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Base states with hamiltonian matrix equations (Feynman Lectures Vol.3 chap. 9-10)

  1. Aug 9, 2012 #1
    It is better for you to have studied "Feynman lectures on Physics Vol.3", because I cannot distinguish whether the words or expressions are what Feynman uses only or not and in order to summarize my questions here, I have to just quote the contents of the book.

    However, one thing I notice is that "base state" that Feynman explains seems to be "basic orthonomal state vector"...

    With a pair of hamiltonian matrix equation

    [itex]i\hbar \frac{d{C}_{1}}{dt} = {H}_{11}{C}_{1} + {H}_{12}{C}_{2}[/itex]
    [itex]i\hbar \frac{d{C}_{2}}{dt} = {H}_{21}{C}_{1} + {H}_{22}{C}_{2}[/itex]

    where [itex]{C}_{x} = <x|\psi>[/itex] , [itex]\psi =[/itex] arbitrary state, the book set the states 1 and 2 as "base states". There are only two base states for some particle. Base states have a condition - [itex]<i|j> = {\delta}_{ij}[/itex].

    I think the "kronecker delta" means that once the particle is in the state of j, we will not be able to find the state i, so if we suppose all the components of hamiltonian are constant, we can say [itex]{H}_{12}[/itex]and[itex]{H}_{21}[/itex] should be zero. ..............(1)

    However the book says that states 1 and 2 are base states and [itex]{H}_{12}[/itex]and[itex]{H}_{21}[/itex] can be nonzero at the same time (if you have the book, refer equ. (9.2) and (9.3) and page 9-3.). There can be probability to transform from state 1 to state 2 and vice versa.....

    Then, the relationship that I think like (1) between the "Kronecker delta" and the components of hamiltonian is not correct at all??
     
  2. jcsd
  3. Aug 9, 2012 #2
    Hi,

    You seem to identify [itex]H_{ij}[/itex] and [itex]<i|j>[/itex] (i and j being base states), but these are not the same things. [itex]<i|j> = \delta_{ij}[/itex] is the amplitude to find something in state i when it is known to be in state j, either at the same time or in circumstances where states are not changing. On the other hand, [itex]H_{ij}[/itex] has different units; It is a derivative (in the calculus sense), proportional to the amplitude per unit time for something in base state i to change into base state j, under circumstances where states can vary with time (the constant of proportionality being [itex]i\hbar[/itex]).

    [itex]H_{ij}[/itex] is related to the amplitude [itex]<i|U(t + \Delta t,t)|j>[/itex] where U is the "waiting" operator discussed in section III:8-4. In the limit as [itex]\Delta t[/itex] goes to 0, [itex]<i|U(t + \Delta t,t)|j> = <i|j> = \delta_{ij}[/itex] . The precise relationship between operators U and H as expressed in Eq. (8.37) is

    [itex]U_{ij}(t + \Delta t,t) = \delta_{ij} - (i/\hbar) H_{ij}(t) \Delta t,[/itex]​

    which can be rewritten

    [itex]H_{ij}(t) = i\hbar(U_{ij}(t + \Delta t,t)-U_{ij}(t,t)) / \Delta t.[/itex]​
     
    Last edited: Aug 9, 2012
  4. Aug 9, 2012 #3
    Re: Base states with hamiltonian matrix equations

    Oh, "find something at the same time" - what I want to know exactly! I have just thought classically that a particle known to be in state A cannot be B simultaneously (improper word...) ignoring the true meaning of probability amplitudes.

    I appreciate your answer.
     
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