Lesson 3 Canonical Quantization of Scalar Field Part IIIB

Transcript

hi everybody welcome back to quantum field theory students perspective i want to um continue today with um a canonical quantization and so we're going to do lesson 3 canonical quantization of scalar fields 3b and the references are the same as before like i said every book does it a little bit differently orkin and drell chapters 11 and 12. and sorry and hatfield sections 3.3 rider section 4.1 coleman where much of this material is taken from this section four point four srineki our main course textbook is chapter three and peskin and schroeder chapter two okay so um we've done canonical quantization from a bottom-up approach let's call it the bottom-up fox-space approach and then um last lecture we worked on the top down call this like a physical approach top down canonical approach sort of like a turn the crank without thinking and we last video we reviewed we viewed this approach at lightning speed for uh classical particle mechanics for quantum mechanics and classical field theory and today today we're gonna fill in the mix missing box you might remember from the last lecture and do uh canonical of quantum field theory in the context now some of the stuff is going to seem like we're repeating stuff that we did earlier and some of the stuff that we did earlier i'm not going to um repeat a lot of this is like a chicken and egg sort of thing before we went forwards now we're sort of going backwards but um the general idea i think is clear and i'll try and um i'll try and do this lecture as if it was just a traditional 1960s canonical quantization approach like borkin and drell do it so um we start with the uh klein gordon equation which is a scalar i'm going to do a real scale here a real scaler lorenzen variant feel so the klein going equation is just and i'm going to work in the barking and drought metric in this lecture so the time time component this positive in minkowski space or the second okay now we already have the lagrangian for this field and that's what we we need but i'm just going to work backwards pretend we don't know a lagrangian and you can do this for a lot of theories it's just going backwards so we define del phi of x our variation in the fields phi prime of x minus v of x this is an infinitesimal variation multiply multiply this by the field equation and integrate over space time so we get whenever we do this we always have finite integrations for time but we integrate over all space some people just like neglect to even say that but it's not really not really that important now as usual the variations vanish at the endpoints of time huh and we're also going to assume localization so that all fields derivative etc go to zero as the absolute value of space goes to infinity plus or minus infinity so um with that we can always integrate we don't have to worry about boundary terms now we can always integrate by parts we either have a spatial thing in which case we use the localization thing or we'll have a time thing in which case we use our usual lagrangian formulation that the variations vanish at 10 points of time and if you carry if you work this backwards for instance um you can get like um let's see like delphi the second partial of phi with respect to t2 is equal to minus one half the partial of del phi with respect to t times the partial of v with respect to t plus the partial phi with respect to t partial of del phi with respect to t and you can see this just by imagining you integrate this by parts the integral of u dv we and we get a minus sign which cancels this minus sign and we'll get del phi times the second partial fee respect to t each of these terms is the same so when we do that when we do that we're going to end up with just um again use doing everything in reverse from what we usually do i don't want to work it out anymore my signs may look like they're off but they're they're really not i don't think so this whole thing is equal to zero and so if you remember our basic lagrangian thing lagrangian is is we have this dell the integral of t1 to t2 d4 x here i'm not gonna of um l of v coming the partial of v with respect to x mu is equal to zero so look comp we bring this del over here all the way out there and so we end up getting our usual lagrangian for the klein gordon equation which like i said you can you can derive this you know just by looking at it okay and this is the same same as before so if you ever have an equation you're trying to figure out what the lagrangian is and you can't find your book or something like that just multiply it by variation and integrate and use some integration by parts okay now from this lagrangian using our usual turn to crank approach i'm going to leave out the vectors a lot of time if but technically you know an x without a vector is has four components so this is our canonical coordinate and then our um and our canonical momentum is given by our usual approach pi again i'm going to define that as pi of x the partial of the lagrangian with respect to v dot x and this is just equal to phi dot x so everything's very simple with the scalar field now to make it quantum we have the equal time canonical commutation relations now the theory we're doing today we're never going to have any problems because it's very simple it's a scale of field theory if you had tried to do this for electromagnetism you'd run into difficulties right away there are ways to canonically kind of quantize the electromagnetic field but we have to change some things i won't have to change anything here but just be aware that not everything we're doing here is going to work all the time anyway our canonical quantizations are simply equal time that's the key thing here so the fees commute with each other the pies community with each other and the usual one that we get in quantum mechanics the coordinates this is the important one okay big box around canonical quantization relations okay [Music] then we you know we do the same procedure to derive the hamiltonian we get a hamiltonian density and the hamiltonian density is defined in our usual way nothing changes here and for the theory we have it's going to be pi squared since v dot is equal to pi if i go back up here so all you have to