- #1

Heidi

- 355

- 35

in fact the answer is given in the book (written by philippe Martin).

we have

$$ (\tau_1| A^{-1} | \tau_2) = 2D \ min(\tau_1 ,\tau_2) = 2D(\tau_1 \theta (\tau_2 -\tau_1)+\tau_2 \theta (\tau_1 -\tau_2))$$

So

$$-1/2D \frac{d^2}{d\tau_1^2} (\tau_1| A^{-1} | \tau_2) = \delta( \tau_1 - \tau_2) $$

the author writes then that we see that we have the operators equality

$$-1/2D \frac{d^2}{d\tau^2} = A$$

is it obvious for you?

thanks

Edit:

does the derivative act in the same way on every members of the

covariant matrix 1/A? in this case we could consider that we have the product ot A and 1/A giving a dirac.

we have

$$ (\tau_1| A^{-1} | \tau_2) = 2D \ min(\tau_1 ,\tau_2) = 2D(\tau_1 \theta (\tau_2 -\tau_1)+\tau_2 \theta (\tau_1 -\tau_2))$$

So

$$-1/2D \frac{d^2}{d\tau_1^2} (\tau_1| A^{-1} | \tau_2) = \delta( \tau_1 - \tau_2) $$

the author writes then that we see that we have the operators equality

$$-1/2D \frac{d^2}{d\tau^2} = A$$

is it obvious for you?

thanks

Edit:

does the derivative act in the same way on every members of the

covariant matrix 1/A? in this case we could consider that we have the product ot A and 1/A giving a dirac.

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