From: Maxime Bagnoud Subject: Re: F = ma in QM Date: 17 Aug 2000 16:47:50 GMT Newsgroups: sci.physics.research Summary: [missing] franz heymann wrote: > Jacques Distler wrote in message > news:8nbvth$jad$1@geraldo.cc.utexas.edu... > > > >If fact, > > >F = m a iff = m , \forall |psi>, > > >perhaps modulo some technicalities. > > > The main "technicality" is that "a" is not an operator in the > > Schroedinger picture. The correct equation (which holds in any picture) > > is > > > > = m d^2/dt^2 > > > > In the Heisenberg picture (where |psi> is time-independent), you can > > just pull the time-derivatives inside the expectation-value, thereby > > obtaining the expectation-value of the operator > > > > a = d^2x/dt^2 > > > > In the Scroedinger picture, there ain't no such operator, a. > > I thought that "for every physical observable there is a corresponding > quantum mechanical operator". Are you implying that acceleration is not a > physical observable? > > Franz Heymann Mmmhhhh....Yes. But in the Schroedinger picture (where operators are time-independant), the quantum-mechanical operator representing the classical observable a is not the constant multiplicative operator 0 (the position operator x is time-independant in the Schroedinger picture, so d^2 x/ dt^2 = 0), but rather a complicated derivation operator, given by: \hat{a} = - hbar^2 * (H^2 * x - 2 * H * x * H + x * H^2) which has a multiplicative action only on energy eigenstates in the position representation. (I let you prove as an exercise that this definition of \hat{a} gives back: = d^2/dt^2 ) To summarize things clearly: If you think of a that way and always adapt its definition to the particular representation chosen, then, yes, F=ma holds as an operator equation. BUT, this seems to be very artificial to me and certainly very misleading. You'd better think that Ehrenfest's theorem is true in all pictures and F=ma only in the Heisenberg picture. As another example, it's certainly true that p = m d/dt q in the Heisenberg picture, but only = m d/dt holds in every picture (as you know very well, \hat{p} = hbar/i d/dx in the Schroedinger picture and \hat{p} = dx/dt doesn't hold there). ===> The true content of Ehrenfest's theorem is that the Hamilton equations: dx/dt = 1/m p dp/dt = -dV/dx only holds on average in QM. Otherwise, our world won't be quantum-mechanical at all. I hope that this was enough to convince Jan and Franz and close the discussion, but if you still have question... Maxime ============================================================================== From: Jacques Distler Subject: Re: F = ma in QM Date: 15 Aug 2000 20:28:09 GMT Newsgroups: sci.physics.research In article <1efdogb.1eu71h31m8kt5aN@de-ster.demon.nl>, jjl@de-ster.demon.nl wrote: >Heisenberg or Schroedinger has nothing to do with it. >Since F = m a is an operator identity in QM, >it holds in all pictures if it holds in one. > >If fact, >F = m a iff = m , \forall |psi>, >perhaps modulo some technicalities. The main "technicality" is that "a" is not an operator in the Schroedinger picture. The correct equation (which holds in any picture) is = m d^2/dt^2 In the Heisenberg picture (where |psi> is time-independent), you can just pull the time-derivatives inside the expectation-value, thereby obtaining the expectation-value of the operator a = d^2x/dt^2 In the Scroedinger picture, there ain't no such operator, a. JD -- PGP public key: http://golem.ph.utexas.edu/~distler/distler.asc ============================================================================== From: baez@galaxy.ucr.edu (John Baez) Subject: Re: F = ma in QM Date: 12 Aug 2000 21:58:50 GMT Newsgroups: sci.physics.research In article <8n48j3$goe$1@geraldo.cc.utexas.edu>, Jacques Distler wrote: >In article <8n1tpi$8bl$1@Urvile.MSUS.EDU>, baez@galaxy.ucr.edu (John >Baez) wrote: >>Newton's law is just as true in QM as it is in CM! >Hmmmm. You were still too easily impressed :-). Because the next thing >that I emphasize to my students is that > > != F() (*) > >so that, even though > > = m d^2/dt^2 > >this does not reduce to classical mechanics (even for ), except in >those cases where we can ignore the difference in (*). Just to rescue my honor: I did not say that Newton's law meant the same thing in quantum mechanics as it does in classical mechanics - nor did I ever commit the sin of thinking that taking expectation values commutes with a nonlinear function F. I just meant that F = ma is a perfectly valid equation among observables in quantum mechanics: the obvious quantization of the corresponding classical equation. THAT's what impressed me. Because some people say silly things like "the concept of `force' no longer exists in quantum mechanics".