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  Axioms for Euclidean Green's functions's paper by Osterwalder and Schrader 2.

+ 4 like - 0 dislike

I have another question from the following text:


On page 91, they define $\hat{T}^t$, and I don't see how did they get (4.7).

I mean if I write it down I get:

$$(f,\hat{T}^tg) = \sum \sigma_{n+m} (\Theta f^*)_n \times (\hat{T}^t g)_m) =  \sum \sigma_{n+m}(\Theta f^*_n \times (g_{(t,1)})_m)$$

Now, by (E1) we get:

$$= \sum \sigma_{n+m}((\Theta f^*_{(-t,1)})_n \times g_m)$$

Now, the $\Theta$ is time inversion, i.e we should get:

$$= \sum \sigma_{n+m}((f^*_{(t,1)})_n\times g_m) = \sum \sigma_{n+m}((\hat{T}^tf^*)_n\times g_m)$$

But on the last equality we should have that it 's equal: $\sum \sigma_{n+m}(\Theta (\hat{T}^t f)^*)_n \times g_m)$, but I don't get it.

Can you help me on this?

asked Jul 23, 2015 in Mathematics by MathematicalPhysicist (205 points) [ no revision ]

1 Answer

+ 2 like - 0 dislike

Remember you have to worry about the temporal inversion by \(\theta\) now. I think it is easiest to see this by explicitly writing it out as seen on page 87.

\[(\theta f)_n (x_1,...,x_n) = f_n(\theta x_1,...,\theta x_n)\] where \(\theta x = (-x^0 , \vec{x})\)

now remember by definition we have \(f_{(a,\mathbf{R})}(x_1,...,x_n)= f(\mathbf{R}x_1 + a,...,\mathbf{R}x_n + a)\) and (E1) states that the Euclidean Green function is invariant under SO(4) rotations and translations (as stated in your previous question).

Using your notation we have

\((f,\hat{T}^tg) = \sum \sigma_{n+m} (\Theta f^*)_n \times (\hat{T}^t g)_m) = \sum \sigma_{n+m}(\Theta f^*_n \times (g_{(-t,1)})_m)\)

Notice the minus sign on the t, for that is how they define the map, \((\hat{T}^t f)_n (x_1,...,x_n) = f_n(x_1 - t,...,x_n - t)\) where \(t = (t,\vec{0})\)

Now explicitly writing out your expression above we have

\[= \sum \sigma_{n+m}(f^*_n(\theta x_1,...,\theta x_n) \times g_m(x_1 - t,..., x_n-t))\]

We then shift the whole thing by a factor of t, giving us

\[= \sum \sigma_{n+m}(f^*_n(\theta x_1+t,...,\theta x_n+t) \times g_m(x_1 ,..., x_n))\]

\[= \sum \sigma_{n+m}(f^*_n(\theta x_1- \theta t,...,\theta x_n-\theta t) \times g_m(x_1 ,..., x_n))\]

\[= \sum \sigma_{n+m}(\theta f^*_n( x_1- t,..., x_n- t) \times g_m(x_1 ,..., x_n))\]

\[= \sum \sigma_{n+m}(\theta (\hat{T}^t f)^*_n( x_1,..., x_n) \times g_m(x_1 ,..., x_n))\]

Where in the second line we used the fact that t is a purely temporal vector. This gives the result you are looking for in equation (4.7)

answered Jul 27, 2015 by Peter Anderson (205 points) [ no revision ]

@PeterAnderson There must be a typo here or in the article, since in the article $(U_s(a)f)_n(x_1,\ldots, x_n) = f_n(x_1-a,\ldots , x_n - a)$, where $a=(0,\vec{a})$, and also $(T^tf)_n (x_1,\ldots , x_n)=f_n(x_1-t,\ldots , x_n-t)$, where $t=(t,\vec{0})$.

In your answer to my previous question in the post called:Axioms for Euclidean Green's functions's paper by Osterwalder and Schrader you wrote:

$(U_s(a)f)_n(x_1,\ldots , x_n) = f_n(x_1+a , \ldots , x_n+a)$ even though that's not how it's written in the text, so you say there must be a typo? (It's all on page 91 of the article).

@MathematicalPhysicist It seems I just missed that fact the first time. I updated the previous answer to align accordingly. Since it is just a definition it doesn't actually affect anything, besides just having a minus sign that needed to be carried through out. Thanks for catching that!

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