- #1
lhcQFT
- 5
- 0
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.
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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.