In physics, an annihilation operator is an operator that lowers the number of particles in a given state by one. A creation operator is an operator that increases the number of particles in a given state by one, and it is the adjoint of the annihilation operator. Depending on the context, the identity of the particles in question varies; for example, in quantum chemistry and many-body theory the creation and annihilation operators often act on electrons. Annihilation and creation operators can also refer specifically to the ladder operators for the quantum harmonic oscillator. In the latter case, the raising operator is interpreted as a creation operator, adding a quantum of energy to the oscillator system. Vice versa for the lowering operator. In many subfields of physics and chemistry, using these operators instead of a wavefunction picture, is known as second quantization.

The mathematics behind the creation and the annihilation operators is identical as the formulae for ladder operators that appear in the quantum harmonic oscillator. For example, the commutator of the annihilation and the creation operator associated with the same state equals one; all other commutators vanish.

While the concept of creation and annihilation operators is well defined for free field theories, in interacting QFTs, they can only be defined in the interaction picture, which does not exist according to Haag's theorem.

Derivation of bosonic creation and annihilation operators

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In the context of the quantum harmonic oscillator, we reinterpret the ladder operators as creation and annihilation operators, adding or subtracting fixed quanta of energy to the oscillator system. Creation/annihilation operators are different for bosons (integer spin) and fermions (half-integer spin). This is because their wavefunctions have different symmetry properties.

Suppose the wavefunctions are dependent on N properties. Then

For bosons: ψ(1,2,3,4,...N) = ψ(2,1,3,4,...N)

For fermions: ψ(1,2,3,4,...N) = -ψ(2,1,3,4,...N)

For now let's just consider the case of bosons because fermions are more complicated.

Start with the Schrödinger equation for the one dimensional time independent quantum harmonic oscillator

$\backslash left(-\backslash frac\{\backslash hbar^2\}\{2m\}\; \backslash frac\{d^2\}\{d\; x^2\}\; +\; \backslash frac\{1\}\{2\}m\; \backslash omega^2\; x^2\backslash right)\; \backslash psi(x)\; =\; E\; \backslash psi(x)$

Make a coordinate substitution to nondimensionalize the differential equation

$x\; \backslash equiv\; \backslash sqrt\{\; \backslash frac\{\backslash hbar\}\{m\; \backslash omega\}\}\; q$.

and the Schrödinger equation for the oscillator becomes

$\backslash frac\{\backslash hbar\; \backslash omega\}\{2\}\; \backslash left(-\backslash frac\{d^2\}\{d\; q^2\}\; +\; q^2\; \backslash right)\; \backslash psi(q)\; =\; E\; \backslash psi(q)$.

Notice that the quantity ħω = hν is the same energy as that found for light quanta and that the parenthesis in the Hamiltonian can be written as

$-\backslash frac\{d^2\}\{dq^2\}\; +\; q^2\; =\; \backslash left\backslash right\backslash leftq\; \backslash right+\; \backslash frac\; \{d\}\{dq\}q\; -\; q\; \backslash frac\; \{d\}\{dq\}$

The last two terms in that equation form the commutator of q with its derivative. So let's calculate that commutator q, ∂/∂q

$\backslash left(q\; \backslash frac\{d\}\{dq\}-\; \backslash frac\{d\}\{dq\}\; q\; \backslash right)f(q)\; =\; q\; \backslash frac\{df\}\{dq\}\; -\; \backslash frac\{d\}\{dq\}(q\; f(q))\; =\; -f(q)$

In other words q, d/dq = - 1 or d/dq, q = 1.

Therefore

$\backslash frac\{1\}\{2\}\; \backslash hbar\; \backslash omega\; \backslash left(\; -\backslash frac\{d^2\}\{dq^2\}\; +\; q^2\; \backslash right)\; =\; \backslash hbar\; \backslash omega\; \backslash left\backslash frac\{-d/\; dq\; +\; q\}\{\backslash sqrt\{2\}\}\backslash right\backslash left/\; dq\; +\; q\}\{\backslash sqrt\{2\}\}\backslash right+\; \backslash frac\{1\}\{2\}\; \backslash hbar\; \backslash omega$

If we define

$a^\backslash dagger\; \backslash equiv\; \backslash frac\{1\}\{\backslash sqrt\{2\}\}\; \backslash left(-\backslash frac\{d\}\{dq\}\; +\; q\backslash right)$ as the "creation operator" or the "raising operator" and

$a\; \backslash equiv\; \backslash frac\{1\}\{\backslash sqrt\{2\}\}\; \backslash left(+\backslash frac\{d\}\{dq\}\; +\; q\backslash right)$ as the "annihilation operator" or the "lowering operator"

the Hamiltonian becomes

$H\; =\; \backslash hbar\; \backslash omega\; \backslash left(\; a^\backslash dagger\; a\; +\; \backslash frac\{1\}\{2\}\; \backslash right)$.

This Hamiltonian is significantly simpler than the original form. Further simplifications of this equation enables one to derive all the properties listed above thus far.

