1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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:
    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
    \left( \cos{\varphi} < v_1 | + \sin{\varphi} < v_2 | \right) \left(\cos{\varphi}\hat{H} | v_2 > - \sin{\varphi} \hat{H}| v_1 > \right)

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

    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
    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 >


    E_{2} sin{\varphi}cos{\varphi} - E_{1} sin{\varphi}cos{\varphi}

    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!

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


    i think i should be looking at

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

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

    \left|v_{e}\right\rangle[/tex] becomes [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?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook