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Eigenvalues of a reduced density matrix

  1. May 5, 2010 #1
    My lecturer keeps telling me that if a density matrix describes a pure state then it must contain only one non-zero eigenvalue which is equal to one. However I can't see how this is true, particularly as I have seen a matrix [tex]\rho_A = \begin{pmatrix} 1/2 & - 1/2 \\ -1/2 & 1/2 \\ \end{pmatrix}[/tex] for which this is not true. He then clarified that if it was in "the diagonal basis" this was true. Can someone clarify this for me or show me a proof please?
  2. jcsd
  3. May 6, 2010 #2
    I guess he means that you can write a density matrix as
    [tex]\rho = | \psi_i \rangle \langle \psi_i |[/tex]
    Which has eigenvalues [itex]\delta_{ij}[/itex] since
    [tex]\rho |\psi_j \rangle = \delta_{ij} |\psi_i \rangle[/tex].
    The example you gave can be written as [itex]\rho = | - \rangle \langle - |[/itex], where [tex]| - \rangle = \frac{1}{\sqrt{2}} \begin{pmatrix}1 \\ -1\end{pmatrix}[/tex] which has [itex]| - \rangle[/itex] as an eigenvector, with eigenvalue 1.
  4. May 7, 2010 #3
    Is that true? I thought $\psi_i \rangle$ and $\psi_j \rangle$ weren't necessarily orthogonal. He also definitely said that when it's pure it only has one non-zero eigenvalue. Maybe he means only in the basis of its eigenvectors, I don't know? That's what I'm trying to find out.
  5. May 8, 2010 #4
    Yeah I think it is the basis of its eigenvectors. I think you can always take a pure state as one possible basis vector. E.g. if you have [itex]|\psi_a \rangle = \alpha |0 \rangle + \beta |1 \rangle[/itex] then the other one is [itex]|\psi_b \rangle = \alpha |0 \rangle - \beta |1 \rangle[/itex]. I can't remember if/how this generalizes to higher dimensions though. :blushing: Maybe Gram-Schmidt?
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