so welcome to election number four of the quantum mechanics course today we talk about projectors bras and ket’s and project display key role in quantum theory and in fact we already saw that in the axioms and they deserve to be studied in detail and today we’ll do that and essentially this is a very cheap linear algebra up to the subtlety that we may deal with infinite dimensional subspaces and there is a little subtlety we have to take care of there so most of the first part of today’s lecture will be rather trivial up to one or two points which I would like to draw your attention to that’s the first topic of the lecture the second topic of the lecture are the so-called bras and ket’s that’s the Dirac bra and ket notation and their definition is very clear and it’s based on the so-called reese lemma that we’re going to prove reese like this on the reese lemma their use in quantum mechanics it’s very intuitive and that’s the reason why people like it because you can write projectors using the bracket notation in irresistable each arming simple form well while they use is intuitive its definition the definition of that use is rather cumbersome so if you really want to say how to use them then you have to invent all kinds of extra rules of how you deal with it and what you really mean by it and at the end of the day all of this is already in our standard notation without inventing lots of standard rule of extra rules so we will not use that notation so apart from an initial charm it carries it only causes trouble and misunderstanding and wrong intuitive conclusions and all of that will not use that notation but obviously I introduced it because the textbooks are covered with this stuff and of course people informally talk using the bras and ket’s but you will always want to translate this into proper notation and then you say oh is that what you mean oh now I understand okay or if that’s what you mean well then I don’t understand okay so we have a fully developed notation we don’t need extra notation with hidden rules okay as intuitive as they may seem so we first start with projections and Pythagoras theorem now obviously in finite dimensional hilbert spaces vector spaces there’s not much to be said nevertheless we will consider a separable Hilbert space so let HB age be a separable Hilbert space forever in the future fail to mention separable I meant to say it that HB a separable Hilbert space with the appropriate structure on it then we have the following definition very simple definition let sigh in age be well let’s say being eight then and II in age a unit vector meaning of course that the norm is one then we define sigh parallel namely parallel to e as that part of sigh which you obtain by projection to e in this fashion so that’s just a definition it’s called

the projection of side to e that’s very simple and whatever is left so if you subtract from sy this projected part you call the orthogonal complement okay so this is really is really very simple now here maybe it’s important to note the order the sigh comes second so that this projection map taking side to side parallel is linear this I was in the first slot it would be anti Linea right so please keep the order in mind okay so now this is very simple this is for one vector and it’s interesting to understand this for an entire well at most countable set of orthogonal vectors II it’s interesting to understand the extension of this elementary or these elementary definitions to at most countable sets II eye where the eye comes from an at most countable set capital I and obviously if we’re dealing with a separable Hilbert space this also normal countable set may well be its basis so this definition comes in the form of a proposition because one cannot simply define here one needs to prove that all the limits exist so proposition again let’s I be in age and let the e is with I from some I be an often normal set in age so not necessarily a basis yet for and at most countable set index set I then we have that a well you can still decompose sigh or you can decompose sigh as sigh orthogonal plus sigh parallel where the siper ll is defined as the sum of over all the I in I of projecting the side to the IEF element of this Oh n set times e I and you say well this is simple enough why is this not just a definition because it comes in form of a definition well here again you have this to be psy – psy parallel well the question is if I is countable and not finite does this limit exist it could simply be that I true such a set but this limit simply does not exist right I need to check hence the whole thing is a proposition rather than a definition okay and and that’s something also to be proven if you take the psy orthogonal and you take the bracket with the jf basis element not basis element from this orthonormal set then you get 0 so this needs to be it’s to be shown then B we have Pythagoras theorem Pythagoras holds how does it look like well it says that if you take the sy norm squared that is the norm squared of the orthogonal complement plus the sum I in I of all possible absolute value squared e I sigh like this and that’s obviously Pythagoras theorem but now we could have countably many summons so it’s a little more than Pythagoras theorem extended to D dimensions its Pythagoras theorem extended to countably well to infinite dimension on a separate

