# Commutator calculus

1. Feb 28, 2014

### Berny

1. The problem statement, all variables and given/known data
|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.)

3. 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.

2. Feb 28, 2014

### hilbert2

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.

3. Feb 28, 2014

### ChrisVer

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/

4. Mar 1, 2014

### Berny

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 ?

5. Mar 1, 2014

### hilbert2

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.

6. Mar 1, 2014

### vela

Staff Emeritus
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 $\langle \phi_n \rvert$?