When should one eigenvector be split into two (same span)?

  • #1

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)
 
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  • #2
Simon Bridge
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You appear to have misunderstood what is going on.
The pdf author starts by finding the operator has 2 eigenvalues ... these are +b and -b.
The next step is to find out the conditions that x,y,z must satisfy if they are to be the components of an eigenvector that points to value +b.
The result is 3 simultanious equations that the author places, for convenience, into a vector form.
This vector is not an eigenvector.

The next step is to solve the equations to get the actual eigenvectors.
There are two vectors that satisfy the conditions: these are the sought-after eigenvectors.

The author has, at not time, split any eigenvectors up.
 
  • #3
The result is 3 simultanious equations that the author places, for convenience, into a vector form.
This vector is not an eigenvector.
...
There are two vectors that satisfy the conditions: these are the sought-after eigenvectors..
Are you sure? The highlighted wording in the question makes me think otherwise.

BBNmCPV.png
 
  • #4
blue_leaf77
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Why is this allowed (separation of one eigenvector into multiple) and when should it be done?
It's about finding the basis for the null space of matrix
$$
\left(\matrix{b&0&0\\0&0&-ib\\0&ib&0}\right) - \left(\matrix{b&0&0\\0&b&0\\0&0&b}\right) = \left(\matrix{0&0&0\\0&-b&-ib\\0&ib&-b}\right)
$$
Using row reduction technique, you should find that any vector in the null space of the above matrix is
$$
\left(\matrix{x\\y\\z}\right) = x\left(\matrix{1\\0\\0}\right) + y\left(\matrix{0\\1\\i}\right)
$$
Which means the basis vectors are ##(1,0,0)^T## and ##(0,1,i)^T## and they also turn out to be already orthogonal.
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]?
You can't do that because the last two vectors are not in the null space of the above matrix.
 
  • #5
andrewkirk
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The solution
[tex]
\hat{x} = \left(\matrix{t\\-iz\\iy}\right)= \left(\matrix{t\\y\\iy}\right)
[/tex]
contains two free variables: ##t## and ##y##. So any vector that can be written in that form, for any ##t,y\in\mathbb{C}##, is an eigenvector with eigenvalue ##b##. That tells us that the space of eigenvectors ('eigenspace') corresponding to eigenvalue ##b## is two-dimensional. To characterize that space we find a basis, which must have two elements. The obvious choice is the pair ## \vec v_1=\left(\matrix{1\\0\\0}\right),\ \vec v_2=\left(\matrix{0\\1\\i}\right)##, so that the solution above is equal to ##t\vec v_1+y\vec v_2##.
Would it be technically acceptable to divide it further into (1,0,0), y(0,1,0) and y(0,0,i)?
You can certainly write the solution as ##t\left(\matrix{1\\0\\0}\right)+y\left(\matrix{0\\1\\0}\right)+y\left(\matrix{0\\0\\i}\right)## but the second and third items are not in the eigenspace, so writing it in that way has no use. The aim is to write the solution as a sum of the basis elements of the eigenspace, and there can be only two basis elements because the eigenspace is two-dimensional.
 
  • #6
Simon Bridge
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I don't see anything in post #3 that counterindicates post #2.
I guess you can think of the intermediate step as an eigenvector if you like... but it is more a collection of valid eigenvectors. Pick 2, any 2. Remember what you need them to be able to do.

The bottom line is you can choose whatever representation you like... so pick the one that makes the maths easier. Hence finding simultanious eigenkets.
 

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