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  Mass, Spin, Internal Energy and 1-Particle States in Galilean Quantum Mechanics

+ 2 like - 0 dislike

I have been reading an article discussing the unitary representation of Galilean group and non-relativistic quantum mechanics. The link to the article is given below.


From equation 2.17a~2.17b, we see three Casimir operators. The authors claimed that the 1-particle states are therefore labeled by three numbers: mass, spin and internal energy.

Unfortunately, I do not understand why $\hat{H}-\frac{1}{2\hat{M}}\hat{P}^{2}$ should be interpreted as internal energy.

As far as I could remember from my bachelor QM course, the 1-particle states have nothing related with thermal dynamics or internal energy.

From the above paper, it seems that the energy (or Hamiltonian) is the internal energy plus kinetic energy. But from my bachelor QM, the Hamiltonian should be kinetic energy plus potential.

Can anyone help understand the 'internal energy'?

asked Aug 10, 2015 in Theoretical Physics by XIaoyiJing (50 points) [ no revision ]
recategorized Aug 11, 2015 by Jia Yiyang

It is talking about free particles, so there's no external potential $V(x)$ involved. The representation theory doesn't care about the "elementarity" of the particles under consideration, so it's rather natural to have an independent "internal energy" parameter, unlike relativistic case where rest mass is intimately related to internal energy. For example, we can have a point particle of mass $M$ and internal energy 0, and we can also have a composite particle with the same mass $M$ but a nonzero internal energy due to internal interactions (again, same rest mass but different internal energies is impossible in relativistic case). In both cases the representation theory applies.  

Thank you so much for the comment @Jia Yiyang. I found the same answer from the lecture notes written by Weinberg last night. Thanks a lot.

No problem. BTW to properly activate the @ function you need to eliminate all the space in the username, for example if I need to @ you I should type @XIaoyiJing. 

@JiaYiyang: Internal interactions are of course also possible in the relativistic case - protons and neutrons in a nucleus, nuclei and electrons in an atom, atoms in a molecule!

Thank you all for your help!

@ArnoldNeumaier, you misunderstood me, I was saying relativistically you can't have two particles with the same rest mass but different internal energies, I'll edit the comment to make it clear.

1 Answer

+ 3 like - 0 dislike

Internal energy is the energy used to build a composite particle. Nonrelativistically, you may think of it as the energy of the bound state making up the particle from its basic constituents. As long as particles do not change their internal state, the internal energy is just a constant and can be ignored.

For a single relativistic particle, the internal energy is $mc^2$, where $m$ is the particle's rest mass. In the limit $c\to\infty$ where one gets the Galilei group from the centrally extended Poincare group, one must subtract this constant part to get finite limiting results. Clearly, one can subtract instead $mc^2-E_0$ for any finite $E_0$ and gets a finite internal energy. As it is constant it has no dynamical effect.

answered Aug 11, 2015 by Arnold Neumaier (15,787 points) [ no revision ]

Thank you very much Arnold. Your explanation about the relativistic particle is really inspiring. 

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