2 equations, 2 unknowns: from hell

  • Thread starter bjnartowt
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  • #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
 

Answers and Replies

  • #2
Office_Shredder
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The first equation is [tex](1-\sqrt{2})C_1=C_2[/tex] and we plug in [tex]C_1=(1+\sqrt{2})C_2[/tex] to get [tex](1-\sqrt{2})(1+\sqrt{2})C_2=C_2[/tex] Either [tex]C_2=0[/tex] or we get [tex]1-2=0[/tex], not 1=1. I'm not sure what you did to get that., but you have an extra 1/(1+sqrt(2)) floating around at the end it seems

So there's only one solution, both C's are zero. This isn't surprising since you just have a matrix equation Ac=0 with A non-singular, so c has to be zero
 
  • #3
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The first equation is [tex](1-\sqrt{2})C_1=C_2[/tex] and we plug in [tex]C_1=(1+\sqrt{2})C_2[/tex] to get [tex](1-\sqrt{2})(1+\sqrt{2})C_2=C_2[/tex]

I'm afraid you dropped a negative sign in the last part. I get C2=C2. The two equations are the same after multiplying the first by 1+sqrt(2).
 
  • #4
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I don't THINK the system is linearly-dependent!

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

I agree.
 
  • #5
vela
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The system is linearly dependent, which is what you want since you're solving for an eigenvector, I believe. If you multiply the first equation by [tex]-(1+\sqrt{2})[/tex], you get

[tex]-(1+\sqrt{2})(1-\sqrt{2})C_1 - (1+\sqrt{2})C_2 = 0[/tex]

which simplifies to

[tex]C_1-(1+\sqrt{2})C_2 = 0[/tex]
 
  • #6
HallsofIvy
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Surely it is not all that difficult! You have, really Ax+ y= 0, x- By= 0.

Multiply both sides ofthe first equation by B to get ABx+ By= 0 and add that to the second equation to eliminate y and leave (AB+ A)x= 0. If [itex]AB+ A\ne 0[/itex] then x= 0 is the only solution to that equation and then y= 0 so that (0, 0) is the unique solution.

If, however, AB+ A= 0 x can be anything and, in that case, from Ax+ y= 0, y= -Ax.

Here, [itex]A= 1+ \sqrt{2}[/itex] and [itex]B= 1- \sqrt{2}[/itex] so that [itex]AB= 1- 2= -1[/itex]. [itex]AB+ A= -1+ 1+ \sqrt{2}= \sqrt{2}\ne 0[/itex] so that x= 0, y= 0 is the unique solution.

If you are looking for "eigenvalues" and "eigenvectors" then something has gone wrong. What was the original eigenvalue problem?
 
  • #7
vela
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Surely it is not all that difficult! You have, really Ax+ y= 0, x- By= 0.

Multiply both sides ofthe first equation by B to get ABx+ By= 0 and add that to the second equation to eliminate y and leave (AB+ A)x= 0. If [itex]AB+ A\ne 0[/itex] then x= 0 is the only solution to that equation and then y= 0 so that (0, 0) is the unique solution.

If, however, AB+ A= 0 x can be anything and, in that case, from Ax+ y= 0, y= -Ax.

Here, [itex]A= 1+ \sqrt{2}[/itex] and [itex]B= 1- \sqrt{2}[/itex] so that [itex]AB= 1- 2= -1[/itex]. [itex]AB+ A= -1+ 1+ \sqrt{2}= \sqrt{2}\ne 0[/itex] so that x= 0, y= 0 is the unique solution.

If you are looking for "eigenvalues" and "eigenvectors" then something has gone wrong. What was the original eigenvalue problem?
The second equation is x-By=0, so when you add the two equations, you're left with (AB+1)x=0, not (AB+A)x=0. Since AB=-1, (AB+1)=0 and x can be anything.
 
  • #8
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Oh, thanks guys! I was silly, and didn't notice that the determinant of the representative matrix was 0.

Well, since this is QM, the sum of the squared magnitudes of C1 and C2 must be unity, and it looks like I'm not without sufficient information after all : )
 

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