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Bacat
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Homework Statement
Given a state [tex]\mid \psi \rangle=\frac{1}{\sqrt{3}}[(i+1)\mid 1 \rangle + \mid 2 \rangle][/tex], find the normalized state [tex]\mid \psi^{'} \rangle[/tex] orthogonal to to it.
Homework Equations
[tex]\langle \psi^{'} \mid \psi \rangle = 0[/tex]
[tex]\langle \psi^{'} \mid \psi^{'} \rangle = 1[/tex]
The Attempt at a Solution
I seek a state such that [tex]\langle \psi^{'} \mid \psi \rangle = 0[/tex] so I am looking for a vector such that [tex]A\frac{1+i}{\sqrt{3}} + B\frac{1}{\sqrt{3}}=0[/tex]
Solving for B...
[tex]B=-A(1+i)[/tex]
Desiring a normalized vector, I let [tex]A=\frac{1}{\sqrt{2}}[/tex], and therefore [tex]B=\frac{-1-i}{\sqrt{2}}[/tex].
Since this is the vector that is dotted with [tex]\mid \psi \rangle[/tex], I find the ket as the transposed conjugate of this vector, hence: [tex]\mid \psi^{'} \rangle = 1 \mid 1 \rangle + (-1 + i) \mid 2 \rangle[/tex].
Now, [tex]\langle \psi^{'} \mid \psi \rangle = \frac{1}{\sqrt{3}}(1+i-1-i) =0[/tex] OK
But, [tex]\langle \psi^{'} \mid \psi^{'} \rangle = 1+(-1-i)(-1+i)=1-i^2=2[/tex] NO
So I found a vector that is orthogonal to [tex]\mid \psi \rangle[/tex] but it is not normalized.
I think one of the following is happening:
1) I need to re-normalized the vector I found.
2) I am approaching this problem in the wrong way entirely. I did try to apply a rotation operator to the given vector (rotating by [tex]\frac{\pi}{2}[/tex] but this gave me nothing better.
How should I proceed?
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