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Time-evolution operator

  1. Mar 14, 2008 #1
    why is the time evolution operator have the form the an imaginary exponential with the hamiltonian in it. I have read some texts on quantum mechanics and so far none of them have given me an explanation for that. If some one can provide me with a proof to it or a website that has the proof I would gladly appriciate it
     
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  3. Mar 14, 2008 #2

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    The simple answer is that the solution of Schrodinger's equation (in abstract notation):

    [tex] i \frac{d}{dt} |\Psi> = H|\Psi> [/tex]

    can be written as:

    [tex] |\Psi(t)> = e^{-iHt} |\Psi(0)> [/tex]

    The operator [itex]e^{-iHt}[/itex] is a unitary operator, and is called the time evolution operator, since it takes a state at time t' to time t+t'. Differentiating the last equation (treating H as an ordinary number) shows it is a solution.

    More generally, any symmetry of the system is represented by a unitary operator on its Hilbert space (here the symmetry is time-translation invariance, ie, the outcome of an experiment is independent of when it is run). This is because we expect a symmetry to have no effect on the transition probabilities between various states, which means it should preserve the inner product on the Hilbert space, and this is precisely what a unitary operator does (you can think of unitary operators as the generalization to Hilbert spaces of orthogonal transformations (ie, rotations)). Then it is just a mathematical fact that unitary operators are of the form [itex]e^{iH}[/itex], where H is a Hermitian operator. In one (complex) dimension, this is just the statement that complex numbers of unit magnitude (1D unitary operators) are all of the form [itex]e^{it}[/itex], where t is a real number (1D Hermitian operator).
     
    Last edited: Mar 14, 2008
  4. Mar 15, 2008 #3

    Fredrik

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    If you describe a system as being in the state [itex]|\alpha\rangle[/itex], then an observer who's translated in time relative to you would describe it as being in another state. We can define a time evolution operator U(t) by saying that the state vector he uses is [itex]U(t)|\alpha\rangle[/itex], where U(t) is unitary. We would expect U(t+s)=U(t)U(s) to hold, and if it does, then U(t) must be an exponential, U(t)=exp(At). (That argument can be made rigorous). It's easy to verify that U(t) is only unitary if A is anti-Hermitian, so it's convenient to write A=-iH, where H is Hermitian. (The minus sign isn't important. It's just a matter of whether t represents how the other guy is translated in time relative to us or how we're translated relative to him).

    You should think of this as the definition of the Hamiltonian. The Schrödinger equation is just what you get when you take the time derivative of the time evolution operator, which exists simply because physics is the same to an observer who's translated in time relative to us. You can think of this as the reason why the Hamiltonian (and therefore energy) exists.
     
  5. Mar 15, 2008 #4

    Fredrik

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    Hmm, don't you need the condition U(t+s)=U(t)U(s) to be sure that it's an exponential?
     
  6. Mar 15, 2008 #5
    Its an applycation of group theory+func analysys to QM

    basically u needo this theorem:

    http://en.wikipedia.org/wiki/Stone's_theorem_on_one-parameter_unitary_groups
     
  7. Mar 17, 2008 #6

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    I'm just talking about a single unitary operator, not a group of them. In the finite dimensional case, this is because a unitary matrix is normal, and so diagonalizable, and by unitarity its eigenvalues must all be complex numbers of norm 1. So, in the basis where its diagonal, it is the exponential of i times a real diagonal matrix, and so in an arbitrary basis, it is the exponential of i times a hermitian matrix. I'm not sure if this works in the infinite dimensional case.
     
    Last edited: Mar 17, 2008
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