remember is pi is equal to phi dot so i have pi squared minus one half pi squared minus del phi squared minus m squared phi squared and so our hamiltonian density ends up being one half pi squared plus del phi squared plus m squared phi squared and it's important to notice that this is great it's great because every term is positive definite so our energy density is bounded from below and we're not going to have any problems with that so now i'm going to do something we did last time in the classical context i'm going to show that the standard heisenberg equations for operators in the heisenberg picture they're the same i could put them either way as the hamiltonian it's hamilton's equations i showed this in the classical context and it should be obvious that it's going to work in the quantum context because the whole idea of the quantum thing is to replace um is to use commutators with del functions which ends up working the same thing as derivatives but i'll show it anyway so oops sorry oh yeah this is the heisenberg equation of motion for any variable any operator that's not um doesn't have explicit time dependence the commutator of the hamilton hamiltonian with the variable so here you just plug in now since we're taking the commutator with phi the only thing we have to worry about is pi's because phi's commute with phi's so i get pi squared y this should be a y comma t okay and now using the usual leibniz rules for commutators i'm going to take a pi out here and have a commutator of pi y with 5x then i'm going to put it on the other side but since we end up getting delta functions it doesn't matter so this ends up being the integral of a d3y there's an i that will come from the delta function which will give me a minus sign but we're taking the commutators in reverse order so we get another minus sign so we're going to get the integral d3y pi of y comma t del 3 x minus y and this is just equal to pi of x comma t so that's our definition that we had earlier as i said i said the only thing we have to remember here is that pi of x comma t is phi dot so we one equation for the derivative of the um of the phi gives us the usual the definition we had before the second equation is a little bit more interesting so we do pi dot and as before we turn the crank now here what we have is we have um [Music] we have two terms here two feet terms which don't compute with phi so we're going to end up getting two terms here so i'm gonna have one half delphi squared with the pi and i'm going to get another term one half phi squared y comma committee comma pi of x comma t now the second term is routine sometimes students are bothered with this term here they don't know how to handle this gradient well it's not really hard [Music] the best thing to do is to integrate by parts well actually you know this is a delphi dot of delphi so let's just write this term i the integral of d3y one-half use the usual liebnitz rule so it's going to be delphi i'll leave out oh i won't leave it out dot commutator del phi y comma t comma pi x comma t and then the other term is going to be the same thing um i'll write it out okay now still a little confused i know people are still a little confused by this but we can just take the gradient out of here so it really becomes i the integral d3 y one half delphi dot del and the dell you should think of this this is a del y obviously del y i'll put on here becomes the commutator fee of y comma t comma pi x comma t and the other term is basically the same now i can evaluate this because i've taken the gradient out of here so i get i or even better you know what i'll see now just do an integration by parts you know the integral udv just do an integration by parts and i get them so i get a minus i the integral of d3y one-half i'm going to have del squared phi of y comma t commutator plus the other term which is the same now i get minus i the integral d3y i'm going to combine both these terms with the same del squared phi of y comma t now i get i del three x minus y and the integral is the i's cancel the integral is trivial to compute so we just get um del squared phi of x comma t then we have to add the term that i didn't bother to evaluate here this is going to be i left the m squared out of there okay it should be an m squared okay so now i'm going to get the total thing i'm going to get pi dot x comma t is equal to del squared phi of x comma t minus m squared phi of x comma t and um let me just doesn't seem like i got that sign right oh because i'm going to be multiplying i have an i out in front and i'm going to get an i in here so that's right okay and now this remember pi dot pi is v dot so this is just v double dot x comma t is equal to del squared phi x comma t minus m squared phi x comma t so this is the kg equation again everything is consistent okay and we are happy i'm sure the first time people did this they weren't 100 sure that it was all gonna work out and so they were probably pretty happy okay now let's just go back to exploring the quantum field theory again the kg equation is linear so we can solve it we can solve by fourier integrals so we're just doing an expansion here so we write phi of x comma t now for this lecture when i use like a and everything i'm using is the non-relativistic normalization so it will differ a little bit from like shinechi but it's um consistent with other books like i said there's two main ways to do it and you either go a a dagger equal one or you go like a a dagger equal two pi cubed two omega k del 3. so just in case you're wondering so i'm just going to write d 3 k 2 pi cubed 2 omega k the one half is the key here that's what makes this non-relativistic normalization and we're going to write it as a p e to the minus i omega p t plus i p dot x plus a dagger p e to the i omega p t minus i p dot x and you can easily see if you substitute this in the klein golden equation um by definition as usual omega k here just plus square root of k squared plus m squared and you could just take the derivatives and substitute it in there and you see that we solved the equation the a's here are just operators and as we'll show they they're independent of time so just for some convenience i'm going to define fk x as being 1 over the square root of 2 pi cubed 2 