- #1

- 5

- 0

## Main Question or Discussion Point

*This question was inspired by 3c) on https://people.phys.ethz.ch/~muellrom/qm1_2012/Solutions4.pdf [Broken]*

Given the operator

[tex]

\hat{B} = \left(\matrix{b&0&0\\0&0&-ib\\0&ib&0}\right)

[/tex]

I find correctly that the eigenvalues are [itex]\lambda = b, \pm b[/itex].

To find the eigenvectors for [itex]b[/itex], I do the following

[tex]

\left(\matrix{b&0&0\\0&0&-ib\\0&ib&0}\right) \left(\matrix{x\\y\\z}\right) = b \left(\matrix{x\\y\\z}\right)

[/tex]

[tex]

bx = bx \hspace{10em} y = -iz \hspace{10em} z = iy\\

[/tex]

[tex]

\hat{x} = \left(\matrix{t\\-iz\\iy}\right)\\

= \left(\matrix{t\\y\\iy}\right)

[/tex]

The pdf then seems to split this into two eigenvectors

[tex]

\hat{x}_1 = \left(\matrix{t\\0\\0}\right) = \left(\matrix{1\\0\\0}\right) \hspace{5em} \text{and} \hspace{5em} \hat{x}_2 = \left(\matrix{0\\y\\iy}\right) = y\left(\matrix{0\\1\\i}\right)

[/tex]

which 'span the eigenspace' of [itex]\lambda = b[/itex].

**Why is this allowed (separation of one eigenvector into multiple) and when should it be done?**

Would it be technically acceptable to divide it further into [itex](1,0,0)[/itex], [itex]y(0,1,0)[/itex] and [itex]y(0,0,i)[/itex]? My current guess is that doing this would be acceptable but just not practical, and the reason the eigenvector here was divided into two was purely because of the [itex]t[/itex] making it difficult to factor the [itex]y[/itex] out. Is this right, or is there a deeper meaning I'm missing? (All these eigenvectors are pre-normalization)

Last edited by a moderator: