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    I Vacuum projection operator and normal ordering

    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...
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    I Vacuum projection operator and normal ordering

    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...
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    I Vacuum projection operator and normal ordering

    Good point, fair enough.
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    I Vacuum projection operator and normal ordering

    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...
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    I Vacuum projection operator and normal ordering

    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.
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    I Vacuum projection operator and normal ordering

    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...
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    I Vacuum projection operator and normal ordering

    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...
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    I Vacuum projection operator and normal ordering

    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 ...
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    I Vacuum projection operator and normal ordering

    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...
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    I Troubles understanding slow-roll conditions in Inflation

    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...
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    I Troubles understanding slow-roll conditions in Inflation

    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)?!
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    I Troubles understanding slow-roll conditions in Inflation

    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)?
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    I Troubles understanding slow-roll conditions in Inflation

    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...
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    I Troubles understanding slow-roll conditions in Inflation

    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##...
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    Velocity correlations and molecular chaos

    Okay, great. Thanks for your time and help!
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    I Killing vectors in isotropic space-times

    You’re right, I got myself mixed up there.
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    I Killing vectors in isotropic space-times

    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?
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    Velocity correlations and molecular chaos

    Is this because the collisions cause particles to scatter, resulting in neighbouring particle velocities becoming more and more randomly directed?
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    I Killing vectors in isotropic space-times

    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}##?
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    I Killing vectors in isotropic space-times

    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.
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    Velocity correlations and molecular chaos

    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...
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    I Killing vectors in isotropic space-times

    What is the intuition for why ##K^{\mu}=0## at a point p, such that it is unchanged?
  23. D

    Velocity correlations and molecular chaos

    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...
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    Velocity correlations and molecular chaos

    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...
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    I Killing vectors in isotropic space-times

    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...
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    I Killing vectors in isotropic space-times

    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...
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    Velocity correlations and molecular chaos

    How should one understand it then? How do velocities become correlated?
  28. D

    Velocity correlations and molecular chaos

    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?
  29. D

    Velocity correlations and molecular chaos

    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)?
  30. D

    Velocity correlations and molecular chaos

    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?
  31. D

    Velocity correlations and molecular chaos

    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...
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    I Diffeomorphism invariance and contracted Bianchi identity

    Thanks for your response. Yeah, I'm still confused over this whole thing. I'm yet to find any satisfactory explanation of it all from any source.
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    I Diffeomorphism invariance and contracted Bianchi identity

    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...
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    I Diffeomorphism invariance and contracted Bianchi identity

    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...
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    I Diffeomorphism invariance and contracted Bianchi identity

    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...
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    A Renormalisation: what are the physical observables?

    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...
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    I A question about momentum integrals and lengths

    Ah ok. I shall heed your advice and read the other books you recommended. Thanks for the link to your notes too!
  38. D

    I A question about momentum integrals and lengths

    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?
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    I A question about momentum integrals and lengths

    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...
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    I A question about momentum integrals and lengths

    The linear mass density of the system would be ##\frac{1}{A}##, right?!
  41. D

    I A question about momentum integrals and lengths

    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?
  42. D

    I A question about momentum integrals and lengths

    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...
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    A Quantum Field Theory - Why quantise fields?

    Thanks for your recommendations, I shall take a look.
  44. D

    A Quantum Field Theory - Why quantise fields?

    Would there be any particular introductory QFT book that you would recommend?
  45. D

    A Quantum Field Theory - Why quantise fields?

    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...
  46. D

    A Quantum Field Theory - Why quantise fields?

    Why is the multiparticle picture inconsistent when relativity is taken into account? Was this the primary motivation for describing particles in terms of fields?
  47. D

    A Quantum Field Theory - Why quantise 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...
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    Particle collisions - a question on angles relative to beam

    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...
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    Particle collisions - a question on angles relative to beam

    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...
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    Particle collisions - a question on angles relative to beam

    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...
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