separable Hilbert space now moreover there is a useful characterization of the parallel component of the projection to this set II so that’s the third part of this proposition for any gamma that lies in the span of all the EIS you remember that the span is the set of all finite linear combinations of such vectors although this might may be countably many the span is the set of all possible finite linear combinations for any gamma in the span of the eye and let me just draw in finite dimensional picture in order to see what is mentioned so if this is e1 and e2 but the point is that this is holds for infinitely many the eyes if I have two E’s and say I’m in a three-dimensional space in total okay and then I can look at a particular vector sy as I usually as we usually do and you can consider the projection of this sy down here into this span okay but now imagine you consider an arbitrary gamma in the span so an arbitrary gamma in the span may be that guy you agree that lies in the span that’s the sy and the red guy here is what you usually call sy parallel right that is the projection in today and if you want to complete the picture although we don’t really need this at this point then this guy here would be psy orthogonal okay so that’s the situation after psy you have the projection into this span that society ll define here and you have an arbitrary gamma now the claim is for any gamma in the span we have the estimate that psy – the gamma so that’s in a sense the distance of the psy from the gamma so that’s well like in colors here let’s take red again now let’s let’s take white so that’s this one here psy minus gamma is this vector here right psy minus gamma is this one here and now compare the length in this intuitive picture of this one to the length of this side of four g”l and then you see ah this is always greater or equal to psy orthogonal length right so this sigh another form psy parallel psy parallel is the one vector the unique vector in the span that is closest to the side because equality holds if and only if the gamma coincides with the side parallel so that’s the idea that’s the picture but of course the pictures here drawn for three dimensions to aid memorizing this but of course the claim is here that this is true for for an arbitrary at most countable index it okay so let’s prove the whole thing let’s start in the beginning and well actually the proof is extremely simple because at first assume I is finite and then of course we may assume that I just goes up to some number n that’s the index set and then the whole thing is really very elementary linear

algebra we very quickly go through this so these two guys are actually definitions it is clear from these two definitions that this one holds so this is the definition this is a definition because of the second definition well the second definition is just this claim so nothing has happened okay nothing has happened if this is a finite sum because then I know it converges well it’s a sum it could it’s well defined the only thing one has to check is this guy here well yeah that’s not too tricky so let’s plug that in we use the definition from here it’s psy – psy parallel but this I parallel is I equals 1 to N it’s a finite sum of the e I comma sy e I and that’s in a bracket with an EJ and the J is arbitrary from I ok so well now it’s just an application of the rules here I can of course this is a finite sum I can pull this apart I can do all possible kinds of things I could write sy comma EJ – now I pull the sum out as I equals 1 to N well I pull this guy out remember that because this is in the first slot pulling out a scalar requires that the scalar gets a bar right because the product is anti linear in the first slot so you pull that one hour and then you have an e I comma and EJ well then we assume that you have an orthonormal set so with this funny symbol Delta IJ you see that the sum only flips only has a contribution if the eye is a J then you can use the property of the inner product that this actually equals psy comma E I bracket like this sorry yeah ooh I like this with this Delta IJ the whole thing is 0 this is what you wanted to show because this one is the sigh orthogonal ok so I mean this is really very very elementary if the eye is finite well similarly I’m not going to do that similarly prove Pythagoras for finite index set I okay in the future if things are that simple I will not show them but here I’ll do that for a certain purpose you’ll see in a second so finally the C part when we just start evaluating sy- gamma now the gamma is supposed to be in the span of the EIS that means there exists coefficients gamma I such if I go over finitely many I have finitely many coefficients I can expand this in since gamma is in the span of the e I by assumption right so I can write this as psy orthogonal plus sy parallel because of step a – well – this this gamma but that amount – if you look at this this is sy orthogonal plus the sum while the site parallel goes from 1 to n e I sigh sorry sy e I – now I put them all under the same sum I can certain do this because to highlight some otherwise I might worry about ordering ambiguities something like this here then the norm the norm closest well distribute it will squares here and then you use Pythagoras that we have just shown you know that everything that has an expansion in E is is in the parallel complement isn’t a parallel part so this is this plus the sum of each individual contribution it’s the e I Sai – gamma I absolute value squared this is from 1 to n and because

