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Normalisation of quantum states

  1. Oct 12, 2012 #1
    Hi,

    Just a little thing that's been puzzling me:

    Consider a state

    [tex] \mid \psi \rangle = \frac{1}{\sqrt{2}} \mid A \rangle + \frac{1}{\sqrt{2}} \mid B \rangle[/tex]

    This is normalised since [tex] [\frac{1}{\sqrt{2}}]^2 + [\frac{1}{\sqrt{2}}]^2 = 1 [/tex]

    Now let A = B:

    [tex] \mid \psi \rangle = \frac{1}{\sqrt{2}} \mid A \rangle + \frac{1}{\sqrt{2}} \mid A \rangle = \frac{2}{\sqrt{2}} \mid A \rangle = {\sqrt{2}} \mid A \rangle[/tex]

    This isn't normalised anymore! What happened?
     
  2. jcsd
  3. Oct 12, 2012 #2

    Matterwave

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    A and B must be normalized and orthogonal for your normalization to work. A and A is obviously not orthogonal.
     
  4. Oct 12, 2012 #3
    You implicitly assumed <A|B> = 0 in your first expression.
     
  5. Oct 12, 2012 #4
    Ah ok.. The reason this came up is that I was looking at latter operators on paired states. So with 2 atoms in a 1,1 state you can get
    [tex] \mid 2,2 \rangle = \mid 1,1 \rangle\mid 1,1 \rangle [/tex]
    And then you can use ladder operators to go down:
    [tex] J_-\mid 2,2 \rangle = (J_-\mid 1,1 \rangle)\mid 1,1 \rangle + \mid 1,1 \rangle(J_- \mid 1,1 \rangle) \\
    \mid 2,1 \rangle = \frac{1}{\sqrt{2}} \mid 1,0 \rangle\mid 1,1 \rangle + \frac{1}{\sqrt{2}} \mid 1,1 \rangle \mid 1,0 \rangle
    [/tex]
    Which is fine, but then the next one:
    [tex]
    \sqrt{6} \mid 2,0 \rangle = \frac{1}{\sqrt{2}} \sqrt{2} \mid 1,-1 \rangle\mid 1,1 \rangle + \frac{1}{\sqrt{2}} \sqrt{2} \mid 1,0 \rangle\mid 1,0 \rangle + \frac{1}{\sqrt{2}} \sqrt{2} \mid 1,0 \rangle \mid 1,0 \rangle + \frac{1}{\sqrt{2}} \sqrt{2} \mid 1,0 \rangle \mid 1,-1 \rangle
    [/tex]
    is actually also fine now I look at it again.. Sorry for wasting your time!
     
  6. Oct 13, 2012 #5

    tom.stoer

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