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Explanation of exponential operator proof

  1. Jun 13, 2013 #1
    Can someone please explain the below proof in more detail?
    Capture_zpsb4f8f1f9.jpg

    The part in particular which is confusing me is Capture2_zpsf444f5b1.jpg

    Thanks in advance!
     
  2. jcsd
  3. Jun 13, 2013 #2
    What don't you understand? That seemed pretty straightforward. Do you know about power series expansions?

    The idea is this:

    We want to show that ##e^{A+B} = e^{A}e^{B}##. If they are, then we can say that the difference, ##e^{A+B} - e^{A}e^{B} = 0##. To demonstrate this, we use a power series expansion.

    The power series for our exponential function is ##e^{A} = \frac{1}{0!}I + \frac{1}{1!}A + \frac{1}{2!}A^2 + \cdots##, implying that ##e^{A}e^{B} = \left(\frac{1}{0!}I + \frac{1}{1!}A + \frac{1}{2!}A^2 + \cdots\right)\left(\frac{1}{0!}I + \frac{1}{1!}B + \frac{1}{2!}B^2 + \cdots\right)##. Matrix multiplication is distributive over addition. In the proof, they shorten it for the benefit of saving space, so they don't show the step between the distributing and the grouping of the terms.

    Is that fairly clear for you?
     
  4. Jun 13, 2013 #3
    Oh ok! So they show the first two terms and "FOIL" it out for simplicity sake. I've been staring at this thing for 20 minutes and can't believe I didn't realize that.

    That was a great explanation, thanks!

    Could you also explain the 1/2!(AB+BA) portion in the last line?
     
    Last edited: Jun 13, 2013
  5. Jun 13, 2013 #4
    Well, the technical term is "distribute," but yes. Sometimes math is silly like that, though, so I wouldn't be too irritated that you didn't see it.

    You're most certainly welcome. :biggrin:
     
  6. Jun 13, 2013 #5

    lurflurf

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    Homework Helper

    (A+B)^2=A^2+AB+BA+B^2
    Which if A and B commute can be written
    (A+B)^2=A^2+2AB+B^2
    In general we cannot assume this so we either include the terms for each ordering ie
    BAA,ABA,AAB
    or we include one ordering and appropriate commutators
    [A,B]=AB-BA
     
  7. Jun 18, 2013 #6
  8. Jun 18, 2013 #7

    DrClaude

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    Staff: Mentor

    You need to expand ##e^{S+T}## further. For instance, a term in ##S^2 T^2## appears only when expanding ##(S+T)^4##.
     
  9. Jun 18, 2013 #8
    Won't you end up with differing coefficients even with further expansion?
    such as [tex]\frac{1}{4}S^2T^2 - \frac{1}{2}S^2T^2[/tex]
     
  10. Jun 18, 2013 #9

    DrClaude

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    Staff: Mentor

    You forgot a factor of ##1/2!## in the last term you wrote for ##e^S e^T##:
    $$
    \frac{1}{2!} S^2 \times \frac{1}{2!} T^2 = \frac{1}{4} S^2T^2
    $$
     
  11. Jun 18, 2013 #10
    Ok, I'll keep working on the proof, but in the mean time I'd like to get some instruction on as to how these exponentials are used.

    For example, if I'm given a matrix A and asked to find the exponential of A these are the steps I take:
    1) Find eigenvalues and then eigenvectors of A
    2) Form a matrix P consisting of eigenvectors of A
    3) Find a matrix B such that [tex]B=PAP^{-1}[/tex]
    4) Match B to a known form
    5) Then [tex]e^{At}=Pe^{Bt}P^{-1}[/tex]

    Is that correct? Here's a screenshot of the notes I'm forming my steps from:
    http://i4.photobucket.com/albums/y117/The0wnage/Capture_zpse86364b1.jpg
     
    Last edited: Jun 18, 2013
  12. Jun 19, 2013 #11

    DrClaude

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    Staff: Mentor

    Strictly speaking, you can only do step 2 as you wrote it if ##A## is diagonalizable. For instance, in the case where you get
    $$
    B = \left[ \begin{array}{cc} \lambda & 1 \\ 0 & \lambda \end{array}\right]
    $$
    then ##P## was not constructed from the eigenvectors of ##A##, since in that case ##A## didn't have two linearly independent eigenvectors. I imagine that in your notes you will have a description of how to construct the three possible matrix forms for ##B##.

    That is ##B=P^{-1}AP##, and see my comment above.

    These steps are actually inverted. Once you have ##B=P^{-1}AP##, using the Taylor expansion you get directly that
    $$
    e^{A} = e^{P B P^{-1}} = P e^B P^{-1}
    $$
    Then, you can solve the exponential by matching the correct result depending on the from of ##B##.
     
  13. Jun 19, 2013 #12
    Ok, so check my logic on this one:

    If you can form a matrix [itex]P[/itex] (ie: [itex]A[/itex] is an [itex]n x n[/itex] matrix and has n eigenvalues with n independent eigenvectors) [itex]B=P^{-1}AP[/itex] will form a diagonolized matrix and then [itex]e^A=P^{-1}e^BP[/itex] so you reach a solution fairly easily.

    If A is an n x n matrix and has n eigenvalues and less than n independent eigenvectors then [itex]P[/itex] won't form a diagonalized matrix and you must match [itex]A[/itex] to a known form of [itex]B[/itex]?
    I'm still hazy on what to do when you can't form a diagonalized matrix [itex]P[/itex]

    As an example, how would you solve the matrix [tex]\begin{matrix} 1 & 2 \\ 0 & -1 \end{matrix}[/tex] as I believe it only has 1 independent eigenvector.

    Sorry for the format, I don't know how to do tex code inline
     
    Last edited by a moderator: Jun 20, 2013
  14. Jun 20, 2013 #13

    DrClaude

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    Staff: Mentor

    Again, be careful that since [itex]B=P^{-1}AP[/itex], you get [itex]e^A=Pe^BP^{-1}[/itex] (note the order of ##P## and ##P^{-1}##).

    In one of your posts, your notes read
    so I guess that the answer to that is shown earlier.

    That one is actually diagonalizable. But if you change the -1 to 1, you get a single eigenvector ##( 1\ 0)^T##, and you construct ##P## using ##(0\ 1)^T## as the second vector, such that
    $$
    P = \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right)
    $$
    so obviously ##P = I## and therefore ##B = P^{-1} A P = A##, which is exactly the second form in your notes.

    Use itex instead of tex.
     
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