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Translation invariant

  1. Jan 7, 2014 #1
    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.
     
  2. jcsd
  3. Jan 7, 2014 #2
    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.
     
  4. Jan 7, 2014 #3

    Jano L.

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    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.
     
  5. Jan 7, 2014 #4

    BruceW

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    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.
     
  6. Jan 7, 2014 #5

    Jano L.

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    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.
     
  7. Jan 8, 2014 #6

    vanhees71

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    Rewrite the Hamiltonian in terms of the CM motion and the relative motion, i.e., in terms of the new variables
    [tex]\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}.[/tex]
    Then you can show that [itex]\vec{P}[/itex], [itex]\vec{H}[/itex], [itex]\vec{l}^2[/itex] and [itex]l_z[/itex] with the orbital angular momentum of the relative motion
    [tex]\vec{l}=\vec{r}\times \vec{p}[/tex]
    form a complete set of observables.
     
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