Phase Space of Two Particles

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  • Thread starter sophiatev
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In Henley and Garcia's Subatomic Physics, they introduce phase space in chapter 10 by considering all the possible locations a particle can occupy in a plot of ##p_x \ vs. x##, ##p_x## being the momentum of the particle in the x direction. They next consider an area pL on this plot, and state that while classically a particle can occupy any of the infinite points in this area, due to Heisenberg's Uncertainty Principle we cannot determine both the momentum and location of the particle to infinite precision. The best we can do is divide the area up into cells of size h and state that the particle resides somewhere inside this cell. In this sense we can only describe ##N = \frac{pL}{h}## states in this phase space. They then generalize this to momentum and spatial coordinates in three-dimensions, so we have a six-dimensional phase space. They confine the particle to a box of volume V and show that ##N_1 = \frac{V}{h^3} \int d^3p##, where the volume of a cell in this phase-space is now ##h^3##. Finally, they consider a situation with two particles where their total momentum is fixed. In this case, if we know the momentum of particle 1, we know the momentum of particle 2, so they claim that "the extra degrees of freedom are not really there". I can see why there are no degrees of freedom for ##p_2##, but what I'm confused about is where the spatial degrees of freedom for particle 2 went. I would think that for a given cell in phase space of particle 1, we have a range that ##p_1## can be. This determines the range of ##p_2##. But can't particle 2 reside spatially anywhere inside the volume ##V##? Drawing an analogy to the 1D case, we would have a whole row of cells for particle 2 given a cell of particle 1. They would all have the same height and width and be centered at the same ##p_{x,2}## value, but be centered at different ##x_2## values. In that case, shouldn't ##N_2 = \frac{V^2}{h^6} \int d^3p_1##? But they claim that ##N_2 = \frac{V}{h^3} \int d^3p_1##, and I can't figure out why the second particle doesn't have any spatial degrees of freedom.
 

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  • #2
anuttarasammyak
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In that case, shouldn't ##N_2 = \frac{V^2}{h^6} \int d^3p_1##? But they claim that ##N_2 = \frac{V}{h^3} \int d^3p_1##, and I can't figure out why the second particle doesn't have any spatial degrees of freedom.
The formula should have V not V^2, spatial degree of freedom for THE SECOND particle not TWO particles.
 
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