Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Heisenberg Picture

  1. Jul 20, 2016 #1
    Can anybody give a natural interpretation of operators and states in the Heisenberg Picture? When I imagine particles flying through space, it seems that the properties of the particles are changing, rather than the position property itself. Is there any way I should be thinking about these time evolving operators, intuitively?
  2. jcsd
  3. Jul 20, 2016 #2
    In Schrodinger's picture the state ket undergoes unitary time evolution dictated by Schrodinger equation. The exepctation value of a operator A is $$ \langle \psi | U A U^\dagger | \psi \rangle $$. In Heissenberg's picture the operator varies with time and is given by $$ A' = U A U^\dagger $$ while the state ket remains stationary in order to keep the expectation value the same as in Schrodinger'es picture. The key point is no matter what picture you are using the expectation value should not change.

    The nice thing about Heissenberg's picture is that the time evolution of the operators resemble that of the dynamical variables in classical mechanics a lot. $$ O_H = \frac{1}{i \hbar} [ O_H, H] $$
  4. Jul 21, 2016 #3


    User Avatar
    Science Advisor

    If you have an observable [itex]A[/itex] with possible measurement outcomes [itex]\{ a_i \}[/itex], the probability to get a particular outcome [itex]a_n[/itex] at a time [itex]t[/itex] is given by the Born rule:
    [tex]p(a_n, t) = \langle a_n | \psi_S(t) \rangle = \langle a_n(t) | \psi_H \rangle[/tex] So in the Heisenberg picture, the time dependence is in the eigenstates of the observable.

    If we take this as a starting point, it suggests the following picture: instead of a dynamical particle and a measurement apparatus which waits for the particle to arrive, the particle state is fixed and the apparatus changes until the measurement takes place. Conceptually, this seems to be similar to the difference between active and passive transformations but it should be taken with a grain of salt because the quantum system and the apparatus are not on equal footing at least in the Copenhagen interpretation.
  5. Jul 21, 2016 #4

    A. Neumaier

    User Avatar
    Science Advisor

    In classical mechanics (Poisson bracket formalism), one distinguishes between a quantity (such as ##q(t)## or ##\phi(x)##) denoting a concept (of something that can in principle be determined by experiment) and its value denoting a number associated with it. Usually this distinction is somewhat blurred when talking about or writing physics, as one tends to identify the concept and its value to keep things simple. But this distincition is essential in classical mechanics when handled in the Poisson bracket formalism (where the identification would lead to nonsense if taken seriously), Here the value map is a homomorphism from the Poisson algebra of quantities to the complex numbers. In stochastic classical mechanics, the value map becomes the expectation map that assigns to ##q(t)## the expectation value ##\langle q(t)\rangle##, etc.; it is no longer a homomorphsm but only a positive linear map, given in terms of an integral ##\langle f\rangle=\int \rho f## involving the phase space density ##\rho##.

    The Poisson formulation of classical mechanics has an immediate extension to the quantum case, with quantities turning into operators ##f##, Poisson brackets turning into commutators, and expectations given by ##\langle f\rangle=## trace ##\rho f## in terms of density operators (or, in the special case of pure states, of wave functions). Thus the operators define the concepts and the state (density operator or wave function) provides the value map.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted