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.