Separation of the centre of mass motion

pat1enc3_17
Messages
8
Reaction score
0
hey

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

So, let's 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 don't understand, why we break the system, cause i don't 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
 
Physics news on Phys.org
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.
 
  • Like
Likes 1 person
vanhees71 said:
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.
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?
 
bump.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!

Similar threads

Replies
3
Views
3K
Replies
7
Views
3K
2
Replies
61
Views
11K
Replies
9
Views
2K
3
Replies
100
Views
11K
Replies
87
Views
7K
Replies
19
Views
4K
Back
Top