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

 Quantum harmonic oscillator Ladder Operators
1.Hamiltonian of the Onedimensional SHO
Let the particle of mass m represents an harmonic oscillator.
Its Hamiltonian is:
H = p^{2}/2m + mω^{2}x^{2}/2
Where x is position operator and p is the momentum operator.
They are given by:
x = x
p =  i ℏ∂/∂x
To find the energy eigenstates and their corresponding energy
levels, we must solve the timeindependent Schrödinger equation
Hψ> = Eψ>.
2. Ladder operator method
The "ladder operator" method is used to find the energy
eigenvalues without directly solving the differential equation.
We define two operators a and its adjoint a^{+} as follows:
a = [mω/2ℏ]^{1/2} [x + (i/mω)p]
a^{+} = [mω/2ℏ]^{1/2} [x  (i/mω)p]
Let's define the product:
N = a^{+}a
N = [mω/2ℏ] [x^{2} + (i/mω)xp  (i/mω)px + (1/mω)^{2}p^{2}]
= [mω/2ℏ] [x^{2} + (i/mω)[x,p] + (1/mω)^{2}p^{2}]
[x,p]= iℏ, so:
N = [mω/2ℏ] [x^{2}  (ℏ/mω) + (1/mω)^{2}p^{2}] =
N = a^{+}a = (mω/2ℏ) x^{2} + (1/2ℏmω) p^{2}  1/2
Now, Let's write:
(N + 1/2) ℏ ω = (mω^{2}/2)x^{2} + (1/2m)p^{2}] = H
Therefore, the Hamiltonian of the quantum harmonic
oscillator can be written in terms of thw ladder
operators:
H = (N + 1/2)ℏω
We have also:
(aa^{+}  1/2)ℏω = H = (N + 1/2)ℏω, so:
(aa^{+})ℏω = (N + 1)ℏω, then: aa^{+} = N + 1
aa^{+} = N + 1
Commutators:
1.
[a, a^{+}] = aa^{+}  a^{+}a =  N + N + 1 = 1
2.
[N,a^{+}] = Na^{+}  a^{+}N = a^{+}aa^{+}  a^{+}a^{+}a =
a^{+}aa^{+}  a^{+}(aa^{+}  1) = a^{+}aa^{+}  a^{+}aa^{+} + a^{+} = a^{+}
3.
[N,a] = Na  aN = a^{+}aa  aa^{+}a = (a^{+}a  aa^{+})a =
[a^{+},a]a =  a
[a, a^{+}] = 1
[N,a^{+}] = a^{+}
[N,a] =  a
Let's write the eigenstate for the quantum harmonic
oscillator ψ_{n}(x)= n>. The number operator N satisfy the
equation:
Nn> = n n>
Na^{+}n> = (a^{+}N + [N,a^{+}])n> = (a^{+}N + a^{+})n> =
a^{+}(N + 1)n> = a^{+}(n + 1)n> = (n + 1) a^{+}n>
Similarly,
Nan> = (aN + [N,a])n>: = (aN  a)n> = (n  1)an>
a^{+} is a raising operator
Na^{+}n> = (n + 1) a^{+}n>
a is a lowering operator
Nan> = (n  1)an>
If the smallest eigen number n is 0 (ground state), then a0> = 0
( we cannot lower less than 0).
Therefore:
H0> = (ℏω/2)0>
For the excited state
Hn> = E_{n} = (ℏω (n + 1/2)n>.
Its corresponding eigenstate is:
n> = [(a^{+})^{n}/(n!)^{1/2}]o>
This expression comes from the following proof:
We force a^{+} to satisfy:
a^{+}n> = C n+1>, so
<naa^{+}n> = C^{2} <nn+1>
n + 1 , then:
C = (n+1)^{1/2}, therefore:
a^{+}n> = (n+1)^{1/2}n+1>
Thus:
By n1 instead of n in a^{+}n> = (n+1)^{1/2}n+1>, we get:
n+1> = (n+1)^{ 1/2} a^{+}n> and
n> = n^{ 1/2} a^{+}n1>
By n2 instead of n in a^{+}n> = (n+1)^{1/2}n+1>, we get:
a^{+}n2> = (n1)^{1/2}n1>
so n1> = a^{+}n2>/(n1)^{1/2}
By substitution n1> in n> = n^{ 1/2} a^{+}n1>, we get:
n> = n^{ 1/2} a^{+} a^{+}n2>/(n1)^{1/2}; = n^{ 1/2} (n1)^{ 1/2}a^{+}^{2}n2>
... and so on ...
=
n> = (n!)^{ 1/2} a^{+}^{n}0>
Finally, without solving the equation of Schrodinger,
we have get the expressions of the eigenvalue and the
eigenstate for an quantum harmonic oscillator:
E_{n} = (ℏω (n + 1/2)
n> = (n!)^{ 1/2} (a^{+})^{n}0>