these are all non-negative we have the estimate as claimed but all of this only for finite in the exits now we need to extend the results to i countable so that’s the essential steps you see the finite case world is finite sums that’s child’s play but how what happens if you go now to countable well you know what you need to show based on the finite formula I find that I mean for finite in the index sets we need to show that all limits exist because you can show all these formulae and then you just wrote them for the partial sums to draw to speak so you definitely need to do this in a finite case you need to do these final calculations but once you did them you just see whether all the the limits exist so here the only thing you need to check is whether this limit here exists whether you need to check whether I equals 1 to n is partial sums over a I sigh e I you need to check whether the limit n to infinity exists because this limit is exactly what we defined a sigh parallel right we wrote sigh parallel is to sum over all I in capital I of this thing so does this limit exist well when do we know whether is sequence so this is the sequence of partial sums we know a sequence converges if it satisfies the criterion for convergence however this is a sequence sequence of partial sums in our Hilbert space right it’s a sequence in Hilbert space so do I need to show convergence or is it sufficient to show that this is a Cauchy sequence well of course it suffices to show that these partial sums are a Cauchy sequence why well because the hilbert space is complete i know if it’s a Cauchy sequence it certainly converges so you see now so far we have to prove something as it is complete but now we’re starting using it right ok so we need to show where this is a cushy sequence so it suffices to show that this sequence of partial sums is cushy yes ok well yes ok so I can go up to here for the camera I don’t want to move the blackboard all the time because in the last row you can get up and get a better angle also follows by this part C of the theorem okay so um well just consider just consider the following estimate we need to check it’s a Cauchy sequence right so just check what happens to if you start with I at some M and you go up to some n so take N greater or equal to M and check what you know about this guy what do we know about it from what we already derived so far well we know that

this is I sorry that this is I equals m to n of the absolute value of e I sigh absolute value squared right this is Pythagoras there’s no parallel orthogonal part here and we know that this guy here is less or equal than the norm of sign well this also Pythagoras right okay well but that means psy is for any fixed sigh this is a number so this is a bounded sequence so from this we conclude this sequence is bounded from the fact that it has only non-negative summons we know that it’s increasing it’s monotonic monotonic Li so it’s bounded from above if you want to be very precise and it’s monotonically increasing what do we know about a sequence that’s monotonically increasing and bounded from above converges okay that’s right if it converges it is certainly Cauchy because any convergent series is Cauchy but if this is Khushi this sequence is Cauchy then this sequence in here is Khushi hence the sum I equals m to n e I sigh e I is Khushi hence it converges hence psy parallel as defined in the theorem is indeed an element of the Hilbert space that’s it that’s it so you see you really do only have to do this stuff one usually does and there’s only a very small extra argument that you have to provide in order to check whether your infinite sum really is an element of the space where you claim it as an element of okay so very often again proofs in the separable hilbert space case that’s infinite-dimensional only require a little extra work but that extra work is required in order to make the theorem true rather than maybe falling into the trap of stating something that’s only true in the finite dimensional case and only very few quantum systems that we actually want to look at in physics would then be treatable there would be a nice exam question right producing that and you will be very careful to not only check the finite case but to check existence of limits okay so um now we look at the relevance of closed linear subspaces of a Hilbert space H now our theorems will very often say something about this set is closed and there’s the closure of this set and so on and that’s a little or maybe at first puzzling why closed closure topological closure comes into the game and what is the relevance of closed linear subspaces well is this not any not any linear subspace let’s call it M of H what is the linear subspace the subspace is a subset but a linear

subspace is one where for all elements M 1 M 2 in the linear subspace and for any complex number it’s true that if M 1 and M 2 are in the subspace that M 1 plus and that’s the plus on the Hilbert space lambda times it’s the times on a Hilbert space right that this linear combination is again in the subspace so it’s a sub vector space a linear subspace is a sub vector space not any linear subspace in an age that is sub vector space is a sub Hilbert space which you might have assumed because while the hilbert space has more structure than a vector space has also in a product it needs to be complete with respect to that inner product for it may fail so this M may fail to be complete aha so this is the problem of just taking a sub vector space and we have a little Emma from topology which we’re going to prove in order not to confine this to mystery lemma a closed subset so you see now we’re no longer looking the linearity is not important at the moment a closed subset M age of a complete space is again complete aha so in order to preserve the completeness property we cannot take any old sub vector space we need to take a topological closed subset okay and the topology we’re talking about is of course the topology on H which is the softball open ball standard topology induced by the norm okay aha proof how can we see this it’s quite instructive to see this well we want to show that a compose subset M is complete if it’s a subset of a complete space what do we need to do we need to consider a Cauchy sequence let’s call it sy n yeah in the subspace if we are able to show that this Cauchy sequence converges in the subspace so that it has a limit that lies in the Subspace then the Subspace is complete right differences seek a Cauchy sequence converges well in the space whose completeness we want to show ok consider a Cauchy sequence we need to show that limit sy n n to infinity exists and is an element of the subspace it it’s not enough if it exists somewhere it needs to exist in M now since the sequence sy n is the sequence in M it’s certainly a sequence in age right it’s a Cauchy sequence in H in the ambient space 8 hence it converges in H so we know that the limit n to infinity sy n exists but we know only that it exists in H right it exists the limit exists in 8 the question is does it not only lie in age but does it even lie in the subset M of H in this subset of which all its members come from but does the limit also lie in there ok well if that is true now since M is closed so now we’re going to use the closure condition the closed subset M what does it mean for m to be closed you tell me

