Wolfram Alpha:
Search by keyword:
Astronomy
Chemistry
Classical Physics
Climate Change
Cosmology
Finance and Accounting
Game Theory
General Relativity
Lagrangian and Hamiltonian Mechanics
Macroeconomics
Mathematics
Microeconomics
Particle Physics
Probability and Statistics
Programming and Computer Science
Quantum Computing
Quantum Field Theory
Quantum Mechanics
Semiconductor Reliability
Solid State Electronics
Special Relativity
Statistical Mechanics
String Theory
Superconductivity
Supersymmetry (SUSY) and Grand Unified Theory (GUT)
The Standard Model
Topology
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.