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Math Help - Factoring exponentials

  1. #1
    Forum Admin topsquark's Avatar
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    Factoring exponentials

    I'm a bit embarrassed to post this, as it might be trivial but what the hey? I'm stuck!

    The problem is to reduce this:
    e^{i \bold{k} \cdot ( \bold{x} - \bold{y} ) } e^{-i k_0 (x_0 - y_0) }  - e^{-i \bold{k} \cdot ( \bold{x} - \bold{y} ) } e^{i k_0 (x_0 - y_ 0 ) }

    into this:
    e^{ i \bold{k} \cdot ( \bold{x} - \bold{y} ) } ~ sin( k_0 (x_0 - y_0 ))

    There is going to be a factor of -2i in the process.

    If it matters, there is an integration over all k in the unreduced form. These (minus a couple of numerical factors) are the intregrands.

    More information upon request.

    -Dan

    Just for completeness, the Minkowski four-vector for k is k^{\mu} = ( -k_0, \bold{k} ). In case you need it.
    Last edited by topsquark; January 22nd 2013 at 06:56 AM.
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    Re: Factoring exponentials

    Quote Originally Posted by topsquark View Post
    I'm a bit embarrassed to post this, as it might be trivial but what the hey? I'm stuck!

    The problem is to reduce this:
    e^{i \bold{k} \cdot ( \bold{x} - \bold{y} ) } e^{-i k_0 (x_0 - y_0) }  - e^{-i \bold{k} \cdot ( \bold{x} - \bold{y} ) } e^{i k_0 (x_0 - y_ 0 ) }

    into this:
    e^{ i \bold{k} \cdot ( \bold{x} - \bold{y} ) } ~ sin( k_0 (x_0 - y_0 ))

    There is going to be a factor of -2i in the process.
    You know that \sin(\theta)=\frac{e^{i\theta}-e^{-i\theta}}{2i}

    Factor -e^{i \bold{k} \cdot ( \bold{x} - \bold{y} ) } out to get the numerator.
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    Forum Admin topsquark's Avatar
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    Re: Factoring exponentials

    Quote Originally Posted by Plato View Post
    You know that \sin(\theta)=\frac{e^{i\theta}-e^{-i\theta}}{2i}

    Factor -e^{i \bold{k} \cdot ( \bold{x} - \bold{y} ) } out to get the numerator.
    I don't see where you are going on this. Sorry. Factoring the e^{i \bold{k} \cdot (\bold{x} - \bold{y} ) } gives
    -e^{i \bold{k} \cdot ( \bold{x} - \bold{y} ) } \left [ e^{-2i \bold{k} \cdot ( \bold{x} - \bold{y} ) } e^{i k_0(x_0 - y_0)} - e^{-i k_0(x_0 - y_0)} \right ]

    (The terms with the ik(x - y) have opposite signs.)

    -Dan
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    Re: Factoring exponentials

    Quote Originally Posted by topsquark View Post
    I don't see where you are going on this. Sorry. Factoring the e^{i \bold{k} \cdot (\bold{x} - \bold{y} ) } gives
    -e^{i \bold{k} \cdot ( \bold{x} - \bold{y} ) } \left [ e^{-2i \bold{k} \cdot ( \bold{x} - \bold{y} ) } e^{i k_0(x_0 - y_0)} - e^{-i k_0(x_0 - y_0)} \right ]

    (The terms with the ik(x - y) have opposite signs.)

    -Dan

    Well e^{i k_0(x_0 - y_0)} - e^{-i k_0(x_0 - y_0)}=2i\sin(k_0(x_0 - y_0))
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    Re: Factoring exponentials

    eia e-ib e-ia eib = ei(a-b) e-i(a-b) = 2isin(a-b)
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    Re: Factoring exponentials

    Okay, I think I know where the problem is in my explanation. I know this is what you are doing Hartlw and I think that this is what you are saying also, Plato. You're actually moving backward, if I'm right. I'm going to give you the whole mess, like I should have in the beginning.

