- #1

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## Homework Statement

This seemingly-innocuous system came up in a quantum mechanics problem:

[tex]\left. \begin{array}{l}

(1 - \sqrt 2 ){C_1} + {C_2} = 0\\

{C_1} - (1 + \sqrt 2 ){C_2} = 0

\end{array} \right][/tex]

I want to find out what C1 and C2 are. Simple, right? Well...

## Homework Equations

usual rules of linear algebra...

## The Attempt at a Solution

[tex]\left. \begin{array}{l}

(1 - \sqrt 2 ){C_1} + {C_2} = 0\\

{C_1} - (1 + \sqrt 2 ){C_2} = 0

\end{array} \right] \to \left. \begin{array}{l}

(1 - \sqrt 2 ){C_1} + {C_2} = 0\\

{\textstyle{1 \over {1 + \sqrt 2 }}}{C_1} - {C_2} = 0

\end{array} \right] \to \left. \begin{array}{l}

(1 - \sqrt 2 ){C_1} + {\textstyle{1 \over {1 + \sqrt 2 }}}{C_1} = 0\\

{\textstyle{1 \over {1 + \sqrt 2 }}}{C_1} - {C_2} = 0

\end{array} \right][/tex]

hmmm...looking at second equation if I substitute:

[tex]{\textstyle{1 \over {1 + \sqrt 2 }}}{C_1} = {C_2}[/tex]

...then i clearly get, in the first equation:

1 = 1

...which is the math blowing a raspberry at me. Perhaps the two equations are linearly-dependent? It doesn't seem so, but let's try finding "angle" between the two row-vectors that come from the coefficients of the equations:

[tex]{\cos ^{ - 1}}{\textstyle{{\left\langle {1 - \sqrt 2 ,1} \right\rangle \bullet \left\langle {1, - (1 + \sqrt 2 )} \right\rangle } \over {\left| {\left\langle {1 - \sqrt 2 ,1} \right\rangle } \right|\left| {\left\langle {1, - (1 + \sqrt 2 )} \right\rangle } \right|}}} = 112.5^\circ [/tex]

hmm...not orthogonal.

Why can't I get explicit values for C[1] and C[2] for this system? I don't THINK the system is linearly-dependent!

problem no longer of interest after: 09/24/10