Redshift Academy


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>

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_pn_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