 to lecture number 11 of the course quantum mechanics and molecular spectroscopy. As usual we will start with a quick recap of the previous lecture and then proceed with the lecture number 11. In the previous lecture we talked about the electric field and magnetic field of the propagating light. So if a light is propagating along a direction R or let us say its vector is represented by R then there is a electric field which is as a direction E and the corresponding electric field is E and there is a magnetic field B with the corresponding direction, okay. Now there are three components of both electric field and the magnetic field. So E will consist of E x, E y and E z. Similarly the magnetic field will have three components that is B x, B y and B z, okay. But in Hamiltonian we have potential and not the field. So we need to convert this into potentials. So there is a transformation that takes the electric and magnetic field to scalar and vector potentials. Phi is the scalar potential and there is a vector potential A which will have three components A x, A y and A z. So totally we have four components, one scalar potential and three vector potentials, so four components and these four components must be reduced from the electric and magnetic fields which have six components. The relationship between the magnetic field or electric field with the vector and scalar potentials are E is equal to del phi minus dA by dt and B is equal to del cross A. Now what we will do is since we have to go from six components to four components of course there is some redundancy. So there are infinite ways one can have this transformation. One of the transformations that we will call it as Coulomb gauge in which we say phi is equal to 0 and del dot A equals to 0. So there are six components, so you are reducing to four components with two criteria that are fixed. So therefore now the transformations becomes easier. Now in a plane wave notation A is given as A is equal to A naught and it will move in the same direction as electric field this one into e to the power of minus i k dot r minus omega t sorry this should be plus e to the power of minus i k dot r minus omega t. Now this is exponential i theta plus exponential minus i theta so this is proportional to cos theta and you can see the electric field is a time derivative with respect to vector potential. So e will be equal to minus i A naught epsilon or plus i rather e to the power of minus i k into omega k dot r minus omega t minus e to the power of minus i so this should be plus k dot r minus omega t. So this should be proportional to sin theta. So which means the vector potential A and the electric field A e are out of phase with respect to each other by pi by 2. So they are out of phase by pi by 2 okay. Now let us get on to the contents of this lecture. For example if you take a charged particle and you put it in the electric field or electromagnetic radiation that is light then the charged particle experiences what is known as Lorentz force okay. This Lorentz force is experienced by charged particle charged particle in electromagnetic field or this electromagnetic field is nothing but light. So if I expose a charged particle to the light then it feels a Lorentz force. So I will write Lorentz force as F L this is equal to Q times electric field plus V into B okay where you know this is electric field, B is magnetic field, V is the velocity of the particle. So of course that will depend on the charge Q. So more the charge more will be the force. And atomic systems of course atomic and molecular system the charge is fixed because the electrons have fixed charge okay. Of course nuclei could have a different charge which could be multiples of multiples of the electronic charge E. Now this is the force that is called Lorentz force but let us take a small teetor and think what is force in terms of classical mechanics, force in classical mechanics okay. What is force is consequential force F equals to mass into acceleration okay. So this is mass and this is acceleration. Now acceleration is nothing but rate of change of velocity. So this is nothing but F is equal to m times dV by dt that is nothing but rate of change velocity okay. Now mass is a fixed quantity it is a number for a given particle. So one can write F is equal to d by dt of mV okay. Now what is mV momentum this is nothing but momentum okay. So now force can be written as d by dt of mV that is nothing but d by dt of P. So force can also be reconciled as rate of change of momentum. So this is nothing but rate of. Now let us get to the Lorentz force okay, FL is equal to Q times E plus V. Now in general the magnetic field its magnitude is going to be is much smaller okay. So one can neglect okay and if you neglect that we get FL is equal to QE okay. This is called electric dipole approximation. You are only looking at the electric part electric field of the electromagnetic radiation okay. So we are neglecting the magnetic field. Of course if you are going to do spectroscopy like NMR or EPR this is not going to be valid but for general spectroscopy terms like you know rotational, vibrational and electronic this is valid okay. Now FL is equal to Q dot E okay. Now I can write this as Q times minus d by dt of A okay because E is equal to del phi minus dA by dt that is what we said you know and in the Coulomb gauge you have d phi is equal to 0. So which means E will turn out to be minus dA by dt. Now if I write this your FL is equal to of course Q is a charge of a system so that can be taken out it is a constant with respect to time okay. So this can be written as minus d by dt of QA. Now but force is this is