# I Squaring a Sum of Ket-Bra Operators

1. Dec 4, 2017

### jstrunk

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 $\alpha$ is real and $\left. {\left| {{\phi _1}} \right.} \right\rangle ,\left. {\left| {{\phi _2}} \right.} \right\rangle$ are normalized eigenstates of an operator ${\hat A}$ that has no degenerate eigenvalues.
Find ${{\hat H}^2}$.

My first step is ${{\hat H}^2} = {\alpha ^2}\left( {\left. {\left| {{\phi _1}} \right.} \right\rangle \left\langle {\left. {{\phi _2}} \right|} \right. + \left. {\left| {{\phi _2}} \right.} \right\rangle \left\langle {\left. {{\phi _1}} \right|} \right.} \right)\left( {\left. {\left| {{\phi _1}} \right.} \right\rangle \left\langle {\left. {{\phi _2}} \right|} \right. + \left. {\left| {{\phi _2}} \right.} \right\rangle \left\langle {\left. {{\phi _1}} \right|} \right.} \right)$.
The first step in the book is ${{\hat H}^2} = {\alpha ^2}\left( {\left. {\left| {{\phi _1}} \right.} \right\rangle \left\langle {\left. {{\phi _2}} \right|} \right. + \left. {\left| {{\phi _2}} \right.} \right\rangle \left\langle {\left. {{\phi _1}} \right|} \right.} \right)\left( {\left. {\left| {{\phi _1}} \right.} \right\rangle \left\langle {\left. {{\phi _1}} \right|} \right. + \left. {\left| {{\phi _2}} \right.} \right\rangle \left\langle {\left. {{\phi _2}} \right|} \right.} \right)$.
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 $\left. {\left| {{\phi _1}} \right.} \right\rangle$ and $\left. {\left| {{\phi _2}} \right.} \right\rangle$ are eigenstates of ${\hat A}$ and ${\hat A}$ is Hermitian,
they must be orthogonal. Since $\left. {\left| {{\phi _1}} \right.} \right\rangle$ and $\left. {\left| {{\phi _2}} \right.} \right\rangle$ are both
normalized and since $\left\langle {\left. {{\phi _1}} \right|} \right.\left. {{\phi _2}} \right\rangle = 0$, we can reduce ${{\hat H}^2}$ to
${{\hat H}^2} = {\alpha ^2}\left( {\left. {\left| {{\phi _1}} \right.} \right\rangle \left\langle {\left. {{\phi _2}} \right|} \right. + \left. {\left| {{\phi _2}} \right.} \right\rangle \left\langle {\left. {{\phi _1}} \right|} \right.} \right)\left( {\left. {\left| {{\phi _1}} \right.} \right\rangle \left\langle {\left. {{\phi _1}} \right|} \right. + \left. {\left| {{\phi _2}} \right.} \right\rangle \left\langle {\left. {{\phi _2}} \right|} \right.} \right)$

2. Dec 4, 2017

### hilbert2

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.