaus Hepp. A more thorough (and maybe complicated) description using semiclassical analysis is given by Pierre-Louis Lions and Thierry Paul. The squeezed coherent state is $C_{\hslash}(q,\xi)=e^{\frac{i}{\sqrt{\hslash}}(\xi x - q p)}\Omega$, where $\Omega$ is the lowest energy eigenvector of the harmonic oscillator. $\endgroup$ – yuggib Jul 4 '16 at 21:39
  • $\begingroup$ With this setting, you have to take $x_{\hslash}=\sqrt{\hslash}x$ and $p_{\hslash}=\sqrt{\hslash}p$ (where $x$ and $p$ are the position and momentum operators, in their $\hslash$-independent version) $\endgroup$ – yuggib Jul 4 '16 at 21:39
  • $\begingroup$ although this is just written for certain cases (and I still have to work through that), basically that's what I want to know about. Thank you :) $\endgroup$ – Quantumwhisp Jul 6 '16 at 16:29

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