Quantum Mechanics
Schrodinger equation
Quantum Mechanics
Propagators : Pg
Quantum Simple Harmonic Oscillator QSHO
Quantum Mechanics
Simulation With GNU Octave
© The scientific sentence. 2010


Hermite polynomials
1. Hermite polynomials
In Physics, Hermite polynomials are used give determine the
eigenstates of the quantum harmonic oscillator. They are the
set of sequences:
The Hermite differential equation is:
y"(x)  2xy'(x) + 2ny(x) = 0
The solution of this nonlinear secondorder ordinary differential
equation is the set of Hermite polynomials.
H_{n} = ( 1)^{n} exp{x^{2}} d^{n}(exp{ x^{2}})/dx^{n}.
H_{n} is the polynomial of degree n.
Here some first sequences:
H_{0}(x) = 1
H_{1}(x) = 2x
H_{2}(x) = 4x^{2}  2
H_{3}(x) = 8x^{3}  12x
H_{4}(x) = 16x^{4}  48x^{2} + 12
H_{5}(x) = 32x^{5}160x^{3} + 120x
The Hermite polynomial are orthogonal with the
weight w(x) = exp{ x^{2}}. Thus:
∫ H_{n}(x) H_{m}(x) exp{ x^{2}} dx = δ(n,m) [x: from  ∞ to ∞]
Recursion relations
H_{n+1}(x) = 2xH_{n}(x) H'_{n}(x)
H'_{n}(x) = 2nH_{n  1}(x)
H_{n+1}(x) = 2xH_{n}(x)  2nH_{n 1}(x)
2. Quantum Simple Harmonic Oscillator (QSHO)
2.1. Schrödinger’s Equation for QSHO
The Schrödinger’s equation for QSHO is:
 ħ ^{2}/2m Ψ"(x) + (1/2) k x^{2} Ψ(x)  E Ψ(x) = 0
or:
Ψ"(x) + [ (2m/ħ^{2})E  (mk/ħ^{2}) x^{2}]Ψ(x) = 0
Where 2πħ is the Planck constant,
m is the mass of the particle, k is the spring constant and
E is the energy of the particle.
Let's write:
x = z/α then:
Ψ'(x) = α Ψ'(z)
Ψ"(x) = α^{2} Ψ"(z)
Therefore
α^{2} Ψ"(z) + [ (2m/ħ^{2})E  (mk/ħ^{2}) (z/α)^{2}]Ψ(x) = 0
Dividing by α^{2} , yields:
Ψ"(z) + [ (2m/ħ^{2} α^{2} )E  (mk/ħ^{2}) (z^{2}/α)^{4}]Ψ(x) = 0
Let's define: km/ħ^{2}α^{4} = 1, so
α^{4} = mk/ħ^{2}
We have then:
Ψ"(z) + [ (2m/ħ^{2} α^{2} )E  z^{2}]Ψ(x) = 0
Now, let's define:
Ψ_{n} (z) = exp{ z^{2}/2} H_{n}(z)
Therefore:
Ψ'_{n} (z) =  z Ψ_{n} (z) + exp{ z^{2}/2} H'_{n}(z)
Ψ"_{n} (z) =  Ψ_{n}  zΨ'_{n} (z) + exp{ z^{2}/2} H"_{n}(z)
 z exp{ z^{2}/2} H'_{n}(z) =
Ψ"(z)_{n} (z) =  Ψ_{n}  z [ z Ψ_{n} (z) +
exp{ z^{2}/2} H'_{n}(z)] + exp{ z^{2}/2} H"_{n}(z)
 z exp{ z^{2}/2} H'_{n}(z)
=  Ψ_{n} + z^{2}Ψ_{n} (z)  z exp{ z^{2}/2} H'_{n}(z) +
exp{ z^{2}/2} H"_{n}(z)  z exp{ z^{2}/2} H'_{n}(z)
= (z^{2}  1)Ψ_{n} (z)  2xexp{ z^{2}/2} H'_{n}(z) + exp{ z^{2}/2} H"_{n}(z)
The Schrodinger's equation becomes:
z^{2}Ψ  Ψ_{n} (z)  2 z exp{ z^{2}/2} H'_{n}(z) +
exp{ z^{2}/2} H"_{n}(z) + [ (2mE/ħ^{2}α^{2})  z^{2}]Ψ(z) = 0
Or:
 2 z H'_{n}(z) + H"_{n}(z) + [ (2mE/ħ^{2}α^{2})  1]H(z) = 0
Rearranging:
H"_{n}(z)  2 z H'_{n}(z) + [(2mE/ħ^{2}α^{2})  1]H(z) = 0
then the Schrodinger's equation becomes the Hermite
equation with:
2mE/ħ^{2}α^{2})  1 = 2n, so
E = (ħ^{2}α^{2}/2m) (2n + 1)
We have α^{2}/2m = (k/m)^{1/2}/2ħ, then
E = (ħ^{2}(k/m)^{1/2}/2ħ) (2n + 1)
= (ħ(k/m)^{1/2}) (n + 1/2)
Let's write:
ω^{2} = k/m
Therefore:
E = ħω (n + 1/2)
E = ħω (n + 1/2)
The wave function becomes:
Ψ_{n} (z) = Ψ_{n} (z) = exp{ z^{2}/2} H_{n}(z)
is the solution of the schrodinger equation, with
z = α x
α^{4} = mk/ħ^{2}.
E = (ħ^{2}α^{2}/2m) (2n + 1)
2.2. Normalization of the function Ψ_{n} (z)
The hermite polynomials are orthogoal as follows:
∫ exp{ z^{2}} H_{n}(z) H_{m} dz = (2^{n} π^{1/2} n!)δ(n,m) [from _∞ to ∞]
Then:
∫ Ψ_{n} (z)dz = 1, so
∫ N^{2} exp{ z^{2}} H^{2}_{n}(z) = 1 [from _∞ to ∞]
Then
N = [n!2^{n} π^{1/2}]^{ 1/2}
Therefore:
Ψ_{n} (z) = [n! 2^{n} π^{1/2}]^{ 1/2} exp{ z^{2}/2} H_{n}(z)
or:
Ψ_{n} (x) = [n! 2^{n} π^{1/2}]^{ 1/2} exp{ (αx)^{2}/2} H_{n}(x)
The results are:
Ψ_{n} (x) = [n! 2^{n} π^{1/2}]^{ 1/2} exp{ (αx)^{2}/2} H_{n}(x)
E = (ħ^{2}α^{2}/2m) (2n + 1) = ħω (n + 1/2)
α^{4} = mk/ħ^{2} ω^{2} = k/m
The energy is not null even at the ground
state where n = 0.
The transition energy between states is ħω.

