Ehrenfest theorem

McLaren Rulez

Hi,

In the Wikipedia derivation of this theorem, there is a step which I do not follow. This uses the Heisenberg picture to get Ehrenfest's theorem. A is an operator at t=0 and A(t) is the evolved operator.

[tex]\frac{d}{dt}A(t) = \frac{i}{\hbar}e^{iHt/\hbar}(HA-AH)e^{-iHt/\hbar} +e^{iHt/\hbar}(\frac{\partial A}{\partial t})e^{-iHt/\hbar} = \frac{i}{\hbar}(HA-AH)+\frac{\partial A}{\partial t}[/tex]

The last step is said to be true because [itex]e^{iHt/\hbar}[/itex] commutes with H. But I don't see how that leads to the equality. Can someone explain?

Secondly, when the Ehrenfest theorem is applied, we get classical laws. For instance, Newton's second law with the original classical variables replaced by expectation values of the quantum operator i.e. [itex]x[/itex] becomes [itex]<x>[/itex] and so on. But in general, is it true that any classical law can be written in terms of QM through Ehrenfest's theorem (provided we can define the relevant operators)? Is Ehrenfest's theorem the thing that actually connects QM with classical mechanics?

Thank you!

olgranpappy

In the last equality the first two "A"s should be written as "A(t)"s... Because they are
e^{iHt}Ae^{-iHt}, since you can pass the e^{iHt} past H.

McLaren Rulez

olgranpappy, thank you. That is something I overlooked.

Regarding the second part of my question, does anyone have any answers? Thank you

kith

I would say yes. Classically, observables are functions on the phase space. The time evolution of these functions can be described by the poisson bracket of the function and the Hamiltonian of the system. This corresponds to the Ehrenfest theorem.

In the Heisenberg picture, this relation is even more explicit. The Heisenberg equation applies not only to expectation values, but to the now time-dependent observables themself.

McLaren Rulez

Thank you kith. I didn't realize how important the Ehrenfest theorem was until now. I wonder why it is hardly emphasized in textbooks like Griffiths which only brings it up in some exercises here and there when it is so fundamental in connecting QM to CM.

dextercioby

For completeness, the Ehrenfest theorem in its general form was only proven in 2010 http://arxiv.org/abs/1003.3372

McLaren Rulez

Thank you dextercioby, but that paper appears to be slightly too technical for me. I would like to clarify another two questions if I may.

1) Would it be correct if I said that a quantum state [itex]\psi[/itex] is an analog of a classical state of a particle if and only if, for every operator [itex]O[/itex], the expectation value [itex]<\psi|O|\psi>[/itex] is the same as the classical measurment that corresponds to that observable [itex]O[/itex] on that particle? For example, if I choose the position operator, x, then [itex]\psi[/itex] is an analog of a classical state of a particle only if [itex]<\psi|x|\psi>[/itex] is the same as the measurement of the position of that classical state. Is this true?

2) I think (do correct me if I am wrong) that Ehrenfest's theorem is often useful in proving the above statement by considering an arbitrary wavefunction to compute the expectation value and thus deriving a classical law from QM. For instance, we consider an arbitrary wavefunction in the relation [itex]\frac{d}{dt}<\psi|x|\psi>[/itex] and after applying Ehrenfest's theorem, we find that this is equal to [itex]\frac{1}{m}<\psi|p|\psi>[/itex]. Since this is true for any wavefunction, we must have the classical equation [itex]\frac{d}{dt}x=p/m[/itex]. Is the above analysis correct?

As always, thank you so much for reading through and helping me out!

dextercioby

You pretty much got the right idea. Indeed, on a superficial level quantum mechanics is not too different from its classical counterpart. The Heisenberg equation of motion strongly resembles the Hamilton equations. Of course, the meaning and the conclusions one reaches are fundamentally different, because one theory is a generalization of the other.

qsa

the problem with Ehrenfest theorem is that it is limited, it cannot explain the classical bohr atom (as this has been discussed in another thread). and cannot generalize QFT to classical. although I do think it has very badly been over looked as for the connection between CM and QM. And that is basically because of the accepted wave-PARTICLE(!) daulity.

Dickfore

wat?

qsa

http://www.physicsforums.com/showthread.php?t=518300

any specific question is welcome.something like what do you mean by ....

Dickfore

Oh, sorry, I confused myself with Nernst Theorem that was propagated by Paul Ehrenfest. As far as I know, Ehrenfest's Theorem is a claim that expectation values of operators satisfy equations of classical mechanics.

kith

I don't think so. Consider a free particle. Position and Momentum are related by the Fourier transform. The expectation values don't change if we change the width of the position distribution. So very delocalized and therefore very "quantum mechanical" states may have the same expectation values like more "classical" states with small uncertainties in both momentum and position.

Btw: In classical mechanics, a measurement simply reads out the value of a observable (which is a function on the phase space). So unlike in QM, there is no need to talk about classical measurements, but only about the observables themselves.

kith

Classical accelerated point charges radiate, so the Bohr model is not classical.

QFT cannot be "generalized" to a classical field theory, since it is the more fundamental theory. And like in non-relativistic QM, the classical equations (Maxwell's) correspond to Heisenberg equations (for the field operators E and B). So there should be no problem in applying Ehrenfest's theorem here.

McLaren Rulez

So what that means is that there is more than one quantum state that corresponds to the classical state. But we only considered the position and momentum operators acting on the state. Isn't it possible that for other operators, say energy, the expectation values for various states differ and only one of those states will give us the same result as the classical energy?

kith

I think this wording is misleading. There is no classical analogon of delocalized states for example, so I would not say that such a state "corresponds" to a classical state just because its expectation value may be the same as for the state of a localized particle.

Or consider a position distribution which is zero at the expectation value <X> and has two widely separated peaks. In measurements, you will never find this particle near <X>. So this state does not at all correspond to a classical state (x, p) in phase space with the classical observables x = <X> and p = <P>.

[Maybe you are being confused by the fact that classically, the position and momentum observables are represented by the trivial functions f(x,p)=x and f(x,p)=p.]

McLaren Rulez

I see your point. Okay, so not every quantum state has a classical analog. But the reverse must be true right? Every classical state must have a quantum analog since QM is a more general theory than CM. And this quantum analog must have expectation values that agree with the classical measurement. Is this statement correct?