SPDC One Arm vs Very Weak Coherent state I know that SPDC(Spontaneous Parametric Down Conversion) is a method to generate heralded single photon source.
So If we do homodyne tomography of single photon fock state, another arm of SPDC is used for trigger signal.
I have question about this point.
What if we observe only one arm without trigger of another arm, What difference does exist comparing with very weak coherent state?
Is it possible to distinguish experimentally SPDC one arm signal with Very weak coherent source which have mean photon number 1??
 A: The state generated by SPDC is a two-mode squeezed vacuum state, which can be written in the Fock basis as:
$$
|\Psi\rangle_{s,i} = 
\sum_{n=0}^{\infty}
\frac{\mathrm{tanh}^n r}{\mathrm{cosh}\;r}  |n,n\rangle_{s,i}
$$
The state in one mode, for example the signal, is given by taking the partial trace over the other mode, in this case the idler:
$$
\rho_s = \mathrm{tr}_i \big[|\Psi\rangle\langle\Psi|_{s,i}\big]
= \sum_{m=0}^{\infty} \langle m|_i
\big(|\Psi\rangle\langle\Psi|_{s,i}\big) |m\rangle_i
$$
Evaluating this gives the state:
$$
\rho_s = (1-\mathrm{tanh}^2 r) \sum_{n=0}^{\infty}
\mathrm{tanh}^{2n} r |n\rangle\langle n|_s
$$
This is simply the thermal state, which isn't surprising because we're tracing out one part of an entangled state. Doing homodyne tomography on one half of an SPDC state without heralding will therefore give you a thermal state.
Identifying $\mathrm{tanh}^2 r = \mathrm{exp}\big[ -\frac{\hbar\omega}{k_B T_s}  \big]$ we can find the temperature of the signal state as:
$$
T_s = \frac{\hbar \omega}{2 k_B \mathrm{ln}(\mathrm{coth}|r|)}
$$
