Find a ket orthogonal to a given ket

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In summary, the conversation discusses finding a normalized state orthogonal to a given state. The solution involves finding a vector such that its dot product with the given state is zero, and then normalizing it by finding the appropriate values for its coefficients. The shortcut for finding the normalized state is also mentioned.
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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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  • #2
Bacat said:
[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].

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

You have found [itex]\langle\psi'|=A\langle 1|+B\langle2|=A\langle 1|-(1+i)A\langle2|[/itex]...To normalize the state, take the norm of that Bra, set it equal to one and solve for [itex]A[/itex].
 
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  • #3
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

[tex]|\psi> = a|1> + b|2>[/tex]

show that

** Edit **

[tex]<\psi '| = -b<1| + a<2|[/tex]

is othonormal to it.
 
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1. How do you find a ket orthogonal to a given ket?

To find a ket orthogonal to a given ket, you can use the Gram-Schmidt process. This process involves taking the given ket and subtracting its projection onto the previous orthogonal kets in the set until you are left with a ket that is orthogonal to all previous kets.

2. Why is finding a ket orthogonal to a given ket important?

Finding a ket orthogonal to a given ket is important in quantum mechanics because orthogonal kets represent distinct quantum states. By finding a ket orthogonal to a given ket, we can determine the possible outcomes of a measurement on the given ket.

3. Can any ket be orthogonal to a given ket?

No, not all kets can be orthogonal to a given ket. In order for two kets to be orthogonal, their inner product must equal 0. This means that the kets must be perpendicular to each other in the vector space.

4. Are there multiple kets that can be orthogonal to a given ket?

Yes, there can be multiple kets that are orthogonal to a given ket. This is because there are infinitely many possible orthogonal kets that can be constructed using the Gram-Schmidt process.

5. How does finding a ket orthogonal to a given ket relate to linear independence?

Finding a ket orthogonal to a given ket is related to linear independence because a set of kets is linearly independent if and only if each ket is orthogonal to all other kets in the set. So, by finding a ket orthogonal to a given ket, we can determine if it is linearly independent from a set of kets.

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