A three-dimensional Hilbert-space has a set of orthonormal basis vectors: $$|0, |1, |2$$ that is $$ = δ_{i j}$$where i, j = 0, 1, 2. In this space there is a ket vector: $$|ψ = i|0 + 2|1 + (1 − i)|2$$ How would I normalize |ψ so that it has unit norm?

Question

A three-dimensional Hilbert-space has a set of orthonormal basis vectors: $$|0>, |1>, |2>$$ that is $$<i|j> = δ_{i j}$$where i, j = 0, 1, 2. In this space there is a ket vector: $$|ψ> = i|0> + 2|1> + (1 − i)|2>$$ How would I normalize |ψ> so that it has unit norm?

turingtest · Answer

whoa, creepy...
first one I've seen with the same name...
let me call on the few that may be able to help:
@across @JamesJ @Zarkon @Mr.Math something hard

anonymous · Answer

Thank you

anonymous · Answer

@No-data do you know how I might go about solving this?

anonymous · Answer

Sorry I've only heard about this.

jamesj · Answer

Suppose $ \{ v_n \} $ were an orthonormal basis for a vector space $ V $ over the complex number.  Suppose you had a vector $ v \in V $ where
$$ v = \sum_{i=1}^n a_n v_n $$
for some some complex numbers $ a_n $.  What is the normal form of $ v $?  Well, it is just
$$ \frac{v}{||v||} $$ where
$$ ||v|| = \left( \sum_{i=1}^n |a_n|^2 \right)^{1/2} $$
This is completely analogous to a vector in $ \mathbb{R}^3 $.  What's the normal form of
$$ v = (1,-1,2) ? $$  Just
$$ \frac{v}{||v||} = \frac{1}{\sqrt{6}}(1,-1,2) $$

jamesj · Answer

See what to do now?

anonymous · Answer

You have no idea how helpful this is. Thank you

jamesj · Answer

good

turingtest · Answer

I actually understood what JamesJ said, but I still don't understand the question.
I think I need to learn about Hilbert spaces, any suggestions where I can learn this stuff?

jamesj · Answer

The essential properties of Hilbert spaces is that they are that they are vector spaces over the complex numbers, and that these vector spaces have an inner product.  This inner product also allows us to define a norm/metric (a function defining distance) on the vector space.  Lastly, using that notion of distance, by definition Hilbert spaces are complete, meaning among many equivalent definitions, that every Cauchy sequence with respect to the metric has a limit in the space.

In other words, they have lots of nice properties.  All of those nice properties make Hilbert spaces very pleasant to work with and avoid all sorts of marginal and at times pathological problems for vector spaces or metric spaces that don't have all of these properties.

The wikipedia article isn't a terrible place to start if you want to read more.  But best of all, take an intermediate/advanced course on analysis.

turingtest · Answer

Thanks James :)