PARTICLE IN A RING

Consider a particle of mass m and rotating in a circle of radius r in the xy plane.

And we will consider If the potential energy (v) of the particle is zero.

                                      The Hamiltonian in the form can be written as, 

H= v- h²v²         ,

         8π²m

Ĥ =   −h²       (∂²/∂x² + ∂²/∂y²) v =0..........(1)

       8π²m      

The Cartesian coordinates x and y are the related to planer coordinates r and φ as, 

x = r cosφ, y = sinφ. 

Ĥ = −   h²         ( ∂²/∂φ²)

       8π²mr²

= −   h²       (∂²/∂φ²)............................(2)

    8π² I  

[ I = mr² ] is the moment of inertia relating to the equation. 

The eigen function F(φ) or simply F will be a function of φ only, 

So, the schrodinger equation is, 

  −h²       ∂²F     = EF

8π² I       ∂φ²

(or)       d²F   + M² F = 0  .....................(3)

             dφ²

where ,    M² =   8π² I E  .............       (3.1)

                              h²

The equation (3) is particle in a 1-dimensional box. It's solution the wavefunction are, 

F = N sin Mφ

F' = N' Cos Mφ  as thecreal set....... (4)

F" = A exp (±iMφ) as the imaginary total sets of numbers as a equation ........ (5)

The trigonometric forms (4) are also called " circular harmonics" 

Equation (4) and (5) are equivalent in view of the theorem, 

exp ( ± iMφ) = Cos Mφ is in Mφ............(6)

The function (4) and (5) are finite and continues for all values of φ and M

They are single − valued too, since the angels φ and φ + 2π represented the same point and the property of single − valued demands that, 

F(φ) = F ( φ +2π) 

Using the real set equation (4), we get 

Sin Mφ = Sin M (φ+2π) (or) 

Cos Mφ = Cos M (φ+ 2π) 

This is possible only if M= 0,±1,±2,.....

For example, 

M = + 1:          sinMφ = Sinφ

                        SinM (ψ+2π) = Sinφ.

Similarly, 

A exp (iMφ) =A exp ( iM (φ+2π)) 

                    = A exp (iMφ)⁺. exp (i2πM) 

exp (i2πM) =1, (or) 

Cos 2πM +isin2πM = 1, only if M = 0, ±1 , ± 2

As regards boundary conditions, there's no barrier to the particles motion as long as it's on the ring. 

So, there's no condition for the wave function to vanish at any point. 

[ For M= 0, N sinMφ = 0 but N' Cos Mφ =N' ]

Normalization and orthogonality :-

                                            The normalization factor N can be determined are follows :

²π₀∫ (N sin Mɸ )² dɸ = 1

[ sin²θ =  1− cos²θ  ]

                      2

(or)

N² ²π₀∫ sin² Mɸ dɸ = N²  ²π₀∫ ( 1−cos²Mɸ )

                                                        2

                  dɸ = 1

           N² =  1       (or)

                    π

N² π = 1 (or) N = 1/√π 

Similarly,

                   N¹ = 1/√π

The normalized real set of wave function are, 

F=  1  sin Mɸ and F =  1  . cos Mɸ .........(7)

     √π                           √π

For   M = 0, and  value F' =  1     , 

                                              √π

For the imaginary set, 

²π₀∫ A* exp ± ( i M ɸ), A exp (± i M ɸ). dɸ= 1

(or)

²π₀∫ ∣ A ∣² dɸ =1 (or) ∣ A ∣².2π = 1(or)

               ∣ A ∣ =  1   

                        √2π

Here F" =  1  exp ( ± i M ɸ ) as a eqn.......(8)

                  √2π

For M = 0 , and vale F" =  1  this equation 

                                         √2π

It's not difficult to show that the function equation (7) and (8), for example, are orthogonal set as. 

For example, 

       1   ²π₀∫ sinɸ . cos ɸ dɸ = 0

       π

It's seen that expect for M = 0, the functions equation (7) and (8) constitute doubly degenerate real and imaginary sets of wave functions respectively. 

Quantization of energy :-

            The expression equation (3.1) can be rearranged to give the expression for energy as, 

               E =  M² h²  

                      8π² I

Thus, the energy is always quantised in units of 

                    h²      

                  8π² I

The parameter M being and it's called " rotational quantum number" 

For  M = 0, E = 0,

i.e., the rotating particle doesn't have zero point energy.