Calculating Energy Distribution of Free Particle - 1D & 2D

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SUMMARY

The discussion focuses on calculating the energy distribution of a free particle in one and two dimensions using the equation (3-12). The user attempts to derive the distribution function g(ε) and initially arrives at g(ε) = √(2m/ε), which is incorrect. The correct approach involves using the Dirac delta function in multiple dimensions, specifically integrating over momentum space in polar or spherical coordinates to simplify the calculations. The final expression for the energy distribution in three dimensions is L³g(ε) ∝ √(ε).

PREREQUISITES
  • Understanding of quantum mechanics concepts, specifically free particle behavior.
  • Familiarity with the Dirac delta function and its properties in integrals.
  • Knowledge of coordinate transformations, particularly polar and spherical coordinates.
  • Proficiency in using the equation (3-12) for energy distribution calculations.
NEXT STEPS
  • Study the derivation of energy distributions for free particles using the equation (3-12).
  • Learn about the application of the Dirac delta function in multiple dimensions.
  • Explore coordinate transformations in quantum mechanics, focusing on polar and spherical coordinates.
  • Investigate the implications of energy distributions in quantum statistical mechanics.
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Students and researchers in quantum mechanics, particularly those studying free particle dynamics and energy distributions in various dimensions.

genericusrnme
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Homework Statement



2. Calculate the energy distribution of the free particle
(a) in one dimension,
(b) in two dimensions,
Use (3-12) to calculate the energy distribution of the state, assuming
(a) V(r) = + K r2
(b) U f r ) = - Z e 2 / r

Homework Equations



(3-12);
L^3 g(\epsilon) = \frac{1}{\hbar ^3} \int d^3x \ d^3 p \ \delta( \epsilon - \frac{p^2}{2m} - U(x))
g is the distribution and L is the length of some container

The Attempt at a Solution



For part 2a) I'm getting g(\epsilon) = \sqrt{\frac{2m}{\epsilon}} which I'm pretty sure is wrong since surely the distribution shouldn't be inversely proportional to the energy..
I arrived at this result by using \delta( f(x)) = \sum_i \frac{\delta(x-x_i)}{|f'(x_i)|} where the xi's are the roots of f(x), I'm pretty sure this is correct so really I'm not sure where I went wrong.

I'm also not sure how to work with the dirac delta when I'm integrating over more than one variable like this, for example in the two dimensional case I'd have (leaving out constants) \int d^2p \ \delta(e - p^2) = \int d^2p \ \delta(e - p_x^2 - p_y^2) and I don't really know (or perhaps remember, it's been about a year since I last looked at a dirac delta) how to deal with this.

This 'energy distribution' thing is kinda confusing me too, the first mention of it is in regards to a particle in a box with all the n_{x,y,z}'s describing the energy and the distribution is given as (again, leaving out constants)

L^3 g(e) = \sum_{n_{x,y,z}} \delta(e - n_x^2 - n_y^2 - n_z^2)

which then goes to an integral
L^3 g(e) = \int d^3n \ \delta(e - n_x^2 - n_y^2 - n_z^2)
and then to an integral over p
L^3 g(e) = \int d^3p \ \delta(e - p^2)
L^3 g(e) \ \alpha \ \sqrt{e}

(as of writing this I realize that if this is correct then I was wrong in what I wrote about the dirac delta earlier)

So I'm not exactly sure what this energy distribution is supposed to be nor am I sure how to work with these dirac deltas over functions of more than one variable. :confused:

Thanks in advance!
 
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Since the argument of the delta function depends only on the magnitude of the momentum vector, go to polar coordinates (2 dimensional problem) or spherical coordinates (3 dimensional problem). The angle integrals will be trivial and you will be left with just an integral over the magnitude of p.
 

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