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I was trying to do this exercise from Cohen-Tannoudji Quantum Mechanics book:

$|\varphi_n\rangle$ are the eigenstates of a Hermitian operator $H$ ( $H$ is, for example, the Hamiltonian of some physical system). Assume that the states $|\varphi_n\rangle$ form a discrete orthonormal basis.

The operator ($U(m,n)$) is defined by:

$U(m,n) = |\varphi_m\rangle \langle \varphi_n|$

a)Calculate the adjoint $U^{\dagger}(m,n)$ of $U(m,n)$

b)Calculate the commutator $[H,U(m,n)]$

c)...

I don't know the right answer, but I've the intuition that this should be $0$, but this is farthest that I got:

Since $|\varphi_n\rangle$ form a basis we have that this obey the closure relation $\sum_{j} |\varphi_j\rangle \langle \varphi_j| = I$.

So we have:

$$[H,U(m,n)] = H|\varphi_m\rangle \langle \varphi_n| - |\varphi_m\rangle \langle \varphi_n|H = IH|\varphi_m\rangle \langle \varphi_n| - |\varphi_m\rangle \langle \varphi_n|HI$$

Using the closure reation:

$$[H,U(m,n)] = \sum_{j} |\varphi_j\rangle \langle \varphi_j|H|\varphi_m\rangle \langle \varphi_n| - \sum_{l}|\varphi_m\rangle \langle \varphi_n|H|\varphi_l\rangle \langle \varphi_l|$$

Since $|\varphi_n\rangle$ are eigenstates of H we have $H|\varphi_k\rangle = a_k|\varphi_k\rangle$, then:

$[H,U(m,n)] = \sum_{j} |\varphi_j\rangle \langle \varphi_j|a_m|\varphi_m\rangle \langle \varphi_n| - \sum_{l}|\varphi_m\rangle \langle \varphi_n|a_l|\varphi_l\rangle \langle \varphi_l| = \sum_{j} a_m|\varphi_j\rangle \delta_{jm} \langle \varphi_n| - \sum_{l} a_l|\varphi_m\rangle \delta_{ln} \langle \varphi_l| = (a_m - a_n)|\varphi_m\rangle \langle \varphi_n|$

So that is what I got, is that right ? So they shouldn't commute unless I've $U(m,m)$? Or is there something more that I didn't see?

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    $\begingroup$ I don't know the right answer, but I've the intuition that this should be 0 - why do you think so? Further: So they shouldn't commute unless I've $U(m,m)$? - or $a_n=a_m$, i.e. there are some degenerate states. $\endgroup$ Nov 13 '21 at 17:48
  • $\begingroup$ I don't know why, it was just an intuition hahaha, but it was really wrong. I had forgotten about the degenerate states, thanks! $\endgroup$ Nov 13 '21 at 23:03

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