1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Find a ket orthogonal to a given ket

  1. Sep 11, 2009 #1
    1. The problem statement, all variables and given/known data

    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.


    2. Relevant equations

    [tex]\langle \psi^{'} \mid \psi \rangle = 0[/tex]

    [tex]\langle \psi^{'} \mid \psi^{'} \rangle = 1[/tex]


    3. 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?
     
    Last edited: Sep 11, 2009
  2. jcsd
  3. Sep 11, 2009 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    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].
     
  4. Sep 11, 2009 #3

    kuruman

    User Avatar
    Homework Helper
    Gold Member

    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.
     
    Last edited: Sep 11, 2009
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Find a ket orthogonal to a given ket
  1. Bra kets (Replies: 3)

  2. Bra - ket? (Replies: 6)

  3. Bra and kets (Replies: 8)

Loading...