Ground state energy of a two particle gravitational atom

quasar_4
Messages
273
Reaction score
0

Homework Statement



Two neutral spinless particles of mass m are gravitationally bound to one another. What is the ground state energy of this two-particle gravitational atom?

Homework Equations





The Attempt at a Solution



So, it's a two particle system, but

H_{total} = H_1 + H_2 and \psi_{12} = \psi_1 \psi_2 implies

H_{total} \psi_{12} = (H_1 + H_2) \psi_1 \psi_2 = (E_1 + E_2) \psi_1 \psi_2.

So I should be able to find the energy eigenvalues for one of the particles since they're identical, then (assuming spinless = zero spin = bosons), I should just be able to add the two ground state energies together.

My Hamiltonian for a single particle is

\frac{p^2}{2m} - \frac{Gm^2}{x^2} = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x ^2} - \frac{Gm^2}{x^2}

so I get an ODE of the form

\frac{\partial^2\psi}{\partial x^2} - \frac{2 G m^3}{\hbar^2 x^2}\psi + \frac{2 mE}{\hbar^2} \psi = 0

What I don't see is how on Earth energy is quantized. I guess I am under the impression that to quantize any observable, we need boundary conditions (i.e., isn't it usually the presence of a boundary condition that causes quantization)? In any case, for a bound state, the total energy must be less than the potential for the system... but how do I fit that into this problem?

Or, option 2: have I set it up totally wrong? I could try to use the center of mass frame, in which things might start to look more like the hydrogen atom. Maybe I'll work on that until this gets some replies.
 
Physics news on Phys.org
What makes you think this is a one-dimensional problem? I would treat it in 3-d, just like the hydrogen atom, except that the constants are different.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top