The semiclassical Rabi problem

1. The semiclassical Rabi problem

2. We have a two level atom,with We look for the solution of the Schrödinger equation as: The atom has a hamiltonian: The field interaction has a corresponding hamilton operator: Our goal is to look for the solution of a two level atom in a classical electric field, described, by, If we are talking about a two level atom,the solution without the field, we can describe by the time dependent Schrödinger equation, If we assume that Now we can apply the time independent schrödinger equation, and we get: Since we have a two level atom, the general solution of the differential equation is

3. This is the two level atom. When there is no radiation there is no transition between states, thus the interaction between field and atom is none. That is why we need to introduce operators. To describe the interactions only when transitions occur We have a classical electromagnetic field: The interaction is only when the electron changes it’s energy level. This is when from state 1 goes to state 2 and from state 2 goes to state 1. Two transitions can be described with two operators: Transition from state 1 to state 2 of the electron.d is the dipole moment vector. The dipole momentum is equal in length: 1. operator: 2. operator: Basically the dipole momentum vector is a 3 dimensional vector, where each of it’s component contains the momentum in the given direction and the classicall electric field contains the electric field strength measured in V/m, divided also in 3 components. Here the two operators decides weather is emitted or absorbed radiation.

4. The operator-matrix of an operator can be written as: The operator matrix of the atom: can be seen: This is true because of the time independent Schrödinger equation: The operator matrix of the field hamiltonian: This is true because the property of the interaction Hamiltonian. (There is no dipole with 1 state, we wolud need Infinite energy)

5. Now we put the atom in an electric field. What happens to ? The state functions remain constant, but the time dependence will change, we can describe this time dependence, by saying that the coefficients are time dependent functions: Now let us write the new equation: Our hamiltonian of the whole system is: This is the total energy. We apply this on our partial solution, but first we write the upper equation in another form, Afterwards we wrote our hamiltonian matrix: Thus…

6. What about the derivative of the equation, we derivate the matrix form and we get: By substituting all this information in the time dependent Schrödinger equation, we get a space invariant set of equations, which will lead us to the Rabi solution: The last two terms are left aside, and by multiplying them back we get two equations: Where:

7. The Rabi frequency is defined as: We know that: The new equations are of the form: The Rabi frequency is an interaction frequency between field and atom. It is a mean frequency.

8. If we neglect the fast oscillations, which are near 2ω , beacause of the near resonance effect, then we apply the rotating wave approximation (RWA). We can do this because the slow oscillations govern the time evolution (the fast oscillations change very fast the sign of the term) The new system is : If we are at perfect resonance (ω0=ω), then te equations are: The solution is: Because we defined the probability of finding the system in a state is, that is why

9. The solution of the Rabi problem

10. Applying inverse Laplace transform: Where we assumed that at t=0 we are in state 1 with probability 1. thus |c10|=1, and |c20|=0.

11. If we assume that in time instant 0 we are in state 1 and we are in perfect resonance, then the probabilities are: In this case the solutions for the probabilities are: We can see on this picture, that the probabilities are inverted in phase, and, they are preserved. If we are at time instant , then the system is in state 1 with ~ 0.7 and in state 2 with ~ 0.3 probability. The final conclusion is that the probability change between is: In this diagram at the peak values we are in state 2 with 1 probability, and at min values we are in state 1 with 1 probability.

12. Density Matrices These are the elements of the thensity matrix:

13. We can fulfill these constraints even if we introduce other variables , to solve the equation with Laplace Transformation: Derivating these: Substituting the derivatives into the original equations:

14. We can write now the system of equations: This is a standard equation of a system with 4 inputs, 4 outputs, and we assume that the system’s initial state is state 1. Initial condition

15. Solutions:

16. In case of perfect resonance again we know much more: 1. We start by the given initial condition, that we are in state 1. then the weight for transition is 0, because we know for shure that we are in state 1. (1) 2. When the electron is halfway on it’s road between states(.5 probability), then the absor- bed energy weight is maximum.(2) 3. If we are in state 2 with 1 probability, then it’s the same story as in 2. (3) 4. If we are again halfway between states but we go from 2 to 1, then the emmitted energy amount reaches it’s maximum. 5. The period closes !! Remark: This is the perfect resonance case.

17. Solutions: Conclusion: The larger the detuning is the the larger the probability of remaining in state 1. We took the Rabi frequency as 1.

