  # Perturbation

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## Everything posted by Perturbation

1. $\langle q_2, t_2|q_1, t_1\rangle =\int_{Q(t_1)=q_1}_{Q(t_2)=q_2}\mathcal{Q(t)}\int\mathcal{P(t)}\exp\left[i\int dt\left(\sum_iP_i\dot{Q}^i-H(Q, P)\right)\right]$
2. Copenhagen is too reliant on the observer, and some aspects are just plain wrong in my opinion. Everett seems pretty superfluous. So neither.
3. Do you have to be involved in research or actually be qualified for them to let you into these sorts of things? I'd expect there's some restriction. I've spent the past three or four years teaching myself mathematics and physics and the thought of being neglected from things like this because I'm not formally educated, or too young, irritates me. York's not too far from me, it's about an hour or so away, If I could get there I'd go.
4. Oh, yeah, axiom of choice. Ignore what I said about separability. http://en.wikipedia.org/wiki/Hilbert_space#Bases
5. It's only an analogy. I don't know any better way of explaining it without going into the mathematics. Space-time is some abstract thing that has non-zero curvature around bodies with mass-energy. If you really hate using the trampoline with gravity just say it's a naturally positively curved trampoline. Maybe some grossly obese person has fallen asleep on it and left an indentation.
6. It depends if the Hilbert space is separable, that is that any state-vector can be written as a sum of elements of a countable basis set. And whether you can expand it into a basis of an operator's eigenvectors depends whether the eigenvectors are linearally independant, i.e. orthogonal. For the eigenvectors of a Hermitian operator this is always the case, it's a simple property of Hermitian operators, at least if they're non-degenerate. It's usually assumed that the Hilbert space is separable, but I do believe the guys working on early quantum gravity came across problems with separability when they tried to build a Hilbert space, and was one reason why they moved to using spin networks. If you wanted to use a set of eigenvectors of some operator as an expansion consisting of purely diagonal terms for some other operator, the two operators would have to commute.
7. Schrodinger quantum mechanics has the time dependence in two things: the state-vector and consequently the state operator. There is no time dependance in the operators that represent dynamical variables. The Schrodinger equation expresses the time dependence of state-vectors as a differential equation, i.e. as a dynamical equation. Start with your state-vector [imath]|\psi (0)\rangle[/imath] defined at initial time t=0 and perform a unitary evolution $|\psi (t)\rangle =e^{-iHt/\hbar}|\psi (0)\rangle$ Differentiate with respect to time $\frac{\partial}{\partial t}|\psi (t)\rangle =\frac{-iH}{\hbar}e^{-iHt/\hbar}|\psi (0)\rangle =\frac{-iH}{\hbar}|\psi (t)\rangle$ Putting this in the usual form gives $H|\psi (t)\rangle =i\hbar\frac{\partial}{\partial t}|\psi (t)\rangle$ This is the Schrodinger equation. For a non-relativistic particle the Hamiltonian H is [imath] H=\frac{P^2}{2M}+V [/imath] And in coordinate representation the wave-function $\psi (\vec{x}, t)=\langle\vec{x} |\psi\rangle$ obeys $\left(\frac{-\hbar^2\nabla^2}{2M}+V(\vec{x}, t)\right) \psi (\vec{x}, t)=i\hbar\frac{\partial}{\partial t}\psi (\vec{x}, t)$ The momentum operator [imath]\vec{P}[/imath] is [imath]-i\hbar\vec{\nabla}[/imath] in coordinate representation. For a relativistic particle we have the Dirac equation $\left(i\gamma^{\mu}\partial_{\mu}-m\right)\psi (\vec{x}, t)=0$ Heinsenberg QM differs in that the state-vectors and state operators have no explicit time depedence; instead the time dependence is in the dynamical operators, such as the Hamiltonian or Momenta.
8. I din't expect you'd get some of it, having already said you don't really know the maths. I was just bored and went for a bit of a ramble. You should be alright with the bit about on- and off-shell if you ignore the mention of four-momentum, which I needn't really have included, and take the equation I gave for the square of the mass as given. I'd edit the reply but it's too late to edit it now.
