state and prove Ehrenfest's theorem
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Answer:
The Ehrenfest theorem, named after Paul Ehrenfest, an Austrian theoretical physicist at Leiden University, relates the time derivative of the expectation values of the position and momentum operators x and p to the expectation value of the force {\displaystyle F=-V'(x)} {\displaystyle F=-V'(x)} on a massive particle moving in a scalar potential {\displaystyle V(x)} V(x),[1]
{\displaystyle m{\frac {d}{dt}}\langle x\rangle =\langle p\rangle ,\;\;{\frac {d}{dt}}\langle p\rangle =-\left\langle V'(x)\right\rangle ~.} {\displaystyle m{\frac {d}{dt}}\langle x\rangle =\langle p\rangle ,\;\;{\frac {d}{dt}}\langle p\rangle =-\left\langle V'(x)\right\rangle ~.}
Although, at first glance, it might appear that the Ehrenfest theorem is saying that the quantum mechanical expectation values obey Newton’s classical equations of motion, this is not actually the case.[2] If the pair {\displaystyle (\langle x\rangle ,\langle p\rangle )} {\displaystyle (\langle x\rangle ,\langle p\rangle )} were to satisfy Newton's second law, the right-hand side of the second equation would have to be
{\displaystyle -V'\left(\left\langle x\right\rangle \right),} {\displaystyle -V'\left(\left\langle x\right\rangle \right),}
which is typically not the same as
{\displaystyle -\left\langle V'(x)\right\rangle .} {\displaystyle -\left\langle V'(x)\right\rangle .}
If for example, the potential {\displaystyle V(x)} V(x) is cubic, (i.e. proportional to {\displaystyle x^{3}} x^{3}), then {\displaystyle V'} V' is quadratic (proportional to {\displaystyle x^{2}} x^{2}). This means, in the case of Newton's second law, the right side would be in the form of {\displaystyle \langle x\rangle ^{2}} {\displaystyle \langle x\rangle ^{2}}, while in the Ehrenfest theorem it is in the form of {\displaystyle \langle x^{2}\rangle } \langle x^{2}\rangle . The difference between these two quantities is the square of the uncertainty in {\displaystyle x} x and is therefore nonzero.
An exception occurs in case when the classical equations of motion are linear, that is, when {\displaystyle V} V is quadratic and {\displaystyle V'} V' is linear. In that special case, {\displaystyle V'\left(\left\langle x\right\rangle \right)} {\displaystyle V'\left(\left\langle x\right\rangle \right)} and {\displaystyle \left\langle V'(x)\right\rangle } {\displaystyle \left\langle V'(x)\right\rangle } do agree. Thus, for the case of a quantum harmonic oscillator, the expected position and expected momentum do exactly follow the classical trajectories.
For general systems, if the wave function is highly concentrated around a point {\displaystyle x_{0}} x_{0}, then {\displaystyle V'\left(\left\langle x\right\rangle \right)} {\displaystyle V'\left(\left\langle x\right\rangle \right)} and {\displaystyle \left\langle V'(x)\right\rangle } {\displaystyle \left\langle V'(x)\right\rangle } will be almost the same, since both will be approximately equal to {\displaystyle V'(x_{0})} {\displaystyle V'(x_{0})}. In that case, the expected position and expected momentum will approximately follow the classical trajectories, at least for as long as the wave function remains localized in position.[3]
The Ehrenfest theorem is a special case of a more general relation between the expectation of any quantum mechanical operator and the expectation of the commutator of that operator with the Hamiltonian of the system [4][5]
{\displaystyle {\frac {d}{dt}}\langle A\rangle ={\frac {1}{i\hbar }}\langle [A,H]\rangle +\left\langle {\frac {\partial A}{\partial t}}\right\rangle ~,} {\frac {d}{dt}}\langle A\rangle ={\frac {1}{i\hbar }}\langle [A,H]\rangle +\left\langle {\frac {\partial A}{\partial t}}\right\rangle ~,
where A is some quantum mechanical operator and ⟨A⟩ is its expectation value. This more general theorem was not actually derived by Ehrenfest (it is due to Werner Heisenberg).
It is most apparent in the Heisenberg picture of quantum mechanics, where it is just the expectation value of the Heisenberg equation of motion. It provides mathematical support to the correspondence principle.
The reason is that Ehrenfest's theorem is closely related to Liouville's theorem of Hamiltonian mechanics, which involves the Poisson bracket instead of a commutator. Dirac's rule of thumb suggests that statements in quantum mechanics which contain a commutator correspond to statements in classical mechanics where the commutator is supplanted by a Poisson bracket multiplied by iħ. This makes the operator expectation values obey corresponding classical equations of motion, provided the Hamiltonian is at most quadratic in the coordinates and momenta. Otherwise, the evolution equations still may hold approximately, provided fluctuations are small.