Quick question: eigenfunctions

  • #1
I'm doing quantum mechanics with only a little experience in linear algebra. I've been working on eigenstates/values/functions/whatever for a couple days but still having a little trouble. Here's a question I had recently, and if anyone can do a quick check of my work and point me in the right direction, I'd appreciate it.

Homework Statement



The Hamiltonian operator for a system in 2-d vector space is iΔ(|w1><w2| - |w2><w1|). |w1> and |w2> are eigenstates of an observable operator Ω(hat).

Homework Equations



The problem asks me to do a bunch of things:
1) Represent H(hat) and Ω(hat) in a matrix in the basis |w1>,|w2>.
2) Find the eigenvalues E1 and E2 and the normalized eigenstates |E1> and |E2>.
3) Express |w1> and |w2> as linear combinations of |E1> and |E2>.
4) Finally, use this linear combination to express H(hat) and Ω(hat) in the basis |E1>,|E2>.

The Attempt at a Solution



Now, first I represented H(hat) as a matrix that looked like 0, iΔ (row 1) and -iΔ, 0 (row 2). I wasn't 100% sure how to do Ω(hat), but I figured the only non-zero values will be on the main diagonal, since we're representing it in the basis of its eigenstates.

Then, I solved for the eigenvalues and got E1 = -Δsqrt(2) and E2 = Δsqrt(2). This is the first area I got stuck -- I sort of "guessed" on how to get the eigenfunction from the eigenvalues. I think it involves solving a system of equations, but again, I'm pretty inexperienced with linear algebra.

In terms of making these eigenstates linear combinations of each other, I'm sure that means transform |w1> and |w2> into an additive combination of the new normalized eigenfunctions |E1>, |E2>. As long as I have |E1> and |E2> this seems straightforward enough.

So, I suppose if anyone can provide a straightforward method to transforming an eigenvalue into an eigenfunction and normalize it, it would be a huge help. Thanks for reading, and for all the help in the past.

EDIT: I guess this wasn't really a "quick" question in the end, sorry. But I feel like most of the helpers on PF can probably shed some light on this in about a sentence or two.
 
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Answers and Replies

  • #2
vela
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Let A be a matrix and x an eigenvector of A. Then you know that Axx. Rearranging a bit, you get (A-λI)x=0, where I is the identity matrix. This is the system you want to solve to find x.

For example, say you have the matrix

[tex]A=\begin{pmatrix}1 & 2 \\ 2 & 1\end{pmatrix}[/tex]

which have eigenvalues λ1=-1 and λ2=3. The first eigenvector satisfies

[tex](A-\lambda I)\vec{x} = \begin{pmatrix}1-(-1) & 2 \\ 2 & 1-(-1)\end{pmatrix}\vec{x} = \begin{pmatrix}2 & 2 \\ 2 & 2\end{pmatrix}\vec{x} = 0[/tex]

This yields one independent equation 2x+2y=0. So just arbitrarily set one of the variables to 1 and solve for the other. In this case, you get x=1 and y=-1. You can verify this indeed is an eigenvector by multiplying it into the original matrix:

[tex]\begin{pmatrix}1 & 2 \\ 2 & 1\end{pmatrix}\begin{pmatrix} 1 \\ -1\end{pmatrix} = \begin{pmatrix} -1 \\ 1 \end{pmatrix} = (-1)\begin{pmatrix}1 \\ -1\end{pmatrix}[/tex]

and verifying the result is the vector multiplied by the eigenvalue.

To normalize the vector, just divide it by its length:

[tex]\|\vec{x}\| = \sqrt{(1)^2+(-1)^2} = \sqrt{2}[/tex]

So

[tex]\hat{x} = \frac{\vec{x}}{\|\vec{x}\|} = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2}\end{pmatrix}[/tex].
 

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