Quantum Mechanics Help

[Ben26]

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:
http://i34.tinypic.com/2j4bpmc.jpg
Any help would be very welcome.

[Ben26]

What entries does \left \langle i \left|\hat{H}\right \left| j \rangle refrer to?

[latentcorpse]

well if i=v_e and j=v_u then i reckon your meant to compute the matrix element

<v_e | \hat{H} | v_\mu >

multiply out those matrices in your first post to get v_e,v_\mu in terms of v1 and v2 and then see what you get...

[Ben26]

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

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

[Ben26]

Still cant see how i get to <v_e|\hat{H}|v_{\mu}>

[Ben26]

Not that i know what <v_e|\hat{H}|v_{\mu}> is or should look like...

[latentcorpse]

well you can write this as

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

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

also, you do know what <i|\hat{H}|j> is - it is the ij^{th} entry in this matrix. as for what it looks like, well, that's going to be the answer to the quesiton.

[Ben26]

\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?

[latentcorpse]

looks fine.
now use orthogonality of the v_i when you multiply out the brackets.

[Jasso]

now use orthogonality of the v_i when you multiply out the brackets.

IE. The fact that <v_a|v_b> is the inner product of states v_a and v_b and that v_1 and v_2 are orthogonal.

[Ben26]

=
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...

[latentcorpse]

ok so, i think i probably could have explained myself better earlier but nonetheless....

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

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

H will look something like this
\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]

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

3 similar calculations will give you the other entries though.

[Ben26]

Finally got there! Thanks for your help!

http://i37.tinypic.com/25qal3k.gif

[Ben26]

...continuing from the same question, here is the next bit which i have tried but cannot do:

http://i36.tinypic.com/21bw3k6.jpg

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\rangl e cos \varphi - \left|v_{1}\right\rangle sin \varphi


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

\left|v_{e}\right\rangle becomes

\left|v_{\mu}\right\rangle

Is this the explanation?

[Ben26]

any ideas? im really stuck...

[latentcorpse]

what's JPARC and T2K?

[Ben26]

Its a place in Japan where they are experimenting with neutrinos, i think its irrelevant to the question.

[Ben26]

JPARC is the accelerator and T2K is the experiment name.

[Ben26]

any ideas on how to go about answering this?