Creation and annihilation operator

Occupation number is a redirect to this article. For the use of the term in statistics, see classification (statistics)

The creation and annihilation operators are the core of a possible solution to the Schrödinger equation of the harmonic oscillator . These operators can also be used to solve problems with quantum mechanical angular momentum more easily (see angular momentum operator ). The creation and annihilation operators are also used in the quantization of fields (the so-called second quantization or occupation number representation).

There are a variety of alternative names such as ladder operators , climbing operators , upgrade - and lowering operators and lifting - and Senkoperatoren . Instead of the "creation operator", the creation operator is sometimes used. In the German-speaking world, the operators and , which change the states of an atom, are also called creation and annihilation operators. ${\displaystyle \sigma _{+}}$ ${\displaystyle \sigma _{-}}$

The problem of the harmonic oscillator in quantum mechanics can be treated using the method of creation and annihilation operators, which is also called the algebraic method . It was mainly developed by Paul Dirac . For this approach, two operators and are defined , which each add or remove an energy quantum from an oscillator . That is why they are called annihilation and creation operators. ${\displaystyle {\hat {a}}}$ ${\displaystyle {\hat {a}}^{\dagger }}$ ${\displaystyle \hbar \omega }$

The circumflex ( symbol) above the symbolizes that this is an operator. This means that the same calculation rules do not apply as for scalars , because the order of operators, for example, generally cannot be exchanged. In the following, the circumflex symbol is omitted for the sake of clarity, unless otherwise stated. All Latin capital letters, with the exception of representing the energy eigenvalue, are operators. ${\displaystyle {\hat {}}}$ ${\displaystyle a}$ ${\displaystyle E_{n}}$

definition

The creation operator and the annihilation operator adjoint to it are defined using the following commutation relations with the occupation number operator (which is also called the particle number operator ) : ${\displaystyle a^{\dagger \,}}$ ${\displaystyle a^{\,}}$ ${\displaystyle N:=a^{\dagger }a}$

${\displaystyle [N,a]=-a\ ,\quad [N,a^{\dagger }]=a^{\dagger }}$

The occupation number operator is a Hermitian operator and therefore has real eigenvalues . The associated eigenvalue equation is: ${\displaystyle N}$ ${\displaystyle n}$

${\displaystyle N\left|n\right\rangle =n\left|n\right\rangle }$

where are Fock states . ${\displaystyle \left|n\right\rangle }$

The occupation number is a nonnegative integer, so . In the case of fermions , there is still a restriction to the values ​​0 and 1. ${\displaystyle n}$ ${\displaystyle n\in \mathbb {N} _{0}}$

By applying or to the state , the state above or below is obtained: ${\displaystyle a^{\dagger }}$ ${\displaystyle a^{\,}}$ ${\displaystyle \left|n\right\rangle }$

${\displaystyle a^{\dagger }\left|n\right\rangle =c_{n}^{+}\left|n+1\right\rangle }$

${\displaystyle a\,\left|n\right\rangle =c_{n}^{-}\left|n-1\right\rangle }$

The constants and depend on whether and satisfy the commutator or anti-commutator exchange relation. ${\displaystyle c_{n}^{+}}$ ${\displaystyle c_{n}^{-}}$ ${\displaystyle a^{\,}}$ ${\displaystyle a^{\dagger }}$

Details

Various properties of are derived below . The eigen-states are normalized. ${\displaystyle N}$ ${\displaystyle \left|n\right\rangle }$

• The occupation number operator is Hermitian, i.e. self-adjoint:

${\displaystyle N^{\dagger }=(a^{\dagger }a)^{\dagger }=a^{\dagger }a^{\dagger \dagger }=a^{\dagger }a=N}$

Thus has real eigenvalues, the occupation numbers . ${\displaystyle N}$ ${\displaystyle n}$

• The eigenvalues ​​are not negative: ${\displaystyle n\geq 0}$

${\displaystyle n=\left\langle n|N|n\right\rangle =\left\langle n|a^{\dagger }a|n\right\rangle =\left\langle an|an\right\rangle =\left\|a\left|n\right\rangle \right\|^{2}\geq 0}$

The inequality follows from the fact that the norm of a vector is non-negative.

• The smallest eigenvalue is 0

The state is a vector in Hilbert space and must not be confused with the zero vector , but is simply numbered with the number 0. ${\displaystyle \left|0\right\rangle }$

${\displaystyle a^{\dagger }a\left|0\right\rangle =0\left|0\right\rangle =0}$   and   ${\displaystyle \left\langle 0|0\right\rangle =1}$

Because must apply: . If one applies the descent operator to the lowest state, one obtains the zero vector. However, this cannot be reversed: By applying to the zero vector, one does not get the ground state, but the zero vector again. ${\displaystyle n\geq 0}$ ${\displaystyle a\left|0\right\rangle =0}$ ${\displaystyle a^{\dagger }}$

• The eigenvalues ​​are integer: ${\displaystyle n\in \mathbb {N} _{0}}$

Assuming the eigenvalues ​​are not integers, then, starting from an eigenstate, eigenstates with negative eigenvalues ​​can be found by repeatedly applying the relegation operator. But this is a contradiction to the condition . In the case of integer eigenvalues, one reaches the ground state at some point and, by applying it again, the zero vector; from here the ladder breaks off automatically. ${\displaystyle n\geq 0}$

