# Tag Info

## New answers tagged quantum-optics

0

Coherent EM field is not characterizable by photon number; it is not an eigenstate of an $a^\dagger a$ operator.

1

It is a superposition of absorbing one photon and not absorbing at all. Say you shine a laser pulse onto atom. $|g\rangle$ will be the state if no photon is absorbed, and $|e\rangle$ if it is absorbed. In quantum picture this light pulse can be described by coherent state $|\alpha_1\rangle ... 2 It is really quite difficult. This gets you to the original paper by Cummings in Phys. Rev. Lett. 140 in 1965, page A1051: http://journals.aps.org/pr/abstract/10.1103/PhysRev.140.A1051 . You must notice first that if$P_e(t)$is given as you wrote then the coefficients$\frac{e^{-\alpha^2}|\alpha|^{2n}}{{n!}}$multiplying$\cos()^2$terms are localized ... 1 Time-dependence of operators when quantizing a free field When quantizing EM field, the QM picture used is about choice. All different pictures of QM are unitary equivalent. But, surely there are some pictures that seems to be more convenient than others. Here, since the ladder operator is the basic bloc of a quantized theory of EM-field, it seems to be ... 0$a$and$a^\dagger$are operators.$\alpha$is a pure imaginary number which we then call$i\alpha$with$\alpha$now real, this is why from eq1 to eq2 we use$\alpha→i\alpha$and$−\alpha^*→i\alpha$. Now from eq2 to eq3 we simply substitute$a+a^\dagger$for$x$operator. I don't understand why you want to write$\alpha=\ldots\$, its just some real constant. ...

0

You are making an incorrect assumption in your question: There is no physical evolution from a number state (aka Fock state). This evolution happened purely inside physicists' heads as it was realized that laser light is not properly described by number states. The problem is your assumption that the particle number ever is well-defined. Lasing action is an ...

Top 50 recent answers are included