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Unitary time evolution - products of consecutive evolutions

  1. Jun 19, 2014 #1
    Hi all,

    I was wondering if anyone could clarify my understanding of unitary time evolution of quantum states, in particular for products of time evolution's:

    Suppose we know state of a quantum system at [itex] t=t_{0}[/itex], given by [itex] \vert\psi\left(t_{0}\right)\rangle[/itex], then to determine its state at a later time [itex]t=t_{2}[/itex] we can first evolve the state from its configuration at [itex] t=t_{0}[/itex] to its configuration at [itex] t=t_{1}[/itex], i.e [tex]\vert\psi\left(t_{1}\right)\rangle= U\left(t_{1},t_{0}\right)\vert\psi\left(t_{0}\right)\rangle[/tex]
    and then subsequently evolve this to its final state at [itex] t=t_{2}[/itex] via a further time evolution, [tex]\vert\psi\left(t_{2}\right)\rangle= U\left(t_{2},t_{1}\right)\vert\psi \left(t_{1}\right)\rangle= U\left(t_{2},t_{1}\right)U \left(t_{1},t_{0}\right)\vert\psi\left(t_{0}\right)\rangle[/tex]
    Alternatively, we could immediately evolve the state from its initial configuration at [itex]t=t_{0}[/itex] to its final state at [itex]t=t_{2}[/itex] in the following manner, [tex]\vert\psi \left(t_{2}\right)\rangle= U\left(t_{2},t_{0}\right)\vert\psi \left(t_{0}\right)\rangle[/tex]
    Given that time is homogeneous (as far as we know), the results of these two evolution's of the system should be equivalent (as the path we take through time in evolving the state should not affect its initial and final configurations), this implies that [tex]U\left(t_{2},t_{0}\right)= U\left(t_{2},t_{1}\right)U\left(t_{1},t_{0}\right)[/tex]

    Is this a correct description?
  2. jcsd
  3. Jun 19, 2014 #2


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  4. Jun 19, 2014 #3
    Ok, thanks.

    On a related note, to prove that time evolution is a unitary operation is it enough to consider a inner product of a quantum state [itex]\vert \psi\left(t\right)\rangle[/itex] that satisfies the Schrodinger equation (with [itex]\hbar=c=1[/itex]) [tex]i \frac{\partial}{\partial t} \left(\vert \psi\left(t\right)\rangle\right) = \hat{H}\vert \psi\left(t\right)\rangle \rightarrow \frac{\partial}{\partial t} \left(\vert \psi\left(t\right)\rangle\right)= -i\hat{H}\vert \psi\left(t\right)\rangle [/tex] and then differentiate it with respect to time, such that [tex]\frac {\partial}{\partial t}\left(\langle \psi\left(t\right) \vert \psi\left(t\right)\rangle \right)=\frac {\partial}{\partial t}\left(\langle \psi\left(t\right) \vert\right) \vert \psi\left(t\right)\rangle + \langle \psi\left(t\right) \vert\frac {\partial}{\partial t}\left(\vert \psi\left(t\right)\rangle \right)= i\langle\psi \left(t\right)\vert\hat{H}^{\dagger} \vert\psi \left(t\right) \rangle - i\langle\psi \left(t\right)\vert\hat{H} \vert\psi \left(t\right) \rangle = 0 [/tex]
    as [itex]\hat{H}^{\dagger} = \hat{H}[/itex]. Hence, the inner product is preserved for all times, i.e. it is preserved under finite time evolution. As a result of this, if we consider a state at time [itex]t=t_{0}[/itex], [itex]\vert\psi \left(t_{0}\right)\rangle [/itex] and its subsequent evolved state at some later time [itex] t[/itex], [itex]\vert\psi \left(t\right)\rangle=U \left(t,t_{0}\right) \vert\psi \left(t_{0}\right)\rangle [/itex], we then have that, [tex] \langle \psi \left(t\right)\vert\psi \left(t\right)\rangle= \langle \psi \left(t_{0}\right)\vert U^{\dagger} \left(t,t_{0}\right) U \left(t,t_{0}\right) \vert\psi \left(t_{0}\right)\rangle= \langle \psi \left(t_{0}\right)\vert\psi \left(t_{0}\right)\rangle [/tex] which is true iff [itex] U^{\dagger} \left(t,t_{0}\right) U \left(t,t_{0}\right) = 1 [/itex], i.e. [itex]U\left( t,t_{0}\right) [/itex] must be a unitary operator?!

    Sorry, I know this should probably be in a new thread, but I thought I'd try and combine the two questions as they are related.
    Last edited: Jun 19, 2014
  5. Jun 19, 2014 #4
    To prove than an operation is unitary you need to show that the inner product of any two states is preserved. It's not enough to show that the norm of each individual state is preserved. There are operations that preserve the norm of all states but are not unitary.

    So you need to consider a general inner product ##\langle \phi | \psi \rangle## in your proof. But actually it should go through in pretty much the same way.
  6. Jun 19, 2014 #5
    Ah ok, so would it be fair to consider two different states, [itex] \vert\psi\left( t\right)\rangle[/itex] and [itex] \vert\phi\left( t\right)\rangle [/itex] which both satisfy the schrodinger equation, i.e. [tex] i\frac{ \partial}{\partial t}\left( \vert\psi\left( t\right)\rangle\right)= \hat{H} \vert\psi\left( t\right)\rangle, \qquad i\frac{ \partial}{\partial t}\left( \vert\phi\left( t\right)\rangle\right)= \hat{H} \vert\phi\left( t\right)\rangle [/tex] and then follow the same procedure from here as before?
  7. Jun 19, 2014 #6


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  8. Jun 19, 2014 #7
    Cool. Thanks for the help guys, much appreciated!
  9. Jun 19, 2014 #8


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    A much easier way to see the same thing perhaps is to notice that the unitary time evolution operator for the Schroedinger equation is simply (for time independent Hamiltonians) ##U(t,t_0)=e^{iH(t-t_0)}##. Since the Hamiltonian is Hermitian, the the time evolution operator is unitary.
  10. Jun 19, 2014 #9
    Yes, you're right. I understood it in that case, but was trying to go for a more general approach, including time-dependent Hamiltonians.
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