• Is eigenvalue, then too ${\displaystyle n}$ ${\displaystyle n+1}$

${\displaystyle Na^{\dagger }\left|n\right\rangle =\left(a^{\dagger }N+[N,a^{\dagger }]\right)\left|n\right\rangle =\left(a^{\dagger }N+a^{\dagger }\right)\left|n\right\rangle =a^{\dagger }\left(N+1\right)\left|n\right\rangle =a^{\dagger }\left(n+1\right)\left|n\right\rangle =\left(n+1\right)a^{\dagger }\left|n\right\rangle }$

If is not equal to the zero vector, a new eigenvalue is obtained . ${\displaystyle a^{\dagger }\left|n\right\rangle }$ ${\displaystyle (n+1)}$

${\displaystyle a^{\dagger }\left|n\right\rangle }$is thus eigenstate to with eigenvalue and thus proportional to : ${\displaystyle N}$ ${\displaystyle (n+1)}$ ${\displaystyle \left|n+1\right\rangle }$

${\displaystyle a^{\dagger }\left|n\right\rangle =c_{n}^{+}\left|n+1\right\rangle }$

• Is eigenvalue, then too ${\displaystyle n>0}$ ${\displaystyle n-1}$

${\displaystyle Na\left|n\right\rangle =\left(aN+\left[N,a\right]\right)\left|n\right\rangle =\left(aN-a\right)\left|n\right\rangle =a\left(N-1\right)\left|n\right\rangle =a\left(n-1\right)\left|n\right\rangle =\left(n-1\right)a\left|n\right\rangle }$

If is not equal to the zero vector, a new eigenvalue is obtained . ${\displaystyle a\left|n\right\rangle }$ ${\displaystyle (n-1)}$

${\displaystyle a^{\,}\left|n\right\rangle }$is thus eigenstate to with eigenvalue and thus proportional to : ${\displaystyle N}$ ${\displaystyle (n-1)}$ ${\displaystyle \left|n-1\right\rangle }$

${\displaystyle a\,\left|n\right\rangle =c_{n}^{-}\left|n-1\right\rangle }$

Bosonic climbing operators

In the bosonic case meet and the commutator -Vertauschungsrelationen: ${\displaystyle a}$ ${\displaystyle a^{\dagger }}$

${\displaystyle [a,a^{\dagger }]=1\ ,\quad [a,a]=[a^{\dagger },a^{\dagger }]=0}$

Consequently

${\displaystyle a^{\dagger }\left|n\right\rangle ={\sqrt {n+1}}\left|n+1\right\rangle }$

${\displaystyle a\,\left|n\right\rangle ={\sqrt {n}}\left|n-1\right\rangle }$

In the bosonic case, the occupation numbers can be arbitrarily large: . ${\displaystyle n}$ ${\displaystyle n\in \mathbb {N} _{0}}$

Details

• First of all, it must be checked whether the above requirements are met:

${\displaystyle [N,a]=[a^{\dagger }a,a]=\underbrace {[a^{\dagger },a]} _{-1}a+a^{\dagger }\underbrace {[a,a]} _{0}=-a}$

${\displaystyle [N,a^{\dagger }]=[a^{\dagger }a,a^{\dagger }]=\underbrace {[a^{\dagger },a^{\dagger }]} _{0}a+a^{\dagger }\underbrace {[a,a^{\dagger }]} _{1}=a^{\dagger }}$

• The next overlying state can be constructed with . The factor results from the following calculation with the commutator : ${\displaystyle a^{\dagger }}$ ${\displaystyle \left|n\right\rangle }$ ${\displaystyle a^{\dagger }\left|n\right\rangle =c\left|n+1\right\rangle }$ ${\displaystyle c}$ ${\displaystyle aa^{\dagger }-a^{\dagger }a=1}$

${\displaystyle \left|c\right|^{2}=\left\Vert c\left|n+1\right\rangle \right\Vert ^{2}=\left\Vert a^{\dagger }\left|n\right\rangle \right\Vert ^{2}=\left\langle a^{\dagger }n|a^{\dagger }n\right\rangle =\left\langle n|aa^{\dagger }|n\right\rangle =\left\langle n|a^{\dagger }a+1|n\right\rangle =\left\langle n|N+1|n\right\rangle =n+1}$

${\displaystyle c={\sqrt {n+1}}e^{i\varphi }}$, but the phase can be neglected, so . ${\displaystyle \varphi }$ ${\displaystyle c={\sqrt {n+1}}}$

• With can the under construct lying state . The factor results from the following calculation: ${\displaystyle a}$ ${\displaystyle \left|n\right\rangle }$ ${\displaystyle a\left|n\right\rangle =c\left|n-1\right\rangle }$ ${\displaystyle c}$

${\displaystyle \left|c\right|^{2}=\left\Vert c\left|n-1\right\rangle \right\Vert ^{2}=\left\Vert a\left|n\right\rangle \right\Vert ^{2}=\left\langle an|an\right\rangle =\left\langle n|a^{\dagger }a|n\right\rangle =\left\langle n|N|n\right\rangle =n\quad \Rightarrow \quad c={\sqrt {n}}}$

• All eigen-states can be constructed starting from the ground state:

${\displaystyle \left|n\right\rangle ={\frac {1}{\sqrt {n}}}a^{\dagger }\left|n-1\right\rangle ={\frac {1}{\sqrt {n!}}}\left(a^{\dagger }\right)^{n}\left|0\right\rangle \ ,\quad n\in \mathbb {N} _{0}}$

