Taking the Hamiltonian operator $H = T + V$, I know that the eigenfunctions are stationary waves, this means that the wave function $\Psi$ is eigenfunction of T + V, this is also true in the case of the ground state of the hydrogen atom, but why is an eigenfunction of T + V? $\Psi$ can be eigenfunction only of the potential energy operator or the KE operator in this case? I mean, what properties have the sum of T + V that leads to an stationary wave as a eigenfunction in this particular case? I try to see what´s the problem on this and I don't find any reason that tells me that the stationary wave can not be an eigenfunction of T or V operators only. Can you help me to understand this?
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1$\begingroup$ So you are essentially asking where the time-independent Schrödinger equation comes from, and why is it the Hamiltonian $H$ appearing there? $\endgroup$– Thomas FritschCommented Dec 3, 2021 at 21:30
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1$\begingroup$ The Hamiltonian represents the total energy of the system, do you think that each energy separately are conserved? They are not as both the kinetic and potential energies change in time, and if each was to have their own stationary state eigenvalues, those eigenvalues would have to be constant in time. $\endgroup$– TriatticusCommented Dec 3, 2021 at 21:40
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$\begingroup$ Let me explain a bit more, taking the above we know that in the ground state of hydrogen atom we have a potential V, this potential is constant (or not?) then we can find a stationary wave function that is the eigenfunction of the V operator, then the problem I see is with the KE, I mean, the real question is: I can find the eigenfunction of the Hamiltonian knowing the eigenfunctions of T and V? $\endgroup$– DoubtDudeCommented Dec 3, 2021 at 22:09
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1$\begingroup$ @DoubtDude If you are asking whether you can obtain the spectrum of $A+B$ from the spectra of $A$ and $B$ separately, the answer is generally no. $\endgroup$– Seth WhitsittCommented Dec 3, 2021 at 22:58
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$\begingroup$ The potential $V\sim 1/r$ is certainly not constant… $\endgroup$– ZeroTheHeroCommented Dec 4, 2021 at 2:51
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1 Answer
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The stationary states of the system are defined as the eigenfunctions of the Hamiltonian operator. $$H|\psi\rangle =E|\psi\rangle $$ These eigenfunctions are necessary to be an eigenfunction of kinetic or potential. This is because these are functions of $x$ and $p$ which don't commute with each other. $$[H,K(p)]\not=0\not=[H,V(x)]$$
Only for free particle when $V(x)=0$ we have $$H=\frac{p^2}{2m}\rightarrow [H,p]=0$$ therefore kinetic energy (or momentum) and Hamiltonian have the same eigenfunctions.
