# Tag Info

1

We start from the equation of time evolution: $$|\psi(t)\rangle = T\left(e^{-i\int_{-\infty}^{t}dt'H(t')}\right)|\psi\rangle$$ Now, we need to evaluate the exponential with the following Hamiltonian: $$H(t) = H_0 + f(t)O_1$$ while paying attention to the time ordering. To do this, for convenience, let us consider the (time-ordered) exponential part alone, ...

0

I think you're on the right path. Just a few things and you're there: Resolution of identity$1=\sum_{k^0} \lvert k^0\rangle\langle k^0\rvert$. That takes you from $\langle k^0| n^2 \rangle$ to $\lvert n^2\rangle$ $\langle k^0\vert n^0\rangle=\delta_{k^0n^0}$ So, when you "cancel the third term", that comes with an assumption that $k^0\ne n^0$. ...

2

Using elementary graph theory identities one can show that the number of loops in a connected diagram is related to the number of external lines and the number of vertices of type $i$ each of which has $n_i$ lines attached to it, is related by $$\sum \left(\frac{n_i}{2}-1\right) V_i -\tfrac{1}{2}E +1= L$$ So you can see that for a fixed process (fixed ...

2

The order of a quantity in general refers to the exponent of the quantity in an expression, ie $$x^3y^2$$ would be 3rd order in $x$ and 2nd order in $y$. According to the Feynman rules, each vertex in a Feynman diagrams contributes a factor of the coupling constant, so the order of each coupling constant is simply the number of vertices of that interaction. ...

0

There are several definitions of nonlinear and each can be somewhat dependent on context. In your question, I would imagine one could say the perturbation was nonlinear if there were higher order terms altering the "shape" of the metric. You could also argue that the perturbations were nonlinear if they made it so the metric could not be expanded in the ...

Top 50 recent answers are included