In this way one obtains a complete discrete set of eigen-states

Fermionic climbing operators

In the fermionic case and satisfy the anti-commutator commutation relations: ${\displaystyle \,a\,}$ ${\displaystyle a^{\dagger }}$

${\displaystyle \{a,a^{\dagger }\}=1\ ,\quad \{a,a\}=\{a^{\dagger },a^{\dagger }\}=0\quad \Rightarrow \quad a^{2}=a^{\dagger \,2}=0}$

Consequently

${\displaystyle a^{\dagger }\left|n\right\rangle =\left(1-n\right)\left|n+1\right\rangle }$

${\displaystyle a\,\left|n\right\rangle =n\left|n-1\right\rangle }$

In the fermionic case, the occupation numbers can only assume the values ​​0 or 1. ${\displaystyle n}$

Details

• With and is : ${\displaystyle a\,a=0}$ ${\displaystyle a^{\dagger }a^{\dagger }=0}$ ${\displaystyle N^{2}=N}$

${\displaystyle N^{2}=a^{\dagger }aa^{\dagger }a=a^{\dagger }(\underbrace {\{a,a^{\dagger }\}} _{1}-a^{\dagger }a)a=a^{\dagger }a-\underbrace {a^{\dagger }a^{\dagger }} _{0}\underbrace {aa} _{0}=a^{\dagger }a=N}$

The occupation number operator has only the eigenvalues ​​0 and 1 and the eigenstates and : ${\displaystyle |0\rangle }$ ${\displaystyle |1\rangle }$

${\displaystyle n^{2}\left|n\right\rangle =N^{2}\left|n\right\rangle =N\left|n\right\rangle =n\left|n\right\rangle \quad \Rightarrow \quad n^{2}=n\quad \Rightarrow \quad n\in \{0,1\}}$

• First of all, it must be checked whether the above requirements are met:

${\displaystyle [N,a]=[a^{\dagger }a,a]=\underbrace {[a^{\dagger },a]} _{2a^{\dagger }a-\{a,a^{\dagger }\}}a+a^{\dagger }\underbrace {[a,a]} _{0}=2a^{\dagger }\underbrace {aa} _{0}-\underbrace {\{a,a^{\dagger }\}} _{1}a=-a}$

${\displaystyle [N,a^{\dagger }]=[a^{\dagger }a,a^{\dagger }]=\underbrace {[a^{\dagger },a^{\dagger }]} _{0}a+a^{\dagger }\underbrace {[a,a^{\dagger }]} _{\{a,a^{\dagger }\}-2a^{\dagger }a}=a^{\dagger }\underbrace {\{a,a^{\dagger }\}} _{1}-2\underbrace {a^{\dagger }a^{\dagger }} _{0}a=a^{\dagger }}$

• The next overlying state can be constructed with . The factor results from the following calculation with the anti-commutator : ${\displaystyle a^{\dagger }}$ ${\displaystyle \left|n\right\rangle }$ ${\displaystyle a^{\dagger }\left|n\right\rangle =c\left|n+1\right\rangle }$ ${\displaystyle c}$ ${\displaystyle aa^{\dagger }+a^{\dagger }a=1}$

${\displaystyle {\begin{aligned}&\left|c\right|^{2}=\left\Vert c\left|n+1\right\rangle \right\Vert ^{2}=\left\Vert a^{\dagger }\left|n\right\rangle \right\Vert ^{2}=\left\langle a^{\dagger }n|a^{\dagger }n\right\rangle =\left\langle n|aa^{\dagger }|n\right\rangle =\left\langle n|1-a^{\dagger }a|n\right\rangle =\left\langle n|1-N|n\right\rangle =1-n\\&\Rightarrow \quad c={\sqrt {1-n}}\end{aligned}}}$

Since there can only be 0 or 1, is (where is the Kronecker delta ). ${\displaystyle n}$ ${\displaystyle c={\sqrt {1-n}}=1-n=\delta _{0,n}}$ ${\displaystyle \delta _{i,j}}$

• With can the under construct lying state . The factor results from the following calculation: ${\displaystyle a}$ ${\displaystyle \left|n\right\rangle }$ ${\displaystyle a\left|n\right\rangle =c\left|n-1\right\rangle }$ ${\displaystyle c}$

${\displaystyle \left|c\right|^{2}=\left\Vert c\left|n-1\right\rangle \right\Vert ^{2}=\left\Vert a\left|n\right\rangle \right\Vert ^{2}=\left\langle an|an\right\rangle =\left\langle n|a^{\dagger }a|n\right\rangle =\left\langle n|N|n\right\rangle =n\quad \Rightarrow \quad c={\sqrt {n}}}$

Since can only be 0 or 1, is . ${\displaystyle n}$ ${\displaystyle c={\sqrt {n}}=n=\delta _{1,n}}$

• All eigen-states can be constructed starting from the ground state:

${\displaystyle \left|n\right\rangle ={\frac {1}{\sqrt {n}}}a^{\dagger }\left|n-1\right\rangle ={\frac {1}{\sqrt {n!}}}\left(a^{\dagger }\right)^{n}\left|0\right\rangle \ ,\quad n\in \{0,1\}}$

