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...
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...
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##...
Thanks for the info. In the case where K is a killing field, is the interpretation of ##K(p)=0## at ##p##, that the point p remains fixed, i.e. its coordinates remain unchanged?
Yes, sorry. I should have stated that I meant infinitesimally. So is it correct to say that, at the infinitesimal level, the reason why the KVF must vanish at the point p is because we require that it remains unchanged, i.e. ##x’^{\mu}=x^{\mu}##?
But is the transformation generated by the KVF of the form ##x’^{\mu}=x^{\mu}+\epsilon K^{\mu}##? If so, it makes sense to me that ##K^{\mu}=0## at p, as then it won’t be transformed.
By this do you mean that if an external force is acting on each particle, but there is no inter-particle interactions, then there will be a perfect correlation?
Why would correlations be present for a gas of particles out of thermal equilibrium? When deriving a Boltzmann equation, one assumes...
If this were the case then the velocities of each molecule would be perfectly correlated (at least in the sense that they will all be pointing in the same direction), right?
Is it correct to say that because it is an average force, not all the particle velocities will be pointing in the same...
Is this because they are all being acted upon by the same external force, causing a net motion in a particular direction, such that the velocities of neighbouring are likely to be similar, i.e. there is a statistical relation between their velocities as a result of them being acted upon by an...
Ah ok. So the Killing vector will move about all other points, but will keep the point p, about which you are applying it, fixed. It must vanish because ##K^{\mu}## generates space-time translations, i.e. ##x'^{\mu}=x^{\mu}+\epsilon K^{\mu}##, right? (And so, at p, we must have...
I've been reading up on Killing vectors, and have got on to the topics of homogeneous, isotropic and maximally symmetric space-times. I've read that for an isotropic spacetime, one can construct a set of Killing vector fields ##K^{(i)}##, such that, at some point ##p\in M## (where ##M## is the...
What if it is out of equilibrium and thermalising?
Also, am I correct in saying that the correlations in the velocities are due to collisions between particles, and momentum conservation relating their velocities?
Ah ok. So, in the abstract case of a gas of particles, do their velocities become correlated via collisions with one another (due to transfer of momentum)?
By that do you mean that the velocities of air molecules at neighbouring points will be similar (in magnitude and direction), and hence are correlated?
I’ve been reading up about Boltzmann transport equations, and the concept of molecular chaos has come up, in which one assumes the velocities of particles are assumed to be uncorrelated. I’m a bit confused about the concept though. In what sense do the velocities become correlated in the first...
I similarly am very confused over diffeomorphisms, and how to interpret passive and active coordinate transformations correctly.
I'm pretty sure, that under an infinitesimal diffeomorphism, tensors (of all rank) transform by a Lie derivative, i.e. ##\delta...
Thanks. I've actually read these notes before, but I was left with some doubts. For example, the author states that one can always carry out a coordinate transformation, such that the coordinate values of the new point are the same as those of the old point. Wouldn't this also introduce a...
I've been reading Straumann's book "General Relativity & Relativistic Astrophysics". In it, he claims that the twice contracted Bianchi identity: $$\nabla_{\mu}G^{\mu\nu}=0$$ (where ##G^{\mu\nu}=R^{\mu\nu}-\frac{1}{2}g^{\mu\nu}R##) is a consequence of the diffeomorphism (diff) invariance of the...
I'm trying to understand renormalisation properly, however, I've run into a few stumbling blocks. To set the scene, I've been reading Matthew Schwartz's "Quantum Field Theory & the Standard Model", in particular the section on mass renormalisation in QED. As I understand it, in order to tame the...
I find it confusing why he claims that ##\int\frac{dp}{2\pi}=\frac{1}{L}## - is he simply making the assumption that one is integrating over a finite region and then formally takes the limit?!
Also, is what I put in the first two parts of my last post (#7) correct at all?
Does the condition ##p_{j}=\frac{2\pi\,n_{j}}{L}## (##j=\lbrace 1,2,3\rbrace## and ##n_{j}\in\mathbb{Z}##) follow from the periodicity of the mode functions, i.e. $$e^{i\mathbf{p}\cdot\left(\mathbf{x}+L\mathbf{e}_{j}\right)}=
e^{i\mathbf{p}\cdot\mathbf{x}}\quad\Rightarrow\quad...
It's not the ##\int\frac{d^{3}p}{(2\pi)^{3}}=\frac{1}{V}## that confuses me, I get that this follows from the one-dimensional case. What I'm not sure about is the one-dimensional case itself, i.e. ##\int\frac{dp}{2\pi}=\frac{1}{L}##. Can one derive this mathematically?
I've been making my way through Matthew Schwartz's QFT book "Quantum Field Theory and the Standard Model". In chapter 6 he derives the differential cross-section for a ##2\rightarrow n## interaction. As part of the derivation, he introduces the Lorentz invariant phase space measure (LIPS), and...
Ok. Is there any credence to what Weinberg wrote in his book "The quantum Theory of Fields: Volume I", about the usage of fields being an inevitable consequence of requiring that a quantum theory satisfies Lorentz invariance and the cluster decomposition principle?
Also, why are particles...
Why is the multiparticle picture inconsistent when relativity is taken into account?
Was this the primary motivation for describing particles in terms of fields?
As I understand it, the need for quantum field theory (QFT) arises due to the incompatibility between special relativity (SR) and "ordinary" quantum mechanics (QM). By this, I mean that "ordinary" QM has no mechanism to handle systems of varying number of particles, however, special relativity...
Ah ok, so should it read
$$\left(E_{+}+E_{-}\right)^{2}-\left(\mathbf{p}_{e^{+}}+\mathbf{p}_{e^{-}}\right)^{2}=4E_{e}^{2}=4m_{e}^{2}+4\mathbf{p}^{2}$$ such that $$\mathbf{p}^{2}=\frac{1}{2}\left(E_{+}E_{-}-m_{e}^{2}-p_{e^{+}}p_{e^{-}}\cos\phi\right)$$ where ##\mathbf{p}## is the momentum of the...
Is there any particular book that you would recommend?
This is my own working so far...
In the CoM frame the initial momenta satisfy ##\mathbf{p}=-\mathbf{p}##, and so by momentum conservation, it must be that the final momenta satisfy ##\mathbf{p}'=-\mathbf{p}'##, which are the momenta of the...
How does one show this mathematically?!
Why is there an angular distribution though? Is it simply because the momentum only needs to satisfy overall momentum conservation, but this doesn't mean that the different momenta of each of the produced particles can't be pointing along directions that...