Photon States

It is now obvious that the integer $n_{k,\alpha}$ is the number of photons in the volume with wave number $\vec{k}$ and polarization $\hat{\epsilon}^{(\alpha)}$. It is called the occupation number for the state designated by wave number $\vec{k}$ and polarization $\hat{\epsilon}^{(\alpha)}$. We can represent the state of the entire volume by giving the number of photons of each type (and some phases). The state vector for the volume is given by the direct product of the states for each type of photon.

$$\bgroup\color{black} \vert n_{k_1,\alpha_1},n_{k_2,\alpha_2},... ...2,\alpha_2}\rangle ...,\vert n_{k_i,\alpha_i}\rangle... \egroup$$

The ground state for a particular oscillator cannot be lowered. The state in which all the oscillators are in the ground state is called the vacuum stateand can be written simply as $\vert\rangle$. We can generate any state we want by applying raising operators to the vacuum state.

$$\bgroup\color{black} \vert n_{k_1,\alpha_1},n_{k_2,\alpha_2},... ...,\alpha_i}}\over\sqrt{n_{k_i,\alpha_i}!}} \vert\rangle \egroup$$

The factorial on the bottom cancels all the $\sqrt{n+1}$ we get from the raising operators.

Any multi-photon state we construct is automatically symmetric under the interchange of pairs of photons. For example if we want to raise two photons out of the vacuum, we apply two raising operators. Since $[a_{k,\alpha}^\dagger,a_{k',\alpha'}^\dagger]=0$, interchanging the photons gives the same state.

$$\bgroup\color{black} a_{k,\alpha}^\dagger a_{k',\alpha'}^\dag... ...{k',\alpha'}^\dagger a_{k,\alpha}^\dagger \vert\rangle \egroup$$

So the fact that the creation operators commute dictates that photon states are symmetric under interchange.