Quick question about matrices & bases?

[jeebs]

Say I was given a 2x2 matrix made from a certain basis {|x\rangle, |y\rangle} , and I split that matrix into two parts, one being the diagonal part and one being the off-diagonal part.

for example, if I had H = H_0 + W = \left(\begin{array}{cc}a&c\\b&d\end{array}\right) = \left(\begin{array}{cc}a&0\\0&d\end{array}\right) + \left(\begin{array}{cc}0&c\\b&0\end{array}\right)

Is it true to say that H₀ is still in the basis {|x\rangle, |y\rangle}, and if it is, is there a way I could determine what |x\rangle and |y\rangle actually look like?

 

[fzero]

Your question is a bit ambiguous. The explicit matrix form that you've written down is in the basis

|e_1\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}, |e_2\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}.

Then

H = a |e_1\rangle \langle e_1 | + d |e_2\rangle \langle e_2 | + b |e_2\rangle \langle e_1 | + c |e_1\rangle \langle e_2 |.

If you want to express this in another basis, you could just rotate |e_{1,2}\rangle into |x,y\rangle.

 

[jeebs]

well to be more specific I'm attempting this:

[PLAIN][URL]http://i52.photobucket.com/albums/g33/long_john_cider/problem.jpg[/PLAIN][/URL]

I've done part one, which involves coming up with a Hamiltonian then splitting it into two parts, an initial part H₀ which has the diagonal elements, and a time dependent part W(t) which has the off diagonal elements.

On the second part I need to mess around with that integral for a(t), and to do that I must need to know what |\uparrow \rangle , | \downarrow \rangle are.

So, I'm wondering how I can find them. I have "constructed the Hamiltonian in the basis |\uparrow \rangle , | \downarrow \rangle " in part one, so now I'm wondering how to find out what this basis actually is...

 

[fzero]

Well |\uparrow\rangle = | e_1\rangle, |\downarrow\rangle = | e_2\rangle, but that doesn't quite answer your question. If you wanted to write an expression for a matrix in another basis, you would need a formula that relates the two bases.

(Edit: Maybe it does answer your question)

 

[jeebs]

I'm not sure I understand how you've decided that |\uparrow\rangle = \left(\begin{array}{c}1&0\end{array}\right) and |\downarrow\rangle = \left(\begin{array}{c}0&1\end{array}\right)

 

[fzero]

I'm not sure I understand how you've decided that |\uparrow\rangle = \left(\begin{array}{c}1&0\end{array}\right) and |\downarrow\rangle = \left(\begin{array}{c}0&1\end{array}\right)

They're the eigenvectors of \sigma_z.

 

[jeebs]

ah, so they are, and I have H_0 = B_0\sigma_z.
I might be getting somewhere.
thanks.