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I Squaring a Sum of Ket-Bra Operators

  1. Dec 4, 2017 #1
    I can't follow the solution given in my text book to the following problem.
    The solution goes right off the rails on the first step.

    Consider a system whose Hamiltonian is given by
    \hat H = \alpha \left( {\left. {\left| {{\phi _1}} \right.} \right\rangle \left\langle {\left. {{\phi _2}} \right|} \right. + \left. {\left| {{\phi _2}} \right.} \right\rangle \left\langle {\left. {{\phi _1}} \right|} \right.} \right)
    where [itex]\alpha[/itex] is real and [itex]\left. {\left| {{\phi _1}} \right.} \right\rangle ,\left. {\left| {{\phi _2}} \right.} \right\rangle[/itex] are normalized eigenstates of an operator [itex]{\hat A}[/itex] that has no degenerate eigenvalues.
    Find [itex]{{\hat H}^2}[/itex].

    My first step is [itex]{{\hat H}^2} = {\alpha ^2}\left( {\left. {\left| {{\phi _1}} \right.} \right\rangle \left\langle {\left. {{\phi _2}} \right|} \right. + \left. {\left| {{\phi _2}} \right.} \right\rangle \left\langle {\left. {{\phi _1}} \right|} \right.} \right)\left( {\left. {\left| {{\phi _1}} \right.} \right\rangle \left\langle {\left. {{\phi _2}} \right|} \right. + \left. {\left| {{\phi _2}} \right.} \right\rangle \left\langle {\left. {{\phi _1}} \right|} \right.} \right)[/itex].
    The first step in the book is [itex]{{\hat H}^2} = {\alpha ^2}\left( {\left. {\left| {{\phi _1}} \right.} \right\rangle \left\langle {\left. {{\phi _2}} \right|} \right. + \left. {\left| {{\phi _2}} \right.} \right\rangle \left\langle {\left. {{\phi _1}} \right|} \right.} \right)\left( {\left. {\left| {{\phi _1}} \right.} \right\rangle \left\langle {\left. {{\phi _1}} \right|} \right. + \left. {\left| {{\phi _2}} \right.} \right\rangle \left\langle {\left. {{\phi _2}} \right|} \right.} \right)[/itex].
    I have no idea how they got that second term. The solution gives the following justification. I understand each of the points given in the justification, but as far as I can see the do nothing to justify the conclusion.
    Since [itex]\left. {\left| {{\phi _1}} \right.} \right\rangle[/itex] and [itex]\left. {\left| {{\phi _2}} \right.} \right\rangle[/itex] are eigenstates of [itex]{\hat A}[/itex] and [itex]{\hat A}[/itex] is Hermitian,
    they must be orthogonal. Since [itex]\left. {\left| {{\phi _1}} \right.} \right\rangle[/itex] and [itex]\left. {\left| {{\phi _2}} \right.} \right\rangle[/itex] are both
    normalized and since [itex]\left\langle {\left. {{\phi _1}} \right|} \right.\left. {{\phi _2}} \right\rangle = 0[/itex], we can reduce [itex] {{\hat H}^2}[/itex] to
    [itex]{{\hat H}^2} = {\alpha ^2}\left( {\left. {\left| {{\phi _1}} \right.} \right\rangle \left\langle {\left. {{\phi _2}} \right|} \right. + \left. {\left| {{\phi _2}} \right.} \right\rangle \left\langle {\left. {{\phi _1}} \right|} \right.} \right)\left( {\left. {\left| {{\phi _1}} \right.} \right\rangle \left\langle {\left. {{\phi _1}} \right|} \right. + \left. {\left| {{\phi _2}} \right.} \right\rangle \left\langle {\left. {{\phi _2}} \right|} \right.} \right)[/itex]
  2. jcsd
  3. Dec 4, 2017 #2


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    Note that for operators, ##(A+B)^2 = A^2 + AB +BA +B^2## which is not necessarily the same as ##A^2 + 2AB + B^2## because ##A## and ##B## don't have to commute. Also, you can add a term like ##\left|\right.\phi_1 \left.\right>\left<\right.\phi_2 |\phi_1 \left.\right>\left<\right.\phi_2 \left.\right|## to any sum without changing the sum because the two vectors are orthogonal.
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