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Normalized state vector for bosons, Shankar problem

  1. Nov 18, 2009 #1
    1. The problem statement, all variables and given/known data
    Two identical bosons are found to be in states [tex]|\phi>[/tex] and [tex]|\psi>[/tex]. Write down the normalized state vector describing the system when [tex]<\phi|\psi>\neq0[/tex].


    2. Relevant equations
    The normalized state vector for two bosons with [tex]<\phi|\psi>=0[/tex], using the fact that [tex]|\psi>\otimes|\phi>=|\psi\phi>[/tex], is:
    [tex]\frac{1}{\sqrt{2}}(|\psi\phi>+|\phi\psi>)[/tex].

    3. The attempt at a solution
    So I thought the new normalization would be to do:
    [tex]1=A^{2}(<\psi\phi|+<\phi\psi|)(|\psi\phi>+|\phi\psi>)
    =A^{2}(<\psi\phi|\psi\phi>+<\phi\psi|\phi\psi>
    +<\psi\phi|\phi\psi>+<\phi\psi|\psi\phi>)
    =A^{2}(2+C+C^{*})[/tex]
    where C is a complex number. Since in general [tex]C=a+bi[/tex] where a is real and b is imaginary, [tex]C+C^{*}=(a+bi)+(a-bi)=2a[/tex]. Thus, the new, normalized state would be
    [tex]\frac{1}{\sqrt{2(1+a)}}(|\psi\phi>+|\phi\psi>)[/tex]. I am not very confident with my answer, could someone please help me?
     
    Last edited: Nov 18, 2009
  2. jcsd
  3. Nov 19, 2009 #2

    lanedance

    User Avatar
    Homework Helper

    are you sure about your calculations steps not shown?

    i'm not 100% sure, but would have gone something like

    [tex] <\psi\phi|\psi\phi>
    = (<\psi|\otimes <\phi|)(|\psi>\otimes|\phi>)
    = (<\psi|\phi><\phi|\psi>)
    = <\phi|\psi>^*<\phi|\psi>
    = |<\phi|\psi>|^2
    [/tex]
     
  4. Nov 19, 2009 #3
    That makes sense. Thanks a lot!
     
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