M is closed topologically what’s the condition for M being closed that’s right age without em is in the topology in views the standard topology induced on H since M is closed H without em in all is open right that’s true now suppose we make an assumption which we’re going to reduce to a contradiction okay now suppose the limits I the limit of sigh n is not an element of Emma so let’s assume it is simply not true that it lies in M which would mean our theorem is wrong so it’s proof by contradiction now suppose the limit is not in M well because H without M is open then there exists an open set you in age without M that contains the limit right because if H without M is open and I have a point that’s not in M its and therefore the limit is an element in age without M if that is open around every point of an open set you find an open set that lies entirely within there right but then there exists in any in the integers such that for all N greater equal to n the ends element of the sequence lies in that open set because it’s an open set around the limit point the sequence converges in H the limit point lies in age even an age without M if you now know yeah well if you know that’s an open set around the limit point then from a certain sequence member onwards all sequence members must lie in this open set okay but this u lies in age without em and at the same time thus I n all I’m a in M why well because the sigh n was the Cauchy sequence in M so it cannot be true that they all lie in M and from short member onwards they do not lie in M contradiction hence our assumption is wrong hence it’s false so hence the limit n to infinity say n is in M but that is what we wanted to show so the Cauchy sequence in M any Cauchy sequence in M because we made no special assumption as to what Cauchy sequence this is converges in M that’s it it’s a very simple topological fact based only on the definition of closed subset and the and the definition of convergence with respect to a / – Paulo G okay so all is fine if in addition we require this linear subspace to be a closed linear subspace in the sense of topology maybe that it’s complement is opening so corollary from this simple fact a closed vector subspace or a linear subspace it’s the same just different terminology terminology for the same thing a closed similar linear subspace I mean H is a sub Hilbert space of course with the inner product induced from eight we just use the same inner product but you only apply it to the elements of M and additionally if h is separable which you might have assumed I mean in this theorem anyway yeah you can assume this I just emphasize it if h is separable then also M is separable obviously it is

it’s you can find the sub bases okay so now you understand why in many places we talk about closed subsets or we’re interested if we construct something we’re interested in checking whether the result of the construction is a closed subset because if then also linear subsets we know are we have a sub Hilbert space why is it so great we have a sub Hilbert space well for instance you know you conclusion of orthonormal basis on it right otherwise you have no justification you know that a separable Hilbert space has an orthonormal basis but if it wasn’t a separable Hilbert space you had no justification assuming so so certain constructions you couldn’t do by the way the converse does not holds it’s not true that only closed subspaces can be sub Hilbert spaces but that virtually never arises in practice okay so let’s immediately proceed – orthogonal projections so again if you have a first for physics I assure your projections are all you need essentially what you need to extract predictions from your calculations in quantum mechanics right to extract expectation values and things of that sort so this is very very important to study this so now we extend this whole thing to the following construction so definition let M in H be a in brackets not necessarily close so we will not assume this from the beginning not necessarily closed however linear subspace of H okay so you know okay it’s a linear subspace but it may not be a sub Hilbert space so definition is then the set M orthogonal which is defined as the collection of all elements in the hilbert space such that for all em em em sy or musai let’s call this mu mu named moose i equals 0 then the set M orthogonal is called the orthogonal complement the orthogonal complement of M in H property or properties M orthogonal is a closed linear subspace of M of H aha so although the set M we started from although the set M we look at whose orthogonal complement reconsider may not be a closed subset once you take the ax form and complement the a formal complement is a closed subspace okay so that that’s great so we know are the orthogonal complement of a linear subspace is certainly a sub Hilbert space let’s see why this is true well it’s first check that M dual is a linear subspace well this is very simple because let’s take M 1 and M 2 in M orthogonal no let’s take sy 1 inside two in M orthogonal right then due to the linearity of the inner product in its second slot however this is not so relevant here because you have a zero here so if it Desai if I had defined the sign the first one you could switch it

