Recent content by "Don't panic!"

Thanks for this. Going back to the normal-ordered stuff, I managed to get hold of a copy of the book I was mentioning in my OP ("Nonequilibrium Quantum Transport Physics in Nanosystems: Foundation of Computational Nonequilibrium Physics in Nanoscience and Nanotechnology" by F. A. Buot). In it...

Possibly, but unfortunately I don't have access to the rest of the book :( This is what leaves me confused too. It is discussed in this paper (right-hand side of page 3) and this paper (right-hand side of page 6) on the arXiv also (admittedly using a different approach), but without further...

Good point, fair enough.

Thanks, this derivation (and stevendaryl’s) makes a lot of sense. It is reasonable to take ##0^0=1## (there doesn’t seem to be a general consensus)? Also, it would interesting to see why the author is able to write: $$: \text{exp}\Big\lbrace\hat{a}^{\dagger}\frac{d}{dZ^{\ast}}\Big\rbrace\vert...

Ah, that's frustrating. I've seen the same argument as the one this author gives in an academic paper too. It would be helpful if they actually explained things in further detail. I can't find anywhere else that gives a detailed derivation.

Okay, so do you think the author is assuming this then? What they are not assuming is that it can be written in the simple form of an exponential of the number operator. What is the point in introducing the ##Z^{\ast}## identity then? One could equally well just use a kronecker delta to write...

But I thought the point of this exercise to show precisely that ##\vert 0\rangle\langle 0\vert## is expressible in terms of ##a## and ##a^{\dagger}##? It is not a priori assumed, at least as far as I can tell. I think it possibly is. If one operates from the left by ##:A^{-1}:##, then we get...

Thanks for your response. I understand this part though. What I don't understand is how one goes from: $$ : \text{exp}\bigg\lbrace\hat{a}^{\dagger}\frac{\mathrm{d}}{\mathrm{d}Z^{\ast}}\bigg\rbrace\,\vert 0\rangle\langle 0\vert\,\text{exp}\big\lbrace\hat{a}Z^{\ast}\big\rbrace ...

I've been reading this book, in which the author expresses the vacuum projection operator ##\vert 0\rangle\langle 0\vert## in terms of the number operator ##\hat{N}=\hat{a}^{\dagger}\hat{a}##, where ##\hat{a}^{\dagger}## and ##\hat{a}## are the usual creation and annihilation operators...

Okay, so can one say that, given the scale factor ##a(t)## at some time ##t##, then over a time interval ##\Delta t## of order the Hubble time (i.e. ##\Delta t\sim H^{-1}##) the universe will change significantly, however, for intervals much less than this (i.e. ##\Delta t\ll H^{-1}##), the...

But I thought the Hubble time set the time-scale over which the scale factor changes appreciably (at least that's what I've read in a set of notes)?!

So the slow roll condition is that the change in ##H## is small per Hubble time then. Ss what I wrote in the couple of sentences at the end of my previous post correct at all (as to why the Hubble time is the appropriate time scale)?

Thanks for your response. I was editing my original post when your answer was posted. I'd appreciate it if you could have a look at the updated version and see if you agree with what I've written. Why is the typical time scale set by ##1/H##? I've read that it's the time scale over which the...

I've been reading up on inflation, and have arrived at the so-called slow roll conditions $$\epsilon =-\frac{\dot{H}}{H^{2}}\ll 1\; ,\qquad\eta =-\frac{\ddot{\phi}}{H\dot{\phi}}\ll 1$$ I have to admit, I'm having trouble understanding a couple of points. First, how does ##\epsilon\ll 1##...

Okay, great. Thanks for your time and help!