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What happens to infinitesimal time in path integral

  1. Aug 4, 2014 #1
    Hi,

    I am studying path integral formulation from Ballentine. Till equation 4.50, I follow quiet well.
    [tex]
    G(x,t;x_0,t_0) = \lim_{N \to \infty}\int\ldots\int\left(\frac{m}{2\pi i\hbar\Delta t}\right)^{\frac{N+1}{2}}\exp{\sum_{j=0}^{N}\left(\frac{im(x_{j+1}-x_j)^2}{2\hbar\Delta t}-V(x_j)\right)}dx_1\ldots dx_N
    [/tex]

    I also follow that in continuum limit, summation converts to integral (argument of exponent). I am wondering what happens to the [itex]\Delta t[/itex] in the expression [itex]\left(\frac{m}{2\pi i\hbar\Delta t}\right)^{\frac{N+1}{2}}[/itex].
     
  2. jcsd
  3. Aug 4, 2014 #2
    It goes into the definition of functional integral
     
  4. Aug 4, 2014 #3
    I am sorry, I was not clear enough. In the continuum limit, [itex]N[/itex] tends to infinity and [itex]\Delta t[/itex] tends to zero. From the given expression, the limit amounts to infinity (there is no indeterminate form) so how come we get a finite number?
     
  5. Aug 4, 2014 #4
    The situation is a little bit more complicated. When you define the functional integral taking the [itex]N\to \infty[/itex] limit (i.e. [itex]\Delta t\to 0[/itex]), you also define a "functional measure":
    $$
    \left(\sqrt{\frac{m}{2\pi i\hbar \Delta t}}\right)^N\int \prod_{i=0}^N dx_i \to \int \mathcal{D}x.
    $$
    The hope is that taking the product of an infinite number of integral divided by an infinitesimal quantity the final result is finite. Anyway, most of the times the actual convergenge of the path integral cannot be proved.
     
  6. Aug 4, 2014 #5

    Avodyne

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    And if you actually carry out the ##N## integrations before taking the limit (which you can do explicitly in special cases such as a free particle or a harmonic oscillator), you find that the final result is finite and well-defined in the limit.
     
  7. Aug 4, 2014 #6

    bhobba

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    Gold Member

    There are deep mathematical issues involved here.

    Rigorously one must go to some some advanced math such as Hida distributions to take care of them:
    http://arxiv.org/abs/0805.3253

    This isn't the only area that has this problem (ie rigorously defining such integrals). Path integrals are the same as the Wiener integral but in imaginary time. Because of that there is a close connection between stochastic white noise theory and path integrals at the mathematical level:
    http://mathlab.math.scu.edu.tw/mp/pdf/S20N41.pdf

    Interestingly both the modern mathematical formalism of QM and Hida distributions make extensive use of Rigged Hilbert Spaces. That may be trying to tell us something important - but exactly what -:confused: :confused::confused::confused:

    Thanks
    Bill
     
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