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Units, Constants and Useful Formulas
Position Creation and Annihilation Operators for Bosons

Notation: Capital Ψ is used to denote field operators and ψ is used
for wavefunctions.
Field operators create or destroy a particle at a particular point in space.
Field operators act by applying the Fourier transform to the creation and
annihilation operators.
Ψ(x) = Σ_{k}α(k)exp(ikx) where α(k) is equivalent to the amplitude term in QM
and,
Ψ^{*}(x) = Σ_{k}α^{*}(k)exp(ikx)
Compare these to the QM case:
ψ(x) = ΣA_{m}ψ_{m}
^{m}
Where ψ_{m} are othornormal basis vectors of the form exp(ikx) and A_{m} are the
Fourier coefficients. Therefore, in QFT the Fourier coefficients are
replaced by the creation and annihilation operators.
Now introduce time:
Ψ(x,t) = Σ_{k}a^{}exp(i(kx  ωt)) ... annihilates a particle at position x.
and,
Ψ^{†}(x,t) = Σ_{k}a^{+}exp(i(kx  ωt)) ... creates a particle at position x.
Ψ(x,t) and Ψ^{†}(x,t) are quantum fields.
Summary:
a^{+}(k)  creation operator for momenta, k
Ψ^{†}(x)  creation operator for position, x
a^{}(k)  annihilation operator for momenta, k
Ψ(x) ^{ } annihilation operator for position, x
Examples:
Preexisting photon, n(5) = 1
0 0 0 0 1 0 0>
Create a new photon in momentum state k = 3 at x. Thus,
n(3) > 1, n(5) = 1
Σ_{k}a^{+}(3)exp(i3x)0 0 0 0 1 0 0> => e^{i3x}0 0 1 0 1 0 0>
Create another new photon in momentum state k = 5 at the same x.
Thus, n(5) > 2
Σ_{k}a^{+}(5)exp(i5x)0 0 1 0 1 0 0> => √2e^{i5x}0 0 1 0 2 0 0>
The factor of √2 comes from a^{+}n> = √(n+1)n+1>
Creation and annihilation operators on bra vectors:
The rule for the creation and annihilation operators when they
operate on bra vectors is the opposite of the case for ket
vectors. Thus, the creation operator acts on a bra as an
annihilation operator and vice versa.
<na^{+} => <n1√n
and
<na^{} => <n+1√n+1
Consider the following scattering processes:
t k_{2}
 \
 \ 
 \ g = coupling contant that measures the strength of
 / the scattering (measured experimentally).
 / 
 / 
 k_{1}

 x
x=0
Ψ^{†}Ψk_{1}> => k_{2}>
The probability that k_{1} results in k_{2} is given by:
<k_{2}Ψ^{†}Ψk_{1}>^{2}
We will just calculate the ...:
<k_{2}g∫dtΣ_{d}a^{+}(d)exp(iω_{2}t)Σ_{e}a^{}(f)exp(iω_{1}t)k_{1}> after setting x = 0
<k_{2}g∫dta^{+}(k_{2})exp(iω_{2}t)a^{}(k_{1})exp(iω_{1}t)k_{1}>
<0g∫dtexp(i(ω_{2}  ω_{1})t)0>
<0gδ(ω_{2}  ω_{1})0>
∴ hω_{2} = hω_{1} ... conservation of energy
k_{2} k_{3}
\ /
\ /
\/
/ g
/
/
k_{i}
Follow the same procedure as above:
Ψ^{†}Ψ^{†}Ψk_{1}> => k_{2}k_{3}>
<k_{2}k_{3}Ψ^{†}Ψ^{†}Ψk_{1}>
<k_{2}k_{3}g∫dxΣ_{d}a^{+}(d)exp(ik_{3}x)Σ_{e}a^{+}(e)exp(ik_{2}x)Σ_{f}a^{}(f)exp(ik_{1}x)k_{1}> after setting t = 0
<k_{2}k_{3}g∫dxa^{+}(k_{3})exp(ik_{3}x)a^{+}(k_{2})exp(ik_{2}x)a^{}(k_{1})exp(ik_{1}x)k_{1}>
<0 = 1 when k_{1}  k_{2}  k_{3} or 0 otherwise 0>
∴ hk_{1}  hk_{2}  hk_{3} = 0 ... conservation of momentum
Time Dependent Schrodinger equation for a field:
Differentiate, Ψ w.r.t. t and x:
∂Ψ/∂t = Σ_{k}(iω)a^{}(k)exp(i(kx  ωt))
∂Ψ/∂x = Σ_{k}(ik)a^{}(k)exp(i(kx  ωt))
∂^{2}Ψ/∂x^{2} = Σ_{k}(ik)^{2}a^{}(k)exp(i(kx  ωt))
So we can write:
∂Ψ/∂t/∂^{2}Ψ/∂x^{2} = Σ_{k}(iω)a^{}(k)exp(i(kx  ωt))/Σ_{k}(ik)^{2}a^{}(k)exp(i(kx  ωt))
^{ }^{ }= (iω)/(ik)^{2}
Which leads to:
^{ }^{ }(ik)^{2}∂Ψ/∂t = (iω)∂^{2}Ψ/∂x^{2}
^{ }^{ }k^{2}∂Ψ/∂t = (iω)∂^{2}Ψ/∂x^{2}
Now E = hω = p^{2}/2m = h^{2}k^{2}/2m
Therefore k^{2} = 2mω/h
(2mω/h)∂Ψ/∂t/∂ = (iω)∂^{2}Ψ/∂x^{2}
∂Ψ/∂t =(ih/2m)∂^{2}Ψ/∂x^{2}
Note: In both cases we could have equally integrated over
position/time to demonstrate conservation of momentum and
conservation of energy respectively.
What is the meaning of ∫Ψ^{†}(x)Ψ(x)dx ?
circle with length L
= ∫_{L}Σ_{m}a^{+}(m)e^{imx}Σ_{n}a^{}(n)e^{inx}dx
= ∫_{L}Σ_{m}a^{+}(m)Σ_{n}a^{}(n)e^{i(nm)x}dx
= LΣ_{m}a^{+}(m)a^{}(m)  occupation number is a^{+}(m)a^{}(m)
= the number of particles
Ψ^{†}(x)Ψ(x) is interpreted as the density of particles at x
Position Creation and Annihilation Operators for Fermions

Bosons can occur in the same state state; Fermions cannot (Pauli
Exclusion Principle). Fermions can only have 0 or 1 in a state. We
define new position creation and annihilation operators as:
Ψ(x,t) = Σ_{k}c^{}exp(i(kx  ωt))
and,
Ψ^{†}(x,t) = Σ_{k}c^{+}exp(i(kx  ωt))
Heisenberg Uncertainty Principle in QFT

Analagous to the relationship in QM, the position creation and
annihilation operators are Fourier transforms of each other.
Thus :
Ψ(x)^{±} = Σ_{k}a(k)^{±}exp(±ikx)
and
a(k)^{±} = ΣΨ(x)^{±}exp(±ikx)
This shows that to create a state of definite position, it is necessary
to sum over many momentum states. Conversely, to create a state
of definite momentum it is necessary to sum over many position
states.