# Quantum Mechanics Help

1. Oct 30, 2009

### 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:

Any help would be very welcome.

2. Oct 30, 2009

### Ben26

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

3. Oct 30, 2009

### 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...

4. Oct 30, 2009

### Ben26

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

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

5. Oct 30, 2009

### Ben26

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

6. Oct 30, 2009

### Ben26

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

7. Oct 30, 2009

### 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.

8. Oct 30, 2009

### 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?

9. Oct 30, 2009

### latentcorpse

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

10. Oct 30, 2009

### Jasso

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.

11. Oct 30, 2009

### 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...

12. Oct 30, 2009

### 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.

13. Oct 31, 2009

### Ben26

Finally got there! Thanks for your help!

14. Oct 31, 2009

### Ben26

...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 $$\varphi$$ by $$\pi /2$$ then $$\left|v_{e}\right\rangle$$ becomes $$\left|v_{\mu}\right\rangle$$

Is this the explanation?

15. Nov 1, 2009

### Ben26

any ideas? im really stuck...

16. Nov 1, 2009

### latentcorpse

what's JPARC and T2K?

17. Nov 1, 2009

### Ben26

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

Last edited: Nov 1, 2009
18. Nov 1, 2009

### Ben26

JPARC is the accelerator and T2K is the experiment name.

19. Nov 1, 2009