Phy5646

Second Quantization

Consider now a wavefunction pertaining to a many-particle system, $\psi (\eta _{1},\eta _{2},\eta _{3},...,\eta _{N})\!$, which is considered to be a field variable. For the many-particle system, this field variable must also quantized by a process known as second quantization.

In order to perform this quantization of the field variable, we must construct special raising and lowering operators, associated with the individual energy levels of the system, ${\hat {a}}_{j}^{\dagger }\!$ and ${\hat {a}}_{j}\!$, which add and subtract particles from the $j^{th}\!$ energy level, respectively. In the presence of spin, an additional subscript is added to separate the creation and annihilation operators for each case of spin, so that each operator only acts on particles with the same spin attributed to said operator. In the simple, although rather non-physical, case of spinless particles, this extra factor can be ignored for simplicity in examining how the operators work on the quantized field:

${\hat {a}}_{j}^{\dagger }|0\rangle =|1\rangle \!$

${\hat {a}}_{j}|1\rangle =|0\rangle \!$

${\hat {a}}_{j}|0\rangle =0\!$

For the case of fermions, an additional constraint on the operators is placed due to the exclusion principle: ${\hat {a}}^{\dagger }|1\rangle =0\!$

Given the two classes of particles, fermions and bosons, two sets of relations result to relate the creation and annihilation operators.

For the case of bosons, the operators obey a commutator relationship of the form:

$[{\hat {a}}_{i},{\hat {a}}_{j}^{\dagger }]=\delta _{ij};[{\hat {a}}_{i},{\hat {a}}_{j}]=[{\hat {a}}_{i}^{\dagger },{\hat {a}}_{j}^{\dagger }]=0\!$

The state of the system, $|n_{0},n_{1},..,n_{N}\rangle \!$, where $n_{j}\!$ refers to the number of particles in the $j^{th}\!$ state, is therefore of the form:

$|n_{0},n_{1},..,n_{N}\rangle ={\frac {({\hat {a}}_{0}^{\dagger })^{n_{o}}}{\sqrt {n_{0}!}}}{\frac {({\hat {a}}_{1}^{\dagger })^{n_{1}}}{\sqrt {n_{1}!}}}...{\frac {({\hat {a}}_{N}^{\dagger })^{n_{N}}}{\sqrt {n_{N}!}}}|0\rangle \!$

Fermions, however, obey anti-commutator relationships, of the following form:

$\{{\hat {a}}_{i},{\hat {a}}_{j}^{\dagger }\}=\delta _{ij};\{{\hat {a}}_{i},{\hat {a}}_{j}\}=\{{\hat {a}}_{i}^{\dagger },{\hat {a}}_{j}^{\dagger }\}=0\!$

For this type of system, the state $|n_{0},n_{1},..,n_{N}\rangle \!$ can be written as:

$|n_{0},n_{1},..,n_{N}\rangle =({\hat {a}}_{0}^{\dagger })^{n_{0}}({\hat {a}}_{1}^{\dagger })^{n_{1}}...({\hat {a}}_{N}^{\dagger })^{n_{N}}|0\rangle \!$

Furthermore, for both classes of particles, we can create an operator that, upon acting on the total state of the system, returns the number of particles in a given $n^{th}\!$ state (for fermions this will obviously be 0 or 1). This operator is of the form ${\hat {a}}_{n}^{\dagger }{\hat {a}}_{n}\!$. Therefore, ${\hat {a}}_{j}^{\dagger }{\hat {a}}_{j}|n_{0},n_{1},..,n_{N}\rangle =n_{j}|n_{0},n_{1},..,n_{N}\rangle \!$. From this it is easy to obtain an operator, ${\hat {N}}\!$ that returns the total number of particles in the system:

${\hat {N}}=\sum _{j=0}^{\infty }{\hat {a}}_{n}^{\dagger }{\hat {a}}_{n}\!$

From this operator, the average number of particles, and therefore the average flux number, can be calculated by the following:

$\langle \Psi |{\hat {N}}|\Psi \rangle =\langle {\hat {N}}\rangle \!$

$\langle \Psi |{\langle {\hat {N}}-{\langle {\hat {N}}\rangle }\rangle }^{2}|\Psi \rangle =\langle {\hat {N}}^{2}\rangle -{\langle {\hat {N}}\rangle }^{2}\!$

To continue the analysis of these second quantization operators, lets consider the projection of ${\hat {a}}^{\dagger }\!$ on the state $|0\rangle \!$, that is to say, define a function $\phi _{j}(\mathbf {r} )\!$ such that $\langle \mathbf {r} |{\hat {a}}_{j}^{\dagger }|0\rangle =\phi _{j}(\mathbf {r} )\!$