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dan815

  • one year ago

State space stuff @kainui http://prntscr.com/8qcmml

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  1. anonymous
    • one year ago
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    horor question :v

  2. dan815
    • one year ago
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    ah its jut an intro question

  3. Kainui
    • one year ago
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    Cool, looks like we just plug and chug into \[|\alpha|^2 + |\beta|^2 = 1\] for that first one.

  4. dan815
    • one year ago
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    i think using alpha=x+iy = r*e^itan^-1(y/x)

  5. dan815
    • one year ago
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    |dw:1444631085836:dw|

  6. Kainui
    • one year ago
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    Well here we're given that \[\alpha = e^{i \gamma} \cos \theta\] and beta as well

  7. dan815
    • one year ago
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    right okay so i just have to show it satisfies that eqn

  8. dan815
    • one year ago
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    it has to because that complex e^itheta is 1 rotated around so modulus of that =1

  9. Kainui
    • one year ago
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    \[|\alpha|^2 = \alpha^* \alpha = e^{i \gamma - i \gamma} \cos^2 \theta = \cos^2 \theta\]

  10. dan815
    • one year ago
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    |dw:1444631358098:dw|

  11. dan815
    • one year ago
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    oo ok

  12. dan815
    • one year ago
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    i didnt think of just using conjugate like that

  13. dan815
    • one year ago
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    is this a property though? |A B| = |A| |B| ?

  14. Kainui
    • one year ago
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    Well we could find out I guess

  15. dan815
    • one year ago
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    |AB|^2 = (AB)* BA = A* B* B A = |A| |B|

  16. Kainui
    • one year ago
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    \[\sqrt{a^*ab^*b} = \sqrt{a^*a}\sqrt{b^*b}\] checks out

  17. dan815
    • one year ago
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    k cool

  18. dan815
    • one year ago
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    next qn

  19. dan815
    • one year ago
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    http://prntscr.com/8qcptc

  20. dan815
    • one year ago
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    what does it mean by same "physical" state

  21. dan815
    • one year ago
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    i can see that multiplying by anything of modulus 1 will not change that state vector property* of the |w> = A|0> + B |1> |A|^2 + |B|^2=1

  22. Kainui
    • one year ago
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    it's the same physical state because a complex phase doesn't change it, \[|e^{i \theta} a|^2 = |a|^2\]

  23. Kainui
    • one year ago
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    So yeah I think that's what you're saying

  24. dan815
    • one year ago
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    hmmm

  25. Kainui
    • one year ago
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    I know this to be true, in fact I know I can always choose my eigenstates to be real and I can prove it.

  26. dan815
    • one year ago
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    so like grapinh PSI has no meaning?

  27. dan815
    • one year ago
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    graphing*

  28. Kainui
    • one year ago
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    Graphing anything has no meaning, however complex phase on a wave function has meaning and it's important to QM

  29. Kainui
    • one year ago
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    It's just when you're looking at pure states that you don't have interference so it doesn't do anything haha

  30. dan815
    • one year ago
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    ok i see i guess multiplying by modulus 1 is just like multiplying some equation by 2 on both sides

  31. dan815
    • one year ago
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    y=x 2y=2x same thing

  32. Kainui
    • one year ago
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    Not quite, the phase is more of a relative thing that gives you the interference patterns from being in two different states simultaneously

  33. dan815
    • one year ago
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    what does that mean

  34. dan815
    • one year ago
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    so since both states are rotated same equally

  35. Kainui
    • one year ago
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    The thing is, the particle has a probability of being in multiple states at once and those PROBABILITIES interfere with each other with the complex phase.

  36. dan815
    • one year ago
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    its just the same sphere with a diff coordinate system

  37. dan815
    • one year ago
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    ah i see since probability didnt change it didnt change anything with the complex phase?

  38. Kainui
    • one year ago
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    No no I think this is something I can't explain you have to hear Alan Adams explain it, I'm just a mortal.

  39. dan815
    • one year ago
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    okay i see this complex phase wrt to prob isnt really makign sense right now, ill come back to that then

  40. Kainui
    • one year ago
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    The 'idea' of where a particle could have traveled is what interferes with the particle's path.

  41. Kainui
    • one year ago
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    and because of this, the complex phase plays a role during this interference, but right now we're not doing anything complicated so let's just ignore this until we come to it and let's move forward.

  42. dan815
    • one year ago
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    oh right i do remember this its kinda weird

  43. dan815
    • one year ago
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    we even calculate the wave collision pattern

  44. dan815
    • one year ago
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    since we assume the particles is all places with some prob

  45. dan815
    • one year ago
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    i am kind of getting an idea, so i should be thinking of the wave patterns that can arise with some changes in this angle

  46. dan815
    • one year ago
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    |dw:1444632321197:dw|

  47. dan815
    • one year ago
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    or maybe a phase just means the same wave pattern just happened somewhere else on the wall now

  48. dan815
    • one year ago
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    |dw:1444632415530:dw|