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Zettili QM Problem on Trace of an Operator

  1. Jan 26, 2016 #1
    1. The problem statement, all variables and given/known data
    In Zettili's QM textbook, we are asked to find the trace of an operator [itex] |\psi><\chi| [/itex]. Where the kets [itex]|\psi> [/itex] and [itex]|\chi> [/itex] are equal to some (irrelevant, for the purposes of this question) linear combinations of two orthonormal basis kets.

    2. Relevant equations
    [itex] Tr(\hat{A}\hat{B})=Tr(\hat{B}\hat{A}) [/itex]

    3. The attempt at a solution
    The solution is given in the textbook. Zettili says the following:

    Since [itex] Tr(\hat{A}\hat{B})=Tr(\hat{B}\hat{A}) [/itex], we know that [itex] Tr(|\psi><\chi|)=Tr(<\chi|\psi>) [/itex].

    From then on calculating the trace is no problem.

    My question is if this is a valid argument. It seems to me that the equation Zettili is starting from talks about the trace of two operators not caring about the order the operators are multiplied in. However, the trace we want to find in this problem is the trace of a single operator, namely [itex] |\psi><\chi| [/itex]. The reversal of the product from ket-bra to bra-ket doesn't seem to be the same concept as changing the order of multiplication of two operators. Can anybody help me out here?

    (If anyone's curious, this is Problem 2.1 c on page 133-134 of the second edition of Quantum Mechanics: Concepts and Applications.)
     
  2. jcsd
  3. Jan 26, 2016 #2
    Simply speaking
    $$ trace(\langle \Psi | \Phi \rangle) = \langle \Psi | \Phi \rangle = \sum_n \langle \Psi | n\rangle \langle n | \Phi \rangle = trace(|\Psi\rangle\langle\Phi|)$$
     
  4. Jan 26, 2016 #3

    samalkhaiat

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    No, this is not correct
    [tex]\mbox{Tr}(\langle \Psi | \Phi \rangle) = (\langle \Psi | \Phi \rangle) \ \mbox{Tr} ( \mbox{identity} ) \neq \langle \Psi | \Phi \rangle [/tex]
     
  5. Jan 26, 2016 #4

    samalkhaiat

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    This is very, very wrong:
    1) You only have one operator. So, you can not use the identity [itex]\mbox{Tr}(AB) = \mbox{Tr}(BA)[/itex].
    2) [itex]\mbox{Tr}( | \Psi \rangle \langle \Phi | ) \neq \mbox{Tr}(\langle \Phi | \Psi \rangle )[/itex]. Because [itex]\langle \Phi | \Psi \rangle [/itex] is a number, you can pull it out of the trace operation, and in n-dimension you get
    [tex]\mbox{Tr}\left( \langle \Phi | \Psi \rangle \right) = \mbox{Tr}\left( \langle \Phi | \Psi \rangle \ I \right) = \langle \Phi | \Psi \rangle \mbox{Tr}( I ) = n \langle \Phi | \Psi \rangle [/tex]
    More similar mistakes some students make are the following
    [tex]\mbox{Tr} \left( | \psi \rangle \langle \phi | \right) = \mbox{Tr} \left( ( | \psi \rangle I ) \langle \phi | \right) = \mbox{Tr} \left( \langle \phi | \psi \rangle \ I \right) ,[/tex]
    [tex]\mbox{Tr} \left( | \psi \rangle \langle \phi | \ I \right) = \mbox{Tr} ( \langle \phi | I | \psi \rangle ) = \mbox{Tr} ( \langle \phi | \psi \rangle )[/tex]
    These are wrong because the objects [itex]( |\psi \rangle I )[/itex], [itex]( \langle \phi | I )[/itex], [itex]( |\psi \rangle )[/itex] or [itex]( \langle \phi | )[/itex] are not operators. Therefore one can not apply the identity [itex]\mbox{Tr}(AB) = \mbox{Tr}(BA)[/itex].
    Now, to solve your problem, you only need to apply the definition of the trace
    [tex]\mbox{Tr}(A) = \sum_{n} A_{nn} = \sum_{n} \langle n | A | n \rangle ,[/tex]
    where [itex]\{ |n\rangle \}[/itex] are complete orthonormal states, i.e., they satisfy
    [itex]\langle m | n \rangle = \delta_{mn}[/itex] and [itex]\sum | n \rangle \langle n | = I[/itex].
    In your case
    [tex]\mbox{Tr}( | \Psi \rangle \langle \Phi | ) = \sum_{n} \langle n | \Psi \rangle \langle \Phi | n \rangle .[/tex]
    Since [itex]\langle n | \Psi \rangle[/itex] and [itex]\langle \Phi | n \rangle[/itex] are numbers, you can write the above as
    [tex]
    \begin{align*}
    \mbox{Tr}( | \Psi \rangle \langle \Phi | ) &= \sum_{n} \langle \Phi | n \rangle \langle n | \Psi \rangle \\
    &= \langle \Phi | \left( \sum_{n} | n \rangle \langle n | \right) | \Psi \rangle \\
    &= \langle \Phi | I | \Psi \rangle \\
    &= \langle \Phi | \Psi \rangle ,
    \end{align*}
    [/tex]
    where the completeness relation [itex]\sum |n \rangle \langle n | = I[/itex] has been used.
     
  6. Jan 26, 2016 #5
    I see your point. I treated the number as a one-by-one matrix because I am not sure whether it is really a scalar times a identity matrix (not clear from the statement of the problem). Except this point, everything is fine.
     
  7. Jan 27, 2016 #6
    Thanks so much for your reply! I originally just calculated it directly when presented with the problem (i.e. I worked straight from the definition of trace) and got the same answer as the author (and afterwards also applied a similar procedure to yours to arrive at the more general expression), but I just didn't understand how Zettili had done it his way. Glad to know it was incorrect, although it confirms my growing suspicion while reading this book that Shankar's text is far superior and more reliable (with so few frickin problems though!).
     
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