Creation and annihilation operators

Creation and annihilation operators are   that  have widespread applications in, notably in the study of s and many-particle systems. An annihilation operator (usually denoted $$\hat{a}$$) lowers the number of particles in a given state by one. A creation operator (usually denoted $$\hat{a}^\dagger$$) increases the number of particles in a given state by one, and it is the of the annihilation operator. In many subfields of and, the use of these operators instead of s is known as.

Creation and annihilation operators can act on states of various types of particles. For example, in and  the creation and annihilation operators often act on  states. They can also refer specifically to the for the. 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.

The mathematics for the creation and annihilation operators for is the same as for the   of the. For example, the of the creation and annihilation operators that are associated with the same boson state equals one, while all other commutators vanish. However, for the mathematics is different, involving  instead of commutators.

Ladder operators for the quantum harmonic oscillator
In the context of the, one reinterprets the ladder operators as creation and annihilation operators, adding or subtracting fixed of energy to the oscillator system.

Creation/annihilation operators are different for s (integer spin) and s (half-integer spin). This is because their s have different.

First consider the simpler bosonic case of the phonons of the quantum harmonic oscillator. Start with the for the one-dimensional time independent ,
 * $$\left(-\frac{\hbar^2}{2m} \frac{d^2}{d x^2} + \frac{1}{2}m \omega^2 x^2\right) \psi(x) = E \psi(x).$$

Make a coordinate substitution to the differential equation
 * $$x \ = \ \sqrt{ \frac{\hbar}{m \omega}} q.$$

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  and that the parenthesis in the  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,
 * $$\frac{d}{dq} q- q \frac{d}{dq} = 1 ,$$

coinciding with the usual canonical commutation relation $$ -i[q,p]=1 $$, in position space representation: $$p:=-i\frac{d}{dq}$$.

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 one defines
 * $$a^\dagger \ = \ \frac{1}{\sqrt{2}} \left(-\frac{d}{dq} + q\right)$$

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", the Schrödinger equation for the oscillator reduces to
 * $$ \hbar \omega \left( a^\dagger a + \frac{1}{2} \right) \psi(q) = E \psi(q).$$

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

Letting $$p = - i \frac{d}{dq}$$, where $$p$$ is the nondimensionalized one has


 * $$ [q, p] = i \,$$

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

The operators $$a\,$$ and $$a^\dagger\,$$ may be contrasted to, which commute with their adjoints. Using the commutation relations given above, the Hamiltonian operator can be expressed as
 * $$\hat H = \hbar \omega \left( a \, a^\dagger - \frac{1}{2}\right) = \hbar \omega \left(  a^\dagger \, a + \frac{1}{2}\right).\qquad\qquad(*)$$

One may compute the commutation relations between the $$a\,$$ and $$a^\dagger\,$$ operators and the Hamiltonian:
 * $$[\hat H, a ] = [\hbar \omega \left ( a a^\dagger - \frac{1}{2}\right ) ,a] = \hbar \omega [ a a^\dagger, a] = \hbar \omega ( a [a^\dagger,a] + [a,a] a^\dagger) =  -\hbar \omega a.$$
 * $$[\hat H, a^\dagger ] = \hbar \omega \, a^\dagger .$$

These relations can be used to easily find all the energy eigenstates of the quantum harmonic oscillator as follows.

Assuming that $$\psi_n$$ is an eigenstate of the Hamiltonian $$\hat H \psi_n = E_n\, \psi_n$$. Using these commutation relations, it follows that
 * $$\hat H\, a\psi_n = (E_n - \hbar \omega)\, a\psi_n .$$
 * $$\hat H\, a^\dagger\psi_n = (E_n + \hbar \omega)\, a^\dagger\psi_n .$$

This shows that $$a\psi_n$$ and $$a^\dagger\psi_n$$ are also eigenstates of the Hamiltonian, with eigenvalues $$E_n - \hbar \omega$$ and $$E_n + \hbar \omega$$ respectively. This identifies the operators $$a$$ and $$a^\dagger$$ as "lowering" and "raising" operators between adjacent eigenstates. The energy difference between adjacent eigenstates is  $$\Delta E = \hbar \omega$$.

The ground state can be found by assuming that the lowering operator possesses a nontrivial kernel: $$a\, \psi_0 = 0$$ with $$\psi_0\ne0$$. Application of the above formula for the Hamiltonian yields
 * $$0=\hbar \omega\, a^\dagger a \psi_0 = \left(\hat H - \frac{\hbar \omega}{2} \right) \,\psi_0.$$

So $$\psi_0$$ is an eigenfunction of the Hamiltonian.

