- #1
jstrunk
- 55
- 2
I can't follow the solution given in my textbook to the following problem.
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]
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]