# Redshift Academy

Wolfram Alpha:

Last modified: May 6, 2020
```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
```