Editing Second quantization (section)

=== Boson creation and annihilation operators ===

The boson creation (resp. annihilation) operator is usually denoted as <math>b_{\alpha}^\dagger</math> (resp. <math>b_{\alpha}</math>). The creation operator <math>b_{\alpha}^\dagger</math> adds a boson to the single-particle state <math>|\alpha\rangle</math>, and the annihilation operator <math>b_{\alpha}</math> removes a boson from the single-particle state <math>|\alpha\rangle</math>. The creation and annihilation operators are Hermitian conjugate to each other, but neither of them are Hermitian operators (<math>b_\alpha\neq b_\alpha^\dagger</math>).

==== Definition ====

The boson creation (annihilation) operator is a linear operator, whose action on a ''N''-particle first-quantized wave function <math>\Psi</math> is defined as
:<math>b_\alpha^\dagger \Psi = \frac{1}{\sqrt{N+1}}\psi_\alpha\otimes_+\Psi,</math>
:<math>b_\alpha\Psi = \frac{1}{\sqrt{N}}\psi_\alpha\oslash_+\Psi,</math>
where <math>\psi_\alpha\otimes_+</math> inserts the single-particle state <math>\psi_\alpha</math> in <math>N+1</math> possible insertion positions symmetrically, and <math>\psi_\alpha\oslash_+</math> deletes the single-particle state <math>\psi_\alpha</math> from <math>N</math> possible deletion positions symmetrically.

===== Examples =====
Hereinafter the tensor symbol <math>\otimes</math> between single-particle states is omitted for simplicity. Take the state <math>|1_1,1_2\rangle=(\psi_1\psi_2+\psi_2\psi_1)/\sqrt{2}</math>, create one more boson on the state <math>\psi_1</math>,
:<math>\begin{array}{rl}b_1^\dagger|1_1,1_2\rangle=&\frac{1}{\sqrt{2}}(b_1^\dagger\psi_1\psi_2+b_1^\dagger\psi_2\psi_1)\\=&\frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{3}}\psi_1\otimes_+\psi_1\psi_2+\frac{1}{\sqrt{3}}\psi_1\otimes_+\psi_2\psi_1\right)\\=&\frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{3}}(\psi_1\psi_1\psi_2+\psi_1\psi_1\psi_2+\psi_1\psi_2\psi_1)+\frac{1}{\sqrt{3}}(\psi_1\psi_2\psi_1+\psi_2\psi_1\psi_1+\psi_2\psi_1\psi_1)\right)\\=&\frac{\sqrt{2}}{\sqrt{3}}(\psi_1\psi_1\psi_2+\psi_1\psi_2\psi_1+\psi_2\psi_1\psi_1)\\
=&\sqrt{2}|2_1,1_2\rangle.\end{array}</math>
Then annihilate one boson from the state <math>\psi_1</math>,
:<math>\begin{array}{rl}b_1|2_1,1_2\rangle=&\frac{1}{\sqrt{3}}(b_1\psi_1\psi_1\psi_2+b_1\psi_1\psi_2\psi_1+b_1\psi_2\psi_1\psi_1)\\=&\frac{1}{\sqrt{3}}\left(\frac{1}{\sqrt{3}}\psi_1\oslash_+\psi_1\psi_1\psi_2+\frac{1}{\sqrt{3}}\psi_1\oslash_+\psi_1\psi_2\psi_1+\frac{1}{\sqrt{3}}\psi_1\oslash_+\psi_2\psi_1\psi_1\right)\\=&\frac{1}{\sqrt{3}}\left(\frac{1}{\sqrt{3}}(\psi_1\psi_2+\psi_1\psi_2+0)+\frac{1}{\sqrt{3}}(\psi_2\psi_1+0+\psi_1\psi_2)+\frac{1}{\sqrt{3}}(0+\psi_2\psi_1+\psi_2\psi_1)\right)\\=&\psi_1\psi_2+\psi_2\psi_1\\=&\sqrt{2}|1_1,1_2\rangle.\end{array}</math>