omega k e to the minus i k dot x okay and then there's another notation that's commonly used in the quantum field theory books if you have like something like a t this is called like a backwards and forwards derivative it can also be with mu i'm we're going to only use it for time here but it can also apply with space as well and this is just defined as a t the partial of b with respect to t minus the partial of a t with this a with respect to t b of t notice we don't change the opera the order here so you can easily establish i think i do it in the notes that i'll provide along with this but i'm not going to do it here you can easily establish that f k star x comma t f k prime x comma t d 3 x is equal to del 3 k minus k prime this is the like orthogonality relation that you're going to have to uh to establish some results i left one thing out here the key thing here is that fee is real so i could use a dagger if you take the complex conjugate of this you see this goes into this so in order for this to be unchanged ap has to go to ap dagger and ap dagger goes to ap sidneki discusses this a little bit more detail showing why that has to be exactly the case if we were doing a complex scalar field which i may do in a few videos that would not be necessarily the case and then you can also show that the integral of um so using um using our definition for fee it's very easy to show it's very easy to do the inversion i'll do part of it here but it's very easy to show that fk star and so i want to check one thing i think i might have left off the uh the omega k there in my notes um that's right okay so um using this it becomes very easy you know it's just two linear equations by the way if you know it's easy to see what's going to happen here when you just substitute in fee and when you substitute in v-dot it's very easy to see what's going to happen um so when you solve this you're going to end up getting a k is equal to i the integral so this is just inverting some books just show it to you i put in a step or two here but i'm not gonna i'm not gonna show it all these back and forths are with the minus signs are actually difficult to do in your head i found out over the years so that's a k and it's um it's pretty easy to show that this is independent of time because each of these is a solution of the um this is like by green's theorem and each of these sort of solves the klein going equation so it's it's not too hard to show that if you take a time derivative here this will vanish okay and obviously a dagger k is the same thing i mean minus i we don't need the fee star because fee is real okay now [Music] and all the books do this suneki does it um everybody does it a certain way we can compute what we want to compute see when we did the bottom up approach we sort of started with knowing what this was here we're going to just compute it i'm just going to write it down i'm not going to do the whole computation but we have like our ak so this is going to end up being um and we can do this at equal times we can take our fields at equal times because any time for these will matter they're all the same okay so do a little crunching do it on a piece of paper so we get our usual commutation relations for the creation operators and similarly we'll find you know a k a k prime is equal to a dagger k a dagger k prime equals zero so now we sort of come full circle we've gotten back our creation and annihilation operators and they're easy to interpret and everything now if you do a detailed calculation which sagnecki does and i really um i really encourage you to go so you do a detailed calculation and you're going to end up getting that h is equal to one half the integral of d3p a dagger p a p these should have vectors on them but you know what they are plus a p a dagger p and there should be an omega p here so this is our hamiltonian and if you use the commutation relations you'll end up getting this is where we started when we had just our relativistic formula and we said why couldn't we just start continue with this the answer is we wouldn't know if we had lorenz covariance we wouldn't know if we had fields that commute outside the light cone and everything so doing it the canonical approach we sort of establish all that on the way this by the way is our usual vacuum energy and [Music] just um we define normal ordering i'm not sure if nikki covers this or not right away eventually he will normal ordering just basically says if you have a string of operators i'll just do it for two here you always put the annihilation operator on the right and the creation operators on the left so in this case that's really all it is annihilation just a way of crossing this out and because once you get that you're always going to get things like 0 a dagger a 0 which is equal to 0 because a annihilates the vacuum so at the end of this right now and i just want to mention what's the advantage of this canonical approach there are other approaches there's a foim and path integral there's um you know there's the fox space ground up approach that we did earlier the answer is very easy and this also applies to the fine and path integral very easy to add interactions so far we've only treated a free theory but if we want we can just go l goes to l minus lambda phi to the fourth this is going to still be lower ends invariant and it's easier to show that we'll still have all those other properties that we want of [Music] commutators vanishing outside the lycone the problem with the bottom up approach that will fail and we can do perturbation theory because we need the we need the solutions to build up our spocs fox space in other words we need the spectrum we don't have the spectrum [Music] because our equations all right now i'm linear so that ends this lecture um the next few videos i'll have some examples not much more to do with canonical quantization but i'll do some examples and um there's still a lot more to do on scalar fields as we'll see in the coming videos we've got things like knowledge theorem and symmetries and um you know um i might have a couple extra videos on non-abelian gauge symmetries just to show you how it works so i'll see you next time and um thank you very much for watching quantum field theory a student's perspective