Question involving eigenvectors

[dingo_d]

Homework Statement

So I have to find the eigenvalues and eigenvectors of

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

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

v_1=t\left(\begin{array}{c}1\\1\end{array}\right)

and

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

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?

 

[Hootenanny]

Homework Statement

So I have to find the eigenvalues and eigenvectors of

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

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

v_1=t\left(\begin{array}{c}1\\1\end{array}\right)

and

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

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

\begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix}\begin{bmatrix}i \\ i\end{bmatrix} = \begin{bmatrix}i \\ i\end{bmatrix}

and

\begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix}\begin{bmatrix}i \\ -i\end{bmatrix} = -\begin{bmatrix}i \\ -i\end{bmatrix}.

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.

 

[dingo_d]

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?

 

[Hootenanny]

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.

 

[dingo_d]

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 ^^