Problem 2
Part a
Consider a cylinder of height h and radius (Subscript[s, 0]). Let the base of the cylinder be at z = 0 and the top end-cap at z = h. We are given that the height of the cylinder is equal to the circumference => h = 2π (Subscript[s, 0])
The Hamiltonian (OverHat[H]) of a particle is given by:

(OverHat[H])= (((Overscript[P, ^ ])^2)/(2m))+V(r)
((OverHat[P])^2):= -(ℏ^2)(((∂^2)/(∂(z^2)))+(1/((Subscript[s, 0])^2))((∂^2)/(∂(ϕ^2))))
Since, V(r) = V((Subscript[s, 0])). Therefore, the potential on the cylinder is a constant. Let’s make our lives easier and set it equal to 0.
With that our Schrodinger’s equation becomes: 

     (OverHat[H]) Ψ( ϕ , z , t ) = i ℏ (∂/(∂t))Ψ( ϕ,z, t)
=> ((-(ℏ^2))/(2m))(((∂^2)/(∂(z^2)))+(1/((Subscript[s, 0])^2))((∂^2)/(∂(ϕ^2))))Ψ( ϕ , z , t ) =  i ℏ (∂/(∂t))Ψ( ϕ,z, t)
We have to solve for the wavefunction Ψ (ϕ, z, t) that satisfies the following boundary conditions:

1. Ψ (ϕ, z = 0, t) = Ψ (ϕ, z ≥ h, t) = 0 as the particle is confined to live on the cylinder and cannot “get out” to an infinite potential region. 
2. Ψ (ϕ, z, t) = Ψ (ϕ + 2π, z, t) (Periodic boundary condition as the points ϕ and ϕ + 2π represent the same location on the cylinder.

Part b

Now, we have to use separation of variables to solve the problem. Let’s do: Ψ (ϕ, z, t) = ρ (ϕ, z) T(t) (i.e. separate the spatial and the temporal parts) 
Plug it into the Schrodinger’s equation:

((-(ℏ^2))/(2m))(((∂^2)/(∂(z^2)))+(1/((Subscript[s, 0])^2))((∂^2)/(∂(ϕ^2))))ρ (ϕ,z) T(t)=i ℏ(∂/(∂t))ρ (ϕ,z) T(t)

Now the (∂/(∂t)) operator only acts on the T(t) part of the wavefunction i.e. ρ (ϕ, ,z) is a “constant” w.r.t the (∂/(∂t)) operator. Similarly, the spatial operators (∂/(∂z)) and (∂/(∂ϕ)) only act on the ρ(ϕ,z) part of the wavefunction i.e. T(t) is  a constant function w.r.t the spatial operators 

((-(ℏ^2))/(2m))T(t)(((∂^2)/(∂(z^2)))ρ (ϕ,z) +(1/((Subscript[s, 0])^2))((∂^2)/(∂(ϕ^2)))ρ (ϕ,z) )=i ℏρ (ϕ,z) (d/dt)T(t)

Finally, let’s divide both sides by the wavefunction Ψ (ϕ, z, t) = ρ (ϕ, z) T(t). 

((-(ℏ^2))/(2m))(1/(ρ (ϕ, z)))(((∂^2)/(∂(z^2)))ρ (ϕ,z) +(1/((Subscript[s, 0])^2))((∂^2)/(∂(ϕ^2)))ρ (ϕ,z) )=((i ℏ)/(T(t))) (d/dt)T(t)

Now, the left hand side is purely a function of spatial “stuff” (i.e. ϕ and z) while the right hand side contains all the time dependence. Therefore, the only way those two “things” can equal one another is if they both equal a constant. Let’s call that (separation) constant E. 

With that we get:
 
   ((i ℏ)/(T(t))) (d/dt)T(t) =E

=>  ((dT(t))/(T(t))) =((-i)/ℏ)E dt 

Solving this yields: 

   ∫((dT(t))/(T(t))) =∫((-i)/ℏ)E dt
   
   => T(t) = (e^(((-i)/ℏ)t))
   
   .......Not a huge surprise as our Hamiltonian is time independent. Therefore, we get the usual time dependence of the exponential form. 
   