This gives the ground state energy $$E_0 = \hbar \omega /2$$, which allows one to identify the energy eigenvalue of any eigenstate $$\psi_n$$ as
 * $$E_n = \left(n + \frac{1}{2}\right)\hbar \omega.$$

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

Consequently,
 * $$\hbar\omega \,\left(N+\frac{1}{2}\right)\,\psi (q) =E\,\psi (q)~.$$

The is then
 * $$U(t)=\exp ( -it \hat{H}/\hbar) = \exp (-it\omega (a^\dagger a+1/2)) ~,$$
 * $$= e^{-it \omega /2} ~ \sum_{k=0}^{\infty}   {(e^{-i\omega t}-1)^k \over k!}  a^{{\dagger} {k}} a^k   ~.$$

Explicit eigenfunctions
The ground state $$\ \psi_0(q)$$ of the can be found by imposing the condition that
 * $$ a \ \psi_0(q) = 0.$$

Written out as a differential equation, the wavefunction satisfies
 * $$q \psi_0 + \frac{d\psi_0}{dq} = 0$$

with the solution
 * $$\psi_0(q) = C \exp\left(-{q^2 \over 2}\right).$$

The normalization constant $C$ is found to be $$1/ \sqrt[4]{\pi}$$  from $$\int_{-\infty}^\infty \psi_0^* \psi_0 \,dq = 1$$,  using the. Explicit formulas for all the eigenfunctions can now be found by repeated application of $$ a^\dagger$$ to $$ \psi_0$$.

Matrix representation
The matrix expression of the creation and annihilation operators of the quantum harmonic oscillator with respect to the above orthonormal basis is
 * $$ 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 & \dots & \sqrt{n} &\dots & \\ \vdots & \vdots & \vdots & \vdots & \vdots &\ddots \end{pmatrix} $$


 * $$ 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}$$

These can be obtained via the relationships $$a^\dagger_{ij} = \langle\psi_i \mid a^\dagger \mid \psi_j\rangle$$ and $$a_{ij} = \langle\psi_i \mid a \mid \psi_j\rangle$$. The eigenvectors $$\psi_i$$ are those of the quantum harmonic oscillator, and are sometimes called the "number basis".

Generalized creation and annihilation operators
The operators derived above are actually a specific instance of a more generalized notion of creation and annihilation operators. The more abstract form of the operators are constructed as follows. Let $$H$$ be a one-particle (that is, any Hilbert space, viewed as representing the state of a single particle).

The (ic) over $$H$$ is the algebra-with-conjugation-operator (called *) abstractly generated by elements $$a(f)$$, where $$f\,$$ranges freely over $$H$$, subject to the relations


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

in.

The map $$a: f \to a(f)$$ from $$H$$ to the bosonic CCR algebra is required to be complex (this adds more relations). Its is $$a^\dagger(f)$$, and the map $$f\to a^\dagger(f)$$ is  in $H$. Thus $$H$$ embeds as a complex vector subspace of its own CCR algebra. In a representation of this algebra, the element $$a(f)$$ will be realized as an annihilation operator, and $$a^\dagger(f)$$ as a creation operator.

In general, the CCR algebra is infinite dimensional. If we take a Banach space completion, it becomes a. The CCR algebra over $$H$$ is closely related to, but not identical to, a.

For fermions, the (fermionic) over $$H$$ is constructed similarly, but using  relations instead, namely


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

The CAR algebra is finite dimensional only if $$H$$ is finite dimensional. If we take a Banach space completion (only necessary in the infinite dimensional case), it becomes a $$C^*$$ algebra. The CAR algebra is closely related to, but not identical to, a.

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

The  is the state $$ with no particles, characterized by
 * $$a(f) \left| 0\right\rangle=0.$$

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

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: $$A+A\to \empty$$. 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 one dimensional lattice. Each particle moves to the right or left with a certain probability, and each pair of particles at the same site annihilates each other with a certain other probability.

The probability that one particle leaves the site during the short time period $dt$ is proportional to $$n_i \, dt$$, let us say a probability $$\alpha n_{i}dt$$ to hop left and $$\alpha n_i \, dt$$ to hop right. All $$n_i$$ particles will stay put with a probability $$1-2\alpha n_i \, dt$$. (Since $dt$ is so short, the probability that two or more will leave during $dt$ is very small and will be ignored.)

