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Expectation value via trace

  1. Jun 11, 2012 #1
    1. The problem statement, all variables and given/known data
    given
    [itex]\mid \psi \rangle = \frac{1}{\sqrt{2}} (\mid1\rangle + \mid2\rangle )[/itex]
    where [itex]\mid1\rangle, \mid2\rangle[/itex] are orthonormal
    calculate
    i)density operator
    ii) [itex]\langle A \rangle[/itex] where A is an observable
    2. Relevant equations

    3. The attempt at a solution
    i) [itex]\rho = \frac{1}{2} (\mid1\rangle\langle1\mid + \mid2\rangle\langle2\mid)[/itex]

    ii) [itex]\langle A \rangle = \frac{1}{2}\langle1\mid A\mid1\rangle + \frac{1}{2}\langle2\mid A\mid2\rangle + \frac{1}{2}\langle2\mid A\mid1\rangle + \frac{1}{2}\langle1\mid A\mid2\rangle[/itex]
    i guess [itex]\frac{1}{2}\langle2\mid A\mid1\rangle + \frac{1}{2}\langle1\mid A\mid2\rangle[/itex] = 0 (because first 2 terms are each half the expected value but surely depends on operator therefore not necessarily zero??)

    also I would like to use the trace to solve this:
    [itex]\langle A \rangle = tr[\rho A] = tr[\frac{1}{2}\mid1\rangle \langle1\mid A + \frac{1}{2}\mid2\rangle \langle2\mid A][/itex] but what does this mean?
     
    Last edited: Jun 11, 2012
  2. jcsd
  3. Jun 11, 2012 #2

    vela

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    This is the density matrix for a mixed state, where half the particles are in state ##\vert 1 \rangle## and half are in state ##\vert 2 \rangle##. The state you were given is a pure state.

    You can't really say anything about those matrix elements since you don't know anything about ##\hat{A}## except that it's an observable.

    Say you have some basis and you find the matrix representing ##\rho\hat{A}##. The trace is simply the sum of the diagonal elements. How would you write that in Dirac notation?
     
  4. Jun 12, 2012 #3
    thanks, I have since worked out that the density matrix I gave is not complete (I assumed that the outer product of two orthogonal vectors would be zero!!)
     
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