9. It's not the basis of, it's a result of the approach of QM. The more general relation is between two self-adjoint operators A and B such that [imath][A, B]=iC[/imath], where C is another self-adjoint operator, then their variances [imath]\Delta_A[/imath] and [imath]\Delta_B[/imath] satisfy $\Delta_A\Delta_B\geq\tfrac{1}{2}|\langle C\rangle |$ Here the angle brackets denote the mean of C in some state ($=Tr(\rho C)/Tr(\rho )$, for some state operator [imath]\rho[/imath]). In the case of position and momentum the commutation relation between the two is [imath][\vec{P},\vec{Q}] =i\hbar[/imath] giving the familiar Heisenberg uncertainty relation between position and momentum $\Delta\vec{P}\Delta\vec{Q}\geq\tfrac{1}{2}\hbar$ For the uncertainty relation between energy and time, the derivation is not the same as this, for there is no "time operator" per se. The time in this uncertainty relation is the "characteristic time" associated with the variance of some dynamical variable, say that of some measuring apparatus. $\Delta E\Delta\tau\geq\tfrac{1}{2}\hbar$ The deal with on/off mass shell is to do with four-momentum, a four-dimensional vector that has the energy and spatial momentum as components. The definition of four-momentum is $p^{\mu}p_{\mu}=E^2-|\vec{p}|^2=m_0^2$ (units with c=1 for simplicity) Here the p's on the left are the four-momentum, E is the energy, [imath]\vec{p}[/imath] is the spatial momentum and [imath]m_0[/imath] is the rest mass/energy (rest mass if we weren't using units with c=1). In quantum field theory it is feasable for this equality to be violated by virtual particles. If it is the case that the equality holds the particle is said to be on-mass-shell. If the identity does not hold then the particle is said to be off-mass-shell. As Severian said, the violation of the equation he gave for energy indicates a particle being off-shell. It's usually called being off-mass-shell so I gave a definition where the relation to mass is more explicit, the two are obviously equivalent. What Severian was saying is that particles being of shell can be seen to imply Heisenberg uncertainty: a particle could have too much energy by being off-shell than it should have, so it has sort of borrowed energy from the uncertainty it is allowed in its energy from Heisenberg. So the norm [imath]p^{\mu}p_{\mu}[/imath] needn't be equal to the square of the mass because of this uncertainty.
10. Original post deleted. Edit: Actually, stuff it. I'm just going to be labelled a dogmatist by this cretin.
11. Go for it. I'm 16, well nearly 17, and in college (UK equivalent of Highschool) and I taught myself quantum mechanics/field theory and all the prerequisite stuff. Reading popular science is definitely the thing to do first. If you need any help any time just give me a shout.
12. Would these texts be mathematically orientated or purely qualitative, popular science and what not? Popular science wise, "QED-The Strange Theory of Light and Matter"-Feynman is a must, but you've probably read that, it's a popular book. Popular science wise for particle physics I don't really know any more than that, I don't read much popular science any more. The only QED books I own and have really read a great deal of are field theoretic, chief amongst which is Schroeder and Peskin's, where the first part is devoted entirely to QED and Feynman diagrams. The second part focuses on the systematics of renormalisation and the final part on non-abelian field theories and the Standard Model of particle physics, i.e. QCD, GSW (electroweak), the Higgs etc. But it's very good, though it requires a fair amount of mathematics/physics as a prerequisite. It would probably be much better if you've been introduced to QED through relativistic quantum mechanics, I'd have thought. Though I never did much relativistic QM before doing field theory, and it didn't take me long to get the hang of the relativistic formulation. I have read a bit of Halzen and Martin, and it doesn't seem too bad. I teach myself so I have to make do with what I can get hold of, so my recommendations aren't as substantial as Severian's or ajb's.
13. They're the same thing. $\mathbf{g}=\frac{\mathbf{F}}{m}$ The force is just $G\frac{mM}{r^2}\hat{\mathbf{r}}$ To get the absolute value just whack | around the things your taking abs of, there should just be a key on your keyboard for it.
14. It's seperable $\frac{dy}{dx}=\sqrt{a-y^2}$ $\frac{1}{\sqrt{a-y^2}}\frac{dy}{dx}=1$ $\int\frac{1}{\sqrt{a-y^2}}\frac{dy}{dx}dx=\int\frac{1}{\sqrt{a-y^2}}dy=\int dx$
15. What he's done is used pythag to get the distance from the origin to (x, y, z) and then sub'ed in what z is from the equation for the surface.
16. They use accepted, well established science to sell pseudo-science under the guise of philosophy.
17. Damn it, stupid accidental post.
18. You're equation for energy, and thus the de' Broglie wavelength, won't work for light, as you've multiplied it by [imath]m_0\gamma[/imath] and [imath]m_0=0[/imath] for real photons.
19. I'm not even going to start on how much I despise "What the Bleep Do We Know".
20. No, I said we don't just whack a Coloumb potential into Schrodinger's equation to get QED. We can get ourselves potentials from QFT if we apply the non-relativistic limit and use the Born approximation, which relates S-matrices to the classical potential (that in the Schrodinger equation). The Coloumb potential comes quite nicely out of QED.
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