==== Action on Fock states ====
Starting from the single-mode vacuum state <math>|0_\alpha\rangle=1</math>, applying the creation operator <math>b_\alpha^\dagger</math> repeatedly, one finds
:<math>b_\alpha^\dagger|0_\alpha\rangle=\psi_\alpha\otimes_+ 1=\psi_\alpha=|1_\alpha\rangle,</math>
:<math>b_\alpha^\dagger|n_\alpha\rangle=\frac{1}{\sqrt{n_\alpha+1}}\psi_\alpha\otimes_+ \psi_\alpha^{\otimes n_\alpha}=\sqrt{n_\alpha+1}\psi_\alpha^{\otimes (n_\alpha+1)}=\sqrt{n_\alpha+1}|n_\alpha+1\rangle.</math>
The creation operator raises the boson occupation number by 1. Therefore, all the occupation number states can be constructed by the boson creation operator from the vacuum state
:<math>|n_\alpha\rangle=\frac{1}{\sqrt{n_\alpha!}}(b_{\alpha}^\dagger)^{n_\alpha}|0_\alpha\rangle.</math>
On the other hand, the annihilation operator <math>b_\alpha</math> lowers the boson occupation number by 1
:<math>b_\alpha|n_\alpha\rangle=\frac{1}{\sqrt{n_\alpha}}\psi_\alpha\oslash_+ \psi_\alpha^{\otimes n_\alpha}=\sqrt{n_\alpha}\psi_\alpha^{\otimes (n_\alpha-1)}=\sqrt{n_\alpha}|n_\alpha-1\rangle.</math>
It will also quench the vacuum state <math>b_\alpha|0_\alpha\rangle=0</math> as there has been no boson left in the vacuum state to be annihilated. Using the above formulae, it can be shown that
:<math>b_\alpha^\dagger b_\alpha|n_\alpha\rangle=n_\alpha|n_\alpha\rangle,</math>
meaning that <math>\hat{n}_\alpha = b_\alpha^\dagger b_\alpha</math> defines the boson number operator.

The above result can be generalized to any Fock state of bosons.
:<math>b_\alpha^\dagger|\cdots,n_\beta,n_\alpha,n_\gamma,\cdots\rangle= \sqrt{n_\alpha+1}|\cdots,n_\beta,n_\alpha+1,n_\gamma,\cdots\rangle.</math>
:<math>b_\alpha|\cdots,n_\beta,n_\alpha,n_\gamma,\cdots\rangle= \sqrt{n_\alpha}|\cdots,n_\beta,n_\alpha-1,n_\gamma,\cdots\rangle.</math>
These two equations can be considered as the defining properties of boson creation and annihilation operators in the second-quantization formalism. The complicated symmetrization of the underlying first-quantized wave function is automatically taken care of by the creation and annihilation operators (when acting on the first-quantized wave function), so that the complexity is not revealed on the second-quantized level, and the second-quantization formulae are simple and neat.

==== Operator identities ====
The following operator identities follow from the action of the boson creation and annihilation operators on the Fock state,
:<math>[b_\alpha^\dagger,b_\beta^\dagger]=[b_\alpha,b_\beta]=0,\quad [b_\alpha,b_\beta^\dagger]=\delta_{\alpha\beta}.</math>
These commutation relations can be considered as the algebraic definition of the boson creation and annihilation operators. The fact that the boson many-body wave function is symmetric under particle exchange is also manifested by the commutation of the boson operators.

The raising and lowering operators of the [[quantum harmonic oscillator]] also satisfy the same set of commutation relations, implying that the bosons can be interpreted as the energy quanta (phonons) of an oscillator. The position and momentum operators of a Harmonic oscillator (or a collection of Harmonic oscillating modes) are given by Hermitian combinations of phonon creation and annihilation operators,
:<math>x_{\alpha}=(b_{\alpha}+b_{\alpha}^\dagger)/\sqrt{2},\quad p_{\alpha}=(b_{\alpha}-b_{\alpha}^\dagger)/(\sqrt{2}\mathrm{i}), </math>
which reproduce the canonical commutation relation between position and momentum operators (with <math>\hbar=1</math>)
:<math>[x_{\alpha},p_{\beta}]=\mathrm{i}\delta_{\alpha\beta},\quad [x_{\alpha},x_{\beta}]=[p_{\alpha},p_{\beta}]=0.</math>
This idea is generalized in the [[quantum field theory]], which considers each mode of the matter field as an oscillator subject to quantum fluctuations, and the bosons are treated as the excitations (or energy quanta) of the field.