In this way one obtains a complete discrete set of eigen-states

Example of boson climbing operators: Harmonic oscillator

The Hamilton operator of the harmonic oscillator is ${\displaystyle H}$

${\displaystyle H={\frac {P^{2}}{2m}}+{\frac {m\omega ^{2}Q^{2}}{2}}}$

${\displaystyle P}$ Momentum operator , position operator , mass , natural frequency ${\displaystyle Q}$ ${\displaystyle m}$ ${\displaystyle \omega }$

In the following, the stationary Schrödinger equation has to be solved:

${\displaystyle H\left|n\right\rangle =E_{n}\left|n\right\rangle }$

${\displaystyle E_{n}}$Energy eigenwert , energy own state ${\displaystyle \left|n\right\rangle }$

Reshape the Hamilton operator

The Hamilton operator can be transformed:

${\displaystyle H={\frac {P^{2}}{2m}}+{\frac {m\omega ^{2}Q^{2}}{2}}=\hbar \omega \left({\frac {P^{2}}{2\hbar m\omega }}+{\frac {m\omega Q^{2}}{2\hbar }}\right)}$

Two new operators are defined:

${\displaystyle {\tilde {P}}:={\frac {P}{\sqrt {2\hbar m\omega }}}}$   and   ${\displaystyle {\tilde {Q}}:={\sqrt {\frac {m\omega }{2\hbar }}}Q}$

The Hamilton operator expressed with the new operators:

${\displaystyle H=\hbar \omega \left({\tilde {P}}^{2}+{\tilde {Q}}^{2}\right)}$

You now try to write the content of the bracket as a product, so ( is the imaginary unit ) ${\displaystyle i}$

${\displaystyle \left(u-iv\right)\left(u+iv\right)=u^{2}+v^{2}+iuv-ivu=u^{2}+v^{2}}$

But since and are operators that do not interchange , the last equal sign does not apply here. ${\displaystyle u}$ ${\displaystyle v}$

The commutator is required to swap two operators : ${\displaystyle {\tilde {Q}}{\tilde {P}}={\tilde {P}}{\tilde {Q}}-\left[{\tilde {P}},{\tilde {Q}}\right]}$

${\displaystyle {\begin{array}{rcl}H&=&\hbar \omega \left({\tilde {Q}}^{2}+{\tilde {P}}^{2}\right)\\&=&\hbar \omega \left({\tilde {Q}}^{2}+{\tilde {P}}^{2}+i\left[{\tilde {Q}},{\tilde {P}}\right]-i\left[{\tilde {Q}},{\tilde {P}}\right]\right)\\&=&\hbar \omega \left({\tilde {Q}}^{2}+{\tilde {P}}^{2}+i{\tilde {Q}}{\tilde {P}}-i{\tilde {P}}{\tilde {Q}}-i\left[{\tilde {Q}},{\tilde {P}}\right]\right)\\&=&\hbar \omega \left(({\tilde {Q}}-i{\tilde {P}})({\tilde {Q}}+i{\tilde {P}})-i\left[{\tilde {Q}},{\tilde {P}}\right]\right)\end{array}}}$

The commutator can be traced back to the commutator of the original operators and : ${\displaystyle \left[{\tilde {Q}},{\tilde {P}}\right]}$ ${\displaystyle Q}$ ${\displaystyle P}$

${\displaystyle \left[{\tilde {Q}},{\tilde {P}}\right]={\sqrt {\frac {m\omega }{2\hbar }}}{\frac {1}{\sqrt {2\hbar m\omega }}}\underbrace {\left[Q,P\right]} _{i\hbar }={\frac {i}{2}}}$

The Hamilton operator looks like this:

${\displaystyle H=\hbar \omega \underbrace {\left({\tilde {Q}}-i{\tilde {P}}\right)} _{a^{\dagger }}\underbrace {\left({\tilde {Q}}+i{\tilde {P}}\right)} _{a}+{\frac {1}{2}}\hbar \omega }$

Now the two ladder operators are defined:

${\displaystyle a^{\dagger }:={\tilde {Q}}-i{\tilde {P}}}$   Creation operator

${\displaystyle a:={\tilde {Q}}+i{\tilde {P}}}$   Annihilation operator

They are also often written as and . Note that the ladder operators are not Hermitian , there . ${\displaystyle a_{+}}$ ${\displaystyle a_{-}}$ ${\displaystyle a\neq a^{\dagger }}$

The ladder operators expressed by the position operator and the momentum operator : ${\displaystyle Q}$ ${\displaystyle P}$

${\displaystyle a^{\dagger }={\sqrt {\frac {m\omega }{2\hbar }}}\left(Q-{\frac {i}{m\omega }}P\right)}$

${\displaystyle a={\sqrt {\frac {m\omega }{2\hbar }}}\left(Q+{\frac {i}{m\omega }}P\right)}$

Conversely, we get for and : ${\displaystyle Q}$ ${\displaystyle P}$

${\displaystyle Q={\sqrt {\frac {\hbar }{2m\omega }}}(a+a^{\dagger })}$

${\displaystyle P={\frac {1}{i}}{\sqrt {\frac {m\omega \hbar }{2}}}(a-a^{\dagger })}$

With the ladder operators, the Hamilton operator is written:

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

Characteristics of the producers and destroyers

The commutator still has to be determined from the two ladder operators:

