Vacuum amplitudes in \phi^4 theory

[Replusz]

TL;DR

I am not sure how to get the second line from the first.
It looks like he is expanding the exponential as a Taylor series, but what is happening to all those integrals?

 

[George Jones]

It looks like he is expanding

To whom does "he" refer? If you ask questions about a specific reference, then you should always explicitly state the reference, because then we might be able to point out something else relevant from the same reference.

Defining the exponential of an operator as a power series, and supposing that the operator $\int dt H\left(t\right)$ exists, what are the first three terms of

$$\exp \left( \int dt H\left(t\right) \right) ?$$

 

[hilbert2]

That looks like the same as when an evolution operator $U(t,t')$ in nonrelativistic QM is written as a Dyson series, but it's not completely the same because the integrals are not only over the time coordinate.

 

[Replusz]

The material we should know is in this PDF:
http://www.damtp.cam.ac.uk/user/tong/qft/qft.pdf

Defining the exponential of an operator as a power series, and supposing that the operator $\int dt H\left(t\right)$ exists, what are the first three terms of

$$\exp \left( \int dt H\left(t\right) \right) ?$$

I assume its just this:

This corresponds to the first three terms of the 2nd line in eq. 7.1
I think I see where I went wrong.
Where it says in the 2nd line in 7.1 "..." those only mean measures right? No integrands. So it is only the very last part that is being integrated n times.
That would make sense, am I correct?

 

[Replusz]

That looks like the same as when an evolution operator $U(t,t')$ in nonrelativistic QM is written as a Dyson series, but it's not completely the same because the integrals are not only over the time coordinate.

I was actually thinking the same thing, I assume the d3^x term is just there to normalize something - I might be completely wrong though.

Thanks to George Jones and hilbert2 for your help!

 

[George Jones]

I assume its just this:
View attachment 260571
This corresponds to the first three terms of the 2nd line in eq. 7.1
I think I see where I went wrong.
Where it says in the 2nd line in 7.1 "..." those only mean measures right? No integrands. So it is only the very last part that is being integrated n times.

I am not quite sure what you mean, as there two instances of "..." in (7.1), one for the integration variables, and one for the integrand. Write the second-order term above as

$$\int dt_1 H\left(t_1\right) \int dt_2 H\left(t_2\right) =\int dt_1 \int dt_2 H\left(t_1\right) H\left(t_2\right) ,$$

so the nth-order term is

$$ \int dt_1 \int dt_2 ... \int dt_n H\left(t_1\right) H\left(t_2\right)... H\left(t_n\right). $$

I was actually thinking the same thing, I assume the d3^x

If $H\left(t\right)$ is a Hamiltonian, what is $\int dt H\left(t\right)$ in terms of a Hamiltonian density?

 

[Replusz]

I am not quite sure what you mean, as there two instances of "..." in (7.1), one for the integration variables, and one for the integrand. Write the second-order term above as

$$\int dt_1 H\left(t_1\right) \int dt_2 H\left(t_2\right) =\int dt_1 \int dt_2 H\left(t_1\right) H\left(t_2\right) ,$$

so the nth-order term is

$$ \int dt_1 \int dt_2 ... \int dt_n H\left(t_1\right) H\left(t_2\right)... H\left(t_n\right). $$

Yes, I understand now. Thank you!

If $H\left(t\right)$ is a Hamiltonian, what is $\int dt H\left(t\right)$ in terms of a Hamiltonian density?

It is the integral with d^4x of the Hamiltonian density. Just what we have!
Thanks!