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** Introduction to Bound States**:

You would remind that a body in the simple harmonic motion bounces backward and forward between the two points where the total or net mechanical energy, 'E', of the body is equivalent to the potential energy. As the kinetic energy becomes zero, the body should turn back. In just similar manner, a quantum-mechanical oscillator is a particle within an infinite potential well. You must observe that certainly, the solutions will be sinusoidal, just the manner it is by a harmonic oscillator. Such a state is an illustration of a bound state. More particularly, we shall state a system is in the bound state if, E < V (-∞). As such the wave function comprised should die at infinity, that is, Ψ(x) → 0 as x → ± ∞. E > V (- ∞) or/and E > V (+ ∞) is termed as a scattering state. The other manner of seeing a bound state is to state that the particle is subjected to the attracting potential.

For a bound state, the given conditions apply:

a) Ψ is continuous across the boundary

b) The first derivative, Ψ', is continuous across the boundary

** Particle in an infinite potential well**:

This is as well termed as a particle in a box. The figure shown below describes a particle in an infinite potential well.

The above figure is the infinite square well potential confining a particle to a region having width 'L'.

Within the well, the potential is zero whereas outside the well, the potential is infinite. We anticipate that the wave function outside the well will be equivalent to zero.

We remember the Schrodinger equation: (∂^{2}Ψ/∂x^{2}) + (8π^{2}m/h^{2}) (E - V) Ψ = 0

If, V = 0,

(d^{2}Ψ/dx^{2}) + (8π^{2}m/h^{2}) EΨ = 0

It can be written as:

(d^{2}Ψ/dx^{2}) + k^{2}Ψ = 0

The common solution of this equation is:

Ψ(x) = c_{1}e^{ikx} + c_{2}e^{-ikx}

Here c_{1} and c_{2} are the constants to be determined, subject to the boundary conditions and i = √-1.

As,

a) Ψ(x) = 0 for x = 0

Ψ(0) = c_{1}e^{ik0} + c_{2}e^{-ik0} = 0

=> c_{1} + c_{2} = 0 => c_{1} = -c_{2} = c

=> Ψ(x) = c (e^{ikx} - e^{-ikx}) = A sin kx

Here A = 2ic

b) Ψ(x) = 0 for x = L

A sin kL = 0 => kL = nπ, where n = 0, 1, 2,........

It obeys that the values of k are quantized, in such a way that k_{n} = nπ/L

Therefore, the corresponding energy E_{n} = (n^{2}h^{2})/(8mL^{2}), and the wave-function corresponding to this energy is Ψ_{n}(x) = A_{n }sin (nπx/L).

A_{n}, termed as the normalization constant is obtained through applying the normalization condition:

_{-∞}∫^{∞} |Ψ(x)|^{2} dx = 1

That is, A_{n}^{2} _{o}∫^{L} sin^{2} (nπx/L) dx = 1

Or A_{n}^{2} x 1/2 = 1,

From which it follows that,

A_{n} = √ (2/L)

We can thus write:

Ψn(x) = √ (2/L) sin (nπx/L) for 0 ≤ x ≤ L; n = 1, 2, 3....

** The Finite Potential Well**:

In case of a finite potential well, the wave-functions 'spill over' to the area outside the potential well.

Region 1:

x ≤ - L/2, V(x) = V_{o}, Putting into time independent Schrodinger equation:

The above can be written in the form:

Ψ_{I }= Ce^{αx} + De^{-αx}

Though, we can't let an exponentially growing term; so we set D equivalent to zero.

Ψ_{I} = Ce^{αx}

Region 2:

- (L/2) ≤ x ≤ (L/2), V(x) = 0, putting it into the time independent Schrodinger equation

The above can be written in the form:

Ψ_{II} = A sin kx + B cos kx

This comprises of an odd and an even solution.

We at first consider the even solution: Ψ_{II} = B cos kx

Region 3:

x ≥ L/2, V(x) = V_{o}

The above can be written in the form:

Ψ_{III} = Fe^{αx} + Ge^{-αx}

Though, we can't let an exponentially growing term; therefore we set F equivalent to zero.

Ψ_{III} = Ce^{-αx}

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## Testing of Means

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