   Let’s look at the other equation: 
   
   ((-(ℏ^2))/(2m))(1/(ρ (ϕ, z)))(((∂^2)/(∂(z^2)))ρ (ϕ,z) +(1/((Subscript[s, 0])^2))((∂^2)/(∂(ϕ^2)))ρ (ϕ,z) ) = E
   
  Why not do separation of variables again to solve for ρ (ϕ,z). Let’s say, ρ (ϕ, z) = Φ (ϕ) Z (z). Substitute that into the above equation (and turn the ∂ into total derivatives (d)). 
  We get, 
 
  ((-(ℏ^2))/(2m))(1/(Φ (ϕ) Z (z)))(((d^2)/(dz^2))Φ (ϕ) Z (z) +(1/((Subscript[s, 0])^2))((d^2)/((dϕ)^2))Φ (ϕ) Z (z) ) = E
  
  Simplifying it yields: 
  
 (1/(Z (z)))((d^2)/(dz^2))Z (z) +(1/(((Subscript[s, 0])^2)Φ (ϕ)))((d^2)/((dϕ)^2))Φ (ϕ) = ((-2m)/(ℏ^2))E 
  
  Now, the “z package”  ((1/(Z(z)))((d^2)/(dz^2))Z(z)) can never equal the “ϕ package” ((1/(Φ(ϕ)))((d^2)/((dϕ)^2))Φ(ϕ))  unless both of these packages equal a constant. 
  Let’s introduce another separation constants : -(Subscript[k, z]^2) and ((-(Subscript[k, ϕ]^2))/4) (the reason for pulling out a factor of (1/4)  from the constant (Subscript[k, ϕ ])will become clearly shortly). We that we get 2 ordinary differential equations. 
  
  (1/(Z (z)))((d^2)/(dz^2))Z (z)  = -(Subscript[k, z]^2)
  (1/(Φ(ϕ)))((d^2)/((dϕ)^2))Φ(ϕ) = ((-(Subscript[k, ϕ]^2))/4)
  
  The (separation) constants will be determined from the boundary conditions. With that our Schrodinger’s equation becomes: 
  
  -(Subscript[k, z]^2) -((Subscript[k, ϕ]^2)/(4(Subscript[s, 0]^2))) = ((-2m)/(ℏ^2))E
  Now we can solve for Energy E. 
  
  E = ((ℏ^2)/(2m)) ((Subscript[k, z]^2)+((Subscript[k, ϕ]^2)/(4(Subscript[s, 0]^2))))
  
  Let’s now solve those 2 differential equations: 
  
  (1/(Z (z)))((d^2)/(dz^2))Z (z)  = -(Subscript[k, z]^2)
  
  => Z(z) = A Sin ((Subscript[k, z])z) + B Cos ((Subscript[k, z])z) 
  Employing the boundary conditions we get: 
  
  1. B = 0 
  2. (Subscript[k, z]) = (((Subscript[n, z])π)/h) = (((Subscript[n, z])π)/(2π (Subscript[s, 0]))) = ((Subscript[n, z])/(2(Subscript[s, 0]))) where (Subscript[n, z]) = {1,2,3,4........}
  Normalizing the wavefunction yields: 
  
  A = (Sqrt[(2/h)]) =(Sqrt[(2/(2π (Subscript[s, 0])))]) =(1/(Sqrt[π (Subscript[s, 0])]))
  
Let’s solve the “ϕ” package:  (Realizing it’s the same system as a “Particle on a Ring”, we can steal the eigenfunctions and energies from that system).
  (1/(Φ(ϕ)))((d^2)/((dϕ)^2))Φ(ϕ) = -(Subscript[k, ϕ]^2)
  
  => Φ (ϕ) = C (e^(i (Subscript[k, ϕ])ϕ))
  
  Applying the (periodic) condition yields: 
  
  (Subscript[k, ϕ ]) = (Subscript[n, ϕ]) where (Subscript[n, ϕ])= {0, ± 1, ±2, ±3.....}
  And normalizing it gives us: C = (1/(Sqrt[2π (Subscript[s, 0])]))
  
  Therefore, our energy E is: 
  