${\displaystyle \left[a,a^{\dagger }\right]=aa^{\dagger }-a^{\dagger }a=\left({\tilde {Q}}+i{\tilde {P}}\right)\left({\tilde {Q}}-i{\tilde {P}}\right)-\left({\tilde {Q}}-i{\tilde {P}}\right)\left({\tilde {Q}}+i{\tilde {P}}\right)=2i\left({\tilde {P}}{\tilde {Q}}-{\tilde {Q}}{\tilde {P}}\right)=2i\underbrace {\left[{\tilde {P}},{\tilde {Q}}\right]} _{-i/2}=1}$

Since it is also true that the climbing operators of the harmonic oscillator are bosonic climbing operators . Thus, all of the above properties apply to boson climbing operators. ${\displaystyle [a,a]=[a^{\dagger },a^{\dagger }]=0}$

The product defines the occupation number operator : ${\displaystyle a^{\dagger }a}$

${\displaystyle N=a^{\dagger }a}$

Solution of the eigenvalue problem

The Hamilton operator can be expressed by the occupation number operator:

${\displaystyle H=\hbar \omega \left(N+{\frac {1}{2}}\right)}$

The eigenvalue problem can be reduced to the eigenvalue equation of the occupation number operator. ${\displaystyle H\left|n\right\rangle =E_{n}\left|n\right\rangle }$ ${\displaystyle N\left|n\right\rangle =n\left|n\right\rangle }$

${\displaystyle \hbar \omega \left(N+{\frac {1}{2}}\right)\left|n\right\rangle =E_{n}\left|n\right\rangle \quad ,\quad \hbar \omega \left(N+{\frac {1}{2}}\right)\left|n\right\rangle =\hbar \omega \left(n+{\frac {1}{2}}\right)\left|n\right\rangle }$

The eigen-states of are also eigen-states of , da . The eigenvalues ​​of the Hamilton operator result from the eigenvalues ​​of the occupation number operator : ${\displaystyle N}$ ${\displaystyle H}$ ${\displaystyle \left[H,N\right]=0}$ ${\displaystyle N}$

${\displaystyle E_{n}=\hbar \omega \left(n+{\frac {1}{2}}\right)}$   and   ${\displaystyle \left|n\right\rangle =\left|n\right\rangle }$

A particularly important property of climbing operators is this:

${\displaystyle a^{\dagger }\left|n\right\rangle ={\sqrt {n+1}}\left|n+1\right\rangle }$

${\displaystyle a\,\left|n\right\rangle ={\sqrt {n}}\left|n-1\right\rangle }$

If the Schrödinger equation is a solution for the energy , then there is a solution for the energy and a solution for the energy . This means that all solutions can be obtained from a solution by simply applying the creation or annihilation operator to that solution. This creates a new solution for the neighboring energy level that is shifted by the energy . ${\displaystyle \left|n\right\rangle }$ ${\displaystyle E_{n}}$ ${\displaystyle a^{\dagger }\left|n\right\rangle }$ ${\displaystyle E_{n}+\hbar \omega }$ ${\displaystyle a\left|n\right\rangle }$ ${\displaystyle E_{n}-\hbar \omega }$ ${\displaystyle \hbar \omega }$

Since the occupation number operator has no negative eigenvalues, no negative energy eigenvalues ​​can exist either. So there is a solution for the minimum occupation number that sits at a minimum energy level ( zero point energy ): ${\displaystyle n=0}$ ${\displaystyle |0\rangle }$

${\displaystyle E_{0}={\frac {1}{2}}\hbar \omega }$

In the state , the energy is composed of the zero point energy and energy quanta of size . The effect of transforms the system into a state with increased energy. This can be interpreted as the creation of an additional energy quantum, which makes the name creation operator understandable. Analogously, the operator converts the system into a state reduced by one energy quantum. So an energy quantum is destroyed, hence the annihilation operator . The eigenvalues ​​of the operator indicate how many energy quanta are excited in an eigenstate. The occupation of a state with energy quanta explains the name occupation number operator . ${\displaystyle \left|n\right\rangle }$ ${\displaystyle E_{n}=\hbar \omega (n+{\tfrac {1}{2}})}$ ${\displaystyle \hbar \omega /2}$ ${\displaystyle n}$ ${\displaystyle \hbar \omega }$ ${\displaystyle a^{\dagger }}$ ${\displaystyle \hbar \omega }$ ${\displaystyle a}$ ${\displaystyle N}$ ${\displaystyle n}$

Eigenfunctions in position representation

If one applies the descent operator to the lowest state, one obtains the zero vector . However, this cannot be reversed: By applying to the zero vector, one does not get the ground state, but the zero vector again . This gives an equation for the ground state : ${\displaystyle a\left|0\right\rangle =0}$ ${\displaystyle a^{\dagger }}$ ${\displaystyle a^{\dagger }a\left|0\right\rangle =0}$

${\displaystyle 0=a\left|0\right\rangle =\left({\tilde {Q}}+i{\tilde {P}}\right)\left|0\right\rangle =\left({\sqrt {\frac {m\omega }{2\hbar }}}Q+i{\frac {P}{\sqrt {2\hbar m\omega }}}\right)\left|0\right\rangle }$

In the position representation, the above operator equation can be represented and solved as a differential equation: and ${\displaystyle \langle x|P|x\rangle =-i\hbar {\frac {\mathrm {d} }{\mathrm {d} x}}}$ ${\displaystyle \langle x|Q|x\rangle =x}$

