Question involving eigenvectors

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Homework Help Overview

The discussion revolves around finding the eigenvalues and eigenvectors of a specific 2x2 matrix, A, given as A=\left(\begin{array}{cc}0&1\\1&0\end{array}\right). The original poster expresses initial confidence in solving for the eigenvalues, which are found to be \lambda_{1,2}=\pm 1, and subsequently derives the eigenvectors. However, a question arises regarding the nature of the parameter t in the eigenvector expressions, specifically whether it must be real or could be complex.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of allowing t to be complex, questioning the assumptions made in linear algebra regarding the nature of eigenvectors. The original poster seeks clarification on whether it is permissible to impose that t is real and what the consequences would be if t were complex.

Discussion Status

Some participants provide insights into the nature of eigenvectors, noting that while real eigenvectors can be guaranteed for symmetric matrices, complex eigenvectors are also possible. The conversation touches on the context of quantum mechanics, where complex eigenvectors are common, leading to further inquiries about their interpretation.

Contextual Notes

The discussion includes considerations of the mathematical properties of eigenvectors in relation to the context of quantum mechanics, where eigenvectors represent states associated with observable values. There is an acknowledgment that the nature of eigenvectors may depend on the specific application or context rather than solely on the matrix itself.

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Homework Statement



So I have to find the eigenvalues and eigenvectors of

[tex]A=\left(\begin{array}{cc}0&1\\1&0\end{array}\right)[/tex]

which is not that special and hard. I solve the characteristic equation and find the eigenvalues: [tex]\lambda_{1,2}=\pm 1[/tex]. So finding the eigenvectors is relatively simple, I just plug that back into characteristic eq and I get:

[tex]v_1=t\left(\begin{array}{c}1\\1\end{array}\right)[/tex]

and

[tex]v_2=t\left(\begin{array}{c}1\\-1\end{array}\right)[/tex].

So that's pretty simple? I can even normalize it and show that they are orthogonal. But!

There was one thing that kinda bugged me, and got my attention. Back on linear algebra class we said that t is some real parameter and we lived our lives happily ever after.

But my professor, now on QM asked this: how do we know that t is real? What if it's complex?

If it is complex then the normalization isn't that trivial.

So my question is: why is it real? Are we free to impose that on the parameter? Or is there some deeper math behind it all?
 
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dingo_d said:

Homework Statement



So I have to find the eigenvalues and eigenvectors of

[tex]A=\left(\begin{array}{cc}0&1\\1&0\end{array}\right)[/tex]

which is not that special and hard. I solve the characteristic equation and find the eigenvalues: [tex]\lambda_{1,2}=\pm 1[/tex]. So finding the eigenvectors is relatively simple, I just plug that back into characteristic eq and I get:

[tex]v_1=t\left(\begin{array}{c}1\\1\end{array}\right)[/tex]

and

[tex]v_2=t\left(\begin{array}{c}1\\-1\end{array}\right)[/tex].

So that's pretty simple? I can even normalize it and show that they are orthogonal. But!

There was one thing that kinda bugged me, and got my attention. Back on linear algebra class we said that t is some real parameter and we lived our lives happily ever after.

But my professor, now on QM asked this: how do we know that t is real? What if it's complex?

If it is complex then the normalization isn't that trivial.

So my question is: why is it real? Are we free to impose that on the parameter? Or is there some deeper math behind it all?
In general, t or your eigenvectors needn't be real. For example, if we let t=i then your eigenvectors are v1 = [i, i]T and v2 = [i, -i]T. Substituting the eigenvectors into the eigenvalue problem yields

[tex]\begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix}\begin{bmatrix}i \\ i\end{bmatrix} = \begin{bmatrix}i \\ i\end{bmatrix}[/tex]

and

[tex]\begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix}\begin{bmatrix}i \\ -i\end{bmatrix} = -\begin{bmatrix}i \\ -i\end{bmatrix}[/tex].

Therefore, it is possible in general to have complex eigenvectors. However, if your matrix is a symmetric (as it is here) nxn square matrix, then there exists n mutually orthogonal, real eigenvectors; but that doesn't mean that there aren't any complex ones!

In general, the context dictates whether the eigenvectors should be real, rather than the matrix itself.
 
And in the context of quantum mechanics? Usually the eigenvectors represent states in which observable has a definite value (the eigenvalue).

What would imaginary eigenvector represent?
 
dingo_d said:
And in the context of quantum mechanics? Usually the eigenvectors represent states in which observable has a definite value (the eigenvalue).

What would imaginary eigenvector represent?
In quantum mechanics, the eigenvectors (or wave functions) belong to a complex Hilbert Space. In other words, quantum mechanical wave functions are, in general, complex.
 
Hootenanny said:
In quantum mechanics, the eigenvectors (or wave functions) belong to a complex Hilbert Space. In other words, quantum mechanical wave functions are, in general, complex.


Oh I see! Cool, thanks on the clearing that out ^^
 

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