Letting p = - i d/dq, where p is the nondimensionalized momentum operator

$a^\backslash dagger\; =\; \backslash frac\{1\}\{\backslash sqrt\{2\}\}(q\; -\; i\; p)$

$a\; =\; \backslash frac\{1\}\{\backslash sqrt\{2\}\}(q\; +\; i\; p)$.

Substituting backwards, the laddering operators are recovered.

Mathematical details

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The operators derived above are actually a specific instance of a more generalized class of creation and annihilation operators. The more abstract (and hence more applicable) form of the operators satisfy the properties below.

Let H be the one-particle Hilbert space. To get the bosonic CCR algebra, look at the algebra generated by a(f) for any f in H. The operator a(f) is called an annihilation operator and the map a(.) is antilinear. Its adjoint is a^†(f) which is linear in H.

For a boson,

$*=*=0$

$*=\backslash langle\; f|g\; \backslash rangle$,

where we are using bra-ket notation.

For a fermion, the anticommutators are

$\backslash \{a(f),a(g)\backslash \}=\backslash \{a*(f),a*(g)\backslash \}=0$

$\backslash \{a(f),a*(g)\backslash \}=\backslash langle\; f|g\; \backslash rangle$.

A CAR algebra.

Physically speaking, a(f) removes (i.e. annihilates) a particle in the state |f> wheareas a^†(f) creates a particle in the state |f>.

The free field vacuum state is the state with no particles. In other words,

$a(f)|0\backslash rangle=0$

where |0> is the vacuum state.

If |f> is normalized so that <f|f>=1, then a^†(f) a(f) gives the number of particles in the state |f>.

Note that the creation and annihilation operators are "generalized complex conjugates" of each other. Usually, the notation is chosen in such a way that the a^†(f) is the creation operator, and a(f) is the annihilation operator. The ^† reminds us that something "extra" is being added to the system. The topic can be misleadingly confusing if this is not done.

Notational caveats and considerations

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In quantum mechanics, Dirac bra-ket notation is often used. However, there is some ambiguity in this notation, particularly when there is the need to differentiate between these things:

• The lowest energy state
• The zero state
• The vacuum state
• The zero ket

Often, these are all interchangeably notated as |0>, or even | >. As a result, it is necessary to read carefully, and consider the context in which the notation is used.

For example, in the quantum harmonic oscillator, the ground state has the property that when the annihilation operator b is applied to it, it satisfies b|0> = 0| > = 0

The intermediate step is rarely indicated as it is considered necessary only when more conceptual/mathematical rigour is needed.

In this example, the lowest energy state is denoted as |0>. It is labeled as the "zero state", but it is important to emphasize that any state can be labeled as the "zero" state. The zero state is often used as a reference state to other quantum states. Therefore, the |0> state need not be the state with the absolutely lowest energy. In the case of the harmonic oscillator, it is due to the particulars of the mathematics that the ground state is chosen to be |0>. The vacuum state is the state where no quanta is available to be extracted. This special null state is denoted by | >. This vacuum state is also known as the "zero ket" because there are zero particles in the state. Unfortunately, the lowest energy state |0> is also known as the "zero ket" for the different reason that the state is labeled as "zero". Care must be taken that the four concepts listed above are not mixed together.

Sometimes, the terms "null state" and "empty state" are used interchangeably for |> and |0>. The meaning for this usage is again dependent on the context.

The vacuum state

The vacuum state is a conceptual state which has no particles. The state is usually denoted as |0>, not the "empty ket" | >. Interestingly enough, no actual function actually represents the |0> state, but for notational purposes, we define the vacuum state as being normalized such that <0|0> = 1 and that |0> is orthogonal to all other states of the form |N>, where N is any indexing of quantum states for a particular system.

Energy spectra

In a quantum mechanical system, the range of discrete energies allowed in a system may be either finite, or infinite, or "semi-infinite".

In a system where the energies are confined to be semi-infinite on the interval [constant, ∞) such as the quantum harmonic oscillator, the vacuum state | >, (different from |0>) needs to be introduced in order to make the theory of creation and annihilation operators consistent. The lowest energy allowed in a semi-infinite energy system is known as the ground state. Since it is often used as the reference state, it is denoted by |0>. However, this state is not empty - the vacuum state | > is introduced to disambiguate these two states.

In a system where the range of energies is (-∞, ∞), the vacuum state is almost always denoted by |0>. There is no need for the "null" state | > as |0> already is sufficient to denote "emptiness". There is also no "ground state" present, which is why the notational ambiguity arises. This interpretation arises directly from the relativistic formalism of quantum mechanics by Paul Dirac, which later became one of the foundations for quantum field theory. One of the shortcomings of quantum field theory however, is its allowance of energy states tending to infinity. The attempt to resolve this problem is very much an active part of quantum mechanical research today.

In summary for infinite and semi infinite systems

COMMON NOTATION FOR STATES infinite semi-infinite ground state none |0> vacuum state |0> | >

There is no | > state needed for infinite-ranged-energy-systems in quantum mechanics.

See also

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• Ladder operators
• Bogolibov transformations - arises in the theory of quantum optics. Also transliterated as Bogolubov transformations'

Quantum mechanics | Quantum field theory