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I'm having some trouble understanding the exact details of the collapse and revival in the Jaynes Cummings model. What I understand is this:

We assume our two level atom to initially be in the excited state, interacting with a single field mode in a coherent state. The combined state can then be written as

$|\psi(t)\rangle = \sum_{n=0}^{\infty}{C_{e,n}\frac{e^{-\frac{\alpha^2}{2}}|\alpha|^n}{\sqrt{n!}}|e,n\rangle} + \sum_{n=0}^{\infty}{C_{g,n+1}\frac{e^{-\frac{|\alpha|^2}{2}}\alpha^n}{\sqrt{n!}}|g,n+1\rangle}$

where in the case of zero detuning we know that the coefficients are given by $C_{e,n} = \cos{(g\sqrt{n+1}t)}, C_{g,n+1} = i\sin{(g\sqrt{n+1}t)}$

Now, if we then look at the probability of being in the excited state as a function of time $P_e(t)$ we can easily find it to be given by

$P_e(t) = \sum_{n=0}^{\infty}{\frac{e^{-|\alpha|^2}|\alpha|^{2n}}{n!}\cos^2{(g\sqrt{n+1}}t)}$

This tells us that the probability varies with various different frequencies, following a poissionian distribution in $|\alpha|^2$. However, this is where I get stuck understanding wise. I want to find the timescale on which this probability collapses, and on which it revives. Most texts I can find write that the collapse is on a timescale given by $1/g$, but personally I don't see this. It has to do with the poissonian nature of the coherent states, I know that much, and an argument involving that the photon numbers lie mostly in the region $|\alpha|^2 - |\alpha|, |\alpha|^2 + |\alpha|$.

So, just to recap, I'm trying to find the collapse and revival times of the abovementioned probability, in the case of no detuning. Could anyone tell me how to do this?

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  • $\begingroup$ It seems like you're just looking for a time $t$ such that $g \sqrt{n+1} t = \pi (k_n+.5)\; \forall n$ where $k_n$ is an integer. Am I misunderstanding your question? $\endgroup$ – user27118 Jan 22 '15 at 23:43
  • $\begingroup$ That would give the collapse time I suppose yes. But in such an infinite sum, I don't see how one would do that? $\endgroup$ – user129412 Jan 22 '15 at 23:49
  • $\begingroup$ They write about it here web.stanford.edu/~rsasaki/AP387/chap6 on page 7 and 8, but I do not understand why the times take on the form they describe. $\endgroup$ – user129412 Jan 22 '15 at 23:51
  • $\begingroup$ In general, you can't. The frequencies of oscillation are mostly irrational numbers. In the link they discuss finding a timescale in the limit $t \ll \frac{|\alpha|}{g}$ (and I assume $\alpha \gg 1$). Is your question how they get 6.49? If so, I would edit your question to ask that specifically. At first glance it's not clear to me either. $\endgroup$ – user27118 Jan 23 '15 at 1:24
  • $\begingroup$ Hm, well I don't specifically need it to be of the form 6.49, there are also sources that come to the same conclusion without using that form. Moreover, they also use a different form to find the revival time, which I don't get either $\endgroup$ – user129412 Jan 23 '15 at 21:12
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It is really quite difficult. This gets you to the original paper by Cummings in Phys. Rev. Lett. 140 in 1965, page A1051: http://journals.aps.org/pr/abstract/10.1103/PhysRev.140.A1051 . You must notice first that if $P_e(t)$ is given as you wrote then the coefficients $\frac{e^{-\alpha^2}|\alpha|^{2n}}{{n!}}$ multiplying $\cos()^2$ terms are localized (spike in values) around $n = n_{av} = |\alpha|^2$. You can get it from the Stirling formula for large $n$ ($n>10$) by approximating

$n! \approx \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n$

and than trying to find the maximum of the function

$\frac{e^{-n_{av}}|n_{av}|^{n}}{{n!}} \approx 1 /\sqrt{2 \pi n}\left(\frac{e}{n}\right)^n e^{-n_{av}}|n_{av}|^{n}$

with respect to $n$ as it was the continuous variable.

Calculating and putting the derivative to zero you get

$e^{-n_{av}} /(2 \sqrt{2 \pi})e^n n_{av}^n n^{-3/2-n}(1 + 2 n_{av} \log n - 2 n \log n) = 0 $

and the approximate solution, neglecting the small $1$ in the bracket, is

$n=n_{av}$.