We can now describe the occupation of particles on the lattice as a `ket' of the form

$$|\dots, n_{-1}, n_0, n_1, \dots\rangle$$. It represents the juxtaposition (or conjunction, or tensor product) of the number states $$\dots, |n_{-1}\rangle$$ $$|n_{0}\rangle$$,  $$|n_{1}\rangle, \dots$$ located at the individual sites of the lattice. Recall


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

$$ a^\dagger$$and
 * $$ a^\dagger \mid\! n\rangle= \sqrt{n+1}\mid\! n+1\rangle,$$

for all $n$ ≥ 0, while
 * $$[a,a^{\dagger}]=1$$

This definition of the operators will now be changed to accommodate the "non-quantum" nature of this problem and we shall use the following definition:

$$a\mid\! n\rangle = n \ |n-1\rangle$$

$$ a^\dagger \mid\! n\rangle=\ \mid n+1\rangle$$

note that even though the behavior of the operators on the kets has been modified, these operators still obey the commutation relation

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

Now define $$ a_i$$ so that it applies $$ a$$ to  $$ |n_i\rangle$$. Correspondingly, define $$ a^\dagger_i$$ as applying $$ a^\dagger$$ to   $$ |n_i\rangle$$. Thus, for example, the net effect of $$ a_{i-1} a^\dagger_i$$ is to move a particle from the $$(i-1)^{\text{th}}$$ to the $$i^{\text{th}}$$site while multiplying with the appropriate factor.

This allows  writing the pure diffusive behavior of the particles as
 * $$\partial_{t}\mid\! \psi\rangle=-\alpha\sum(2a_i^\dagger a_i-a_{i-1}^\dagger a_i-a_{i+1}^\dagger a_i)\mid\!\psi\rangle=-\alpha\sum(a_i^\dagger-a_{i-1}^\dagger)(a_i-a_{i-1}) \mid\! \psi\rangle, $$

where the sum is over $$i$$.

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 $$\lambda n(n-1)dt$$, yielding a term
 * $$\lambda\sum(a_i a_i-a_i^\dagger a_i^\dagger a_i a_i)$$

where number state n is replaced by number state n − 2 at site $$i$$ at a certain rate.

Thus the state evolves by
 * $$\partial_t\mid\!\psi\rangle=-\alpha\sum(a_i^\dagger-a_{i-1}^\dagger)(a_i-a_{i-1}) \mid\!\psi\rangle+\lambda\sum(a_i^2-a_i^{\dagger 2}a_i^2)\mid\!\psi\rangle $$

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 and s one works with creation and annihilation operators of quantum states, $$a^\dagger_i$$ and $$a^{\,}_i$$. These operators change the eigenvalues of the ,
 * $$N = \sum_i n_i = \sum_i a^\dagger_i a^{\,}_i$$,

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

The commutation relations of creation and annihilation operators in a multiple- 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 and $$\delta_{i j}$$ is the.

For s, the commutator is replaced by the $$\{\ \, \ \ \}$$,
 * $$\{a^{\,}_i, a^\dagger_j\} \equiv a^{\,}_i a^\dagger_j +a^\dagger_j a^{\,}_i = \delta_{i j},$$
 * $$\{a^\dagger_i, a^\dagger_j\} = \{a^{\,}_i, a^{\,}_j\} = 0.$$

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

If the states labelled by i are an orthonormal basis of a Hilbert space H, then the result of this construction coincides with the CCR algebra and CAR algebra construction in the previous section but one. If they represent "eigenvectors" corresponding to the continuous spectrum of some operator, as for unbound particles in QFT, then the interpretation is more subtle.

Normalization
While Zee obtains the normalization $$[\hat a_{\mathbf p},\hat a_{\mathbf q}^\dagger] = \delta(\mathbf{p} - \mathbf{q})$$ via the  for Fourier transforms, Tong and Peskin & Schroeder use the common asymmetric convention to obtain $$[\hat a_{\mathbf p},\hat a_{\mathbf q}^\dagger] = (2\pi)^3\delta(\mathbf{p} - \mathbf{q})$$. Each derives $$[\hat \phi(\mathbf x), \hat \pi(\mathbf x')] = i\delta(\mathbf x - \mathbf x')$$.

Srednicki additionally merges the Lorentz-invariant measure into his asymmetric Fourier measure, $$\tilde{dk}=\frac{d^3k}{(2\pi)^3 2\omega}$$, yielding $$[\hat a_{\mathbf k},\hat a_{\mathbf k'}^\dagger] = (2\pi)^3 2\omega\,\delta(\mathbf{k} - \mathbf{k}')$$.