I Normalization of an Eigenvector in a Matrix

Dwye
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I am working through some linear algebra questions in the Griffith's Book "Introduction to Quantum Mechanics", and I am unsure why a constant of 1/sqrt(2) is added into the answer.
I understand that the question is detailing a rotation about axis x & y, and that 1/sqrt(2) is the value of 45 degrees for both Sin and Cos, is this the reason for the addition; a generalization?
In fact I have seen this number quite a lot in Quantum Mechanics, is there something more to this number?
ProblemA.18.JPG
AnswerA.18.JPG
 
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Do you know, how to get the norm of a vector in ##\mathbb{C}^2## or, more generally, how to define a scalar product on a complex vector space? It's very important to get these concepts right, before starting to study quantum theory, for which you need the "infinite-dimensional version" of these ideas, the socalled (separable) Hilbert space (more precisely what physicists do with this is rather the extension to a "rigged Hilbert space").
 
Dwye said:
Summary:: I am working through some linear algebra questions in the Griffith's Book "Introduction to Quantum Mechanics", and I am unsure why a constant of 1/sqrt(2) is added into the answer.
I understand that the question is detailing a rotation about axis x & y, and that 1/sqrt(2) is the value of 45 degrees for both Sin and Cos, is this the reason for the addition; a generalization?
In fact I have seen this number quite a lot in Quantum Mechanics, is there something more to this number?

The vector ##(1, i)## has magnitude ##\sqrt 2##, so Griffiths decided to normalise it. The question doesn't actually ask for a normalised eigenvector, so ##(1, i)## would be just as valid an answer.

In QM it's generally a good idea to normalise vectors. The factor here has nothing directly to do with being the sine of ##45°##. If the eigenvector were ##(1, 2i)##, then the normalisation factor would be ##\frac 1 {\sqrt{5}}##.
 
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PeroK said:
The vector ##(1, i)## has magnitude ##\sqrt 2##, so Griffiths decided to normalise it. The question doesn't actually ask for a normalised eigenvector, so ##(1, i)## would be just as valid an answer.

In QM it's generally a good idea to normalise vectors. The factor here has nothing directly to do with being the sine of ##45°##. If the eigenvector were ##(1, 2i)##, then the normalisation factor would be ##\frac 1 {\sqrt{5}}##.
PeroK said:
The vector ##(1, i)## has magnitude ##\sqrt 2##, so Griffiths decided to normalise it. The question doesn't actually ask for a normalised eigenvector, so ##(1, i)## would be just as valid an answer.

In QM it's generally a good idea to normalise vectors. The factor here has nothing directly to do with being the sine of ##45°##. If the eigenvector were ##(1, 2i)##, then the normalisation factor would be ##\frac 1 {\sqrt{5}}##.
Thank you very much!
 
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