Unique to within a constant factor?

[cks]

unique to within a constant factor?

we always listen to the phrase in quantum mechanics textbook that says this eigenvector is unique to within a constant factor.

What does it actually mean?

unique means there is one and only one??
constant factor: a constant 2 or 3 or... maybe??

I don't really catch the physical picture of it, I just hope for some mathematical examples that can illustrate this. Thank you

 

[CompuChip]

Well, suppose that $$\left| a \rangle$$ is an eigenvector to an operator  with eigenvalue $a$, that is:
$$\hat A \left| a \rangle = a \left| a \rangle$$.
Then of course, since scalars commute with any operator,
$$\hat A \left( \lambda \left| a \rangle \right) = \lambda \left( \hat A \left| a \rangle \right) = \lambda \left( a \left| a \rangle \right) = a \left( \lambda \left| a \rangle \right)$$.
So as you see, $$\lambda \left| a \rangle$$ is also an eigenvector for the same eigenvalue, for any (complex) number lambda. But other than that, the eigenvector is really unique, that is, if
$$\hat A \left| a' \rangle = a \left| a' \rangle$$
then |a'> cannot be anything different than a multiple of |a>.

This is what is meant by "unique up to a constant factor", which is enough since in QM, the pre-factor doesn't carry any physical meaning anyway (we normally just use it for convenience, e.g. to normalize eigenkets).

 

[malawi_glenn]

There is a complex (global-)phase: exp(i*phi), cos' we deal with complex numbers. And as convention we often we often taken phi = 0, so exp(i phi) = 1.

Is that what you looked for? You can see an example of this in Sakurai "modern QM" chapter 1 in discussion of Sx and Sy and their eigen-vectors.

 

[cks]

Thank you CompuChip,

I understand now. Really appreciate that.

 

[dextercioby]

quote: >>the pre-factor doesn't carry any physical meaning anyway (we normally just use it for convenience, e.g. to normalize eigenkets).<<

Actually this phase freedom of vectors is of crucial importance when it comes to describing symmetries in QM.