  E = ((ℏ^2)/(2m)) ((Subscript[k, z]^2)+((Subscript[k, ϕ]^2)/(4(Subscript[s, 0]^2)))) = ((ℏ^2)/(2m))(((Subscript[n, z]^2)/(4(Subscript[s, 0]^2)))+((Subscript[n, ϕ]^2)/(4(Subscript[s, 0]^2)))) = ((ℏ^2)/(8m (Subscript[s, 0]^2)))((Subscript[n, z]^2)+(Subscript[n, ϕ]^2))  
  
  (Now you see why we pulled out a factor of (1/4) from the constant (Subscript[k, ϕ]). If we didn’t do it, our expression for energy wouldn’t have been this nice. There’s no way I knew that pulling out a (1/4) would make it nicer. I solved it without pulling it out and the expression for energy was not as nice. This motivated me to pull all the necessary constants to make E look nice. It is purely a matter of convenience. You could have not pulled out (1/4) and your expression for Energy would have been same). 
  
 So putting it all together. 
 
 Our eigen-energies are: (Subscript[E, (Subscript[n, z]),(Subscript[n, ϕ])]) = ((ℏ^2)/(8m (Subscript[s, 0]^2)))((Subscript[n, z]^2)+(Subscript[n, ϕ]^2))  where (Subscript[n, z]) = {1,2,3,4,5....} and (Subscript[n, ϕ]) = {0, ±1, ±2, ±3......}
 Our eigenfunctions are: Ψ (ϕ, z, t) = Φ (ϕ) Z(z) T(t) = ((1/(Sqrt[2π (Subscript[s, 0])]))(e^(i (Subscript[n, ϕ])ϕ)))((1/(Sqrt[π (Subscript[s, 0])]))Sin(((Subscript[n, z])/(2(Subscript[s, 0])))z))((e^(((-i)/ℏ)(Subscript[E, (Subscript[n, z])]),(Subscript[n, ϕ])t))) = (1/((Sqrt[2])π (Subscript[s, 0])))Sin(((Subscript[n, z])/(2(Subscript[s, 0])))z)(e^(i (Subscript[n, ϕ ])ϕ))(e^(((-i)/ℏ)(Subscript[E, (Subscript[n, z]),(Subscript[n, ϕ])])t))
 We can represent them as kets as well: 
 Therefore, wavefunctions can be written as: (Ket[(Subscript[n, z]), (Subscript[n, ϕ])]) := (∑_((Subscript[n, z])=1)^∞)(∑_((Subscript[n, ϕ])=-∞)^∞)(Subscript[C, (Subscript[n, z]),(Subscript[n, ϕ])])(1/((Sqrt[2])π (Subscript[s, 0])))Sin(((Subscript[n, z])/(2(Subscript[s, 0])))z)(e^(i (Subscript[n, ϕ])ϕ))

  Part C
  For the ground state, our quantum numbers are (Subscript[n, z])=1 and (Subscript[n, ϕ])=0 
  =>Ground state =  (Ket[1,0])
  The Energy will be (Subscript[E, 1,0])= ((ℏ^2)/(8m (Subscript[s, 0]^2)))((1^2)+(0^2)) =((ℏ^2)/(8m (Subscript[s, 0]^2))) 
  The full wave function is: Ψ(Subscript[(ϕ,z,t), ground]) = (1/((Sqrt[2])π (Subscript[s, 0])))Sin((1/(2(Subscript[s, 0])))z)(e^(-i(ℏ/(8m (Subscript[s, 0]^2)))t))
   
   
 The 1st excited state has an energy greater than the ground state or (Subscript[E, 1st excited state]) has the next lowest energy (after the ground state). 
 =>(1^st) excited state = (Ket[1,1]) 
 (Subscript[E, 1,1])=((ℏ^2)/(8m (Subscript[s, 0]^2)))((1^2)+(1^2)) =((2(ℏ^2))/(8m (Subscript[s, 0]^2)))
 
But hey, if (Subscript[n, ϕ])=-1, then we still get the same energy.  
(Subscript[E, 1,-1])= ((ℏ^2)/(8m (Subscript[s, 0]^2)))((1^2)+(((-1))^2)) =((2(ℏ^2))/(8m (Subscript[s, 0]^2)))
=> The first excited state is degenerate! as we have two states ((Ket[1,1]) and (Ket[1,-1])) with the same energy. 
The wavefunctions associated with the 1st excited state are: (Ket[1,±1]):=(1/((Sqrt[2])π (Subscript[s, 0])))Sin((1/(2(Subscript[s, 0])))z)(e^(±i ϕ))(e^(-i((2ℏ)/(8m (Subscript[s, 0]^2)))t)) 

