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n-point Wightman functions (notation confusion)

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!

Reply 1

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...

Reply 2

Gregorius

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?

Reply 3

Gregorius

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?

Reply 4

Gregorius

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?

Reply 5

Implication

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 :biggrin:

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 :biggrin:

Reply 6

Gregorius

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 :biggrin: