Finite potential well
Unlike the infinite potential well, there is a probability associated with the particle being found outside the box.
The quantum mechanical interpretation is unlike the classical interpretation, where if the total energy of the particle is less than the potential energy barrier of the walls it cannot be found outside the box.
In the quantum interpretation, there is a non-zero probability of the particle being outside the box even when the energy of the particle is less than the potential energy barrier of the walls (cf quantum tunnelling).
For the one-dimensional case on the x-axis, the time-independent Schrödinger equation can be written as: where[1] For the case of the particle in a one-dimensional box of length L, the potential is
For the region inside the box, V(x) = 0 and Equation 1 reduces to[3]
resembling the time-independent free schrödinger equation, hence
Thus, the solution is never square integrable; that is, it is always a non-normalizable state.
This does not mean, however, that it is impossible for a quantum particle to have energy greater than
, i.e., the non-normalizable states still contribute to the continuous part of the spectrum as generalized eigenfunctions of an unbounded operator.
Now in order to find the specific solution for the problem at hand, we must specify the appropriate boundary conditions and find the values for A, B, F, G, H and I that satisfy those conditions.
In this case, the finite potential well is symmetrical, so symmetry can be exploited to reduce the necessary calculations.
In order for the wave function to be square integrable, we must set
In other words, the values of the functions and their derivatives must match up at the dividing points: These equations have two sorts of solutions, symmetric, for which
What we have found is that the continuity conditions cannot be satisfied for an arbitrary value of the energy; because that is a result of the infinite potential well case.
Nevertheless, we will see that in the symmetric case, there always exists at least one bound state, even if the well is very shallow.
[7] Graphical or numerical solutions to the energy equations are aided by rewriting them a little and it should be mentioned that a nice approximation method has been found by Lima which works for any pair of parameters
, solutions exist where the blue semicircle intersects the purple or grey curves (
Each purple or grey curve represents a possible solution,
On the right we show the energy levels and wave functions in this case (where
But this is just the energy of the bound state of a Delta function potential of strength
In the previous equations only the positive derivative parts of the functions have to be considered.
Existence of root to above equation is not always guaranteed, for example, one can always find a value of
The results of symmetrical well is obtained from above equation by setting
satisfying the condition This equation does not always have a solution indicating that in some cases, there are no bound states.
The minimum depth of the potential well for which the bound state first appears at
case) is given by[12] The results above can be used to show that, as to the one-dimensional case, there is two bound states in a spherical cavity, as spherical coordinates make equivalent the radius at any direction.
The ground state (n = 1) of a spherically symmetric potential will always have zero orbital angular momentum (ℓ = n−1), and the reduced wave function
Notice that for (n = 1) angular part is constant (ℓ = 0).
It fulfils the condition where the wave does not find any potential inside the sphere:
Different differential equation lay on when ℓ ≠0, so as above titles, here it is: The solution can be rationalized by some changes of variable and function to rise a Bessel like differential equation, which solution is: where
