Quantum mechanics bound states

The result will be the number of states.In summary, the conversation discusses the problem of determining the number of bound states quantum mechanically. The approach suggested is to consider the Hamiltonian function and the set of states within a specific energy range. It is stated that the number of states is determined by the 2n-dimensional volume of the set, and in the case of a finite volume, there are only a finite number of distinguishable quantum mechanical states. The conversation then moves on to discuss the specific Hamiltonian function provided and provides a hint for solving the problem. The method mentioned involves calculating the n-dimensional surface of a unit sphere and using this to find the number of states. Ultimately, the goal is to find the volume of the 4
  • #1
45
0

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
 
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  • #2
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!
 
  • #3
quZz said:
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?
 
  • #4
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].
 

1. What are bound states in quantum mechanics?

Bound states in quantum mechanics refer to the states of a system in which the particles are confined to a certain region of space due to the potential energy of the system. These states have discrete energy levels and cannot escape the potential well, unlike unbound states that have continuous energy levels.

2. How do bound states differ from unbound states in quantum mechanics?

Bound states have discrete energy levels and are confined within a specific region of space, while unbound states have continuous energy levels and are not confined to a specific region. Bound states are also typically more stable than unbound states.

3. What is the significance of bound states in quantum mechanics?

Bound states play a crucial role in understanding the behavior of atoms, molecules, and other quantum systems. They help explain the stability of matter and the properties of materials, and are important in applications such as quantum computing and nuclear physics.

4. How are bound states described in quantum mechanics?

Bound states are described by the Schrödinger equation, which is a fundamental equation in quantum mechanics that describes the behavior of particles in a potential well. The solutions to this equation give the energy levels and wavefunctions of the bound states.

5. Can bound states exist in all potential energy wells in quantum mechanics?

No, bound states can only exist in potential wells that are deep enough to confine the particles. If the potential well is too shallow, the particles can escape and the system will not have any bound states. The shape and depth of the potential well determine the existence and properties of bound states in a system.

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