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If we consider a system made out of two subsystems (i.e particles etc) and we do not consider interaction between the two subsystems. Then we have:

$H_1=-\frac {\hbar^2}{2m_1}\Delta_1 + V(r_1)$ (Hamiltonian of the first subsystem)

$H_2-\frac {\hbar^2}{2m_2}\Delta_2 + V(r_2)$ (Hamiltonian of the second subsystem)

$H=H_1 + H_2$ (The total Hamiltonian of the joint system)

$\vec L_1$, $\vec L_2$ are the angular momentum vector operators of each subsystem.

The the following is said:

$[\vec L_1,H]=[\vec L_2,H]=0$ (I was able to prove both statements, so I am perfectly clear here.)

But then this is what confuses me:

If this holds true: $[\vec L_1,H]=[\vec L_2,H]=0$ then:

The eigenfunctions of the total system are products of the eigenfunctions of each subsystem. Now how do we end up with this conclusion and should it say tensor product instead of simply product?

Edit:

It is obvious and logical to think that the eigenstate of a joint system of two subsystems is a tensor product of the eigenstates of each of them. It is not obvious (at least to me) how this being true : $[\vec L_1,H]=[\vec L_2,H]=0$ implies that the joint eigenstate is a tensor product of eigenstates of each system.

imbAF
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1 Answers1

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The wavefunctions are numerical products while the states are tensor products. If we label the eigenstates of the two systems by $n_1$ and $n_2$, the eigenstates of the combined system are $$ |n_1,n_2\rangle \stackrel{\rm def}{=} |n_1\rangle\otimes |n_2\rangle $$ and the eigenfunctions are $$ \langle x_1,x_2| n_1,n_2\rangle = (\langle x_2|\otimes \langle x_2|)(|n_1\rangle\otimes |n_2\rangle) =\langle x_2|n_1\rangle \langle x_2|n_2\rangle $$

mike stone
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