Conservative physical quantities

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If a physical quantity like the angular momentum operator vector L is conservative, indicated by the commutation relation [H, L] = 0, then its components Lx, Ly, and Lz are also conserved. This principle applies to all conservative vector operators, including the spin vector S. The conservation of these components is determined by the Hamiltonian, which dictates whether an operator is conserved over time. Therefore, if [H, L] = 0, it can be inferred that [H, Lx], [H, Ly], and [H, Lz] will also equal zero without needing to check each component individually. Understanding these relationships is crucial for analyzing constants of motion in quantum mechanics.
KostasV
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Hello ppl !
If i find that a physical quantity (lets say angular momentum operator vector L) is conservative (this means [H,L]=0 - H=hamiltonian ) then its 3 components Lx , Ly and Lz are being conserved too ?
That happens with every conservative vector operator ? Like spin vector S and his components?
I am confused ... :S
 
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The constants of the motion is dependent on the Hamiltonian, for simple Hamiltonian such as hydrogen atom whose potential is symmetric, angular momentum L and its components are indeed conserved in time.
KostasV said:
That happens with every conservative vector operator ?
Its the Hamiltonian which decides whether an operator is conserved in time or not.
 
blue_leaf77 said:
The constants of the motion is dependent on the Hamiltonian, for simple Hamiltonian such as hydrogen atom whose potential is symmetric, angular momentum L and its components are indeed conserved in time.

Its the Hamiltonian which decides whether an operator is conserved in time or not.
i am concerned about the components !
Ok, let's say that i find that [H,L]=0 (so L -angular momentum vector- is being conserved) ! Do i have to find the commutators [H,Lx] , [H,Ly] , [H,Lz] or i am sure that they will all be zero due to the fact that [H,L]=0 ?
 
In that case, where the one which commutes the Hamiltonian is the vector L, the components all commute with the Hamiltonian as well.
 
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blue_leaf77 said:
In that case, where the one which commutes the Hamiltonian is the vector L, the components all commute with the Hamiltonian as well.
Thank you very much for the help ;)
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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