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Comments:
<0> ok I'm off
<0> good luck all
<1> cya
<2> thanks microacg
<1> thanks for the help man
<0> wish me luck grading AP exams XD first time ever, kinda strange
<3> haha
<0> maybe not grading, we'll see ugh
<4> microacg: do you have a job as teacher?
<1> got it right micro
<0> student teacher
<1> thanks for the help
<0> XPX1, of course :D
<1> also one more question with it
<1> you said a = 0
<1> that makes sense but
<1> where did it come into play in this equation
<4> student teacher?? what is that?
<0> the equation came from the fact that a = 0
<1> Ooh otherwise we would put a in there?
<0> in USA student teacher means 4th year college student or graduate student teaching cl***es in school
<0> XPX1, the original equation was F1+F2 = m * a
<0> but a=0 so F1 = -F2
<0> that's where my equation came from generallys peaking
<1> Aaah
<0> the first equation was just Fnet = ma
<0> the 2 tensions in the string cancelled
<1> F1+f2 = m * 0 so F1+F2=0
<0> only left the 2 gravity forces, one straight down, and one along the ramp
<1> F1+F2=0 so F1=-F2
<0> XPX1, yeah
<0> F1 and F2 are in oppiste directions
<1> alright
<0> so the negative sign goes away
<0> and you just use the size
<4> microacg: do you get paid for it?
<0> thus m2 g = m1 g sin (theta)
<1> man for physics being a hard cl*** atleast it makes you think lol
<0> Manyfold, no, it is part of teaching certification
<4> :(
<0> physics makes you think more than
<0> most cl***es
<0> in high school or college
<4> and what is this ap stuff all about
<0> that's okay, I'll get a job for sure next school year
<1> also I think I am able, with my awesome ascii skills... to make a picture of the 1st problem
<0> ap is high school cl***es for advanced students
<0> look up advanced placement
<4> are this courses thaught in high school?
<0> the college board offers it
<0> you can read about it easily
<0> yeah
<0> XPX1 I'm leaving, good luck
<4> gimme a link
<0> later Manyfold
<0> www.thecollegeboard.com
<0> woops
<0> http://www.collegeboard.com/student/testing/ap/about.html
<4> naura: still there??
<1> ok heres a test..
<1> 0
<1> -|- \
<1> \ _________
<1> 21 | BOX |
<1> ----- | |
<1> |________|
<1> now that didnt work out as I planned
<3> Manyfold: YEP!
<3> oops, capslock
<4> ok when you measure something in qm you apply an operator to a state
<1> got time for 1 more question micro?
<4> XPX1: he is gone
<1> d'oh
<1> lol
<1> Manyfold: Do you know stuff about pulleys and newtons laws?
<4> i am occupied
<1> alright no prob
<4> now that operator splits the state into orthonormal states which have an eigenvalue asigned to it
<4> the eigenvalue is the value you actually measure
<4> naura: everything clear till now?
<3> Manyfold: one sec.
<3> Manyfold: yes. but to clarify, a state is an element of a complex hilbert space, yes?
<4> yeah
<3> okay. crystal clear, then
<3> oh, one more thing
<3> linear operator?
<3> differential operator?
<4> a linear operator is a matrix
<4> or better a matrix is a linear operaor
<1> #conference
<1> oops
<3> i thought operators were functions, or are you referring to the function whose matrix is the subject?
<4> no i refer to matrices
<3> er, how is a matrix an operator, then? application of the matrix?
<0> hm ap c part 2 is cake this year :D
<1> oh hey micro!
<4> lets look at H|psi> =E|psi> where H is the operator |psi> is an element of a complex hilbertspace and E is an eigenvalue of H
<3> okay.
<1> Microacg: you busy?
<4> we see that H maps |psi> to an vector that is parallel to |psi>
<4> XPX1: he is gone
<3> Manyfold: indeed.
<1> hes like the wind.. lol
<4> as E is a real number
<3> and H is an operator. gotcha.
<3> so the linear transformation denoted by a matrix would be a linear operator.
<4> okay where jave we been so fa?
<4> got messed up over here
<3> haha. well, i understand the time-independent equation H|psi> = E|psi>, where H is the Hamiltonian (a linear 1-adic operator), |psi> is an element of a complex hilbert space, and E is an eigenvalue of the H operator.
<3> amirite?
<4> ok as |psi> is an element of a complex vector space it surely can be brocken down to components of some orthonormal basis
<4> so |psi> = sum_i a_i <a_i|psi>
<3> where <a_i|psi> denotes the inner/dot product of a_i (an element of the orthonormal basis) and psi, yeah?
<4> yes
<3> okay, so essentially we're taking the dot product of psi with each element of the orthonormal basis and adding them up (they're scalar quantities, right?)
<4> so we can write our operator as A|psi> = sum_i a_i|a_i><a_i|psi>
<4> right
<4> did you see how this works?
<3> well, one sec. when we take the dot product of psi with each element of the orthonormal basis and add them up, wouldn't we get a scalar quantity? that doesn't seem to work, as psi is a vector.
<3> am i misunderstanding the summation here?
<4> use A|psi>=a|psi>
<3> then that follows, yes, as a_i would be the elements of A
<4> argh i did a mistake above
<4> |psi> = sum_i |a_i> <a_i|psi>
<4> now it's correct
<3> ah, yes. now |psi> is the sum of (the elements of the orthonormal basis times (the dot product of the elements of the orthonormal basis and psi))
<3> aka it's the sum of multiples of the elements of the orthonormal basis
<4> now we see that what is happening our operator A decomposes psi into an orthonormal basis
<3> indeed
<4> time for another joint please stand by
<3> haha. i smoke every morning before school, wake 'n bake is delightful.
<4> naura: what school are you attending too?
<3> high school
<3> in texas :\
<3> (public)
<4> the red neck state :)
<3> gah, don't get me started. our school cut the number of cl***es we can take to fund the athletic program.
<3> you're in high school too?
<4> eh no i am already finished with school
<3> ah, sweet.
<4> ok back to our measurement process
<3> righto
<4> now our operator has split up |psi> in handy components
<3> 'k.
<4> of these one is picked at random with probabillity |<a_i|psi>|^2
<4> the ***iociated value we measure is a_i
<4> why do you do qm?
<4> that's for a discrete spectrum
<4> if you want a continous spectrum just replace the sum with qa
<4> an integral
<4> for example position and momentum have a continous spectrum
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