Find a ket orthogonal to a given ket

[Bacat]

Homework Statement

Given a state \mid \psi \rangle=\frac{1}{\sqrt{3}}[(i+1)\mid 1 \rangle + \mid 2 \rangle], find the normalized state \mid \psi^{'} \rangle orthogonal to to it.

Homework Equations

\langle \psi^{'} \mid \psi \rangle = 0

\langle \psi^{'} \mid \psi^{'} \rangle = 1

The Attempt at a Solution

I seek a state such that \langle \psi^{'} \mid \psi \rangle = 0 so I am looking for a vector such that A\frac{1+i}{\sqrt{3}} + B\frac{1}{\sqrt{3}}=0

Solving for B...

B=-A(1+i)

Desiring a normalized vector, I let A=\frac{1}{\sqrt{2}}, and therefore B=\frac{-1-i}{\sqrt{2}}.

Since this is the vector that is dotted with \mid \psi \rangle, I find the ket as the transposed conjugate of this vector, hence: \mid \psi^{'} \rangle = 1 \mid 1 \rangle + (-1 + i) \mid 2 \rangle.

Now, \langle \psi^{'} \mid \psi \rangle = \frac{1}{\sqrt{3}}(1+i-1-i) =0 OK

But, \langle \psi^{'} \mid \psi^{'} \rangle = 1+(-1-i)(-1+i)=1-i^2=2 NO

So I found a vector that is orthogonal to \mid \psi \rangle 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 \frac{\pi}{2} but this gave me nothing better.

How should I proceed?

 

[gabbagabbahey]

B=-A(1+i)

Desiring a normalized vector, I let A=\frac{1}{\sqrt{2}}, and therefore B=\frac{-1-i}{\sqrt{2}}.

What makes you think that A=\frac{1}{\sqrt{2}} produces a normalized vector?

You have found \langle\psi'|=A\langle 1|+B\langle2|=A\langle 1|-(1+i)A\langle2|...To normalize the state, take the norm of that Bra, set it equal to one and solve for A.

 

[kuruman]

This is easier than most problems of this kind because the dimensionality is 2. Prove first then use the following shortcut

Given a normalized state

|\psi> = a|1> + b|2>

show that

** Edit **

<\psi '| = -b<1| + a<2|

is othonormal to it.