Wolfram Alpha:

```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) = Σkα(k)exp(ikx) where α(k) is equivalent to the amplitude term in QM

and,

Ψ*(x) = Σkα*(k)exp(-ikx)

Compare these to the QM case:

ψ(x) = ΣAmψm
m

Where ψm are othornormal basis vectors of the form exp(ikx) and Am are the
Fourier coefficients.  Therefore, in QFT the Fourier coefficients are
replaced by the creation and annihilation operators.

Now introduce time:

Ψ(x,t) = Σka-exp(i(kx - ωt))  ... annihilates a particle at position x.

and,

Ψ†(x,t) = Σka+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:

Pre-existing 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

Σka+(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

Σka+(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.

<n|a+ => <n-1|√n

and

<n|a- => <n+1|√n+1

Consider the following scattering processes:

t        k2
|         \
|          \ |
|           \| g = coupling contant that measures the strength of
|           /|       the scattering (measured experimentally).
|          / |
|         /  |
|       k1
|
-------------------- x
x=0

Ψ†Ψ|k1> => |k2>

The probability that k1 results in k2 is given by:

|<k2|Ψ†Ψ|k1>|2

We will just calculate the |...|:

<k2|g∫dtΣda+(d)exp(iω2t)Σea-(f)exp(-iω1t)|k1> after setting x = 0

<k2|g∫dta+(k2)exp(iω2t)a-(k1)exp(-iω1t)|k1>

<0|g∫dtexp(i(ω2 - ω1)t)|0>

<0|gδ(ω2 - ω1)|0>

∴ hω2 = hω1 ... conservation of energy

k2     k3
\    /
\  /
\/
/ g
/
/
ki

Follow the same procedure as above:

Ψ†Ψ†Ψ|k1> => |k2k3>

<k2k3|Ψ†Ψ†Ψ|k1>

<k2k3|g∫dxΣda+(d)exp(-ik3x)Σea+(e)exp(-ik2x)Σfa-(f)exp(ik1x)|k1> after setting t = 0

<k2k3|g∫dxa+(k3)exp(-ik3x)a+(k2)exp(-ik2x)a-(k1)exp(ik1x)|k1>

<0| = 1 when k1 - k2 - k3 or 0 otherwise |0>

∴ hk1 - hk2 - hk3 = 0 ... conservation of momentum

Time Dependent Schrodinger equation for a field:

Differentiate, Ψ w.r.t. t and x:

∂Ψ/∂t = Σk(-iω)a-(k)exp(i(kx - ωt))

∂Ψ/∂x = Σk(ik)a-(k)exp(i(kx - ωt))

∂2Ψ/∂x2 = Σk(ik)2a-(k)exp(i(kx - ωt))

So we can write:

∂Ψ/∂t/∂2Ψ/∂x2 = Σk(-iω)a-(k)exp(i(kx - ωt))/Σk(ik)2a-(k)exp(i(kx - ωt))

= (-iω)/(ik)2

(ik)2∂Ψ/∂t = (-iω)∂2Ψ/∂x2

k2∂Ψ/∂t = (iω)∂2Ψ/∂x2

Now E = hω = p2/2m = h2k2/2m

Therefore k2 = 2mω/h

(2mω/h)∂Ψ/∂t/∂ = (iω)∂2Ψ/∂x2

∂Ψ/∂t =(ih/2m)∂2Ψ/∂x2

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

= ∫LΣma+(m)e-imxΣna-(n)einxdx

= ∫LΣma+(m)Σna-(n)ei(n-m)xdx

= LΣma+(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) = Σkc-exp(i(kx - ωt))

and,

Ψ†(x,t) = Σkc+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)± = Σka(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.  ```