Calculating Commutator [H,U(m,n)] with Homework Statement

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The discussion revolves around calculating the commutator [H, U(m,n)] where H is a Hermitian operator and U(m,n) is defined as |phi(m)><phi(n)|. The participants derive the expression for the commutator, leading to the conclusion that [H, U(m,n)] = (E(m) - E(n))U(m,n), where E(m) and E(n) are the eigenvalues corresponding to the states |phi(m)> and |phi(n)>, respectively. It is emphasized that the commutator's value is independent of the specific state of the quantum system. The conversation also highlights the importance of precise notation in quantum mechanics to avoid errors in calculations. Overall, the calculation confirms the relationship between the commutator and the energy eigenvalues of the system.
Berny
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Homework Statement


|phi (n)> being eigen states of hermitian operator H ( H could be for example the hamiltonian
of anyone physical system ). The states |phi (n)> form an orthonormal discrete basis.
The operator U(m,n) is defined by:

U(m,n)= |phi(m)><phi(n)|
Calculate the commutator:
[H,U(m,n)]

( this is part of the first problem in Cohen, Tannoudji, Diu, Laloe textbook in quantum mechanics.)

The Attempt at a Solution

\

[HU-UH] (ψ) = H|phi(m)><phi (n)|ψ> - |phi (m)><phi(n)| H| ψ>

= <phi(n)|ψ> H |phi (m>) - |phi (m><phi(n)| <ψ | H

and then ? i did not find symbol phi.
 
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Note that the operator ##H## can be written as ##H=\sum\limits_{k}E_{k}\left|\phi(k)\right>\left<\phi(k)\right|##, where the ##E_{k}## are its eigenvalues. Also, note that the vectors ##\left|\phi(k)\right>## form an orthonormal set.
 
you should use an eigenstate of the Hamiltonian instead of \psi (if you ask how you can do that, you can expand \psi as a superposition of the Hamiltonian's eigenstates)
Then in general, following correct paths you will reach the desired result/
 
Berny said:

Homework Statement


|phi (n)> being eigen states of hermitian operator H ( H could be for example the hamiltonian
of anyone physical system ). The states |phi (n)> form an orthonormal discrete basis.
The operator U(m,n) is defined by:

U(m,n)= |phi(m)><phi(n)|
Calculate the commutator:
[H,U(m,n)]

( this is part of the first problem in Cohen, Tannoudji, Diu, Laloe textbook in quantum mechanics.)

The Attempt at a Solution

\

[HU-UH] (ψ) = H|phi(m)><phi (n)|ψ> - |phi (m)><phi(n)| H| ψ>

= <phi(n)|ψ> H |phi (m>) - |phi (m><phi(n)| <ψ | H

and then ? i did not find symbol phi.

many thanks for help. I find a commutator value depending upon the system energy :

commutator = E(m) U(m,n) if E= E(m) and - E(n) U(m,n) if E=E(n).

in other cases it's zero.
Is this correct ?
 
I got something like [H,U(m,n)] = (E(m)-E(n))U(m,n). The commutator does not depend on what state the quantum system is in.
 
Berny said:

Homework Statement


|phi (n)> being eigen states of hermitian operator H ( H could be for example the hamiltonian
of anyone physical system ). The states |phi (n)> form an orthonormal discrete basis.
The operator U(m,n) is defined by:

U(m,n)= |phi(m)><phi(n)|
Calculate the commutator:
[H,U(m,n)]

( this is part of the first problem in Cohen, Tannoudji, Diu, Laloe textbook in quantum mechanics.)

The Attempt at a Solution

\

[HU-UH] (ψ) = H|phi(m)><phi (n)|ψ> - |phi (m)><phi(n)| H| ψ>

= <phi(n)|ψ> H |phi (m>) - |phi (m><phi(n)| <ψ | H
You can say that ##\hat{H}\lvert \phi_m \rangle \langle \phi_n \vert \psi \rangle = \langle \phi_n \vert \psi \rangle \hat{H}\lvert \phi_m \rangle## because ##\langle \phi_n \vert \psi \rangle## is a number, though it doesn't really help you in this case. What you can't do is say ##\lvert \phi_m \rangle \langle \phi_n \lvert \hat{H} \rvert \psi \rangle## equals ##\lvert \phi_m \rangle \langle \phi_n \lvert \langle \psi \rvert \hat{H}## because ##\hat{H}\lvert \psi \rangle## and ##\langle \psi \rvert \hat{H}## aren't the same. One's a bra; the other, a ket. You need to be a bit more precise with your notation, otherwise you're invariably going to make errors.

You have, so far,
\begin{align*}
[\hat{H},\hat{U}] &= \hat{H}\hat{U} - \hat{U}\hat{H} \\
&= \hat{H}\lvert \phi_m\rangle\langle\phi_n\rvert - \lvert \phi_m\rangle\langle\phi_n\rvert\hat{H}
\end{align*} Now in the first term, apply ##\hat{H}## to the ket ##\lvert \phi_m \rangle##. What do you get? Similarly, in the second term, what do you get when ##\hat{H}## acts on the bra ##\langle \phi_n \rvert##?
 

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