The 2nd excited state has a higher energy than the 1st excited state. 
(Ket[2,0]) has energy: (Subscript[E, 2,0]) = ((ℏ^2)/(8m (Subscript[s, 0]^2)))((2^2)+(0^2))=((4(ℏ^2))/(8 m (Subscript[s, 0]^2)))
There’s no degeneracy in the 2nd excited state.
Wavefunction: |2,0⟩: = (1/((Sqrt[2])π (Subscript[s, 0])))Sin((2/(2(Subscript[s, 0])))z)(e^(-i((4ℏ)/(8 m (Subscript[s, 0]^2)))t))

3rd excited state has the next highest energy. 

The energy of (Ket[1,+±2]) is:
 (Subscript[E, 1,± 2])= ((ℏ^2)/(8m (Subscript[s, 0]^2)))((1^2)+(((±2))^2))=((5(ℏ^2))/(8 m (Subscript[s, 0]^2)))  
But the state (Ket[+2,+±1]) has the same energy as (Ket[1,+±2]) as:
(Subscript[E, 2, ±1])= ((ℏ^2)/(8m (Subscript[s, 0]^2)))((2^2)+(((±1))^2)) = ((5(ℏ^2))/(8 m (Subscript[s, 0]^2))) 
Wavefunction associated with the ket (Ket[+1,±2]): = (1/((Sqrt[2])π (Subscript[s, 0])))Sin((1/(2(Subscript[s, 0])))z)((e )^(±2iϕ))(e^(-i ((5ℏ)/(8π (Subscript[s, 0]^2)))t))
Wavefunction associated with the ket (Ket[+2,±1]): = (1/((Sqrt[2])π (Subscript[s, 0])))Sin((2/(2(Subscript[s, 0])))z)((e )^(±iϕ))(e^(-i ((5ℏ)/(8π (Subscript[s, 0]^2)))t))
Part D
 Let’s do the 2nd excited state: 
 |2,0⟩: = (1/((Sqrt[2])π (Subscript[s, 0])))Sin((2/(2(Subscript[s, 0])))z)(e^(-i((4ℏ)/(8 m (Subscript[s, 0]^2)))t))
 Probability density is given by: P = |(BraKet[2,0, 2,0])((|)^2)
 * Doing that get’s rid of the time dependence, (which is to be expected as an eigenstate is a stationary state that does not evolve/ change with time). Therefore, the probability density does not change with time. 
 * Further, the state will definitely evolve/ change with time as given by the Schrodinger’s equation. The time evolution is simply: (e^(((-i)/ℏ)E t))
 

 P= |(BraKet[2,0, 2,0])((|)^2) = (1/(2(π^2) (Subscript[s, 0]^2)))(Sin^2)((z/(Subscript[s, 0])))    (The dimensions make sense, as to obtain probability we integrate the probability density over a finite region on the cylinder)
 Doing that gives us (∫_0^(2π))(∫_0^h)P (z,ϕ) (Subscript[s, 0])dzdϕ = (∫_0^(2π))(∫_0^(2π (Subscript[s, 0])))(1/(2(π^2) (Subscript[s, 0]^2)))(Sin^2)((z/(Subscript[s, 0]))) (Subscript[s, 0])dzdϕ = 1, thus ensuring that our wavefunction is properly normalized. 
 At t = 0, (Ket[2,0])= (1/((Sqrt[2])π (Subscript[s, 0])))Sin((z/(Subscript[s, 0])))
 There are no imaginary parts. It’s simple to plot it in Mathematica.

Plot3D[(1/((Sqrt[2])π))Sin[(z/1)],{z,0,2π},{ϕ,0,2π},PlotLabel->"Wavefunction (Ket[2,0]) at t=0",AxesLabel->{z,ϕ,ψ}]

Plot3D[(1/(2(π^2)))(((Sin[(z/1)]))^2),{z,0,2π},{ϕ,0,2π},PlotLabel->"Probability density for (Ket[2,0])",AxesLabel->{"z","ϕ",(("|ψ|")^2)},ImageSize->Medium]