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Homework Help: Square of density matrix

  1. Dec 29, 2009 #1
    1. The problem statement, all variables and given/known data
    to prove : square of density matrix= the density matrix itself (for a pure ensemble)

    2. Relevant equations
    density matrix=sum over P(i) ket(i) bra(i) where Pi = probability that random chosen system from ensemble shows state i.
    summed over i , where P=1 for pure ensemble

    3. The attempt at a solutioni thought it is giving the identity matrix, whose square is also I , hence the proof. but i am not sure about this.:frown:
    Last edited: Dec 29, 2009
  2. jcsd
  3. Dec 29, 2009 #2


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    No, the density matrix is NOT in general the identity matrix! There are many matrices with the property that [itex]A^2= A[/itex].
  4. Dec 29, 2009 #3
    then how will i prove it?
  5. Dec 29, 2009 #4
    What are the P_i for a pure state?
  6. Dec 29, 2009 #5
    Ok, first we assume that the set [tex] \left \lbrace \left| i \right \rangle \right \rbrace [/tex] forms a complete orthonormal base (otherwise if we have a general state [tex] \left| \psi \right \rangle [/tex] we have to expand it into such a base). Further we just deal with the discrete case, the continuous one can be done similarly.

    If [tex] \rho = \sum_{i} \left| i \right \rangle \left \langle i \right| [/tex] is the density matrix, the square is simply given by:

    [tex] \rho^{2} = \sum_{i,j} \left| i \right \rangle \langle i \vert j \rangle \left \langle j \right| [/tex]

    Now we use [tex] \left \langle i \vert j \right \rangle = \delta_{ij} [/tex] and arrive at:

    [tex] \rho^{2} = \sum_{i} \left| i \right \rangle \left \langle i \right| = \rho [/tex].
  7. Dec 29, 2009 #6
    Pi for pure ensemble=1 since all are in same state, isnt it?
  8. Dec 29, 2009 #7
    i can just say -'COOL.' MANY thanks.
    ......sorry, just one more thing
    how can i just keep them side by side, i mean[tex]\rho[/tex] is a density matrix and to use it as a matrix i need to find its elements in some basis. how can i simply keep the abstract notation?
    Last edited: Dec 29, 2009
  9. Dec 29, 2009 #8
    This proof is wrong. What you've done is computing the square of the identity matrix which obviously is the identity matrix.
  10. Dec 29, 2009 #9
    No, for a pure state all the P_i are zero, except for one particular value of i (say i = r). So, the state is with certainty some particular state |r>.
  11. Dec 29, 2009 #10
    yes. i was wrong.
    but does that mean there will be no summation?(for pure ensemble)
    then how will i proceed?
  12. Dec 29, 2009 #11
    Sorry, the proof is of course wrong, such ugly things should not happen, very embarrassing...

    In the pure case there is indeed no summation and we end up with one certain state [tex] \vert r \rangle [/tex].

    If this state is normed, i.e. [tex] \langle r \vert r \rangle = 1 [/tex] and your density matrix is [tex] \rho = \vert r \rangle \langle r \vert [/tex] than you have the (trivial) relation:

    [tex] \rho^{2} = \vert r \rangle \langle r \vert r \rangle \langle r \vert = \vert r \rangle \langle r \vert = \rho [/tex]
  13. Dec 29, 2009 #12
    parton, to err is human.
    i learnt from your mistake. that will not alter my belief that u are a genius. thanks very much for helping me.
    doing mistakes is a regular job for confused people like me.:biggrin:
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