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While studying introduction to statistical mechanics ,I came across a new idea phase space where we use both position and momentum coordinates to denote a system .In my book the author calculates the number of quantum states within the energy range between E to E+dE.But after a while he writes that the number of quantum states in both momentum and energy states are equal but isn't energy E=p^2/2m and so the results are not equal as there will be a constant multiplied. Why the number of quantum states under these circumstances be equal is unclear to me...

Any help will be appreciated

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  • $\begingroup$ The number of states will be equal, not the actual eigenvalues of the energy or momentum operators. Does that answer your question? If not you might want to clarify your question. $\endgroup$ – Martin C. Dec 28 '17 at 15:41
  • $\begingroup$ If I calculate g(E) dE and then calculate g(p)dp by substituting E= p^2/2m how do we know that they are equal? i.e,g(E)=g(p)...I'm sorry but I am new to this sort of phase space and hence a little elaboration will be appreciated $\endgroup$ – Sam95 Dec 28 '17 at 15:46
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    $\begingroup$ Ah. That seems to be a question about integration rather than the number of states. I suggest you read about changing integration variables. Or you could post a concrete example of an integral that you are having difficulty with. $\endgroup$ – Martin C. Dec 28 '17 at 15:57
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The number of states with an energy between $E$ and $E+dE$ is: $$dN_E=D(E)dE$$ where $D(E)$ is the density of energy states. The number of available states in a system is one, and this number is the same no matter what you consider, whether it's energy or momentum. And it's from here that you impose that the number of energy states is equal to the number of momentum states. $$dN_E=D(E)dE=dN_p=D(p)dp$$ where $D(p)$ and $dN_p$ are respectively the density and number of momentum states. So since $dN_E=dN_p=dN$ you would have $$D(p)=D(E)\frac{dE}{dp}$$ and since $E=\frac{p^2}{2m}$ so $\frac{dE}{dp}=\frac{p}{m}$ you finally get $$D(p)=D(E)\frac{p}{m}$$ So yes, it's the density of energy and momentum states that is different and is multiplied by a constant as you said, but the number of states $dN$ is the same for momentum and energy.

Note: the $g(E)$ and $g(p)$ that you intend are just the integral of my $dN$ over the range of energy/momentum considered.

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