# Creation and annihilation operators

(Redirected from Annihilation operator)

Creation and annihilation operators are mathematical operators that have widespread applications in quantum mechanics, notably in the study of quantum harmonic oscillators and many-particle systems.[1] An annihilation operator lowers the number of particles in a given state by one. A creation operator increases the number of particles in a given state by one, and it is the adjoint of the annihilation operator. In many subfields of physics and chemistry, the use of these operators instead of wavefunctions is known as second quantization.

Creation and annihilation operators can act on states of various types of particles. For example, in quantum chemistry and many-body theory the creation and annihilation operators often act on electron states. They 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 (similarly for the lowering operator). They can be used to represent phonons.

The mathematics for the creation and annihilation operators for bosons is the same as for the ladder operators of the quantum harmonic oscillator.[2] For example, the commutator of the creation and annihilation operators that are associated with the same boson state equals one, while all other commutators vanish. However, for fermions the mathematics is different, involving anticommutators instead of commutators.[3]

## Ladder operators for quantum harmonic oscillator

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.

First consider the simpler bosonic case of the phonons of the quantum harmonic oscillator.

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

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

Make a coordinate substitution to nondimensionalize the differential equation

$x \ = \ \sqrt{ \frac{\hbar}{m \omega}} q$.

and the Schrödinger equation for the oscillator becomes

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

Note that the quantity $\hbar \omega = h \nu$ is the same energy as that found for light quanta and that the parenthesis in the Hamiltonian can be written as

$-\frac{d^2}{dq^2} + q^2 = \left(-\frac{d}{dq}+q \right) \left(\frac{d}{dq}+ q \right) + \frac {d}{dq}q - q \frac {d}{dq}$

The last two terms can be simplified by considering their effect on an arbitrary differentiable function f(q),

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

which implies,

$\frac{d}{dq} q- q \frac{d}{dq} = 1$

Therefore

$-\frac{d^2}{dq^2} + q^2 = \left(-\frac{d}{dq}+q \right) \left(\frac{d}{dq}+ q \right) + 1$

and the Schrödinger equation for the oscillator becomes, with substitution of the above and rearrangement of the factor of 1/2,

$\hbar \omega \left[\frac{1}{\sqrt{2}} \left(-\frac{d}{dq}+q \right)\frac{1}{\sqrt{2}} \left(\frac{d}{dq}+ q \right) + \frac{1}{2} \right] \psi(q) = E \psi(q)$.

If we define

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

then the Schrödinger equation for the oscillator becomes

$\hbar \omega \left( a^\dagger a + \frac{1}{2} \right) \psi(q) = E \psi(q)$

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

Letting $p = - i \frac{d}{dq}$, where "p" is the nondimensionalized momentum operator then we have

$[q, p] = i \,$

and

$a = \frac{1}{\sqrt{2}}(q + i p) = \frac{1}{\sqrt{2}}\left( q + \frac{d}{dq}\right)$
$a^\dagger = \frac{1}{\sqrt{2}}(q - i p) = \frac{1}{\sqrt{2}}\left( q - \frac{d}{dq}\right)$.

Note that these imply that

$[a, a^\dagger ] = \frac{1}{2} [ q + ip , q-i p] = \frac{1}{2} ([q,-ip] + [ip, q]) = \frac{-i}{2} ([q, p] + [q, p]) = 1$

in contrast to the so-called "normal operators" of mathematics, which have a similar representation (e.g. $A= W_1 + i\, W_2)\,,$ with self-adjoint $W_i\,.$ But in the case of normal operators one would be dealing with commuting $W_i\,,$ i.e. with $W_1W_2=W_2W_1\,,$ so that the 1 at the extreme r.h.s. of the previous equation would be replaced by 0, which would have the consequence of one-and-the-same set of eigenfunctions (and/or eigendistributions) for both $W_1$ and $W_2$, whereas here common eigenfunctions or eigendistributions of the operators p and q don't exist.

