# 2 equations, 2 unknowns: from hell

1. Sep 17, 2010

### bjnartowt

1. The problem statement, all variables and given/known data
This seemingly-innocuous system came up in a quantum mechanics problem:
$$\left. \begin{array}{l} (1 - \sqrt 2 ){C_1} + {C_2} = 0\\ {C_1} - (1 + \sqrt 2 ){C_2} = 0 \end{array} \right]$$

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

2. Relevant equations
usual rules of linear algebra...

3. The attempt at a solution

$$\left. \begin{array}{l} (1 - \sqrt 2 ){C_1} + {C_2} = 0\\ {C_1} - (1 + \sqrt 2 ){C_2} = 0 \end{array} \right] \to \left. \begin{array}{l} (1 - \sqrt 2 ){C_1} + {C_2} = 0\\ {\textstyle{1 \over {1 + \sqrt 2 }}}{C_1} - {C_2} = 0 \end{array} \right] \to \left. \begin{array}{l} (1 - \sqrt 2 ){C_1} + {\textstyle{1 \over {1 + \sqrt 2 }}}{C_1} = 0\\ {\textstyle{1 \over {1 + \sqrt 2 }}}{C_1} - {C_2} = 0 \end{array} \right]$$

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

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

...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:
$${\cos ^{ - 1}}{\textstyle{{\left\langle {1 - \sqrt 2 ,1} \right\rangle \bullet \left\langle {1, - (1 + \sqrt 2 )} \right\rangle } \over {\left| {\left\langle {1 - \sqrt 2 ,1} \right\rangle } \right|\left| {\left\langle {1, - (1 + \sqrt 2 )} \right\rangle } \right|}}} = 112.5^\circ$$

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

2. Sep 17, 2010

### Office_Shredder

Staff Emeritus
The first equation is $$(1-\sqrt{2})C_1=C_2$$ and we plug in $$C_1=(1+\sqrt{2})C_2$$ to get $$(1-\sqrt{2})(1+\sqrt{2})C_2=C_2$$ Either $$C_2=0$$ or we get $$1-2=0$$, 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. Sep 17, 2010

### Phrak

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. Sep 17, 2010

### rs1n

I agree.

5. Sep 18, 2010

### vela

Staff Emeritus
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 $$-(1+\sqrt{2})$$, you get

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

which simplifies to

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

6. Sep 18, 2010

### HallsofIvy

Staff Emeritus
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 $AB+ A\ne 0$ 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, $A= 1+ \sqrt{2}$ and $B= 1- \sqrt{2}$ so that $AB= 1- 2= -1$. $AB+ A= -1+ 1+ \sqrt{2}= \sqrt{2}\ne 0$ 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. Sep 18, 2010

### vela

Staff Emeritus
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. Sep 18, 2010

### bjnartowt

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 : )