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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)x2 + (1/2)ω2x2 where ω = √(K/m)
Euler-Lagrange:
.
d(∂L/∂x)/dt - ∂L/∂x = 0
Leads to
..
x = -ω2x
with solution x = cosωt
Hamiltonian:
. .
p = ∂L/∂x = x
.
H = px - L
. .
= px - [(1/2)x2 + (1/2)ω2x2]
.
= (1/2)x2 + (1/2)ω2x2
= (1/2)p2 + (1/2)ω2x2 (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)ω2x2ψ(x)
= (-1/2)∂2ψ(x)/∂x2 + (1/2)ω2x2ψ(x) (set h = 1)
Schrodinger Equation:
i∂ψ/∂t = Hψ
= (-1/2)∂2ψ(x)/∂x2 + (1/2)ω2x2ψ(x)
Energy Eigenvectors:
H|ψ = E|ψ>
-(1/2)∂2ψ/∂x2 + ω2x2ψ(x) = Eψ(x)
ψ(x) = Aexp(-ω/2)x2) 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)(p2 + ω2x2) where p = -i∂/∂x
= (1/2)(ωx + ip)(ωx - ip)
= (1/2)(p2 + ω2x2 + iω(px - xp)) - x and p do not commute in QM
= (1/2)(p2 + ω2x2 + 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>
ap†a= np 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 = Σpnp
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: ap† increases quanta (hω) in momentum
state p by 1; ap 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:
[ap,aq] = [a†p,a†q] = 0
[ap,a†q] = δpq (1 when p = q, 0 when p ≠ q)
[a†p,aq] = -δpq
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:
{cp,cq} = {c†p,c†q} = 0
{cp,c†q} = δpq (1 when p = q, 0 when p ≠ q)
{c†p,cq} = -δpq