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19

What you observe is the general phenomenon that in relativistic theories time translation is replaced by "affine-parameter-translation" or "wordline translation symmetry" and hence the corresponding Hamiltonian becomes a constraint, the constraint that states must be invariant under this symmetry. Yes, this works for the relativistic spinning particle and ...


16

Gravity has to be subject to quantum mechanics because everything else is quantum, too. The question seems to prohibit this answer but that can't change the fact that it is the only correct answer. This proposition is no vague speculation but a logically indisputable proof of the quantumness. Consider a simple thought experiment. Install a detector of a ...


15

Dear Calvin, if any portion of the world is described by probabilistic wave functions, then the whole world has to be. It's easy to show it. Take a decaying nucleus, connect it to a hammer that kills a cat a that also makes the Sun explode into 2 pieces. The nucleus is evolving into a linear superposition of "decayed" and "not yet decayed" states. ...


13

Space and time are continuous, in quantum mechanics or otherwise. In particular, whenever our theories of any kind talk about time, it is always a real continuous parameter. Similarly, spatial positions of particles in ordinary quantum mechanics are operators $\hat x$ whose eigenvalues are continuous, too. This fact is related to the continuity of time by ...


13

One more answer against “second quntization”, because I think it is a good demonstration of how a lame notation can obscure a physical meaning. The first statement is: there is no second quantization. For example, here is citation from Steven Weinberg's book “The Quantum Theory of Fields” Vol.I: It would be a good thing if ...


12

First, you are right in that non-Minkowski solutions to string theory, in which the gravitational field is macroscopic, it should be thought of as a condensate of a huge number of gravitons (which are one of the spacetime particles associated to a degree of freedom of the string). (Aside: a point particle, corresponding to quantum field theory, has no ...


12

Let me add two references to points already mentioned in this discussion: Today, there is no reason known why the electric charge has to be quantized. It is true that the quantization follows from the existence of magnetic monopoles and the consistency of the quantized electromagnetic field, which was shown first by Dirac, you'll find a very nice exposition ...


12

A classical1 theory of (relativistic) $p$-dimensional membranes exists for any non-negative integer $p$. Such classical membrane-like objects appears in the full non-perturbative formulation of string theory. The problem arises if one tries to quantize a membrane (in a first quantized sense) using standard perturbative quantization methods, which have ...


10

Same as @SamRoelants answer but not restricted to gravitational waves. Given $$G_{\mu\nu}=8\pi T_{\mu\nu}$$ $T_{\mu\nu}$ is constructed from the matter fields (Klein Gordon, Dirac or whatever). These are operators (or operator-valued distributions if you like), hence so is the gravitational source $T_{\mu\nu}$. So the right hand side obeys the rules of ...


9

Charge comes from discrete symmetries and is countable and additive. Mass comes from continuous 4d space, is exchangeable with energy and, in quantum mechanical dimensions not linearly additive, thus not countable. Suppose you have an elementary quantum of mass, $m_q$. In the world we know two such quanta would not end up as $2m_q$. One would add the ...


8

Dear asmailer, the reason is simple and completely understood: the electric charge is the generator of a $U(1)$ symmetry which is compact and may be parameterized by an angle, $\phi$. So wave functions may only depend on the angle $\phi$ in a periodic way, $\exp(iQ\phi)$ where $Q$ is integer (or an integer multiple of $e/3$, if I look at the elementary ...


8

You seem to be talking about the "old covariant quantization" in which $L_n$ for positive $n$ and $(L_0-a)$ annihilate physical ket states $|\psi\rangle$, right? It's analogous to the Gupta-Bleuler quantization http://en.wikipedia.org/wiki/Gupta-Bleuler_quantization which was a standard procedure used already in electromagnetism. The idea is that the ...


8

I have written an answer to Mathoverflow in which explicit formulas for the classical and quantum Hamiltonians of a spin system (Generators of $SU(2))$ were written explicitely. The classical Hamiltonians are given by means of functions on the two sphere and the quantum Hamiltonians by means of holomorphic differential operators (which act on the sections of ...


8

If I'm only allowed to use one single word to give an oversimplified intuitive reason for the discreteness in quantum mechanics, I would choose the word 'compactness'. Examples: The finite number of states in a compact region of phase space. See e.g. this Phys.SE post. The discrete spectrum for Lie algebra generators of a compact Lie group, e.g. angular ...


8

The word quantization surface is not standard terminology. It apparently refer to a (generalized) Cauchy surface. A (generalized) Cauchy surface is a hypersurface on which the initial conditions are given for a well-posed initial value problem. Phrased differently, for given initial conditions on the Cauchy surface, there exists a unique solution for the ...


7

One way to see the validity of the background field method (BFM) lies in the proof of the equivalence of the effective action calculated with the BFM to the standard effective action. Let $\Gamma[v]$ be the effective action (Legendre transform of the connected generating function $W[J]$) where $v=v(J)=\frac{\delta W[J]}{\delta J}$ is the "classical" field ...


7

First, quite generally, it is not true that the action $S$ is ever required to be a multiple of Planck's constant $h$. What quantum mechanics implies is pretty much exactly the opposite thing. The formulation of quantum mechanics (any quantum mechanical theory) that uses the action is the Feynman path integral where the action enters via the exponent $$ ...


7

The ordering ambiguity is the statement – or the "problem" – that for a classical function $f(x,p)$, or a function of analogous phase space variables, there may exist multiple operators $\hat f(\hat x,\hat p)$ that represent it. In particular, the quantum Hamiltonian isn't uniquely determined by the classical limit. This ambiguity appears even if we require ...


7

The core of perturbative string theory has a mathematically rigorous formulation. In fact much of mathematical physics and mathematical insight into quantum field theory as such has been gained from the study of the low-dimensional QFTs that constitute the worldvolume theories of the string and the various branes. For instance the axiomatization of QFT in ...


6

Discrete eigenvalues for measurable observables in QM occur more broadly than just for bound states. For example angular momentum, whether orbital or spin, is always restricted to discrete eigenvalues for all systems, free and unbound as well as bound systems. It is the case, however, that discrete spectra is 'associated' with compact geometries. Thus the ...


6

Your example shows that you may use symmetry to get a Hamiltonian (which should be invariant) and for classification of its solutions: it is convenient to choose wavefunctions in a way that they form the basis of irreducible representations of the symmetry group. To get the numbers you need to solve the equations, their symmetry is not enough. Symmetry may ...


6

This answer addresses the geometrical origin of the Bohr-Sommerfeld condition. In geometric quantization, the additional structure required beyond the symplectic data of the phase space is a polarization. The quantization spaces are constructed as spaces of polarized sections with respect a polarization. The most "obvious" type of a polarization is the ...



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