Thus, although in the present case one is explicitly dealing with non-normal operators, by the commutation relation given above, the Hamiltonian operator can be expressed as

$\hat H = \hbar \omega ( a \, a^\dagger - \frac{1}{2}).$
$\hat H = \hbar \omega ( a^\dagger \, a + \frac{1}{2}).$

And $a$ and $a^\dagger\,,$ operators give the following commutation relations with the Hamiltonian[4]

$[\hat H, a ] = -\hbar \omega \, a.$
$[\hat H, a^\dagger ] = \hbar \omega \, a^\dagger .$

These relations can be used to find the energy eigenstates of the quantum harmonic oscillator. Assuming that $\psi_n$ is an eigenstate of the Hamiltonian $\hat H \psi_n = E_n\, \psi_n$. Using these commutation relations it can be shown that[5]

$\hat H\, a\psi_n = (E_n - \hbar \omega)\, a\psi_n .$
$\hat H\, a^\dagger\psi_n = (E_n + \hbar \omega)\, a^\dagger\psi_n .$

This shows that $a\psi_n$ and $a^\dagger\psi_n$ are also eigenstates of the Hamiltonian with eigenvalues $E_n - \hbar \omega$ and $E_n + \hbar \omega$. This identifies the operators $a$ and $a^\dagger$ as lowering and rising operators between the eigenstates. Energy difference between two eigenstates is $\Delta E = \hbar \omega$.

The ground state can be found by assuming that the lowering operator will collapse it, $a\, \psi_0 = 0$. And then using the Hamiltonian in terms of rising and lowering operators,

$a^\dagger a \psi_0 = 0 = \left(\frac{\hat H}{\hbar \omega} - \frac{1}{2} \right) \,\psi_0 = \left(\frac{E_0}{\hbar \omega} - \frac{1}{2} \right) \,\psi_0.$

the wave-function on the right is non-zero, thus term in brackets must be. This gives the ground state energy $E_0 = \hbar \omega /2$. This allows to identify the energy eigenvalue of any eigenstate $\psi_n$ as[6]

$E_n = (n + \frac{1}{2})\hbar \omega$

Furthermore it can be shown that the first-mentioned operator, the number operator $N=a^\dagger a\,,$ plays a most-important role in applications, while the second one, $a \,a^\dagger\,,$ can simply be replaced by $N +1\,.$ So one simply gets

$\hbar\omega \,(N+\frac{1}{2})\,\psi (q) =E\,\psi (q)$.

### Applications

The ground state $\ \psi_0(q)$ of the quantum harmonic oscillator can be found by imposing the condition that

$a \ \psi_0(q) = 0$.

Written out as a differential equation, the wavefunction satisfies

$q \psi_0 + \frac{d\psi_0}{dq} = 0$

which has the solution

$\psi_0(q) = C \exp(-{q^2 \over 2}).$

The normalization constant C can be found to be  $1\over \sqrt[4]{\pi}$  from $\int_{-\infty}^\infty \psi_0^* \psi_0 \,dq = 1$,  using the Gaussian integral.

### Matrix representation

The matrix counterparts of the creation and annihilation operators obtained from the quantum harmonic oscillator model are

$\hat{a}^{\dagger}=\begin{pmatrix} 0 & 0 & 0 & \dots & 0 &\dots \\ \sqrt{1} & 0 & 0 & \dots & 0 & \dots\\ 0 & \sqrt{2} & 0 & \dots & 0 & \dots\\ 0 & 0 & \sqrt{3} & \dots & 0 & \dots\\ \vdots & \vdots & \vdots & \ddots & \vdots & \dots\\ 0 & 0 & 0 & 0 & \sqrt{n} &\dots & \\ \vdots & \vdots & \vdots & \vdots & \vdots &\ddots \end{pmatrix}$

$\hat{a}=\begin{pmatrix} 0 & \sqrt{1} & 0 & 0 & \dots & 0 & \dots \\ 0 & 0 & \sqrt{2} & 0 & \dots & 0 & \dots \\ 0 & 0 & 0 & \sqrt{3} & \dots & 0 & \dots \\ 0 & 0 & 0 & 0 & \ddots & \vdots & \dots \\ \vdots & \vdots & \vdots & \vdots & \ddots & \sqrt{n} & \dots \\ 0 & 0 & 0 & 0 & \dots & 0 & \ddots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix}$

Substituting backwards, the laddering operators are recovered. They can be obtained via the relationships $a^\dagger_{ij} = \langle\psi_i | \hat{a}^\dagger | \psi_j\rangle$ and $a_{ij} = \langle\psi_i | \hat{a} | \psi_j\rangle$. The wavefunctions are those of the quantum harmonic oscillator, and are sometimes called the "number basis".

### Mathematical details

The operators derived above are actually a specific instance of a more generalized class of creation and annihilation operators. The more abstract 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,

$[a(f),a(g)]=[a^\dagger(f),a^\dagger(g)]=0$
$[a(f),a^\dagger(g)]=\langle f|g \rangle$,

where we are using bra-ket notation.

For a fermion, the anticommutators are

$\{a(f),a(g)\}=\{a^\dagger(f),a^\dagger(g)\}=0$
$\{a(f),a^\dagger(g)\}=\langle f|g \rangle$.

