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Matrix exponential in QM

  1. Apr 7, 2013 #1
    In class, my teacher was motivating Schrodinger's Equation, but there is one step that I do not understand or even have intuition for. I'll give the argument leading up to the step I do not understand for context.
    Let [itex] \hat{U}(t) [/itex] be the operator that gives the wave function after time t, given the initial wave function. [itex] \langle\Psi(t) |=\hat{U}(t) |\Psi(0) \rangle [/itex]

    Then [itex] \langle \Psi(t) | \Psi(t) \rangle = \langle \Psi(0) | \hat{U}^\dagger(t) \hat{U}(t)| \Psi(t)\rangle =1 [/itex]

    This implies that [itex] \hat{U} [/itex] is unitary and we can write any unitary operator in the form [itex] e^{i\hat{A}} [/itex], where [itex] \hat{A} [/itex] is Hermitian. Now this is the step I do not understand: In this case, [tex] \hat{A} = \frac{-t}{\hbar} \hat{H} [/tex] where [itex] \hat{H} [/itex] is the Hamiltonian. Once we got past this step, the derivation of Schrodinger's Equation was easy, but I am confused as to how we know what A is.

    Now, there are a couple math concepts used that are not completely clear, but they are at least logical. The derivative of a state is defined in the natural way and the ability to write a unitary operator as an exponential makes sense once you realize what Hermitian conjugation does to an exponential.
    But how do we know the correct operator is related to the Hamiltonian in this way? Is this an axiom or was the reasoning not presented because it is too complicated? Part of the reason why this is confusing is that the only other time we have seen [itex] \hbar [/itex] used was in the commutator relation [itex] [\hat{x},\hat{p}]=i\hbar [/itex], which does not seem to be very relevant to this problem.

    Edit: I said matrix exponential in the title because I was thinking of the Hamiltonian as a diagonal matrix with its energy eigenvalues on the diagonal, but I probably should have said operator.
     
    Last edited: Apr 7, 2013
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  3. Apr 7, 2013 #2
    I think you should view this as the definition of the Hamiltonian H. The Hamiltonian is, by definition, the operator that generates time evolution of the system.

    Also, you are missing a t somewhere; we want ##\hat{U}(t) = e^{-i \hat{H} t / \hbar}##.

    The ##\hbar## appears basically for dimensional reasons: the exponent has to be dimensionless. In QM, energy is the same thing as frequency and momentum is the same thing as wavenumber. ##\hbar## is just the conversion factor between these pairs, with units of J/Hz = (energy)/(frequency) = (kg m/s)/(1/m) = (momentum)/(wavenumber).
     
  4. Apr 7, 2013 #3
    Thanks, I fixed that typo. Hmm, that is not how we defined the Hamiltonian in class. We defined [itex] \hat{H} [/itex] as an operator whose eigenstates are all allowable energy states and whose eignevalues are the values of those states. We made no reference to time evolution when defining the Hamiltonian. I guess that my other confusion is just not really understanding what [itex] \hbar [/itex] is. This was just our second lecture in QM and we have not used it before. Thanks for your help!
     
  5. Apr 7, 2013 #4

    atyy

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    It is a definition that the Hamiltonian governs time-evolution, so it cannot be derived.

    Another way of saying this us that the Schroedinger's equation is postulated, not derived.
     
  6. Apr 7, 2013 #5
    OK, I'll take this for granted. Hopefully, it will become clear to me next class.
     
  7. Apr 7, 2013 #6

    vanhees71

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    One should also state that the above is valid for the socalled Schrödinger picture of time evolution, where the state vector obeys the equation
    [tex]\mathrm{i} \hbar \partial_t |\psi,t \rangle=\hat{H} |\psi,t \rangle.[/tex]
    If [itex]\hat{H}[/itex] itself is not time dependent, which is the case for most problem in introductory quantum mechanics, then this equation is solved by
    [tex]|\psi,t \rangle = \hat{U}(t) |\psi,t=0 \rangle,[/tex]
    where
    [tex]\hat{U}(t)=\exp \left (-\frac{\mathrm{i}}{\hbar} t \hat{H} \right ).[/tex]
    It is very important that this time-evolution operator acts from the left on the ket and not on the bra vector as indicated in one of the previous postings.
     
  8. Apr 7, 2013 #7
    Thanks for your response. You are right that I made the mistake of using a bra for a ket (now fixed). In the case that the Hamiltonian is dependent on time, the teacher said that [tex] \hat{U}(t) = e^{\frac{-i \int^t_0 \hat{H}(t)\,dt}{\hbar}} [/tex]
    but that we would never need to use it. I am aware that what you posted is the solution to Schrodinger's equation, but we did it the other way around in class. We used this relation to derive Schrodinger's equation (just differentiate |ψ>=U|ψ(0)> wrt t and move [itex] \hbar [/itex] and [itex] i [/itex] to the other side.
     
    Last edited: Apr 7, 2013
  9. Apr 7, 2013 #8

    vanhees71

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    The more general case is more complicated than the formula your teacher gave suggests. This is, because the Hamiltonian need not commute at different times. Thus you have to take the socalled time-ordered integral. For a complete treatment, see my qft script which contains a brief summary about non-relativistic quantum mechanics in the first chapter:

    http://fias.uni-frankfurt.de/~hees/publ/tft.pdf
     
  10. Apr 7, 2013 #9
    I looked up time-ordered integral :bugeye: I think that is beyond the scope of the course (at least what I can tell from the syllabus ). Thanks everyone for your help, I appreciate it!
     
  11. Apr 7, 2013 #10

    atyy

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    The German word for "boost" is "boost"?
     
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