Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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 $H_2$?

share|improve this question

1 Answer 1

up vote 2 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 $\zeta$.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.