Physically speaking, a(f) removes (i.e. annihilates) a particle in the state | f$\scriptstyle \rangle$ whereas a(f) creates a particle in the state | f$\scriptstyle \rangle$.

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

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

where | 0 $\scriptstyle \rangle$ is the vacuum state.

If | f$\scriptstyle \rangle$ is normalized so that $\scriptstyle \langle$f | f$\scriptstyle \rangle$ = 1, then a(f) a(f) gives the number of particles in the state | f$\scriptstyle \rangle$.

### Creation and annihilation operators for reaction-diffusion equations

The annihilation and creation operator description has also been useful to analyze classical reaction diffusion equations, such as the situation when a gas of molecules A diffuse and interact on contact, forming an inert product: A + A → ∅ . To see how this kind of reaction can be described by the annihilation and creation operator formalism, consider $n_{i}$ particles at a site $i$ on a 1-d lattice. Each particle diffuses independently, so that the probability that one of them leaves the site for short times $dt$ is proportional to $n_{i}dt$, say $\alpha n_{i}dt$ to hop left and $\alpha n_{i}dt$ to hop right. All $n$ particles will stay put with a probability $1-2\alpha n_{i}dt$.

We can now describe the occupation of particles on the lattice as a `ket' of the form | n1, n2, ...$\scriptstyle \rangle$. A slight modification of the annihilation and creation operators is needed so that

$a|n\rangle= \sqrt{n} \ |n-1\rangle$

and

$a^{\dagger}|n\rangle= \sqrt{n+1} \ |n+1\rangle$.

This modification preserves the commutation relation

$[a,a^{\dagger}]=1$,

but allows us to write the pure diffusive behaviour of the particles as

$\partial_{t}|\psi\rangle=-\alpha\sum(2a_{i}^{\dagger}a_{i}-a_{i-1}^{\dagger}a_{i}-a_{i+1}^{\dagger}a_{i})|\psi\rangle=-\alpha\sum(a_{i}^{\dagger}-a_{i-1}^{\dagger})(a_{i}-a_{i-1})|\psi\rangle$

The reaction term can be deduced by noting that $n$ particles can interact in $n(n-1)$ different ways, so that the probability that a pair annihilates is $\lambda n(n-1)dt$ and the probability that no pair annihilates is $1-\lambda n(n-1)dt$ leaving us with a term

$\lambda\sum(a_{i}a_{i}-a_{i}^{\dagger}a_{i}^{\dagger}a_{i}a_{i})$

yielding

$\partial_{t}|\psi\rangle=-\alpha\sum(a_{i}^{\dagger}-a_{i-1}^{\dagger})(a_{i}-a_{i-1})|\psi\rangle+\lambda\sum(a_{i}^{2}-a_{i}^{\dagger 2}a_{i}^{2})|\psi\rangle$

Other kinds of interactions can be included in a similar manner.

This kind of notation allows the use of quantum field theoretic techniques to be used in the analysis of reaction diffusion systems.

## Creation and annihilation operators in quantum field theories

In quantum field theories and many-body problems one works with creation and annihilation operators of quantum states, $a^\dagger_i$ and $a^{\,}_i$. These operators change the eigenvalues of the number operator,

$N = \sum_i n_i = \sum_i a^\dagger_i a^{\,}_i$,

by one, in analogy to the harmonic oscillator. The indices (such as $i$) represent quantum numbers that label the single-particle states of the system; hence, they are not necessarily single numbers. For example, a tuple of quantum numbers $(n, l, m, s)$ is used to label states in the hydrogen atom.

The commutation relations of creation and annihilation operators in a multiple-boson system are,

$[a^{\,}_i, a^\dagger_j] \equiv a^{\,}_i a^\dagger_j - a^\dagger_ja^{\,}_i = \delta_{i j},$
$[a^\dagger_i, a^\dagger_j] = [a^{\,}_i, a^{\,}_j] = 0,$

where $[\ \ , \ \ ]$ is the commutator and $\delta_{i j}$ is the Kronecker delta.

For fermions, the commutator is replaced by the anticommutator $\{\ \ , \ \ \}$,

$\{a^{\,}_i, a^\dagger_j\} \equiv a^{\,}_i a^\dagger_j +a^\dagger_j a^{\,}_i = \delta_{i j},$
$\{a^\dagger_i, a^\dagger_j\} = \{a^{\,}_i, a^{\,}_j\} = 0.$

Therefore, exchanging disjoint (i.e. $i \ne j$) operators in a product of creation of annihilation operators will reverse the sign in fermion systems, but not in boson systems.