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Change of basis of density matrix

  1. Nov 24, 2012 #1
    I have a density matrix in one basis and need to change it to another. I know the eigenvectors and eigenvalues of the basis I want to change to. How do I do this?

    Any help really appreciated- thanks!
     
  2. jcsd
  3. Nov 25, 2012 #2
    All I can say is that TRACE(ρA) is basis independent.
     
  4. Nov 25, 2012 #3
    Thanks- but in this case I need to know more than the trace?
     
  5. Nov 25, 2012 #4
    Any other base state can be represented as a linear combination of original base states(complete).So may be by equality of traces some relation can be obtained for it.
     
  6. Nov 25, 2012 #5

    Bill_K

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    Science Advisor

    If your density matrix is ρ = Σ|i>pij<j| in the original basis with states |i> and probabilities pij, and the new basis states are |α>, then expand the old basis in terms of the new: |i> = Σ|α><α|i>. This gives you ρ = ΣΣΣ|α><α|i>pij<j|β><β| = Σ|α>Pαβ<β| where Pαβ = ΣΣ<α|i>pij<j|β>.
     
  7. Nov 25, 2012 #6
    By using the matrix identities:
    [tex]
    \mathbf{\rho} \cdot \mathbf{U} = \mathbf{U} \cdot \mathbf{\Lambda}
    [/tex]
    where U is a unitary matrix ([itex]\mathbf{U}^{\dagger} \cdot \mathbf{U} = \mathbf{U} \cdot \mathbf{U}^{\dagger} = 1[/itex]) whose columns are the normalized eigenvectors of the density matrix, and [itex]\Lambda[/itex] is a diagonal matrix with the corresponding eigenvalues along the main diagonal.
     
  8. Nov 26, 2012 #7

    tom.stoer

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    Science Advisor

    I think we should distinguish between the density matrix ρmn and the density operator ρ.

    Using a basis {|n>} this reads

    [tex]\rho = \sum_{mn}\rho_{mn}|m\rangle\langle n|[/tex]

    Now let's introduce a different basis {|k'>}

    [tex]\rho = \sum_{k^\prime} |k^\prime\rangle\langle k^\prime| \;\;\sum_{mn}\rho_{mn}|m\rangle\langle n|\;\; \sum_{l^\prime} |l^\prime\rangle\langle l^\prime| = \sum_{k^\prime l^\prime} \left(\sum_{mn}\langle k^\prime|m\rangle\,\rho_{mn}\,\langle n|l^\prime\rangle \right) |k^\prime\rangle\langle l^\prime| = \sum_{k^\prime l^\prime}\rho_{k^\prime l^\prime} |k^\prime\rangle\langle l^\prime|[/tex]
     
  9. Nov 26, 2012 #8
    The change-of-basis identity:
    [tex]
    \rho_{k' l'} = \sum_{m n}{\langle k' \vert m \rangle \, \rho_{m n} \, \langle n \vert l' \rangle}
    [/tex]
    with the identification of the matrix:
    [tex]
    U_{n l'} \equiv \langle n \vert l'\rangle
    [/tex]
    can be rewritten as:
    [tex]
    \mathbf{\rho}' = \mathbf{U}^{\dagger} \cdot \mathbf{\rho} \cdot \mathbf{U}
    [/tex]
    where we used the fact that:
    [tex]
    \hat{U}^{\dagger}_{k' m} = U^{\ast}_{m k'} = (\langle m \vert k' \rangle)^{\ast} = \langle k' \vert m \rangle
    [/tex]

    The unitarity of the similarity transformation matrix U is an expression of the orthonormality of the old and new bases:
    [tex]
    \langle m \vert n \rangle = \sum_{k'} {\langle m \vert k' \rangle \, \langle k' \vert n \rangle} = \sum_{k'} {U_{m k'} \, U^{\dagger}_{k' n}} = \left[\mathbf{U} \cdot \mathbf{U}^{\dagger}\right]_{m n} = \delta_{m n} \Rightarrow \mathbf{U} \cdot \mathbf{U}^{\dagger} = \mathbf{1}
    [/tex]
    [tex]
    \langle k' \vert l' \rangle = \sum_{m} \langle k' \vert m \rangle \, \langle m \vert l' \rangle = \sum_{m} U^{\dagger}_{k' m} \, U_{m l'} = \left[ \mathbf{U}^{\dagger} \cdot \mathbf{U} \right]_{k' l'} = \delta_{k' l'} \Rightarrow \mathbf{U}^{\dagger} \cdot \mathbf{U} = \mathbf{1}
    [/tex]
     
  10. Nov 26, 2012 #9

    tom.stoer

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    Science Advisor

    exactly - I hope Bizarre123 is still interested ;-)
     
  11. Nov 27, 2012 #10
    Thanks for the replies- they've been really helpful! Yes, I was getting the density operator and the density matrix confused.
     
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