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Time evolution of operators as projectors - confusion

  1. Mar 25, 2013 #1
    I have a confusion regarding expressing operators as projectors in Schrodinger and Heisenberg pictures. Please help.

    Consider a two-state system with |1> and |2>
    We know that e.g. a raising operator can be expressed as: [itex]\hat{\sigma}_+=|2><1|[/itex]

    But here's my line of thought:

    In the Schrodinger picture:

    [itex]\hat{\sigma}_+[/itex] is supposed to remain constant in time, while the two stationary states evolve as:
    [itex]|1(t)>=e^{-\frac{iE_1 t}{\hbar}}|1(0)>[/itex] and [itex]|2(t)>=e^{-\frac{iE_2 t}{\hbar}}|2(0)>[/itex]

    But this seems to suggest that [itex]\hat{\sigma}_+(t) = e^{-\frac{i(E_2-E_1)t}{\hbar}}\hat{\sigma}_+(0)[/itex], so the operator seems to be evolving, which it shouldn't be.

    Similarly in the Heisenberg picture:

    From the Heisenberg equation of motion we expect:
    [itex]\hat{\sigma}_+(t) = e^{\frac{i(E_2-E_1)t}{\hbar}}\hat{\sigma}_+(0)[/itex]
    And |1> and |2> are expected to be constant.

    But if so, then the above equation states that:
    [itex]|2><1| = e^{\frac{i(E_2-E_1)t}{\hbar}} |2><1| [/itex]
    Which is paradoxical.

    Where am I making a mistake?
     
  2. jcsd
  3. Mar 26, 2013 #2
    If I understand your question right, the problem is: suppose ##|n \rangle## is an eigenstate of ##\hat{H}## with eigenvalue ##E_n##. Then after ##t## seconds, the new state is
    ##
    e^{-\frac{\imath}{\hbar}E_n t} | n \rangle
    ##
    ...which does not necessarily equal ##| n \rangle##. But that number in front
    ##
    e^{-\frac{\imath}{\hbar}E_n t}
    ##
    is the exponential of ##\imath## times a real number - so it's just a phase factor. Multiplying a state vector by a phase factor produces a new state vector which represents the same physical state.

    This is one of my favorite things about density-matrix QM. Instead of a vector ##| \Psi \rangle##, a state is represented by a projection operator ##| \Psi \rangle \langle \Psi |##. Projection operators are more complicated, but they have a nice side effect: the meaningless (and distracting) phase factor goes away. More precisely: any two state vectors which differ by a phase factor map to the same projection operator.
     
  4. Mar 26, 2013 #3
    Thanks for your reply.

    Yes, that's what I was asking about.

    I agree that it's just a phase factor, which we usually don't worry much about. But isn't it the case, that e.g. in the Heisenberg picture states shouldn't evolve at all? (i.e. [itex]\frac{d}{dt}|\psi>=0[/itex])? - so not even time-dependent phase factors?

    Because this rotating phase factor next to a stationary state follows from the Schrodinger equation, which doesn't hold in the Heisenberg picture?

    And also this is a simple example. I should perhaps think a better one up, but e.g.:
    [itex]\hat{\sigma}_x = |2><1| + |1><2| [/itex]
    So in this case in the Schrodinger picture this operator wouldn't just get a phase factor, but the two projectors would get counter-rotating phases, which surely can't be just ignored so easily?
     
  5. Mar 26, 2013 #4

    kith

    User Avatar
    Science Advisor

    You have to distinguish between state kets and base kets.

    Let's have a look at a simple transition amplitude, where we start in the state |α> at time 0 and ask for the overlap with state |β> at time t. Independent of the picture, this amplitude can be written as <β|U(t)|α>.

    In the Schrödinger picture, the state kets carry the time dependence, so the amplitude is <β|α(t)>. <β| is a base bra and does not carry a time dependence. Just like it is wrong to use some <β(t)| here, it is wrong to use |1(t)> and |2(t)> to express an observable in the Schrödinger picture. Also note that in the Heisenberg picture, the base kets/bras do carry a time dependence because the amplitude is <β(t)|α> there.

    This is explained in more detail in Sakurai, chapter 2.2, paragraph "Base Kets and Transition Amplitudes".
     
    Last edited: Mar 26, 2013
  6. Mar 26, 2013 #5
    I almost never use the Heisenberg picture, so I'm a little rusty with it. If I remember right, state vectors in the Heisenberg picture really are constant - they don't even pick up a meaningless phase factor. Instead, observables transform. If ##\hat{A}## is an observable and ##\hat{H}## is the system's Hamiltonian operator, the Heisenberg equation is:

    ##
    \hbar \frac{d}{dt}\hat{A} = \imath [ \hat{H}, \hat{A} ] + \partial_t \hat{A}
    ##

    For practice, suppose ##\hat{H}## and ##\hat{A}## have no explicit time-dependence: ##\partial_t \hat{H} = \partial_t \hat{A} = 0##. Then (assuming I don't screw up any minus signs, etc) ##\hat{A}## evolves like this:

    ##
    \hat{A}(t) = \exp[-\tfrac{\imath}{\hbar}\hat{H} t] \hat{A}(0) \exp[\tfrac{\imath}{\hbar}\hat{H} t]
    ##

    where ##\exp## means matrix exponential. If ##\hat{H}## is diagonal (with respect to whatever basis we decided to use), then the time evolution operator ##\exp(-\tfrac{\imath}{\hbar}\hat{H} t)## is a diagonal matrix whose diagonal elements are phase factors.

    If ##\hat{A}(0)## and ##\hat{H}## commute, then a matrix-exponential trick lets us move the time-evolution operators around so they cancel each other out:

    ##
    \hat{A}(t) = \exp(-\tfrac{\imath}{\hbar}\hat{H} t) \exp(\tfrac{\imath}{\hbar}\hat{H} t) \hat{A}(0)
    = \hat{A}(0)
    ##

    If ##\hat{A}(0)## and ##\hat{H}## don't commute, then we might have to solve a group-theory mess instead.
     
  7. Mar 26, 2013 #6
    Thanks for the replies.

    I think kith answered that!

    So, since I don't have Sakurai at hand now, could you just confirm that I understood correctly?:

    There's a distinciton between state kets, and base kets - in the Schrodinger picture state kets evolve, while base kets don't. And state kets are used to describe states, density matrices, while base kets are used in e.g. the resolution of identity, projectors, etc.

    (I have yet to think about how it is in the Heisenberg picture)

    Thanks again
     
  8. Mar 26, 2013 #7

    kith

    User Avatar
    Science Advisor

    Exactly.
     
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