18. After this new form,we can see, that if we irradiate, by a pulse of length then we put the electron in state 2. if we Irradiate by then we get back in state 1. Relaxation: Dapming is because of the spontaneous emission of Photons. Thus the probability of being in state 1 increases after a given amount of time, and that of being in state 2 decreases. Furthermore the electron’s probability reaches a steady state. (It’s value is constant after a time.) We introduce two coefficients. A passes the probability from state 2 to 1. (spont. emmission) ץ – is the coherent interaction

19. The reasons of decay (A): Thermal reservoir, electronic discharges (noise), and everything that affects the whole system – this is the spontaneous emmission coefficient. The reasons of decay (ץ): This represens the the interactions between the coherent states (|1><2|, and |2><1|) due to the interaction between electrons. We must not forget, that: … and do our calculations accordingly.

20. <-the probability of finding the electron in state 1. <-the probability of finding the electron in state 2.

21. In this figure we can see, that after the steady state of the ρ11 and ρ22 probabilities The coherence of the is constant too. These facts have theoretical background too. Let’s assume, that:

22. The quantized electromagnetic field

23. We know from quantummechanics the Hamilton equations: We also know the global definition for impulse: The Lagrange density function can be defined as the difference between the kinetic and pot- ential energy density functions. In case of electromagnetic fields, we define the Hamilton density function as: From classical electromagnetic field theory (where A is the vector potential):

24. If we want to write the electromagnetic field in a unitary cube, we should introduce two orthogonal vector functions and the wave vector, which make an orhogonal system. We introduce two orthogonal unit vectors: The sinusoidal and the orthonormal property:

25. This is because p,and q functions were arbitrary chosen so the coeff. can be arbitrary chosen too. We introduce the space and impulse operaors: From here the emission and absorption operators are:

29. Gaussian pulses

30. In practice one frequency of an electric field cannot be radiated, it is always a gaussian pulse, which in case it is narrow enough it approximates very good the discrete frequency. Instead of scatterings we want to use the full width half maximum ( ) length, because we can measure it. It is at the half of a gaussian function. (which is the monochromatic light spectrum) We want the Gaussian function to be: Our gaussian function is applied in the power spec- trum description of light: Where Ep is the pulse energy We get Ep by integrating P(t). The peak value, at Time instant t0 :

31. This pulse width is Gaussian, because, when the pulse begins, it’s power is minimal, and increases exponentially. At peak value the device that makes the pulse is shut down, thus the pulse’s power decayes exponentially in time. In general if we don’t know the shape of the pulse we can transform it in a Gaussian, by calculating the sqare scatterig and transforming it into the fwhm and expected value in time: length: and by calculating the expected value. Here the beam spectrum has it’s peak value at: These integrals are done on the whole pulse length.

32. The caracteristics of Gaussian beams: A laser beam, can be caracterized by it’s wave vector, which tells us in which direction the wave goes to at the given coordinate. Because it’s intensity usually in one direction is higher, that is why we assume that it’s intensity in the x, y direction is decaying exponentially. We „cut” the beam in mind and realise that in the core the intensity is higher: So let’s assume at z=0 and t = 0 the electric field is in the form of: Where w0 is the beam radius. (the time is separable). The intensity is The beam’s electric field goes in three directions, so We can expand it. This way the wavenumber is preserved in the length of wave vectors. Where: This way:

33. The electric field is built up from different amplitudes corresponding to different wave vectors, by the Fourier Integral: The amplitude depends only from p and q, because going in the z direction the amplitude is approximatly always the same. (the waves going in x direction have high p, because the weight of the x direction is p) By inverse Fourier integral of E(x,y,0), we get that at a given wavevector of x and y what is the electirc field amplitude. If we assume that the wavenumbers in the z direction are the highest then we can say that in the equation, this way we can get k.

34. Now we deal with the Fourier integral, by substituting k. So the the derived equation. <- The result Finally we get:

35. First we assumed, that the beam rad ius was w0, but we know that the elec- tric field is Re{E(x,y,z,t)} This way the beam radius as a function of z is in the exponent : This is the Rayleigh length which is also called the focus depth of a beam. • The diameter is increasing linearly in function • of z, at large distances.

36. Is the electric field of the beam We usually say that the radius of a gaussian beam is where the intensity reaches it’s 1/e^2 value. We know that: It was easy to rewrite it in polar coordinates: 86.5% of power is inside the beam radius.