${\displaystyle \left({\sqrt {\frac {m\omega }{2\hbar }}}x+{\sqrt {\frac {\hbar }{2m\omega }}}{\frac {\mathrm {d} }{\mathrm {d} x}}\right)\langle x|0\rangle =0}$   returns normalized   ${\displaystyle \langle x|0\rangle =\left({\frac {m\omega }{\pi \hbar }}\right)^{\frac {1}{4}}\exp \left(-{\frac {m\omega }{2\hbar }}x^{2}\right)}$

By applying the ascending operator to the solution of the ground state, all higher eigenfunctions are obtained:

${\displaystyle \left|n\right\rangle ={\frac {1}{\sqrt {n}}}a^{\dagger }\left|n-1\right\rangle ={\frac {1}{\sqrt {n!}}}(a^{\dagger })^{n}\left|0\right\rangle }$

In the representation of the location you get:

${\displaystyle \langle x|n\rangle ={\frac {1}{\sqrt {n!}}}\left({\sqrt {\frac {m\omega }{2\hbar }}}x-{\sqrt {\frac {\hbar }{2m\omega }}}{\frac {\mathrm {d} }{\mathrm {d} x}}\right)^{n}\langle x|0\rangle }$

Matrix representation of bosonic climbing operators

The eigen-states of the occupation number operator form a complete orthonormal system . With the help of this Hilbert space basis , a matrix representation of the ladder operators is to be determined. Note that here all indices run from 0 (not from 1) to infinity. The eigen-states can be represented as vectors : ${\displaystyle \left|n\right\rangle }$

${\displaystyle \left|0\right\rangle =\left({\begin{matrix}1\\0\\0\\0\\\vdots \\\end{matrix}}\right),\quad \left|1\right\rangle =\left({\begin{matrix}0\\1\\0\\0\\\vdots \\\end{matrix}}\right),\quad \left|2\right\rangle =\left({\begin{matrix}0\\0\\1\\0\\\vdots \\\end{matrix}}\right)}$    etc.

The completeness of this basis provides a representation of the unit operator:

${\displaystyle \sum \limits _{n=0}^{\infty }{\left|n\right\rangle \left\langle n\right|}=1}$

Creation operator

A 1 (unit operator) is inserted before and after the creation operator:

${\displaystyle a^{\dagger }=\sum \limits _{m,n=0}^{\infty }{\left|m\right\rangle \underbrace {\left\langle m\right|a^{\dagger }\left|n\right\rangle } _{a_{mn}^{\dagger }}\left\langle n\right|}}$

The matrix element is calculated as

${\displaystyle a_{mn}^{\dagger }=\left\langle m\right|a^{\dagger }\left|n\right\rangle ={\sqrt {n+1}}\left\langle m|n+1\right\rangle ={\sqrt {n+1}}\ \delta _{m,n+1}}$

The creation operator represented by the basis vectors

${\displaystyle a^{\dagger }=\sum \limits _{n=0}^{\infty }{\left|n+1\right\rangle {\sqrt {n+1}}\left\langle n\right|}}$

This results in the matrix representation of the creation operator with respect to the occupation eigenbase (all unspecified elements are equal to 0):

${\displaystyle a^{\dagger }=\left({\begin{matrix}0&{}&{}&{}&{}\\{\sqrt {1}}&0&{}&{}&{}\\{}&{\sqrt {2}}&0&{}&{}\\{}&{}&{\sqrt {3}}&0&{}\\{}&{}&{}&\ddots &\ddots \\\end{matrix}}\right)}$

Annihilation operator

By analogous calculation we get for the annihilation operator:

${\displaystyle a=\sum \limits _{m,n=0}^{\infty }{\left|m\right\rangle \underbrace {\left\langle m\right|a\left|n\right\rangle } _{a_{mn}}\left\langle n\right|}=\sum \limits _{m,n=0}^{\infty }{\left|m\right\rangle {\sqrt {n}}\ \delta _{m+1,n}\left\langle n\right|}=\sum \limits _{n=0}^{\infty }{\left|n\right\rangle {\sqrt {n+1}}\ \left\langle n+1\right|}}$

The matrix element has already been used:

${\displaystyle a_{mn}=\left\langle m\right|a\left|n\right\rangle ={\sqrt {n}}\left\langle m|n-1\right\rangle ={\sqrt {n}}\ \delta _{m,n-1}={\sqrt {n}}\ \delta _{m+1,n}}$

Matrix representation of the annihilation operator with regard to the occupation own base:

${\displaystyle a=\left({\begin{matrix}0&{\sqrt {1}}&{}&{}&{}\\{}&0&{\sqrt {2}}&{}&{}\\{}&{}&0&{\sqrt {3}}&{}\\{}&{}&{}&0&\ddots \\{}&{}&{}&{}&\ddots \\\end{matrix}}\right)}$

You can see that the matrix is exactly the transpose of . This is understandable, since the two operators are adjoint to one another (= transposed + complex conjugate). ${\displaystyle a^{\dagger }}$ ${\displaystyle a}$

Simple example

Examples with orthonormal bases matrix forms of and ${\displaystyle \left|0\right\rangle ={\begin{pmatrix}1\\0\\0\\\vdots \end{pmatrix}},\ \left|1\right\rangle ={\begin{pmatrix}0\\1\\0\\\vdots \end{pmatrix}},\ \ldots }$ ${\displaystyle a}$ ${\displaystyle a^{\dagger }}$

