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Tensor Product

  1. Jul 14, 2010 #1
    Hi,

    I'm trying to understand something:
    Supposing I have two states:
    |[tex]\uparrow[/tex]> which is an eigen state of H1
    and [tex]\psi[/tex] which is the space rep. of an eigen state of H2

    Now, as far as I understand |[tex]\uparrow[/tex]> [tex]\otimes[/tex] [tex]\psi[/tex] will be an eigen state of H1 [tex]\otimes[/tex] H2 (is that true or is it the sum of Hamiltonians?)

    But supposing I knew that |[tex]\uparrow[/tex]> [tex]\otimes[/tex] [tex]\psi[/tex] was an eigen state of a known H and I knew |[tex]\uparrow[/tex]> and H1 how could I find H2?

    Thank you
     
    Last edited: Jul 14, 2010
  2. jcsd
  3. Jul 15, 2010 #2
    It is the sum of the Hamiltonians and the product of the wavefunctions so H2 = H - H1 :smile:
     
  4. Jul 15, 2010 #3
    Thank you for you reply but it doesn't quite fit with something:

    https://www.physicsforums.com/showthread.php?t=416185" is a post I posted with all the details of the question and as you can see it doesn't seem to apply (does it?)
     
    Last edited by a moderator: Apr 25, 2017
  5. Jul 16, 2010 #4
    Actually, H2 is exactly H - H1 provided it is
    suitably interpreted. However what you want from your other post
    is not H2 at all. You want an effective Hamiltonian.

    I'll try to explain.
    Given vector spaces X and Y we can form a NEW vector space
    X[itex]\otimes[/itex]Y. For any pair of operators (A,B) with
    A:X[itex]\rightarrow[/itex]X, B:Y[itex]\rightarrow[/itex]Y there
    exists a unique map A[itex]\otimes[/itex]B.

    Why is this important?
    When you're looking at your hamiltonian as acting on the product
    what you start with are the maps A and B above but you need to
    be carefull about how they lift. Specifically H1 lifts
    to H1[itex]\otimes[/itex]id2. Where id2
    is the identity map on the second factor.

    Similarly H2 lifts to id1[itex]\otimes[/itex]H2.

    So that your hamiltonian is actually
    H = H1[itex]\otimes[/itex]id2 + id1[itex]\otimes[/itex]H2.

    Now how can we use this to "factor out" a piece, or find an effective hamiltonian
    Suppose we know the eigenbasis of H2. And further suppose we've
    orthonormalized it, then we can look at the matrix elements of H, which stay
    in a fixed H2 subspace.
    Let [itex]|m>, |n>[/itex] denote vectors in the first space, and [itex]|\mu>, |\nu>[/itex]
    denote orthonormal eigenvectors of H2.

    Then
    [tex] <m \otimes \mu| H |n \otimes \nu>
    = <m \otimes \mu| (H_1 \otimes \text{id}_2) |n \otimes \nu>
    + <m \otimes \mu| (\text{id}_2 \otimes H_2) |n \otimes \nu>
    [/tex]
    [tex]
    = <m|H_1|n><\mu|\nu> + <m|n>\lambda_\nu<\mu|\nu>
    [/tex]

    where [itex]\lambda_\nu[/itex] is the appropriate eigenvalue. Now
    if we are only interested in a given state [itex]\nu[/itex] we
    can put
    [tex]
    <m \otimes \nu| H |n \otimes \nu>
    = <m|H_1|n> + <m|n>\lambda_\nu
    = <m| H_1 + \lambda_\nu \text{id}_1 |n>
    [/tex]

    This looks then like a hamiltonian only on the 1st factor.
     
  6. Jul 17, 2010 #5
    Wow,
    thank you for your detailed post now all I need to do is understand it. I think it will take me some time to properly understand it.

    Thank you very much!
     
  7. Jul 18, 2010 #6
    Your unswer seems to be very usefull since I did find in a related article something like this:
    [tex]
    <\uparrow (\phi)| \otimes <\phi| \frac{\Pi^2}{2M} |\phi '> \otimes |\uparrow (\phi ')>
    = <\phi| \frac{Q^2} {2M} |\phi '> \\ where\ Q=\Pi -const.
    [/tex]

    I think I am begining to understand what you mean evan though there are a few terms you used which are unfamiliar to me.
    1) An identity map: I think is an operater that sends a vector to the unit vector in that space?
    2) I don't quite understand what you mean by 'lifting'
    3) I gues that [itex]\lambda_\nu[/itex] is the eigenvalue we get by applying [itex]H_2|\nu>[/itex] (i.e. [itex]H_2|\nu>=\lambda_\nu|\nu>[/itex])


    And finally, at the end you said that:
    dosn't that necessarilly mean that: [itex]<\mu|\nu>=0[/itex] ?
     
    Last edited: Jul 18, 2010
  8. Jul 18, 2010 #7
    An identity map sends a vector to itself.

    With lifting he means what the operator looks like on the larger space.
     
  9. Jul 19, 2010 #8
    thank you, the identity map was a stupid mistake of me
    But after having a better look in Cohen-Tannoudji I understood that qbert made a slite typo mistake
    and it should be

    [tex] <m \otimes \mu| H |n \otimes \nu>
    = <m \otimes \mu| (H_1 \otimes \text{id}_2) |n \otimes \nu>
    + <m \otimes \mu| (\text{id}_1 \otimes H_2) |n \otimes \nu>
    [/tex]

    What I still don't understand (even though thanks to the both of you I am understanding much more) is how to get this:
    [tex]<\uparrow (\phi)| \otimes <\phi| \frac{\Pi^2}{2M} |\phi '> \otimes |\uparrow (\phi ')> = <\phi| \frac{Q^2} {2M} |\phi '> \\ where\ Q=\Pi -const.[/tex]

    and why isn't [itex]<m|n>=<\mu|\nu>=0[/itex] ?

    Thanks
     
  10. Jul 20, 2010 #9
    Can I assume that a spinor (spin state that depends on location) state and an eigenstate of momentum (which also depends on location) are in separate Hilbert spaces or does the momentum operator act on the spinor as well?
     
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