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n-point Wightman functions (notation confusion) Watch

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    The problem I am trying to solve is trivial, but I am struggling with notation. I have a recursive formula for the nth Wightman function,

    \displaystyle W_n(x_1,...,x_n)= \sum_{k=2}^n \Delta_+(x_1-x_k)W_{n-2}(x_1,...,\hat{x}_k,...,x_n),

    where \hat{x}_k indicates that the kth argument has been 'cancelled'. I also know that W_0=1 and W_1=0.

    The question is to find W_{2n+1}. Intuitively it looks like the answer should be zero, as it will be some product of the n-point Wightman functions with odd n. This includes W_1=0, and so W_{2n+1}=0.

    However, syntactically the recursion formula doesn't actually seem to work. If we let n \rightarrow 2n+1 in the formula, we get

    \displaystyle W_{2n+1}(x_1,...,x_{2n+1})= \sum_{k=2}^{2n+1} \Delta_+(x_1-x_k)W_{2n-1}(x_1,...,\hat{x}_k,...,x_{2n+1  }).

    The next step would be to use the recursion formula again to compute

    W_{2n-1}(x_1,...,\hat{x}_k,...,x_{2n+1  }),

    but this doesn't seem to be possible since in the formula the subscripts on W and the final argument are equal, and this isn't the case here.

    Specifically, if we let n \rightarrow 2n-1 in the formula, we get an expression for

    W_{2n-1}(x_1,...,x_{2n-1}),

    and if we let n \rightarrow 2n+1 in the formula, we get an expression for

    W_{2n+1}(x_1,...,x_{2n+1}).

    Neither of these is what we want: the former has the wrong arguments, and the latter is the wrong function!
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    As an example I have looked at the 4-point functions. Using the formula,

    \displaystyle W_4(x_1,x_2,x_3,x_4) = \Delta_+(x_1-x_2)W_2(x_1,x_3,x_4)+\Delta_+(x_  1-x_3)W_2(x_1,x_2,x_4) +\Delta_+(x_1-x_4)W_2(x_1,x_2,x_3).

    But then we can't use the formula because the formula tells us how to compute the nth function with n arguments, where we now have nth functions with n+1 arguments...
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    (Original post by Implication)
    The problem I am trying to solve is trivial, but I am struggling with notation. I have a recursive formula for the nth Wightman function,

    \displaystyle W_n(x_1,...,x_n)= \sum_{k=2}^n \Delta_+(x_1-x_k)W_{n-2}(x_1,...,\hat{x}_k,...,x_n),
    Eek! It's over thirty years since I studied quantum field theory and the neurons where it was kept have long since been retired. However...

    There does seem to be a serious problem with this formula as the left hand side implies that W_n is a function of n variables, whereas the right hand side is a sum of W_{n-2}'s that appear to be functions of n-1 variables. Unless that \Delta_+ is doing something weird. Remind me, what is it? Something nasty with distributions?
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    (Original post by Implication)
    As an example I have looked at the 4-point functions. Using the formula,

    \displaystyle W_4(x_1,x_2,x_3,x_4) = \Delta_+(x_1-x_2)W_2(x_1,x_3,x_4)+\Delta_+(x_  1-x_3)W_2(x_1,x_2,x_4) +\Delta_+(x_1-x_4)W_2(x_1,x_2,x_3).

    But then we can't use the formula because the formula tells us how to compute the nth function with n arguments, where we now have nth functions with n+1 arguments...
    Exactly. Where have you got the formula from?
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    (Original post by Implication)
    The problem I am trying to solve is trivial, but I am struggling with notation. I have a recursive formula for the nth Wightman function,

    \displaystyle W_n(x_1,...,x_n)= \sum_{k=2}^n \Delta_+(x_1-x_k)W_{n-2}(x_1,...,\hat{x}_k,...,x_n),
    I've found something like this formula in Roepstorff's "Path Integral Approach to Quantum Physics", and there the formula is:

    \displaystyle W_n(x_1,...,x_n)= \sum_{k=1}^{n-1} \Delta_+(x_k-x_n)W_{n-2}(x_1,...,\hat{x}_k,...,x_{n-1}),

    So is this just a typo situation?
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    (Original post by Gregorius)
    I've found something like this formula in Roepstorff's "Path Integral Approach to Quantum Physics", and there the formula is:

    \displaystyle W_n(x_1,...,x_n)= \sum_{k=1}^{n-1} \Delta_+(x_k-x_n)W_{n-2}(x_1,...,\hat{x}_k,...,x_{n-1}),

    So is this just a typo situation?
    Yes! The formula I posted was lifted straight from the teacher's lecture notes for the course. I compared it to the textbook and thought it was the same, but I've just quadruple checked and discovered that the arguments on the (n-2)th function should start from x_2, not x_1:

    \displaystyle W_n(x_1,...,x_n)= \sum_{k=2}^n \Delta_+(x_1-x_k)W_{n-2}(x_2,...,\hat{x}_k,...,x_n)

    That means that the first x is omitted as well as the kth and right number of arguments are there after all and everything works out fine

    The formula you've posted looks like it should work out to be equivalent to this (corrected) version. I've checked for n=4 and it gives the same so thank you very much! This is like a weight off my shoulders haha, I went to chat to my lecturer about it and she just didn't understand what my problem was so it's good to know I wasn't going crazy
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    (Original post by Implication)
    This is like a weight off my shoulders haha, I went to chat to my lecturer about it and she just didn't understand what my problem was so it's good to know I wasn't going crazy
 
 
 
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