force okay Lorentz force is given by d by dt of minus QA okay. Now force as I said is equal to d by dt of P that is nothing but rate of change of momentum okay. So if you compare these two forces of course force is a force you know whether it is of some mechanical origin or of electromagnetic origin it is a force that is fine okay. Now you can clearly see by comparing these two equation you can see momentum P is equal to minus QA okay. This is the momentum that a charge particle will gain in a vector potential okay. So in your Hamiltonian H what you have we have P square by 2 m plus your potential V of x. Now but P in this case is the velocity or the momentum that particle has on its own but the momentum you put the moment you put the charge particle in the electric field then its momentum is going to get modified. So your P prime will be equal to P plus okay the newly attained momentum because of the electric field P of the because of the electric field and that is the perturbation that you have. So your P prime will now be equal to P minus QA. So your Hamiltonian H will be nothing but actually it should be P prime square by 2 m plus Vx. Now let us write the in terms of okay this is not really the so your Hamiltonian so you write the corresponding operators. So this will be equal to P minus QA square by 2 m plus V of x why V of x because we told you I told you that for this course okay the potential is generally independent of time okay. So this is equal to H into now this will be nothing but what is your P this is nothing but minus ih bar T by dx that is the P minus QA whole square by 2 m plus V of x that is my Hamiltonian. So when I write that so your Hamiltonian H is equal to minus ih bar T by dx minus I mean only looking in the one dimension case okay one can generalize to three dimensions okay minus QA whole square P square by 2 m plus V of x. So this is nothing but A square plus B square plus 2 AB. So A square will be minus ih bar square by 2 m T square by Tx square A square that is B square will be Q square A square by 2 m okay but 2 AB okay or it cannot be 2 AB because they are operator AB minus PA okay. In fact I can take a negative sign outside so this is equal to AB this is minus and this is minus so just AB. So H bar plus H bar bear by H bar by 2 m H bar by 2 sorry ih bar by 2 m okay d by dx of QA plus and if I multiply this will be QA times ih bar by 2 m QA times d by dx this potential V of x okay. So your H will now be equal to minus H bar square by 2 m d square by dx square okay minus H bar by 2 m so this is Q square by A square this will be plus ih bar by 2 m let us say d by dx of QA plus ih bar by 2 m QA d by dx plus Q square A square by 2 m plus V of x. Now we will see this equation if you take the first term and the last term this is nothing your is normal Hamiltonian so H is equal to minus H bar square by 2 m d square by dx square plus V of x plus ih bar by 2 m d by dx of QA minus sorry plus ih bar by 2 m QA d by dx plus Q square A square by 2 m. Now this is nothing but your H naught the Hamiltonian of the original molecule so your H will now be equal to H naught plus ih bar by 2 m d by dx QA plus ih bar by 2 m QA d by dx plus Q square A square by okay let us evaluate the term so let me write down again the Hamiltonian H is equal to H naught plus ih bar by d by dx of Q is a constant so you can take it out of A plus ih bar by 2 m QA d by dx minus Q sorry plus Q square by 2 m A square okay. Now let us evaluate the term d by dx of A but this is Hamiltonian it can always it has to be always evaluated with respect to some function okay some function psi of x so this is equal to okay now there are product of so they will get dA by dx okay into psi of x plus A times d by dx of psi of x okay now this is nothing but del dot A plus A this is nothing but del now according to Coulomb gauge del dot A is equal to 0 so this is nothing but A or this is nothing but A d by dx so your H is equal to H naught plus ih bar Q by 2 m d by dx of A that is nothing but A d by dx A d by dx and there is another term which is similar to that it is equal to ih bar Q by 2 m A d by dx plus Q square by 2 m A square so this is nothing but H naught plus these two terms are equals or two times so this will be ih Q A d by dx plus Q square by 2 m A square so this is nothing but H naught plus ih bar Q by m sorry by m A del plus Q square by 2 m A square okay. So your Hamiltonian now comes out to be equals to H naught plus ih bar Q by m A dot del minus or plus Q square by 2 m A square okay so that is the Hamiltonian that you are going to get so which means your H prime of T should be equal to ih bar Q by m A dot del plus Q square by 2 m okay. Now in spectroscopy the amount of light is very feeble so that is called weak perturbation in the weak perturbation limit we will ignore the A square term therefore H prime of T will be equal to ih bar Q by m A dot del that is your Hamiltonian or the perturbation Hamiltonian so your total Hamiltonian nothing but H naught plus H prime T so H naught is something that is the molecular Hamiltonian and your perturbation Hamiltonian time dependent perturbation is just given by ih bar Q by m A dot del okay. Now we know the now I can slightly modify this now we know the Hamiltonian perturbation Hamiltonian so this is nothing but Q by m times minus ih bar del okay what is minus ih bar del minus ih bar del is nothing but P okay remember minus ih bar D by DX is equal to operator P so this is nothing but minus Q by m A that is your H prime of T okay so this is my time dependent okay. Now we know what is the time dependent perturbation okay we will stop it here and continue in the next class.