Editing Second quantization (section)

==== 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>