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I have a very basic question that is confusing me even though I keep turning it over in my head, that I've come across in trying to convert a Hamiltonian into a form which uses normal modes of oscillation. I took Linear Algebra last semester so I know that this equation should be convertible to matrix form, I'm just stumped as to how exactly I can do that; the equation I have is this:

$$ A(x_1 + x_2 + x_3 + x_4) + B(x_1 - x_2 + x_3 - x_4) + C(x_1 + x_2 - x_3 - x_4) + D(x_1 - x_2 - x_3 + x_4) = -2x_1 - 2x_4$$ as well as $$ x_1( A + B + C + D + 2) + x_2(A - B + C - D) + x_3(A + B - C - D) + x_4(A - B - C + D + 2)=0$$

Now, I've converted this into a matrix of the form $$ \begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & -1 & 1 & -1 \\ 1 & 1 & -1 & -1 \\ 1 & -1 & -1 & 1 \\ \end{pmatrix} \begin{pmatrix} A \\ B \\ C \\ D \\ \end{pmatrix} = \begin{pmatrix} 2 \\ 0 \\ 0 \\ 2 \end{pmatrix} $$

with the goal being to solve for an appropriate A,B,C, and D. I figured that in order to do this, I would just have to solve the matrix here, but I'm skeptical that this will return a correct answer, given that the 2's are connected with $x_1$ and $x_4$ and I'm not particularly certain if I constructed this matrix correctly. Any advice is of course, greatly appreciated.

As a final note, I already took a look at the questions Converting Single Equation To A System Of Equations and How would I find a non-zero matrix that fits the following equation? which don't seem to have the answer I'm looking for. I specifically do not really know if I've constructed this system correctly from this single equation, or if that is really even possible based on what I've been given (although I am fairly certain that it is). Many thanks

Edit: if there is some mathematical nuanace here that I am missing as well, I would greatly appreciate any explanation

Dylan
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Brandon
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  • You're getting downvotes because readers believe you have 1 equation and 8 unknowns (that is $A,B,C,D,x_1,x_2,x_3,x_4$). – Mikkel Rev Jun 19 '18 at 19:38
  • Even if $A,B,C,D$ are supplied parameters you have only one equation in four unknowns. There should be a three dimensional solution space. Seeing the regularities in your equation I suspect you really have four separate equations and did not transcribe it properly. As written there is no solution. – Ross Millikan Jun 26 '18 at 05:42
  • @RossMillikan I'm assuming this is supposed to be true for all $x_1,x_2,x_3,x_4$, then we're just comparing coefficients. Of course, this is something OP should clarify. – Dylan Jun 26 '18 at 06:46

2 Answers2

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It is possible to make a system of equations. I am assuming that you mean $A, B, C, D$ are constants and $x1,x2,x3,x4$ can be anything with the equation given is always satisfied. If this is not what you mean, then you can tell me what you mean Although I can't rigorously prove it, the point is to set the $x1$ parts, $x2$ parts, $x3$ parts, and $x4$ parts to be equal to each other. This is from my intuition after dealing with complex numbers and the floor/fractional functions. It's probably true, but I'm not skilled enough with math yet. This makes the four equations: $A+B+C+D=-2,$ $A-B+C-D=0,$ $A+B-C-D=0,$ and $A-B-C+D=0.$

Jason Kim
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  • Well, I guess I can't do a full solution because I had to assume some stuff in it. After all, setting them equal is only necessary when it is true for all $x1,x2,x3,x4.$ – Jason Kim Jun 26 '18 at 04:22
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If you are solving it for all $x_1,x_2,x_3,x_4$ you can separate it into four equations, one for each one. Your matrix equation is then correct. You can solve this by Gaussian elimination or by simple substitution, which are really the same approach. $$ A + B + C + D=- 2\\ A - B + C - D=0\\ A + B - C - D=0\\A - B - C + D =-2\\ A-D=0\\B-C=0\\A+B=-2$$ where I added and subtracted the second and third to get the fifth and sixth. I then plugged into the first to get the seventh. You have a dependent set of equations, so really only three equations in four unknowns. You can choose any variable to be anything you want, then compute the rest. I will choose $A$ to be the independent variable. $$D=A\\B=-2-A\\C=-2-A$$

Ross Millikan
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