anyway and then the complex conjugate of zero is zero okay but in its second slot psy 1 plus I 2 is also in the complement and also lambda times Sai 1 is an element of the complement for all lambda in the complex numbers so that’s trivial because this inner product is linear in its second slot ok well it remains to show closure to throw that M orthogonal is closed well the first thing we note is that M orthogonal is the intersection overall mu in M of the preimage of the 0 in the complex numbers under maps Omega mu I’ll say a word about this in a second where the Omega nu Maps right so the MU comes from M is a map from H to the complex numbers that assigns to each alpha in H it assigns the complex number mu alpha so why is this true well I the M orthogonal as you see from here is all the size such depth is each side such that for all new condition is satisfied well this condition is that Omega mu of Sai zero said right so maybe to make this clearer I call this instead of alpha I call this sign so this is this zero and one map picks a particular mu but the M orthogonal is all the size where this is true for all the MU so it’s the intersection of all those sets and the pre-image of Omega mu of the zero is exactly this condition that this I belongs in here okay so this is just by definition this is just a rewriting of the definition of M orthogonal nothing more nothing less okay now I’m whenever you deal with topology and you deal with unions and and intersections you know that the Union the arbitrary and arbitrary union of open sets is open but only a finite intersection of open sets is open however we want to show that this is closed and if we could show that each individual pre individual pre image for a fixed mu was a closed set then we know arbitrary intersections of closed sets are again closed okay so we already have we almost we would already be there if we have it we will already be able to show that this is closed if we only knew that this is closed forever emu in M because then an arbitrary intersection of closed sets is closed okay so this is why this is it’s clever to rear a rewrite it like this remain to show that M orthogonal is closed at suffices to show that for each new in em the preimage of the map Omega mu of the set containing only the complex zero is closed okay so consider this set is a subset of the Hilbert space right because the map goes from H to C so we consider the entire Hilbert space without this pre image Omega mu of the complex set containing complex 0 obviously we want to show that this is open if we can show that this is open we showed that this is closed what is this this is the set of all the sigh in age for which the condition that qualifies you to be in

here is not true well such that for all know not form you there’s no mu we have a fixed mu such that mu comma sy is not 0 right because we take the complement if you didn’t have the complement would be those where it is 0 this is where it’s not 0 well but this can be rewritten cyan H such that mu sy is an element of any complex not at the set of any complex number without 0 it can lie anywhere but there this years of course Omega mu of sign but hence this is the pre image of Omega mu of C without the zero by the definition of the pre image aha well but we know that this year is open because the one element set is closed in the standard topology on C so this complement is open in the standard topology on C which were of course always employing on C but now Omega mu being just a map that takes a sigh to new sigh this is the inner product and the inner product is continuous in the second slot thus Omega mu is continuous thus we know here that the preimage of a contain the preimage of an open set with respect to a continuous map is open because of continuity hence we know this is open in the topology on edge since this is continuous and this continuity goes into here done that’s we showed that any such premature Omega mu of the zero is closed and here we have an arbitrary intersection of closed subsets well that is closed that was our claim any questions but if that’s not the case let’s have a seven minute break to a half past and we continue so let’s continue so definition the map P taking H down to M that sends sigh – sigh parallel where the PI sigh parallel is defined by the set M here let me see do I need a closed set I think I want the closed set yes so H to M and let this be a closed linear subspace so now I do no longer allow M to be just any linear subspace now I want M itself to be a closed linear subspace if it’s a closed linear subspace it’s a sub Hilbert space that means there exists an or n basis for the M itself agree because it’s a Hilbert space so there is a separable Hilbert space there’s an or any basis for m and then this parallel is defined with respect to this or any basis you remember we defined the siper ll as this sum I in I where I is now the space could be

