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Quantum Mechanics Help

  1. Oct 30, 2009 #1
    1. The problem statement, all variables and given/known data

    I am working through past paper questions because i am finding the quantum mechanics module im taking very hard. I dont know how to go about this question:
    2j4bpmc.jpg
    Any help would be very welcome.
     
  2. jcsd
  3. Oct 30, 2009 #2
    What entries does [tex]\left \langle i \left|\hat{H}\right \left| j \rangle[/tex] refrer to?
     
  4. Oct 30, 2009 #3
    well if i=v_e and j=v_u then i reckon your meant to compute the matrix element

    [itex] <v_e | \hat{H} | v_\mu >[/itex]

    multiply out those matrices in your first post to get [itex]v_e,v_\mu[/itex] in terms of v1 and v2 and then see what you get...
     
  5. Oct 30, 2009 #4
    [tex]\left|v_{e}\right\rangle=\left|v_{1}\right\rangle cos \varphi + \left|v_{2}\right\rangle sin \varphi[/tex] for i

    [tex]\left|v_{\mu}\right\rangle=\left|v_{2}\right\rangle cos \varphi - \left|v_{1}\right\rangle sin \varphi[/tex] for j
     
  6. Oct 30, 2009 #5
    Still cant see how i get to [tex]<v_e|\hat{H}|v_{\mu}>[/tex]
     
  7. Oct 30, 2009 #6
    Not that i know what [tex]<v_e|\hat{H}|v_{\mu}>[/tex] is or should look like...
     
  8. Oct 30, 2009 #7
    well you can write this as

    [itex] \left( \cos{\varphi} < v_1 | + \sin{\varphi} < v_2 | \right) \hat{H} \left(\cos{\varphi} | v_2 > - \sin{\varphi} | v_1 > \right)[/itex]

    see what happens after you apply the Hamiltonian on the second bracket

    also, you do know what [itex]<i|\hat{H}|j>[/itex] is - it is the [itex]ij^{th}[/itex] entry in this matrix. as for what it looks like, well, that's going to be the answer to the quesiton.
     
  9. Oct 30, 2009 #8
    [itex]
    \left( \cos{\varphi} < v_1 | + \sin{\varphi} < v_2 | \right) \left(\cos{\varphi}\hat{H} | v_2 > - \sin{\varphi} \hat{H}| v_1 > \right)
    [/itex]

    [itex]=
    \left( \cos{\varphi} < v_1 | + \sin{\varphi} < v_2 | \right) \left(\cos{\varphi}\ E_{2} | v_2 > - \sin{\varphi} \ E_{1}| v_1 > \right)
    [/itex]

    before i continue, is this right?
     
  10. Oct 30, 2009 #9
    looks fine.
    now use orthogonality of the [itex]v_i[/itex] when you multiply out the brackets.
     
  11. Oct 30, 2009 #10
    IE. The fact that [itex]<v_a|v_b>[/itex] is the inner product of states [itex]v_a[/itex] and [itex]v_b[/itex] and that [itex]v_1[/itex] and [itex]v_2[/itex] are orthogonal.
     
  12. Oct 30, 2009 #11
    [itex]
    =
    E_{2} cos{\varphi}^{2} < v_1 |v_2 > - E_{1} cos{\varphi}sin{\varphi} < v_1 |v_1 > + E_{2} cos{\varphi}sin{\varphi} < v_2 |v_2 > - E_{1} sin{\varphi}^{2} < v_2 |v_1 >

    [/itex]

    [itex]
    =
    E_{2} sin{\varphi}cos{\varphi} - E_{1} sin{\varphi}cos{\varphi}
    [/itex]

    Is this right? I still need to get to a matrix somehow...
     
  13. Oct 30, 2009 #12
    ok so, i think i probably could have explained myself better earlier but nonetheless....

    ok so this entry we have [itex](E_2-E_1) \sin{\varphi} \cos{\varphi}[/itex]

    so you're trying to get this matrix H where the entries in H are given by [itex]<i|\hat{H}|j>[/itex] and [itex]i,j \in \{ v_e , v_\mu \}[/itex]

    H will look something like this
    [itex] \left[ \begin {array}{cc} \left[ \begin {array}{ccc} < v_{{e}}& | \hat{H} |&v_{{e
    }} > \end {array} \right] & \left[ \begin {array}{ccc} < v_{{e}}& | \hat{H} | &v_{{\mu}} >
    \end {array} \right] \\ \noalign{\medskip} \left[ \begin {array}{ccc}
    < v_{{\mu}}& | \hat{H} | &v_{{e}} > \end {array} \right] & \left[ \begin {array}{ccc} < v_{
    {\mu}}& | \hat{H} | & v_{{\mu}} > \end {array} \right] \end {array} \right] [/itex]

    so we have computed the entry that goes in the first row,2nd column

    3 similar calculations will give you the other entries though.
     
  14. Oct 31, 2009 #13
    Finally got there! Thanks for your help!

    25qal3k.gif
     
  15. Oct 31, 2009 #14
    ...continuing from the same question, here is the next bit which i have tried but cannot do:

    21bw3k6.jpg

    i think i should be looking at

    [tex]
    \left|v_{e}\right\rangle=\left|v_{1}\right\rangle cos \varphi + \left|v_{2}\right\rangle sin \varphi
    [/tex]
    [tex]
    \left|v_{\mu}\right\rangle=\left|v_{2}\right\rangle cos \varphi - \left|v_{1}\right\rangle sin \varphi
    [/tex]

    and i can kind of see that if you translate the [tex]
    \varphi
    [/tex] by [tex]\pi /2 [/tex] then [tex]

    \left|v_{e}\right\rangle[/tex] becomes [tex]

    \left|v_{\mu}\right\rangle[/tex]

    Is this the explanation?
     
  16. Nov 1, 2009 #15
    any ideas? im really stuck...
     
  17. Nov 1, 2009 #16
    what's JPARC and T2K?
     
  18. Nov 1, 2009 #17
    Its a place in Japan where they are experimenting with neutrinos, i think its irrelevant to the question.
     
    Last edited: Nov 1, 2009
  19. Nov 1, 2009 #18
    JPARC is the accelerator and T2K is the experiment name.
     
  20. Nov 1, 2009 #19
    any ideas on how to go about answering this?
     
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