Worked Problems in Physics

Help Needed with Nielsen and Chuang Exercise 2.73

Posted in Nielsen/Chuang by rpmuller on June 1, 2010

Exercise 2.73 Given density matrix {\rho}, the minimal ensemble {\{p_k,\left|k\right>\}} contains a number of elements equal to the rank of {\rho}. Let {\left|i\right>} be any state in the support of {\rho}, where the support is defined as being spanned by the eigenvectors of {\rho} with nonzero eigenvalues. Show there exists a minimal ensemble for {\rho} that contains {\left|i\right>}, and that {\left|i\right>} appears with probability

\displaystyle  p_i = \frac{1}{\left<i|\rho^{-1}|i\right>}

where {\rho^{-1}} is the inverse that acts only on the support of {\rho}.

The density matrix {\rho} has the eigendecomposition {\rho=p_k\left|k\right>\left<k\right|}. We are given state {\left|i\right>} in the support of {\rho}, meaning that it is spanned by the eigenvectors with nonzero eigenvalues. This means that we can expand

\displaystyle \left|i\right> = \sum_{k'}\left<k'|i\right>\left|k'\right>= \sum_{k'}C_{k'i}\left|k'\right>

where the primed summation is only over those vectors in the support of {\rho}. Among other things, this means that the rank of {\left|i\right>} is no greater than {\rho}. Since we can expand {\left|i\right>} in the eigenvectors of {\rho}, this also means that {\left|i\right>} we can create a minimal ensemble for {\rho} than contains {\left|i\right>}.

Given {\rho} with rank {m}, consider the {m\times m} matrix {\sum_{k'}p_k\left|k\right>\left<k\right|}. We can write the projection of {\rho} onto {\left|i\right>} via

\displaystyle  \rho\left|i\right>=\sum_{kl'} C_{li}p_k\left|k\right>\left<k|l\right>=\sum_{k'}C_{ki}p_k\left|k\right>.

I need help here, if anyone has any suggestions. Just can’t push the equations home.

3 Responses

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  1. gf said, on January 17, 2011 at 5:30 pm

    Basically we want to prove that =1.
    Using theorem 2.6 we can rewrite this in terms of the eigenfunctions of ρ as
    =Sum over j and j’ of (Uij’*Uij). It’s not that hard to show that = δj’j (so we’ve solved the problem in this special case where the pure states are eigenstates of ρ.)
    We then have that
    =Sum over j of (Uij*Uij)=1. The second equality follows since the U’s are unitary.

  2. an3 said, on May 27, 2015 at 11:40 am

    I don’t know, but this is what I think about this:

    for a unique ensamble ρ = pi | ψi > ρ ρ-1 = pi | ψi > tr( ρ ρ-1 ) = pi tr( | ψi > 1 = pi => pi = 1 / .

    • an3 said, on May 27, 2015 at 11:49 am

      Sorry, I said

      for a unique ensamble
      ρ = pi | ψi > < ψi | ρ-1 )
      1 = pi
      so
      pi = 1 /


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