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Creation and Annihilation Operators
-----------------------------------
Quantum Harmonic Oscillator
---------------------------
Consider unit mass hanging on a spring.
----------
/
\ ^
/ | x
\
unit mass
.
Lagrangian, L = (1/2)x^{2} + (1/2)ω^{2}x^{2} where ω = √(K/m)
Euler-Lagrange:
.
d(∂L/∂x)/dt - ∂L/∂x = 0
Leads to
..
x = -ω^{2}x
with solution x = cosωt
Hamiltonian:
. .
p = ∂L/∂x = x
.
H = px - L
. .
= px - [(1/2)x^{2} + (1/2)ω^{2}x^{2}]
.
= (1/2)x^{2} + (1/2)ω^{2}x^{2}
= (1/2)p^{2} + (1/2)ω^{2}x^{2} (set m = 1)
In CM H can be zero because x and p can be set to zero. Not so in
QM because of uncertainty principle.
H|ψ> = (1/2)(-ih∂/∂x)(-ih∂/∂x)ψ(x) + (1/2)ω^{2}x^{2}ψ(x)
= (-1/2)∂^{2}ψ(x)/∂x^{2} + (1/2)ω^{2}x^{2}ψ(x) (set h = 1)
Schrodinger Equation:
i∂ψ/∂t = Hψ
= (-1/2)∂^{2}ψ(x)/∂x^{2} + (1/2)ω^{2}x^{2}ψ(x)
Energy Eigenvectors:
H|ψ = E|ψ>
-(1/2)∂^{2}ψ/∂x^{2} + ω^{2}x^{2}ψ(x) = Eψ(x)
ψ(x) = Aexp(-ω/2)x^{2}) is a solution to this equation where A
is the normalization factor.
If we plug this into the above equation we get,
ω/2 = E - The Ground State
H = (1/2)(p^{2} + ω^{2}x^{2}) where p = -i∂/∂x
= (1/2)(ωx + ip)(ωx - ip)
= (1/2)(p^{2} + ω^{2}x^{2} + iω(px - xp)) - x and p do not commute in QM
= (1/2)(p^{2} + ω^{2}x^{2} + ω) <- replace (px - xp) by [p,x] = -i
Therefore, the corrected Hamiltonian is:
H = (1/2)(ωx + ip)(ωx - ip) + ω/2
Define:
a^{†} = √(ω/2)(x + ip/√(2ω) or (1/√(2ω)(ωx + ip)
and
a = √(ω/2)(x - ip/√(2ω) or (1/√(2ω)(ωx - ip)
Therefore,
x = (1/√(2ω))(a + a^{†})
and
p = -i√(ω/2)(a - a^{†})
The -i makes the operator Hermitian. Consider 2 the addition
and subtraction of a complex number and its conjugate:
x + iy + (x - iy) = 2x = real
x + iy - (x - iy) = 2iy = imaginary
The resulting commutator is:
[a,a^{†}] = 1
Returning to the Hamiltonian,
H = ω(a^{†}a + 1/2)
Let us go further with this. You might think:
a^{†}|n> => |n+1>
a|n> => |n-1>
So,
a^{†}a|n> = a^{†}|n - 1> = |n>
and
aa^{†}|n> = a|n + 1> = |n>
However, this is not correct because [a^{†},a] = 0 and not -1. What
we need instead is:
a^{†}|n> = √(n+1)|n+1> ... A.
a|n> = √n|n-1> ... B.
Now,
a^{†}a|n> = a^{†}√n|n-1> = √n√n|n> = n|n> ... using A.
and
aa^{†}|n> = a√(n+1)|n+1> = √(n+1)√(n+1)|n> = n+1|n> ... using B.
Now [a,a^{†}] = aa^{†} - a^{†}a = (n+1) - n = 1 as before
Number Operator
---------------
a^{†}a|n> = √na^{†}|n-1>
^{ } = √n√n|n> (after replacing n with n+1)
^{ } = n|n>
a_{p}^{†}a= n_{p} counts the number of quanta in state p and is
referred to as the OCCUPATION NUMBER. These operators change
the eigenvalues of the NUMBER OPERATOR, N. Thus,
N = Σ_{p}n_{p}
Knowing that a^{†}a = n, we can rewite the Hamiltonian as:
H = hω(N + 1/2) (after putting h back in)
with eigenvalues of:
|0> = (1/2)ω|0>
a^{†}|0> = (3/2)ω|0>
a^{†}a^{†}|0> = (5/2)ω|0>
a^{†}a^{†}a^{†}|0> = (7/2)ω|0>
Note regarding interpretation: The energy eigenvales of the harmonic
oscillator could be interpreted in 2 ways. The increase in energy of a
single quanta or the addition of quanta with energy hω. It is this
second idea that we want focus on. Thus, the creation and annihilation
operators add and subtract quanta: a_{p}^{†} increases quanta (hω) in momentum
state p by 1; a_{p} decreases quanta in momentum state p by 1.
Commutation Rules for Bosons
----------------------------
From before we have shown that:
[a,a^{†}]|n> = ((n + 1) - n))|n>
^{ } = |n>
Thus, for example:
a^{†}|0> = |1>
a^{†}|1> = |2>
a|2> = |1>
a|1> = |0>
a|0> = not allowed
Therefore,
[a,a^{†}]|0>(aa^{†} - a^{†}a)|0> = |0>
[a,a^{†}]|1>(aa^{†} - a^{†}a)|1> = |1>
Summarizing:
[a_{p},a_{q}] = [a^{†}_{p},a^{†}_{q}] = 0
[a_{p},a^{†}_{q}] = δ_{pq} (1 when p = q, 0 when p ≠ q)
[a^{†}_{p},a_{q}] = -δ_{pq}
Commuation Rules for Fermions
-----------------------------
Bosons can occur in the same state state. However, because
of the Pauli Exclusion Principle, Fermions cannot. Fermions
can only have 0 or 1 in a state. We define new creation and
annihilation operators c and c^{†} such that:
c^{†}|0> = |1>
c|1> = |0>
c^{†}|1> = not allowed
c|0> = not allowed
c^{†}c|0> = not allowed
cc^{†}|0> = |0>
c^{†}c|1> = |1>
cc^{†}|1> = not allowed
Now consider combinations of c^{†} and c:
(cc^{†} + c^{†}c)|0> = |0>
(cc^{†} + c^{†}c)|1> = |1>
Therefore
{c,c^{†}}|0> = |0>
and,
{c,c^{†}}|1> = |1>
Conclusion: c and c^{†} follow the ANTICOMMUTATOR relationship.
The anticommutator can be written as:
{c,c^{†}}|n> = ((n + 1) + n))|n>
^{ } = (2n + 1)|n>
For this to work, n has to equal 0. Therefore, the anticommutation
relationship allows a state to be filled only if it is empty.
Summarizing:
{c_{p},c_{q}} = {c^{†}_{p},c^{†}_{q}} = 0
{c_{p},c^{†}_{q}} = δ_{pq} (1 when p = q, 0 when p ≠ q)
{c^{†}_{p},c_{q}} = -δ_{pq}