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Quantum mechanics bound states

  • Thread starter jc09
  • Start date
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1. Homework Statement
How many bound states are there quantum mechanically ?


We are told to approach the problem semi classically.
Consider the Hamiltonian function
H : R 2n → R
(whose values are energies), and for E0 < E1 the set
{(p, x) ∈ R 2n |H(p, x) ∈ [E0 , E1 ]} ⊆ R 2n
,
which we assume to have the 2n-dimensional volume V (2n) . It is a fact that when-
ever V (2n) is finite, then there are only finitely many (distinguishable) quantum
mechanical states. More precisely, one has
V (2n) hn ≈ ♯{states of energy E ∈ [E0 , E1 ]},
where h = 2π. Moreover, strict equality holds provided the l.h.s. is an integer.

Asked to consider
asked to consider the Hamiltonian function
H(p, x) = p1^2 2m1 + p2^2 2m2 + 1 /2 m1 ω 1^2 x1^2 + 1/ 2 m2 ω 2^ 2 x2^2 ,
and to determine the approximate number of states of energy E
≤ Etotal .
Hint: This is the equation of an el lipsoid in 4-dimensional phase space with
coordinates (p1 , p2 , x1 , x2 ). The volume of the ellipsoid with radii a, b, c, d is abcd
times the volume of the 4-dimensional unit sphere


I'm stuck trying to find a starting point for the problem
 
125
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If I recall this method correctly, [tex]$ N \approx \int d^np d^n x/(2\pi\hbar)^n$[/tex], where N is the number of states - it is just the amount of "cells" in phase space. The integral should be taken over the whole possible space in it. In your case this space is 4-dimensional (p1, x1, p2, x2) and is bounded by the energy conservation law:
[tex]$ E = p_1^2/2m_1 + m_1 \omega_1^2 x_1^2 /2 + p_2^2/2m_2 + m_2 \omega_2^2 x_2^2 /2 \le E_{total}$[/tex]
This is an equation of a 4 dimensional ellipsoid.
To calculate the n-dimensional surface of a unit sphere do the following:
consider the integral [tex]$\int \exp(-r^2)d^n r$[/tex] over the whole space. Now write [tex]$ d^n r = \Omega_n r^{n-1} dr$[/tex] where [tex] $\Omega_n$ [/tex] is the surface of a unit n-dimensional sphere. Substituting [tex]$ r = \sqrt{z}$[/tex] you get gamma function
[tex] $\int \exp(-r^2)d^n r = \Omega_n \Gamma(n/2)/2$ [/tex].
On the other hand, the integral can be represented as
[tex] $\int \exp(-r^2)d^n r = \int_{-\infty}^{+\infty}\exp(-x_1^2)dx_1 \times \int_{-\infty}^{+\infty}\exp(-x_2^2)dx_2 \times ... \times \int_{-\infty}^{+\infty}\exp(-x_n^2)dx_n$ [/tex].
Each of these integrals is Poisson integral and is equal to [tex]$\sqrt{\pi}$[/tex]. Comparing both sides you get at last
[tex]$\Omega_n = 2 \pi^{n/2}/\Gamma(n/2)$[/tex].
For instance, when n = 2 ("surface area" of a circle) this gives [tex]$\Omega_2 = 2 \pi/\Gamma(1)=2\pi$[/tex],
for n = 3 taking into account that [tex]$\Gamma(3/2) = 1/2 \times \Gamma(1/2) = \sqrt{\pi}/2$[/tex] you get (as it should be) [tex]$\Omega_3 = 4\pi$[/tex]
in your case n = 4 and so [tex] $\Omega_4 = 2\pi^2$ [/tex].

Good luck!
 
45
0
If I recall this method correctly, [tex]$ N \approx \int d^np d^n x/(2\pi\hbar)^n$[/tex], where N is the number of states - it is just the amount of "cells" in phase space. The integral should be taken over the whole possible space in it. In your case this space is 4-dimensional (p1, x1, p2, x2) and is bounded by the energy conservation law:
[tex]$ E = p_1^2/2m_1 + m_1 \omega_1^2 x_1^2 /2 + p_2^2/2m_2 + m_2 \omega_2^2 x_2^2 /2 \le E_{total}$[/tex]
This is an equation of a 4 dimensional ellipsoid.
To calculate the n-dimensional surface of a unit sphere do the following:
consider the integral [tex]$\int \exp(-r^2)d^n r$[/tex] over the whole space. Now write [tex]$ d^n r = \Omega_n r^{n-1} dr$[/tex] where [tex] $\Omega_n$ [/tex] is the surface of a unit n-dimensional sphere. Substituting [tex]$ r = \sqrt{z}$[/tex] you get gamma function
[tex] $\int \exp(-r^2)d^n r = \Omega_n \Gamma(n/2)/2$ [/tex].
On the other hand, the integral can be represented as
[tex] $\int \exp(-r^2)d^n r = \int_{-\infty}^{+\infty}\exp(-x_1^2)dx_1 \times \int_{-\infty}^{+\infty}\exp(-x_2^2)dx_2 \times ... \times \int_{-\infty}^{+\infty}\exp(-x_n^2)dx_n$ [/tex].
Each of these integrals is Poisson integral and is equal to [tex]$\sqrt{\pi}$[/tex]. Comparing both sides you get at last
[tex]$\Omega_n = 2 \pi^{n/2}/\Gamma(n/2)$[/tex].
For instance, when n = 2 ("surface area" of a circle) this gives [tex]$\Omega_2 = 2 \pi/\Gamma(1)=2\pi$[/tex],
for n = 3 taking into account that [tex]$\Gamma(3/2) = 1/2 \times \Gamma(1/2) = \sqrt{\pi}/2$[/tex] you get (as it should be) [tex]$\Omega_3 = 4\pi$[/tex]
in your case n = 4 and so [tex] $\Omega_4 = 2\pi^2$ [/tex].

Good luck!



So This means that I have my surface area which is 2*pi(squared) and what I should do is integrate over this within the region of the 4 dimensional ellipsoid???
 
125
0
Well, if I recall this method correctly :wink: all you have to do is just find the volume of the 4-dimensional ellipsoid given it's radii a,b,c,d (you can extract them from the energy conservation law). As it is said in your first post this volume is equal to a*b*c*d*2*pi^2. Then divide this volume by [tex]$ (2\pi\hbar)^4$[/tex].
 

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