${\displaystyle a={\begin{pmatrix}0&{\sqrt {1}}&0&\ldots \\0&0&{\sqrt {2}}&\ldots \\0&0&0&\ldots \\\vdots &\vdots &\vdots &\ddots \end{pmatrix}}}$

${\displaystyle a^{\dagger }={\begin{pmatrix}0&0&0&\ldots \\{\sqrt {1}}&0&0&\ldots \\0&{\sqrt {2}}&0&\ldots \\\vdots &\vdots &\vdots &\ddots \end{pmatrix}}}$

${\displaystyle a\left|1\right\rangle ={\begin{pmatrix}0&{\sqrt {1}}&0&\ldots \\0&0&{\sqrt {2}}&\ldots \\0&0&0&\ldots \\\vdots &\vdots &\vdots &\ddots \end{pmatrix}}{\begin{pmatrix}0\\1\\0\\\vdots \end{pmatrix}}={\begin{pmatrix}1\\0\\0\\\vdots \end{pmatrix}}=\left|0\right\rangle }$

${\displaystyle a^{\dagger }\left|0\right\rangle ={\begin{pmatrix}0&0&0&\ldots \\{\sqrt {1}}&0&0&\ldots \\0&{\sqrt {2}}&0&\ldots \\\vdots &\vdots &\vdots &\ddots \end{pmatrix}}{\begin{pmatrix}1\\0\\0\\\vdots \end{pmatrix}}={\begin{pmatrix}0\\1\\0\\\vdots \end{pmatrix}}=\left|1\right\rangle }$

(the same applies to the dual representation)

Occupation number operator

Matrix element of the occupation number operator with regard to the occupation eigen-basis:

${\displaystyle N_{mn}=\left\langle m\right|N\left|n\right\rangle =n\left\langle m|n\right\rangle =n\ \delta _{m,n}}$

alternatively with the ladder operators:

${\displaystyle N_{mn}=\sum \limits _{k=0}^{\infty }{a_{mk}^{\dagger }a_{kn}}=\sum \limits _{k=0}^{\infty }{\underbrace {{\sqrt {k+1}}\ \delta _{m,k+1}} _{a_{mk}^{\dagger }}\underbrace {{\sqrt {n}}\ \delta _{k+1,n}} _{a_{kn}}}=n\ \delta _{m,n}}$

Matrix representation of the staffing number operator with regard to the self-staffing base :

${\displaystyle N=\left({\begin{matrix}0&{}&{}&{}\\{}&1&{}&{}\\{}&{}&2&{}\\{}&{}&{}&\ddots \\\end{matrix}}\right)}$

Hamilton operator of the harmonic oscillator

Matrix element of the Hamilton operator for the harmonic oscillator with regard to the occupation eigen-basis or the energy eigen-basis:

${\displaystyle H_{mn}=\left\langle m\right|H\left|n\right\rangle =\left\langle m\right|\hbar \omega \left(N+{\frac {1}{2}}\right)\left|n\right\rangle =\hbar \omega \left(n+{\frac {1}{2}}\right)\left\langle m|n\right\rangle =\hbar \omega \left(n+{\frac {1}{2}}\right)\delta _{m,n}}$

Matrix representation of the Hamilton operator for the harmonic oscillator with regard to the occupation eigen-basis or the energy eigen-basis:

${\displaystyle H=\hbar \omega \left({\begin{matrix}{\frac {1}{2}}&{}&{}&{}\\{}&1+{\frac {1}{2}}&{}&{}\\{}&{}&2+{\frac {1}{2}}&{}\\{}&{}&{}&\ddots \\\end{matrix}}\right)}$

Since the operators and are Hermitian , it follows that the associated matrices are symmetric with respect to the eigenbases . ${\displaystyle N}$ ${\displaystyle H}$

Eigenstates of bosonic climbing operators ("coherent states")

The eigen-states of the annihilation operator are the coherent states . The annihilation operator  (for clarification, the symbols for the operators are explicitly reintroduced here) satisfies the following eigenvalue equation: ${\displaystyle \left|\alpha \right\rangle }$ ${\displaystyle {\hat {a}}}$ ${\displaystyle {\hat {}}}$

${\displaystyle {\hat {a}}\left|\alpha \right\rangle =\alpha \left|\alpha \right\rangle }$

For the creation operator, with a left-eigen state ( Bra- eigen state):

${\displaystyle \left\langle \alpha \right|{\hat {a}}^{\dagger }=\alpha ^{*}\left\langle \alpha \right|}$

The annihilation operator can - in contrast to the creation operator - have right- eigen-states ( Ket- eigen-states). The creation operator increases the minimum particle number of a state in Fock space by one; the resulting condition cannot be the original. In contrast, the annihilation operator reduces the maximum number of particles by one; but since a state in the Fock space can contain components of all particle numbers (including any number of particles), it is not forbidden to have eigen-states. These are the coherent states: ${\displaystyle {\hat {a}}}$ ${\displaystyle {\hat {a}}^{\dagger }}$ ${\displaystyle {\hat {a}}}$

The "coherent state" results as a certain linear combination of all states with a fixed number of particles according to the formula: ${\displaystyle |\alpha \rangle }$ ${\displaystyle |n\rangle \,,}$

