Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Separation of the centre of mass motion

  1. Jul 25, 2014 #1
    hey

    I got a questions and appreciate obv any reply. Im regarding to Capitel 12, page 128ff. from Gustafson, Sigal; Mathematical Concepts for Quantum Mechanics.

    So, lets consider on $L^2(\mathbb R^n)$
    \begin{align}
    H_n= \sum_{j=1}^n \frac{1}{2 m_j} p_j^2 +V(x)
    \end{align}
    where $$p_j=-i\hbar\nabla_{x_j},\qquad V(x)=\frac{1}{2} \sum_{i\neq j} V_{ij}(x_i-x_j)$$ is teh momentum operator.

    Separation of the centre-of-mass

    the latter Hamiltonian has purely essential spectrum, cause it commutes with the total translation of the system
    $$ T_h: \psi(x_1,\ldots,x_n) \mapsto \psi(x_1+h,\ldots,x_n+h) $$
    First Question: is there a explanation or physical Interpretation behind, why this follows from translationinvariance?

    we are now on page 129 middle.

    So, now we "break" the translation invariance, with fixing the centre of mass at the origin, to get some interessting spectral information.
    Second Question: I dont understand, why we break the system, cause i dont get if we then got some interessting informatino about the spectrum, the old Hamiltonian still have purely essential spectrum. Arent we looking for his Spectrum. I think, I need some hit in the right direction, to understand this.

    if further information for the system/ or sth else is needed pls let me know.

    bests, tks in adv
    pat1enc3_17
     
  2. jcsd
  3. Jul 25, 2014 #2

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    Ad Question #1: The total momentum is the generator of translations of the system as a whole, and it commutes with the Hamiltonian. Thus the energy eigenfunctions do not change under translations.

    Ad Question #2: Of course, you don't break any symmetries just by introducing new variables that are more adequate to describe your system. Thus the formulation that you break translation invariance is misleading. What happens is that the center-of-mass motion separates from the relative motions of the particles. According to Noether's theorem, applied to the invariance under Galilei boosts, the center of mass motion is that of a free particle.
     
  4. Jul 25, 2014 #3
    tks for the quick reply!

    to answer #1: "The total momentum is the generator of translations of the system as a whole" can you outline this a bit pls.
    If i get that right, if I have a point $b$ in the spectrum, so cause the corresponding eigenfunctions d not change under translation, so i can perturb it a bit and its still an eigenfunction, so $b+h$ is also an eigenvalue, so its essential?

    to answer #2: ok, if free particle, I got a Hamiltonian only with Kinetic Energy($H_1$ relative motion)
    \begin{align}
    H = H_1 \otimes 1 + 1\otimes H_{com},
    \end{align}
    and as a result i got a part of the centre of mass motion, tho i can apply: long range -> infinite boundstates(min-max-principle) and for short range -> finite (at infinity kinetic energie dominates cause of hardys-inequality the boundstates favoring potential), right?
    Still, if i break, i have a broken system, with probably boundstates in the Clusters (depending on the potential, and esp on the ionization threshold), but still if i go back to my old system, I still have purely essential spectrum. But then, why I do this?
     
  5. Jul 29, 2014 #4
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Separation of the centre of mass motion
  1. Separation constant (Replies: 1)

Loading...