Than you can Taylor-expand $g(n+1)^{1/2}$ up to the first order around $n_{av}$ i.e. $g(n+1)^{1/2} \approx g (n_{av}+1)^{1/2} + g/[2 (n_{av}+1)^{1/2}](n-n_{av})$ to care only about the most contributing terms in the sum. The energies get linear in $n$ now like for the harmonic oscillator. You get both the revival time and the short time decay using this approximation after some page of further math (which is not in the paper as it was the Letter). Revivals are much simpler since now all cosines oscillate with the multiple of the same frequency $g/[2 (n_{av}+1)^{1/2}]$ and therefore they are seen directly to rephase. The revival time is therefore $2 \pi (n_{av}+1)^{1/2}/g$ while you get the decay time $1/g$ using small time expansion of the sum terms for short times to sum them up to the real dropping exponent. Paradoxically revivals were not noticed or ignored by Cummings in his original paper deriving $1/g$ collapse rate even if the approximate formula he derived has them in it and they were discovered later by Eberly in Phys. Rev. lett. 44, 1980, page 1323: http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.44.1323 . He derives a complicated compact formula for the arbitrary case of the nonzero detuning that is long time periodic behavior of the full revival after the time $2 \pi (n_{av})^{1/2}/g \approx 2 \pi (n_{av}+1)^{1/2}/g$ that resembles slightly the oscillatory motion of the weakly spreading free particle wave packet if one calculated the overlap of it with itself from the time $0$ (the autocorrelation function) and with the revival period while the fast oscillations are some phase.

To obtain the decay time: Now we have:

$P_e(t) \approx \sum_{n=0}^{\infty}{\frac{e^{-n_{av}}n_{av}^{n}}{n!}} \times$ $\cos^2{[g (n_{av}+1)^{1/2} + g/[2 (n_{av}+1)^{1/2}](n-n_{av})]}$

or

$P_e(t) \approx \sum_{n=0}^{\infty}{\frac{e^{-n_{av}}n_{av}^{n}}{n!}} \times$ $[\cos{[2 g (n_{av}+1)^{1/2} t + g/[ (n_{av}+1)^{1/2}](n-n_{av}) t]}/2+1/2]$

Now we use to formula for the $\cos (a+b)=\cos a \cos b - \sin a \sin b$ twice and drop all terms that have $\sin (...t) \approx 0$ in it since they they are small and terms with $\cos (...t)$ except the most oscillatory with $2g$ not having the running $n$ we put to $1$ for the same reason. The result is:

$P_e(t) \approx \sum_{n=0}^{\infty}{\frac{e^{-n_{av}}n_{av}^{n}}{n!}} \times$ $[\cos [2 g (n_{av}+1)^{1/2} t] \cos{[g/[ (n_{av}+1)^{1/2}]n t]}/2+1/2]$

Now using the formula for $\cos(x)= (e^{ix}+e^{-ix})/2$

the series can be readily summed up from the series exponent expansion

$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$

(note the terms containing the exponent of the exponent):

$P_e(t) \approx (1/4)\cos [2 g (n_{av}+1)^{1/2} t] e^{-n_{av}} [e^{n_{av}e^{i g/[ (n_{av}+1)^{1/2}]t}}$ $+e^{n_{av}e^{-i g/[ (n_{av}+1)^{1/2}t]}}]$ +1/2$

The one more step is to expand again for small $t$ (we care only about the real parts while the imaginary have the small $\sin (...t)$ again):

$e^{-i g/[ (n_{av}+1)^{1/2}t} \approx 1 - t^2 g^2 /2/ (n_{av}+1)$

and so

$e^{i g/[ (n_{av}+1)^{1/2}t} \approx 1 - t^2 g^2 /2/ (n_{av}+1)$

approximating

$n_{av}+1 \approx n_{av}$ we obtain $P_e(t) \approx (1/2) \cos [2 g (n_{av}+1)^{1/2} t] e^{-g^2 t^2/2}+ 1/2$ so the decay time is $1/g$

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  • $\begingroup$ Hi matt, could you please stop making minor edits to this answer? Every time you make an edit it bumps the post to the top of the list on the main site, taking attention from other deserving questions. It's quite reasonable to edit a post roughly 3 or 4 times, but if you find yourself editing more than that, you should take it as a sign you need to be more conservative with how you edit. $\endgroup$ – David Z Mar 16 '15 at 2:20
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    $\begingroup$ Sorry, I did not know that. This should be fixed software-wise or so. I should be able to edit as much as I want to make it perfect. Besides someone asked me to nice-latex it before. $\endgroup$ – matt Mar 17 '15 at 17:17
  • $\begingroup$ Well you're certainly not the only one who thinks that way, and I'd encourage you to check out some of the relevant feature requests on Meta Stack Exchange (e.g. asking for the ability to make minor edits which don't bump the post). But for now, just make sure to fix everything you can find that needs fixing when you edit a post. $\endgroup$ – David Z Mar 23 '15 at 11:29

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