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Probability amplitude in Heisenberg e Schrodinger pictures

  1. Jul 24, 2010 #1
    Hi, this is my question:

    suppose that at time t' our system is in the state [tex]| \psi(t')\rangle[/tex]
    The probability for the system to be in the state [tex]| \phi\rangle[/tex] at the time t'' is the norm of
    [tex]\langle \phi| \psi(t'')\rangle[/tex]
    This in the Schrodinger picture. But how i can write the same thing in the Heisenberg picture?
     
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  3. Jul 26, 2010 #2

    dextercioby

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    The probability is actually the squared modulus of the expression you wrote. It's supposed to be a number independent of description/picture. You need to know how to connect the 2 pictures and then it's easy to get the result.
     
  4. Jul 26, 2010 #3

    vanhees71

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    In the Schrödinger picture, the state ket evolves in time via

    [tex] |\psi,t \rangle_S=\exp(-\mathrm{i} \hat{H} t) |\psi,0\rangle [/tex] .

    Here, I assume that the Hamiltonian of the system is not explicitly time dependent.

    The observables like position, momentum, etc. are time-independent by definition. Thus, for any observable operator we have

    [tex]\hat{O}_S(t)=\hat{O}(0)[/tex].

    Thus, also the eigenvectors are time-independent

    [tex]|o,t \rangle_S=|o,0 \rangle.[/tex]

    The probability to measure the eigenvalue [tex]o[/tex] is then given by

    [tex]P(t,o)=|_S\langle o,t|\psi,t\rangle_S|^2=\langle o,0|\exp(-\mathrm{i} \hat{H} t) |\psi,0 \rangle.[/tex].

    In the Heisenberg picture, the states are constant in time:

    [tex]|\psi,t \rangle_H=|\psi,0 \rangle [/tex].

    The operators representing observables move with the full Hamiltonian,

    [tex]\frac{d}{d t} \hat{O}_H(t)=\frac{1}{\mathrm{i}} [\hat{O}_H(t),\hat{H}][/tex].

    The solution is

    [tex]\hat{O}_H(t)=\exp(\mathrm{i} \hat{H} t) \hat{O}(0) \exp(-\mathrm{i} \hat{H} t)[/tex],

    and thus for the eigenvectors we have

    [tex]|o,t \rangle_H=\exp( \mathrm{i} \hat{H} t) |o,0\rangle[/tex].

    Thus again, we have

    [tex]P(o,t)=|_H\langle o,t|\psi,t \rangle_H|^2=|\langle \exp( \mathrm{i} \hat{H} t) o,0|\psi,0 \rangle|^2 = |\langle o,0|\exp(- \mathrm{i} \hat{H} t)|\psi,0 \rangle|^2[/tex],

    which is the same result as in the Schrödinger picture. The observable quantities do not depend on the picture of time evolution used, as it must be.
     
  5. Jul 27, 2010 #4
    vanhees pretty much explained it.

    You can also interpret it as follows:

    The overlap

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

    is basically what you want, modulus squared. Then

    [tex]\hat{A}_S = | \phi\rangle \langle \phi| [/tex]

    is a projection operator onto the state [itex]| \phi\rangle [/tex] in the Schroedinger picture. So switching to the Heisenberg picture gives

    [tex]\hat{A}_H(t) = e^{iHt} \hat{A}_S e^{-iHt} = e^{iHt} | \phi\rangle \langle \phi|e^{-iHt} [/tex]

    But [itex]e^{iHt} | \phi\rangle = |\phi(t)\rangle[/itex] is just the time-evolved version of the state you are projecting on:

    [tex]\hat{A}_H(t) = | \phi(t)\rangle \langle \phi(t)|[/itex]

    In effect, the time evolution switched from the state to the (projection) operator. The time dependence of the projection operator is hidden in the time-evolution of the state you are projecting on.
     
  6. Jul 27, 2010 #5
    To make the probability measurable, one of the two vectors must be some dynamical quantity's eigenvector. As eigenvectors change with time in an opposite way to state vector, the product is identical in both picture.
     
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