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Homework Help: Commutator algebra

  1. Oct 12, 2012 #1
    1. The problem statement, all variables and given/known data
    Analytic functions of operators (matrices) A are defined via their Taylor expansion about A=0 .Consider the function

    g(x) = exp(xA)Bexp(-xA)

    Compute : dng(x) /dxn |x=0
    for integer n

    and then show that :exp(A)Bexp(-A)= B+[A,B] +1/2 [A,[A,B]] +1/6[A,[A,[A,B]]]+ ...

    2. Relevant equations

    for the first part of the question i don't understand what x=0 means in the formalism and it is unclear how one should proceed ,becuase we are dealing with operators.

    about the second part i would like to note that :

    g(1) =g(0) +g'(0) +1/2 g''(0)+ ...
    if A=0 e^0 =1 and we start with B in the expansion formula

    3. The attempt at a solution

    About the first part i cannot imagine how i could proceed ,but about the second question i think we must use somehow a taylor expansion to resemble the monsterous expression of the RHS. But what i can't see is how one should manipulate [A,B], how could this come from a taylor expansion of the exponential ?
  2. jcsd
  3. Oct 12, 2012 #2

    D H

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    x is a scalar. Some hints:
    • Matrix multiplication is transitive, but not necessarily commutative. That the latter is the case is the point of the commutator.
    • The chain rule still works as expected with your dg(x)/dx. You will either need to show this or cite the relevant theorem from your text/lectures.
    • For any square matrix A, exp(A) commutes with An for all nonnegative integers n. Once again, you will either need to show this or cite the relevant passage.
    • What is [itex]\frac{d}{dx} \exp(xA)[/itex]?
  4. Oct 12, 2012 #3
    If chain rule is correct ,then i would have that the derivative is equal to 0 ,right?
    about the second part do i just take the taylor expansions of both exponentials and mmultiply them ?
  5. Oct 12, 2012 #4

    D H

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    No! You apparently don't understand the notation.

    This means you are to compute the value of the nth derivative of g with respect to x at x=0. It says nothing about the derivative being zero.

    Ooops. My bad. Sorry about that. I meant product rule, not chain rule.
  6. Oct 12, 2012 #5
    I understand the notation ,but im very tired becuase i dont sleep overnight and i make a mistake ,
    so we have :
    dng(x) /dxn = -2xAg(x) ,do you agree with this ?
  7. Oct 24, 2012 #6

    D H

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    Sorry for the very late response. I didn't see that you had added a post until just now.

    No, I don't agree with that. Let's start with [itex]dg/dx[/itex]. Applying the product rule (which is still valid because matrix multiplication is associative),
    \frac {d\,g(x)}{dx} &= \frac d {dx} \bigl(\exp(xA)\,B\,\exp(-xA)\bigr) \\
    &= \frac{d\exp(xA)}{dx}\,B\,\exp(-xA) +
    The key challenge is to evaluate [itex]d\exp(xA)/dx[/itex]. Using the fact that [itex]A[/itex] and [itex]\exp(xA)[/itex] commute (show this!), you can reduce the derivative to something of the form
    \frac {d\,g(x)}{dx} = \exp(xA)\,C\,\exp(-xA)
    where [itex]C[/itex] is some matrix. The question itself suggests a form for this matrix: It has to somehow involve the commutator. (Hint: It is the commutator.) See if you can get to that. With this, getting [itex]d^ng/dx^n[/itex] should be a snap.
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