I can't follow the solution given in my text book to the following problem.(adsbygoogle = window.adsbygoogle || []).push({});

The solution goes right off the rails on the first step.

Consider a system whose Hamiltonian is given by

[itex]

\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)

[/itex],

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]

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# I Squaring a Sum of Ket-Bra Operators

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