Translation Invariant: Seeing it Intuitively & Mathematically

[aaaa202]

How do I see that when my hamiltonian is translation invariant i.e. H = H(r-r') it means that it is diagonal in the momentum basis? I can see it intuitively but not mathematically.

 

[The_Duck]

If you can show that $[\hat p, \hat H] = 0$ then you have shown that $\hat p$ and $\hat H$ can be simultaneously diagonalized, which is what you are after.

 

[Jano L.]

With your definition of translational invariance (if I understood it, it is invariance with respect to coordinate shifts) the hydrogen atom Hamiltonian is translationally invariant, because the potential energy term is function of $|\mathbf r_1 - \mathbf r_2|$. But this Hamiltonian is not diagonal in the momentum basis.

 

[BruceW]

the 'true' hydrogen atom Hamiltonian should be diagonal in the momentum basis. But usually people just talk about a central potential, and ignore the nucleus, since it is much heavier than the electron. Once we make this approximation of a central potential and ignore the nucleus, the potential energy term now depends on absolute spatial position, so the Hamiltonian is translationally variant.

 

[Jano L.]

Even if the Hamiltonian
$$
\hat{H} = \frac{\hat{\mathbf{p}}_1^2}{2m_1} + \frac{\hat{\mathbf{p}}_2^2}{2m_2} - \frac{Kq^2}{4\pi} \frac{1}{|\mathbf r_1 - \mathbf r_2|},
$$
is translationally invariant, I do not think it is diagonal in the momentum basis $\mathbf p_1, \mathbf p_2$. Only the sum of the two momentum operators commutes with the Hamiltonian, but the momentum operators individally do not.

 

[vanhees71]

Rewrite the Hamiltonian in terms of the CM motion and the relative motion, i.e., in terms of the new variables
$$\vec{R}=\frac{m_1 \vec{r}_1 + m_2 \vec{r}_2}{m_1+m_2}, \quad \vec{P}=\vec{p_1}+\vec{p}_2, \quad \vec{r}=\vec{r}_1-\vec{r}_2, \quad \vec{p}=\frac{m_2 \vec{p}_1-m_1 \vec{p}_2}{m_1+m_2}.$$
Then you can show that $\vec{P}$, $\vec{H}$, $\vec{l}^2$ and $l_z$ with the orbital angular momentum of the relative motion
$$\vec{l}=\vec{r}\times \vec{p}$$
form a complete set of observables.