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I Proof of unitarity of time evolution in Susskind's book

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  1. May 17, 2016 #1
    In "The Theoretical Minimum" of Susskind (p.98) it says that if we take any two basisvectors [itex]|i \rangle[/itex] and [itex]|j \rangle[/itex] of any orthonormal basis, and we take any linear time-development operator [itex]U[/itex], that the inner product between [itex]U(t)|i \rangle[/itex] and [itex]U(t)|j \rangle[/itex] should be 1 if [itex]|i \rangle=|j \rangle[/itex]. Why is this so? (Why is the product normalized?) I can see how it is demonstrated that the inner product of [itex]U(t)|i \rangle[/itex] and [itex]U(t)|j \rangle[/itex] is 0 if [itex]|i \rangle \neq |j \rangle[/itex] (in fact, he assumes it). The reasoning is aimed to show that time evolution is unitary.
     
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  3. May 17, 2016 #2

    A. Neumaier

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    <##Ux|Uy##>=<##x|U^*Uy##>=<##x|y##> since U is unitary.
     
  4. May 17, 2016 #3
    No, no, the reasoning is aimed to prove unitarity of U.
     
  5. May 17, 2016 #4

    A. Neumaier

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    Unitarity of ##U(t) =e^{-itH}## is trivial since ##U(t)^*=e^{-i^*tH^*}=e^{itH}=U(t)^{-1}##.
     
  6. May 17, 2016 #5
    Does this hold for any linear time-development operator [itex]U[/itex]? Are they all expressed as ##U(t) =e^{-itH}##?
     
  7. May 17, 2016 #6

    A. Neumaier

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    Yes, in units where ##\hbar=1##. Without this restriction, the Schroedinger equation
    ##i \hbar \dot\psi(t)= H\psi(t)## has the solution ##\psi(t)=e^{-itH/\hbar}\psi(0)=U(t)\psi(0)##
    where ##U(t):=e^{-itH/\hbar}##.
    If the Hamiltonian depends on time, the formula is more complicated (with Texp), and unitarity is easier to see by differentiating ##\langle \phi(t)|\psi(t)\rangle##, showing that it is constant.
     
  8. May 17, 2016 #7
    Thank you. But still I don't get the proof Susskind is trying to give, so if anyone understands his line of reasoning on this, my question still stands! (peruse my OP!) :biggrin:
     
  9. May 17, 2016 #8

    Strilanc

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    I'm guessing that you're muddling the argument somehow. Unitary matrices preserve preserve the inner product / perpendicularity (i.e. ##\langle a | b \rangle = \langle Ua | Ub \rangle##). Knowing that the linear time-development is unitary would allow you conclude length-preserving, and vice versa knowing the time-development was length-preserving would be important to showing unitarity, but from your quote it's not clear which is being proved from the other.
     
  10. May 17, 2016 #9
    The postulate of U=unitary is not made! U=unitary is to follow from the postulates he gave in his book, which I tried to describe in my OP. (That is: it is to be proven that U=unitary)
     
  11. May 17, 2016 #10
    The core of my question is:
     
  12. May 17, 2016 #11

    Strilanc

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    Then you missed some of the postulates. Does the book give a definition for "linear time development operator"?
     
  13. May 17, 2016 #12
    No, none at all...
     
  14. May 17, 2016 #13

    DrClaude

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    Since ##U(t)|i\rangle## and ##U(t)|j\rangle## are the same state, the condition imposed is simply conservation of the norm.
     
  15. May 17, 2016 #14
    Exactly! :biggrin: But why is the norm conserved?! (it doesn't follow from the proof)
     
  16. May 17, 2016 #15

    Strilanc

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    I'm not sure I believe you. Can you take a picture of the page?
     
  17. May 17, 2016 #16

    DrClaude

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    Because of the Born rule. You want probabilities to stay probabilities.
     
  18. May 17, 2016 #17

    Strilanc

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  19. May 17, 2016 #18
    Unfortunately the pictures are out of focus... :frown:
     
  20. May 17, 2016 #19
    But the form [itex]\langle i|M|j \rangle[/itex] isn't a probability, is it? :wideeyed: At this point in the book, probabilities are formulated as the product of the inproduct of the eigenvector with the state and the the conjugate of the inproduct of the eigenvector with the state.
     
  21. May 17, 2016 #20

    DrClaude

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    No, but the evolution operator returns a state.

    Take a quantum system in state ##|\psi\rangle = |i\rangle##. What is the probability that it is in state ##|j\rangle##?
    ##\langle j | \psi \rangle = \langle j | i \rangle = \delta_{ij}##.

    Now, after some time ##t##, what is the probability that the system is in state ##| \phi\rangle \equiv U(t)|j\rangle##?
    ##\langle \phi | U(t) | \psi \rangle = \langle \phi | U(t) | i \rangle = \langle j | U^\dagger(t) U(t) | i \rangle##

    Now, if ##| i \rangle = | j \rangle##, then ##U(t) | \psi \rangle = | \phi \rangle##, so setting
    ##\langle \phi | U(t) | \psi \rangle = \langle \phi | \phi \rangle = 1##
    is the same as taking ##U(t)## to be unitary.
     
  22. May 17, 2016 #21
    Susskind writes:
    And then:
    Then he mentions ##U(t)## requires to obey the conservation of distinctions, which is that ##U(t)## operating on orthogonal states remains orthogonal. He illustrates this by taking two different vectors ##|i \rangle## and ##|j \rangle## of an arbitrary orthonormal basis.

    He writes:
    Take note of the "should be 1" part!
     
    Last edited: May 17, 2016
  23. May 17, 2016 #22
    I fail to see why ##\langle \phi | \phi \rangle = 1## if ##| \phi\rangle = U(t)|j\rangle##. Why is the norm conserved? For instance: ##U(t)|j \rangle## could be twice as long as ##|j \rangle##!
     
    Last edited: May 17, 2016
  24. May 17, 2016 #23

    DrClaude

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    The choice of normalizaton is arbitrary, but we should expect that once that choice is made, the equations are consistent and that the norm doesn't change (especially since the norm is related to probabilities).
     
  25. May 17, 2016 #24
    If the norm is conserved, I can see why ##U## is unitary. It seems to me Susskind hasn't made that conclusive. I have to take your word on that it is necesary to preserve probabilities (perhaps it should be a postulate then?). Thanks for your replies! (and the others :wink: )
     
    Last edited: May 17, 2016
  26. May 17, 2016 #25

    bhobba

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    To make the math easier so the Born rule can be used it is desirable that states of unit length remain unit length. This implies it must be unitary.

    There is a deep theorem associated with this Susskind doesn't mention because its advanced (you will learn of it in Ballentine) called Wigners Theroem:
    https://en.wikipedia.org/wiki/Wigner's_theorem

    But don't worry about it for now. As I often say in QM the full story sometimes isn't told at the start - you need to build up to it.

    Thanks
    Bill
     
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