How Does Quantum Mechanics Explain Nearly Equal Energy States in Ammonia?

[noblegas]

Homework Statement

In a streamlined model for the low energy states of an ammonia atom, (NH3), imagine that a nitrogen atom moves in one dimension in the potential V(x) sketched in figure I.1(found in Peebles textbook on p.86); The potential has two minima, one on each side of the triangle defined by three hydrogen atoms, and a relatively high peak between the minima, at the plane of the three hydrogen atoms. Thus in classical physics the nitrogen atom in the ground state would sit in one of the minima. Work is required to pull the nitrogen atom away from the molecule , and in classical physics work is required to push the nitrogen atom into the plane of the three hydrogen atoms at x=0. Sketched shapes of the waves functions [tex]\varphi[/tex]₀ and [tex]\varphi[/tex]₁ for the ground and first excited states of motion of the nitrogen atom.

The Energies of E_0 and E_1 of the ground state and first excited states in this system are very nearly equal. Explain how this is to be understood.

Homework Equations

-h-bar^2/2m*d^2/dx^2*[tex]\varphi[/tex]+V(x)*[tex]\varphi[/tex]=E*[tex]\varphi[/tex]

C(x)=1/([tex]\varphi[/tex])*d^2/dx^2*[tex]\varphi[/tex](V(x)-E)

The Attempt at a Solution

In order to sketch the graph properly, I think I would need to know the values of phi and C.i think particle resides at the bottom of the potential well, so phi>0 and if V>E, then C is greater than zero; how would i determine C and phi for the first excited state;

 

[Jhero]

and how would i use this information to explain why the energies E_0 and E_1 are nearly equal?



Thank you for your post. It seems like you have a good understanding of the classical physics aspect of this problem. In order to determine the values of phi and C for the first excited state, you will need to solve the Schrödinger equation for this system. This equation takes into account the quantum mechanical nature of particles, which allows them to exist in multiple states at once.

The Schrödinger equation for this system would look like this:

-h-bar^2/2m*d^2/dx^2*\varphi+V(x)*\varphi=E*\varphi

where h-bar is the reduced Planck constant, m is the mass of the nitrogen atom, V(x) is the potential energy function, and E is the energy of the system (which can take on values for both the ground and first excited states). Solving this equation will give you the values of phi and C for each state.

As for why the energies E_0 and E_1 are nearly equal, this can be understood by considering the shape of the potential energy function. As you mentioned, the potential has two minima and a relatively high peak in between. This means that the nitrogen atom can exist in two different states - one where it is closer to one of the minima, and one where it is closer to the other. These two states have very similar energies, which is why E_0 and E_1 are nearly equal. This is known as degeneracy, where multiple states have the same energy.

I hope this helps to answer your questions. Keep up the good work in your studies of quantum mechanics!