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Heisenberg vs Schrodinger Picture

  1. Jul 21, 2010 #1
    Please tell me where my understanding of the Heisenberg and/or the Schrodinger picture falls apart.

    -Schrodinger says the state vector of a system changes with time according to a unitary operator that doesn't change with time.
    -Hesienberg says the state vector of a system doesn't change with time but the operator acting on that state vector has time dependence.
    -In the heisenberg equation of motion http://en.wikipedia.org/wiki/Heisenberg_picture , the second term (partial A with respect to t) dissapears (A doesn't change with time) when the Hamiltonian is autonomous.
    - But if the Hamiltonian is autonomous and we are in the Heisenberg picture then A(t)=A(0) and the system will not evolve with time because the operator A is the only time dependent variable.
    -However, in the schrodinger picture, the phase of a stationary state can evolve with time.
    Conclusion:
    -Therefore, the two pictures do not seem entirely equal.
     
  2. jcsd
  3. Jul 21, 2010 #2

    olgranpappy

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    Homework Helper

    No. A(t) does not equal A(0) in the Heisenberg picture in general.

    In general,
    [tex]
    A(t)=e^{i H t}A(0)e^{-i H t}
    [/tex]

    A(t) only equals A(0), for all t, if A(0) commutes with the Hamiltonian.

     
  4. Jul 22, 2010 #3

    tom.stoer

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    Science Advisor

    Forget about explicit time dependence for a moment (no time dependent force or something like that.

    You have an operator H, an evolution operatur U(t) = exp(-iHt) and state vector |state>

    Now you can calculate

    |state, t> = U(t) |state>

    and look at

    A |state, t>;

    this is the Schrödinger picture for some operator A.

    Or you can look at

    U*(t) A |state, t> = U*(t) A U(t) |state> = A(t) |state>

    with A(t) = U*(t) A U(t);

    this is the Heisenberg picture for A(t).

    It's just playing around with A.
     
  5. Jul 22, 2010 #4
    First of all, I don't really see why you draw the conclusion that A(t) = A(0).. This is certainly not true if A does not commute with the Hamiltonian while not having some explicit time dependence.

    Anyways, it's best to look at the one thing that matters in quantum mechanics: the amplitudes. Written in the Schroedinger picture they look like:

    [tex]\langle \psi(t)| A | \psi(t)\rangle[/tex]

    where the time evolution of the state is dictated by some unitary evolution operator [tex]U(t,t_0)[/tex] acting on some reference state [tex]| \psi(t_0)\rangle [/tex], so

    [tex]| \psi(t)\rangle = U(t,t_0) | \psi(t_0)\rangle[/tex]

    In the Heisenberg picture the time evolution is switched, such that it is located within the operators. This simply amounts to letting the unitary evolution operators act on the operators instead (I'm dropping the t_0 label on the reference state).

    [tex]\langle \psi(t)| A | \psi(t)\rangle = \langle \psi|U^\dag(t,t_0) AU(t,t_0) | \psi\rangle = \langle \psi|A_H(t) | \psi\rangle [/tex]

    where [tex]A_H(t) = U^\dag(t,t_0) AU(t,t_0)[/tex] is now the Heisenberg representation of the operator A. Strictly speaking you should always attach a subscript to each operator and state to label in which picture they are in.

    Now the moral of the story is that the amplitude, which is usually denoted as [tex]\langle A \rangle[/tex], is the same in both pictures. This is pretty clear from the above since I just moved around the time evolution operators. If for instance the states only have some phase factor time dependence in the Schroedinger picture (and [tex]A_H(t) = A(0)[/tex] like you mentioned), then this still doesn't matter for a quantum amplitude -- the phases just cancel.

    Now, in the following notation (which is neither the Schroeding or Heisenberg picture)

    [tex]\langle \psi|U^\dag(t,t_0) AU(t,t_0) | \psi\rangle[/tex]

    all time dependence is put explicitly into the time evolution operators U. This is, for me, the clearest way to look at both quantum mechanics and QFT. You start with some state [tex]| \psi\rangle[/tex] at some reference time t0. You let the system evolve until it hits A at time t. Then you evolve backwards in time, back to the reference state at t0 again. And this manipulation will give you the quantum amplitude [tex]\langle A(t)\rangle[/tex]. It turns out to be unpractical to work with this notation directly -- the operator U(t,t_0) is problematic -- and that is why you switch to either the Schroedinger or Heisenberg picture, depending on what type of problem you are dealing with.


    ( It will get 'worse' by the way. In the Dirac picture (also interaction picture) the time evolution is split! Both the operators and the states have their own time evolution operator -- both evolving with respect to a different part of the Hamiltonian. This comes down to writing U(t,t') as a product of two unitary operators, U_0 and U_I -- one then acts on the states, and the other on the operators. )
     
    Last edited: Jul 22, 2010
  6. Jul 22, 2010 #5
    OK I get it now, just because A doesn't change with time (partial of A with respect to t = 0) doesn't meant that A(t) doesn't change with time because the time dependence for A(t) is actually in U.
     
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