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I'm having trouble following the proof of the adiabatic theorem (apparently due to Messiah) on Wikipedia.

At one stage we have:

$U(t_1,t_0)=1+{1\over i}\int_{t_0}^{t_1}H(t)dt+{1\over i^2}\int_{t_0}^{t_1}\int_{t_0}^{t'}dt'dt''H(t')H(t'')+\ldots$

which I'll write as $1+H_1+H_2+\ldots$.

The argument then goes

$\zeta=\left<0|(1+iH_1)(1-iH_1)|0\right>+$ other terms.

So $\zeta = \left<0|H_1^2|0\right>+$ other terms.

But if we're computing to second order in $H$, shouldn't we keep terms to second order all the way through the computation? In which case we really need:

$\zeta=\left<0|(1+iH_1+H_2)(1-iH_1+H_2)|0\right>+$ other terms.

So $\zeta = \left<0|H_1^2+2H_2|0\right>$+other terms?

Why is apparently OK to drop $H_2$?

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up vote 3 down vote accepted

Yes, you are right that you should keep all terms to a given order. However if you look at the full expression that is being calculated:

$\zeta= \left<0|(1+iH_1+H_2)(1-iH_1+H_2)|0\right>-\left<0|(1+iH_1+H_2)|0\right> \left<0|(1-iH_1+H_2)|0\right>$

You will find that the $H_2$ terms cancel out identically.

Surprisingly, this answer is controversial, so let's go through this more slowly. From Wikipedia:

$\zeta = \langle 0|\hat{U}^\dagger(t_1,t_0)\hat{U}(t_1,t_0)|0\rangle - \langle 0|\hat{U}^\dagger(t_1,t_0)|0\rangle\langle 0|\hat{U}(t_1,t_0)|0\rangle$

Dan expands $U$ as $U \approx 1 + iH_1 + H_2$.

Then Dan correctly expands the first term for $\zeta$ to second order in H as:

$\zeta = \langle 0| H_1^2 + 2H_2|0\rangle +...$

and asks why it is OK to drop $H_2$? The answer is that the second part of $\zeta$ is, to the same order in H,

$...-\langle 0|H_1|0\rangle \langle 0|H_1|0\rangle - 2\langle 0|H_2|0\rangle$

Therefore, the $\langle 0|H_2|0\rangle $ terms cancel exactly from $\zeta$.

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Do they cancel? $\left<0|H_2^2|0\right>$ isn't the same as $\left<0|H_2|0\right>\left<0|H_2|0\right>$. – Dan Piponi May 22 '12 at 20:40
Dan is being nice. They don't cancel. Can you delete the answer? – Ron Maimon May 22 '12 at 20:55
The $<0|H_2|0>$ terms cancel. The $H_2^2$ terms are of higher order than $H_1^2$ terms. From the question it appeared the questioner understood $H_2$ was of the same perturbative order as $H_1^2$. – user1631 May 22 '12 at 21:02
$<0|H_2^2|0>$ would be 4th order in H. The question specifically referenced terms up to second order in H. What is the confusion? – user1631 May 22 '12 at 21:05
@user1631: The confusion was due to two things 1. the Wikipedia person not making clear that H_1 is an infinitesimal perturbation. 2. the fact that you write "the H_2 terms cancel exactly" in your answer, when you mean H_1 terms. – Ron Maimon May 22 '12 at 21:19

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