Help a Physics Student with Two Dim. Oscilator Problem

[noamriemer]

Hello! I am new here...
I am a physics student, seeking for help. There is something I just can't seem to understand...
Your help will be much appreciated...

In my exercise, I have the following Hamiltonian:
[tex]H=\frac{p_x^2}{2m}+\frac{p_y^2}{2m}+\frac{1}{2}m \omega^2 (x^2+y^2)+\lambda xy[/tex]

I was supposed to find a transformation for which there will be no "linked" coordinates. This is what I came up with:
[tex]H=\frac{p_a^2}{2m}+\frac{p_b^2}{2m}+\frac{a^2}{2}(m\omega^2+\lambda)+\frac{b^2}{2}(m\omega^2-\lambda)[/tex]

I need to find the transition matrix. And here I am facing my problems:
The matrix is given by:
[tex]H=\left \langle u_i|H|u_j \right \rangle= \left \langle u_i|v_j \right \rangle\left \langle v_j|H|v_l \right \rangle\left \langle v_l \right |u_j\rangle[/tex]

But I can't understand... what are the [tex]u_j[/tex] ?
Does it refer to the coordinates- so in my case it will be:
[tex]H=\left \langle x|H|y \right \rangle= \left \langle x|b \right \rangle\left \langle b|H|a \right \rangle\left \langle a \right |y\rangle[/tex]

If it is so: what does [tex]|x\rangle[/tex] mean?

Thank you all so much!

 

[singhvi]

In the first part you need to add and subtract terms such that there is no x*y term, or in other words, the equation can be separated into two independent equations, each dependent on only one co-ordinate.
So if you can write the equation in a new co-ordinate system a and b, like you have, then yes, it is what is required as it can be separated into two independent equations.

As for the second part, I am not a 100% about this, but ui and uj should not be the co-ordinates.

 

[haael]

If it is so: what does [tex]|x\rangle[/tex] mean?

This is a purely localized state. A particle (wavefunction) that has an exact, 100% accurate position, but totally undefined momentum. You can imagine it as a delta function with a peak at position x. I like to think that it's the image of a drop just hitting the surface of water.

 

[noamriemer]

Thank you, guys... you've been great help!