Jump to content


Senior Members
  • Posts

  • Joined

  • Last visited

Profile Information

  • Location
    in the Lab, mostly
  • Favorite Area of Science
    Materials/Solid State Physics


  • Molecule

DQW's Achievements


Molecule (6/13)



  1. As Yggdrasil points out, the First Law is nothing but a restatement of energy conservation.
  2. 4p < 5s < 4d got missed. Also, note that these are the order of filling orbitals in isolated gaseous atoms (not, for instance, in a crystalline solid, where near-neighbor interactions slightly alter the energy levels). The "order" can be generated from the following two rules : 1. An orbital with a lower value of n+l fills first, 2. In case of a tie, the orbital with the lower n fills first.
  3. Illegal use of units : 10P * 10P = 100 P2 not 100P
  4. Sorry to be blunt, but this is entirely gibberish.
  5. At 3 minutes, is the bulb on or off ?
  6. The canonical conjugate variable for angular momentum is angular position, not position.
  7. Right. And you "did that" by writing out H = p^2/2m + V = total energy (at least when the Lagrangian is not explicitly time-dependent). The idea was to essentially "modify" the Classical Hamiltonian by using the momentum operator (as "derived") instead of the classical canonical momentum.
  8. Severian's approach postulates the wavefunction of a spinless free particle to be such-and-such, and derives the SE from there (but only for a free-particle). My (incomplete) approach starts with the classical mechanics and quantizes the various elements of it. On the other hand, the SE can itself be postulated (as in Dirac's formulation of NRQM) and things calculated from there.
  9. Sound is purely longitudinal, only in a medium that can not support shear (ie: a fluid, like air or water). In a solid you have both longitudinal and transverse modes of propagation. Metals are sonorous because (i) they are crystalline, and have little or no means to disprerse the sound energy (ie: they have very small damping coefficients), and (ii) their elastic modulii have values that are right to make everday sized metal objects possess a natural frequency that is in the audible range.
  10. I thought Polignac stated that any even number can be written as a difference of primes in an infinite number of ways (a more general version of the twin-prime conjecture) ? <after quickly Googling> But this conjecture (odd = prime + power of 2) also appears to be attributed to Polignac. There's another counterexample at 877, but I'm not sure if this is the next one. The smallest counterexample is 127. It looks like many of these numbers are themselves primes.
  11. At what level of math preparation ?
  12. It's not trivial. You must arrive at the general result that the momentum operator (in position representation) is given by the gradient. This requires going over a similar process as above with spatial translations... but I decided to skip that. Or alternatively, I could say "But you already did that for me !"
  13. A nice way of arriving the the SE is by looking at time evolution of a quantum state' date=' and modeling this through a unitary operator. Let the state [imath']| \alpha \rangle [/imath] evolve with time as : [math]| \alpha \rangle ~ \xrightarrow{time~evolution} | \alpha,t \rangle [/math] We create an opertor [imath]{\cal U} (t,0) [/imath] that provides this time-evolution, so that : [math] | \alpha ,t \rangle = {\cal U} (t,0) ~| \alpha \rangle [/math] Now, this time evolution operator is required to satisfy a whole bunch of properties. For instance, one must have [math] \lim _{dt \rightarrow 0} {\cal U} (dt,0) = \mathbf {1} [/math] All these requirements are satisfied by writing [imath] {\cal U} (dt,0) = \mathbf {1} - i \mathbf{\Omega} dt[/imath], where [imath]\mathbf{\Omega} [/imath] is a Hermitian operator. Now borrowing from classical mechanics, the idea that the Hamiltonian is the generator of time-translation (just as momentum is the generator of spatial translations), we choose [imath]\mathbf{\Omega} [/imath] to be [imath] {\cal H} / \hbar [/imath]. And the Schrodinger Equation follows automatically from playing with the composition property of the time-evolution operator. This property requires [math] {\cal U} (dt_1 + dt_2, 0) = {\cal U} (dt_2,dt_1) ~{\cal U} (dt_1, 0) [/math] The need for this property is obvious : if you evolve a state through time [imath]dt_1[/imath] and then through [imath]dt_2[/imath], you expect the final state to be the same as that caused by a time-evolution through [imath]dt_1 + dt_2[/imath]. Now, we can use this composition property to write [math] {\cal U} (t+dt, 0) = {\cal U} (t+dt,t) ~ {\cal U} (t,0) = \left( \mathbf{1} - \frac{i {\cal H} dt}{\hbar} \right) {\cal U} (t,0) [/math] Multiplying out, and rearranging terms, this can be written in the differential form : [math]i \hbar \frac {\partial}{\partial t} {\cal U} (t,0) = {\cal H} {\cal U} (t,0) [/math] Oprating this on the initial state ket [imath] | \alpha \rangle = |\alpha, t=0 \rangle [/imath] gives [math]i \hbar \frac {\partial}{\partial t} {\cal U} (t,0)~| \alpha \rangle = {\cal H} {\cal U} (t,0)~| \alpha \rangle [/math] But from the definition of the time-evolution operator, this gives [math]i \hbar \frac {\partial}{\partial t} | \alpha , t \rangle = {\cal H} | \alpha ,t \rangle [/math] ta dah !
  14. Most high Tc cuprates are not very "chemically stable", primarily, I think, due to the possibility of oxygen diffusion into the material altering the doping level. I don't know if there's any problem in an oxygen-free atmosphere, but IO suspect there are. Recently, people have been working on DyBCO as an alternative to YBCO, mostly because of its better chemical stability. Also, folks have been looking into inert coating with good lattice matching (primarily MgO, I think). Give it a look see.
  15. I like this quote : --John Allen Paulos, author of Innumeracy: Mathematical Illiteracy and its Consequences Ummm...that's all.
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.