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Integral involving exponentials and the Time ordering operator

  1. Feb 25, 2010 #1
    How can I show that:

    [tex]\sum^{\infty}_{n=0} \frac{(-i)^n}{n!} \int^{t}_{t'} dt_{1} dt_{2}...dt_{n} T(H_I(t_1)...H_I(t_n)) \equiv Texp[-i\int^{t}_{t'} dsH_I(s)][/tex]

    My concern is that the integral

    [tex]\int^{t}_{t'} dt_{1} dt_{2}...dt_{n} T(H_I(t_1)...H_I(t_n))[/tex]

    is not raised to the power of 'n' so we can't really manipulate the expression on the LHS to fit:

    [tex]exp(x)=\sum^{n=0}_{\infty}\frac{x^n}{n!}[/tex].

    Also, why can you write a product of Hamiltonians as a single one - physically that makes no sense (to me anyway, a big qualification there!) ie why is HI(t1)HI(t2)...HI(tn)=HI(s)?

    Thanks in advance...
     
  2. jcsd
  3. Feb 25, 2010 #2

    CompuChip

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    All your questions relate to the power of the integral.

    Can you write out what
    [tex]\left( \int f(x) \, dx \right)^3 [/tex]
    is (for some function f)?
     
  4. Feb 25, 2010 #3
    I wish I could say:

    [tex]\left( \int f(x) \, dx \right)^3 =\int dx f(x) \int dy f(y) \int dz f(z) [/tex]

    as this would clearly solve the problem. But the variables in the integral in the OP aren't dummy variables - I would guess each hamiltonian is a different function of its respective variable, so I can't see how one can do what you've suggested?
     
  5. Feb 25, 2010 #4

    CompuChip

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    Err, why not? I see an integral over t1 up to tn?

    In principle (if H(t) was an ordinary function) you could immediately write

    [tex]
    \sum^{\infty}_{n=0} \frac{(-i)^n}{n!} \int^{t}_{t'} dt_{1} dt_{2}...dt_{n} H_I(t_1)...H_I(t_n) = \sum_{n = 0}^\infty \frac{(-i)^n}{n!} \left( \int^t_{t'} ds H(s) \right)^n
    [/tex]
    and define the exponential by the formal sum on the right hand side.

    Exactly the same is happening here. The only thing that you need to take care of is the non-commutativity of H with itself at different time instances.

    Anyway, it's not like (what you seem to imply) there are different definitions for H. For example, H may be defined as [itex]\hat H = \hat p^2 / 2m + m \omega^2 \hat x^2[/itex] such that H(t)H(t') is not H(t')H(t). But it is not true that H(t) = f(t) and H(s) = g(s) for different functions f and g.
     
  6. Feb 25, 2010 #5
    Thanks CompuChip:) - that's a convincing explanation. I guess the the time ordering operator takes care of the non-commutativity of H with itself at different time instances.
     
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