String Theory

Sunday, 9 June 2013

Bosons in Second qantizations

Given the occupation number, we introduce the annihilation $b_{\nu_j}$ and creation ${b^{\dagger}}_{\nu_j}$ operators that lowers(raises) the occupation number in the state $| \nu_j \rang$ by 1,

$b_{\nu_j}|\dots,n_{\nu_{j-1}}, n_{\nu_j}, n_{\nu_{j+1}},\dots \rang=\sqrt{n_{\nu_j}}|\dots,n_{\nu_{j-1}}, n_{\nu_j}-1, n_{\nu_{j+1}},\dots \rang$

${b^{\dagger}}_{\nu_j}|\dots,n_{\nu_{j-1}}, n_{\nu_j}, n_{\nu_{j+1}},\dots \rang=\sqrt{n_{\nu_j}+1}|\dots,n_{\nu_{j-1}}, n_{\nu_j}+1, n_{\nu_{j+1}},\dots \rang$

Since bosons are symmetric in the single-particle state index $\nu_j$ we demand that $b_{\nu_j}$ and ${b^{\dagger}}_{\nu_j}$ commute, So, we can obtain the mean properties of these operators:

$\begin{matrix} [{b^{\dagger}}_{\nu_j},{b^{\dagger}}_{\nu_k}] = 0 & [b_{\nu_j},b_{\nu_k}]=0 & [b_{\nu_j},{b^{\dagger}}_{\nu_k}]=\delta_{\nu_j\nu_k}\\ {b^{\dagger}}_{\nu_j}|n_{\nu_j}\rang=\sqrt{n_{\nu_j}+1}|n_{\nu_j}+1 \rang & b_{\nu_j}|n_{\nu_j}\rang=\sqrt{n_{\nu_j}}|n_{\nu_j} -1\rang & b_{\nu_j}|0\rang=0\\ {b^{\dagger}}_{\nu_j}b_{\nu_j}|n_{\nu_j} \rang=n_{\nu_j}|n_{\nu_j} \rang &\left({b^{\dagger}}_{\nu_j}\right)^{n_{\nu_j}}|0 \rang=\sqrt{(n_{\nu_j})!}|n_{\nu_j} \rang & n_{\nu_j}=0,1,2,\dots\\ \end{matrix}$

and therefore identify the first and second quantized states,

$\hat{S}_+|\psi_{n_{\nu_1}}(\bold{r}_1)\rang|\psi_{n_{\nu_2}}(\bold{r}_2)\rang\dots |\psi_{n_{\nu_1}}(\bold{r}_N)\rang= {b^{\dagger}}_{n_{\nu_1}}{b^{\dagger}}_{n_{\nu_2}}\dots{b^{\dagger}}_{n_{\nu_N}}|0\rang$

with $\hat{S}$ the symmetrization operator. Here, both contain N-particle state-kets completely symmetric in the single-particle state index $\psi_{\nu_j}$ . Because the creation and annihilation operators of the quantum harmonic oscillator obey these properties, one can classify the field associated to it as bosonic.

String Theory

Sunday, 9 June 2013

Bosons in Second qantizations

No comments:

Post a Comment