    So. I'm trying to find the commutator \[ \phi (x), \phi ^{ \dagger } (y) \]

    where
    \phi (x) = \int \frac{d^3 k}{(2 \pi )^3 2k_0} \[ a(k) e^{-ikx} + b^{ \dagger } (k) e^{ikx} \]

    and thus
    \phi ^{ \dagger }(x) = \int \frac{d^3 k}{(2 \pi )^3 2k_0} \[ b(k) e^{-ikx} + a^{ \dagger } (k) e^{ikx} \]

    Some definitions. k, a, and b are 4-vectors in Minkowski space, k = (k_0, -\nabla ^3 \bold{k}) and likewise for operators a and b. The 4-vector product is defined as kx = k_0 x_0 - \bold{k} \cdot \bold{x}.

    We also have the commutators
    \[ a(k),a^{ \dagger } (k') \] = \[ b(k),b^{ \dagger } (k') \] = (2 \pi )^3 2k_0 \delta ^3 (\bold{k} - \bold{k'} )

    So on to the commutator:
    \[ \phi (x), \phi ^{ \dagger } (y) \] = \frac{1}{(2 \pi )^6 } \int \int \frac{d^3k d^3k'}{2k_0~2k_0'} \{ \[ a(k),a^{ \dagger } (k') \] e^{-ikx + ik'y} + \[ b^{ \dagger }(k),b(k') \] e^{ikx - ik'y} \}

    Using the a and b commutators, integrating over k', and using the 4-vector product to separate the components, for example, kx = k_0 x_0 - \bold{k} \cdot \bold{x} we get:

    \[ \phi (x), \phi ^{ \dagger } (y) \] = \frac{1}{(2 \pi )^3} \int \frac{d^3k}{2k_0} \[ e^{i \bold{k} \cdot ( \bold{x} - \bold{y} ) } e^{-i k_0 (x_0 - y_0)} - e^{-i \bold{k} \cdot ( \bold{x} - \bold{y} ) } e^{i k_0 (x_0 - y_0)} \]

    I understand all of this and this is where I took the integrand out. We didn't use e^{ikx} - e^{-ikx} because we needed to separate out the 3-vector k to do the last integration over k. The next line should read

    \[ \phi (x), \phi ^{ \dagger } (y) \] = \frac{-i}{(2 \pi )^3} \int \frac{d^3k}{k_0} e^{i \bold{k} \cdot ( \bold{x} - \bold{y} ) }sin(k_0(x_0 - y_0))

    But I don't know how to get from the penultimate line to the final.

    -Dan
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    Re: Factoring exponentials

    Quote Originally Posted by topsquark View Post
    The problem is to reduce this: e^{i \bold{k} \cdot ( \bold{x} - \bold{y} ) } e^{-i k_0 (x_0 - y_0) }  - e^{-i \bold{k} \cdot ( \bold{x} - \bold{y} ) } e^{i k_0 (x_0 - y_ 0 ) }into this: e^{ i \bold{k} \cdot ( \bold{x} - \bold{y} ) } ~ sin( k_0 (x_0 - y_0 ))
    (Reference above for a and b)

    Approach: What is X st A = B + X, and X integrates out?

    eiae-ib - e-iaeib = -eia(e-ib + eib) -2X, after substituting for sinb.

    X = ieibsina

    Works if sina term integrates out because of symmetry.
    Last edited by Hartlw; January 23rd 2013 at 12:49 PM.
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    Re: Factoring exponentials

    Well that took a while to verify. (Which means I tried to keep each step verifiable, ie. I actually made sure each term was Mathematically correct, not necessarily Physically correct.)

    Hartlw: Thank you. It worked out nicely. I had thought that there may be a way to extract an essentially 0 integral but I couldn't figure out how to do it.

    Thanks to you and Plato. Couldn't have done it without both of you. And thanks for putting up with slow to understand student that I occasionally am.

    -Dan
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