# In the interacting scalar field theory, I have a question.

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1. Jun 24, 2015

### lhcQFT

First of all, I copy the text in my lecture note.
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In general, $$e^{-iTH}$$ cannot be written exactly in a useful way in terms of creation and annihilation operators. However, we can do it perturbatively, order by order in the coupling $$\lambda$$. For example, let us consider the contribution linear in $$\lambda$$. We use the definition of the exponential to write:

$$e^{-iTH} = [1-iHT/N]^N = [1-i(H_0 + H_{\text{int}})T/N]^N$$ - - - (1)

for $$N \rightarrow \infty$$. Now, the part of this that is linear in $$H_{\text{int}}$$ can be expanded as:

$$e^{-iTH} = \sum_{n=0}^{N-1} [1-iH_0T/N]^{N-n-1}(-iH_{\text{int}}T/N)[1-iH_0T/N]^n$$ - - - (2)

(Here, we have dropped the 0th order part, $$e^{-iTH_0}$$, as uninteresting; it just corresponds to the particles evolving as free particles.)
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So, my question is how do I derive from eq. (1) to (2)? If you teach me the method, I really thank you.

p.s. In this environment, inline math mode is not worked. Sorry for inconvenient.

2. Jun 24, 2015

### Orodruin

Staff Emeritus
Just write out the product of the N terms and keep terms linear in the interaction Hamiltonian.

3. Jun 24, 2015

### Staff: Mentor

Set off your equations using ## instead of  and they will display inline.

4. Jun 24, 2015

### ChrisVer

Take eg $N=2$ and $N=3$ and try then a general N...
I think from the N=2 and N=3 you will be able to see what is going on...