${\displaystyle \left|\alpha \right\rangle :=e^{-{\frac {\left|\alpha \right|^{2}}{2}}}\sum \limits _{n=0}^{\infty }{\frac {\alpha ^{n}}{\sqrt {n!}}}\left|n\right\rangle =e^{-{\frac {\left|\alpha \right|^{2}}{2}}}\sum \limits _{n=0}^{\infty }{\frac {\alpha ^{n}}{\sqrt {n!}}}{\frac {\left({\hat {a}}^{\dagger }\right)^{n}}{\sqrt {n!}}}\left|0\right\rangle =e^{-{\frac {\left|\alpha \right|^{2}}{2}}}e^{\alpha {\hat {a}}^{\dagger }}\left|0\right\rangle }$

This state is therefore the eigenstate of the annihilation operator, specifically to the eigenvalue, while the associated creation operator only has left eigenstates. Here is a non-vanishing complex number that fully defines the coherent state and may also explicitly depend on time.  is the expected value of the occupation number of the coherent state. ${\displaystyle \alpha \,,}$ ${\displaystyle \alpha }$ ${\displaystyle \left|\alpha \right|^{2}}$

Coherent states (like the ground state of the harmonic oscillator) have minimal fuzziness and remain coherent with time evolution. They are the best way to describe the - generally explicitly time-dependent - electromagnetic wave of a laser mode (so-called Glauber states ).

Creation and annihilation operators in quantum field theories

In quantum field theory and many-particle physics , expressions of the form where the complex numbers are used, while the creation and annihilation operators represent: These increase or decrease the eigenvalues ​​of the number operator by 1, analogous to the harmonic oscillator. The indices take into account the degrees of freedom of spacetime and generally have several components. If the creation and annihilation operators depend on a continuous variable, instead of discrete quantum numbers, to write them as field operators , . The number operators are self-adjoint ( “Hermitian” ) and take on all non-negative integer values: The nontrivial commutation relations are, like with the harmonic oscillator: where [.,.] Represents the so-called commutator bracket, while the Kronecker symbol is. ${\displaystyle \gamma _{i}a_{i}^{\pm }\,,}$ ${\displaystyle \gamma _{i}}$ ${\displaystyle a_{i}^{\pm }}$ ${\displaystyle \sum a_{i}^{+}a_{i}^{-}}$ ${\displaystyle i}$ ${\displaystyle \phi ({\vec {x}})}$ ${\displaystyle \phi ^{+}({\vec {x}})}$ ${\displaystyle n_{i}=a_{i}^{+}a_{i}^{-}}$ ${\displaystyle n_{i}\in \{0,\,\,1,\,\,2,\,\,\dots \}\,.}$
${\displaystyle [a_{i}^{-},a_{j}^{+}]:=a_{i}^{-}a_{j}^{+}-a_{j}^{+}a_{i}^{-}=\delta _{ij}\,,}$ ${\displaystyle \delta _{ij}}$

The above applies to bosons , whereas for fermions the commutator has to be replaced by the anti-commutator . In the fermionic case, the consequence is that the number operators only have the eigenvalues ​​0 and 1. ${\displaystyle \{a_{i}^{-},a_{j}^{+}\}:=a_{i}^{-}a_{j}^{+}+a_{j}^{+}a_{i}^{-}=\delta _{ij}\,.}$ ${\displaystyle n_{i}=a_{i}^{+}a_{i}^{-}}$

Relation to diagram techniques

Concrete calculations using the creation or annihilation operators can usually be supported by diagram techniques (→  Feynman diagrams ). So you can z. B. illustrate three-particle interactions of the shape with three lines, of which the first two enter a vertex and are "destroyed" there, while a third line is "created" at this vertex and runs out of it. The energy and momentum rates must be explicitly taken into account in the associated rules. ${\displaystyle \gamma _{1,2;3}\,a_{1}^{-}a_{2}^{-}a_{3}^{+}}$

The specified term, which describes a so-called "confluence process", has at low temperatures i. A. lower probability than the inverse so-called "splitting process" that belongs to the adjoint term . This is because the transition rate corresponds to every creation operator , analogously to the harmonic oscillator , while   the term is missing for the associated annihilation operator . In this way, the last-mentioned terms are usually more important than the first-mentioned at low temperatures. ${\displaystyle \propto |\gamma _{1,\,2;\,3}|^{2}\cdot \{\langle n_{1}\rangle \,\langle n_{2}\rangle \,(1+\langle n_{3}\rangle \}\,,}$ ${\displaystyle \gamma _{1,2;3}^{*}\,a_{3}^{-}a_{2}^{+}a_{1}^{+}\,,}$ ${\displaystyle a_{i}^{+}}$ ${\displaystyle \propto |\gamma _{\,i,\dots }^{*}|^{2}\cdot (1+\langle n_{i}\rangle )}$ ${\displaystyle \propto 1}$

literature

• Cohen-Tannoudji, Diu, Laloë: Quantum Mechanics 1/2 . de Gruyter, Berlin
• Nolting: Basic course in theoretical physics. Vol. 5/1: Quantum Mechanics . Springer, Berlin

See also

• Jib room

supporting documents

1. ↑ See the chapter on time-dependent perturbation calculations in most of the standard textbooks on Quantum Mechanics II
2. ↑ See e.g. B. Hermann Haken : Quantum field theory of the solid , Teubner 1973, ISBN 3-519-03025-X