(a) The total potential energy for a particle in a one-dimensional rectangular well. (b) The ground state wave function for this all potential (c) The first excited - state wave function.
First, we can consider E< V₀. The schrodinger equation.
is regions (l) and (lll) is d²ψ/dx² +(2m/ħ²) (E −V⁰) ψ= 0.This is a linear homogeneous differential Equation with constant of coefficient, and auxiliary equation
is regions (l) and (lll) is d²ψ/dx² +(2m/ħ²) (E −V⁰) ψ= 0.This is a linear homogeneous differential Equation with constant of coefficient, and auxiliary equation
Is s² +(2m/ħ²) (E−V₀)= 0
with roots s=± (2m/ħ²)¹/². Therefore,
ψ(l) = C exp[(2m/ħ²)¹/² (V₀−E)¹/²x] + Dexp [−2m/ħ²)¹/² (V₀−E)¹/²x]
ψ(lll) = F exp [(2m/ħ²) ¹/² (V₀−E)¹/²x] +Gexp [ −(2m/ħ²)¹/² (V₀−E) ¹/²x]
Where C,D, F, G are constants.
Then we must prevent ψ₁ from becoming infinite as x →−∞, since we are assumed that E<V₀, the quantity (V₀−E)¹/₂ is a real and positive number, and to keep ψ₁ finite as x → ∞, we must have D=0.similarly,to keep ψ (lll) finite as x →+ ∞, we must have F=0.Therefore,
ψ₁ = Cexp [(2m/ħ²)¹/² (V₀−E) ¹/₂, ψlll = Gexp [−(2m/ħ²) ¹/₂ (V₀−E)¹/₂x]
In region ll, V=0, the Schrodinger equation is d²ψll/dx² + 2m/ħ² E ψll =0 in solution
ψll = Acos [ħ⁻¹ (2mE)¹/₂x ] +B sin [ħ⁻¹ (2mE) ¹/₂x] so,
Ψll = Acos [(2m/ħ²)¹/₂ E¹/₂x] + B sin [2m/ħ²)¹/₂ E¹/₂x ].....(1)
To complete this problem, we can must apply for the boundary conditions. As with the particles in a box with infinite walls, we can required that wave function to be continuous at x= 0,and at x= l: so ψ₁ (0) = ψ ll (0) and ψ ll (l) = ψ lll (l). The wave function have four arbitrary constant, so more than this two boundary conditions are needed. As well a require to ψ to be continously, we can require that and it's derivatives dψ /dx be continuous everywhere. To justify these requirements, we note that if the dψ/ dx changed dis continously at the point, then it's derivatives is d²ψ/ dx² would become the infinite at that same point. However, for the particle in a rectangular well, the Schrodinger equation d²ψ/ dx² = (2m/ħ²) (V−E) ψ doesnot contain anything infinite on the right side, so d²ψ/ dx² is can't become infinite. Therefore dψ₁/dx = dψll/dx at x= 0 and dψ ll /dx = dψ lll/ dx at x=l.
From ψl (0) = ψ ll(0) we get C = A. From ψl' (0) = ψll' (0), and we get,
B= (V₀ −E) ¹/₂ A /E¹/₂ from ψ (1) = ψ lll (l) and we get the a complicated equation that the allows G to be found in terms of A. The constant A is found by normalization.
Taking ψ'll (l) = ψ' lll (l) dividing it's by ψll (l) and expressing B in terms of A, we get the following equation for the energy levels,
( 2E− V₀) sin [(2mE) ¹/₂ l/ħ] = 2(V₀E−E²) cos [(2mE) ¹/₂ l/ħ].....equation (2). Although E= 0,satisfied equation (2) it's not allowed energy values, but since it gives ψ= 0, Defining the dimensionless at constants ε and b as
ε≡ E/V₀, and b≡ (2mV₀)¹/₂ l/ħ.........equation (3)
We can divide equation (3) by V₀ to get,
(2ε-1) sin (bε¹/₂) − 2 (ε−ε ²)¹/₂ cos (bε¹/₂) = 0........eqution (4)
Only the particular values of E that satisfied equation (2) give a wave function that's continuous derivatives, so that the energy levels are quantized for E<V₀. To find the allowed energy levels we can putting the plot in the left side equation (5) versus ε for 0< ε <1
And find that points where the curve crossed the horizontal axis. The detailed study shows that number of allowed energy levels with E< V₀ is N, where is N satisfied
N−1<b/π ≤N, where b≡ (2mV₀)¹/₂ l/ħ.......equation (6)
For example,if V₀= h²/ml², then b/π =2(2¹/₂) = 2.83, and N= 3.
See above the figure (top) shows that ψ for the lower two energy levels. The wave function is oscillatory inside the bo and dies off exponentially outside the box. It will be turned out of the number of nodes, increasing by one for each higher level.
So far we can consider only states with E< V₀. For E>V₀, the quantity (V₀−E)¹/₂ is a imaginary, and instead of dying off the zero as x goes to ± ∞, ψl and ψ lll oscillating similar to the free particles ψ. We no longer have any reasons to set D in ψl and F in ψ lll equal to the zeros, and with these additional constant available to satisfied the boundary conditions on ψ and ψ', one finds that E needs not be restricted to obtain the properly behaviour of the wave function. Therefore, all the energies above the V₀ are allowed.
A states in which ψ →0 as x →−∞ is called a bound states. For a bound states, significant probability for the finding that particles exists in only a finite region of space. For an unbound states, ψ doesn't go to zero as x → ± ∞ and it's not normalizable. For the particles in rectangular well, states that with E < V₀ are bound and states with E> V₀ are unbound. For the particles in a box and infinitely higher walls, all the states are bound. For the free particles, all the states are unbound.
Good explanation
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