Question involving eigenvectors

  • Thread starter Thread starter dingo_d
  • Start date Start date
  • Tags Tags
    Eigenvectors
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
4 replies · 2K views
dingo_d
Messages
199
Reaction score
0

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?
 
Physics news on Phys.org
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 ^^