the spaces a I sigh AI of em so this map P is called the affordable projector the a formal projector to the SAP Hilbert space M and now you see if this was enclosed you understand our disclosure criterion if it wasn’t closed you were not guaranteed that you have an O n basis then you wouldn’t know how to define this I fell in the first place okay it’s called the authorial projection projector to the sub Hilbert space M it has two following properties and let’s call this map P down M because it obviously depends on the choice of M the first property is that if you take pm and you execute it after itself again you still obtain the same map and sometimes this is written p.m. squared this p.m. of course the square is here the square with respect to composition the second property of this map is that for all sigh and Phi in the hilbert space the inner product P of p.m. acting on side with v you can actually take the PM over here assai p.m. v all this follows from the definition and the third property of this guy is that if you take the P not of em but of its orthogonal complement why may we do that well because if M is a linear subspace which certainly is then the complement is a linear subspace the closed linear subspace so where I’ve wrote the M here I could also write the M orthogonal this guy of course goes from age to M orthogonal and it takes the cyan h2 thus I orthogonal we had before that’s a simple consequence okay and fourth the p.m. map is bounded it’s a bounded map from age to em you remember the symbol for the bounded maps this curly L is a bounded map and the proof is immediate and will be found on the problem sheet okay so there’s not it’s not a big thing quite interesting is the converse quite interesting and heavily used is the converse so theorem let P naught P dot sup in but just appeal map that’s a bounded map on H have the properties first property P after P is P second property for all sci fi in H the inner product P sy comma Phi is the same as IP Phi later on we’ll call this self adjoint miss and we will call em so equals be call this the projection property and the self adjoint Ness what that will be more subtle is simple here the self adjoint is because this defined on all of H we will come to that

nevertheless at this point we need like this have the properties then the range or some people call it the image of H under PI of P is closed and the P is the P from before so you have to P from before where you have a set down here where you put a closed set down here so the range the image is of course the P apply to the whole domain I claim this is closed and then the P you have must be the P that has been constructed in the way we did in the previous statement so for any map that’s bounded and it has this property in that property you find a closed subset may be the range of the map such that it’s the projector through that closed subspace okay so any projector can be written like this right so the moral is any projector is the orthogonal projector to some sub Hilbert space and the sum is a slight understatement because we even know which one in with the range of the map and it’s good to practice this sort of proofs on the problem sheet so this will be quiet heavily used so final section is three Slama bras and ket’s so the Ries lemma is saying about how you can understand the dual space of a Hilbert space but let’s start differently let’s say consider a lambda in the Hilbert space so lambda is usually a complex number now lambda is an element of the urban space in this case so consider lambda an element of the Hilbert space then you may define linear map let’s call it L sub lambda it’s constructed from the lambda it takes you from H to the complex numbers in the following fashion alpha is being mapped to lambda comma alpha well this is clearly linear since the inner product is linear here in the second entry so this is a linear map it’s also a bounded map how do we find out well we calculate the operator norm of the sky do you remember what the operator law of a map was it was the supreme M over all elements in its domain of the quotient you apply the map and in the resulting space you take the norm well because the complex number is the absolute value and

you divide by the norm of the element right now this is the soup alpha in H of lambda comma a of course we just defined now you know I’m sorry and here we have an absolute value and now we know from vessels inequality now I’d nonsense from cauchy-schwarz we know that this is less or equal to the soup alpha in H we take the norm of lambda times the norm of alpha divided by alpha so this is of course just the norm of lambda but because you have a fixed lambda this is of course less than infinity thus the L lambda is a bounded map from H to C right and how do we write the set of bounded maps we have another name for that the space of bounded linear maps from H to C so dual space H star so again I emphasize H star it’s not all linear functionals from H to C but in the context of Banach spaces and hints of Hilbert spaces it’s only the bounded linear maps from edge to see they also emphasize as shown one of the problem sheets the bounded linear maps are always continuous maps as well bounded implies continuous it’s actually if and only if so okay so that means you can take an element of the hilbert space and you can construct from it using the inner product a bounded linear functional big deal well the question is can le any L in the dual space so can any bounded linear functional from H to C be uniquely written as L is the L of some lambda so the L sub lambda is the construction we just made and can any L any bounded linear functional be written like this uniquely and then we have the Reece lemma very short lemma it says yes okay that can be done and to the yes that belongs a why so proof so let’s do this we have to find for any L in any bound functional we have to find the lambda that does it well that’s one way to show it constructively and that’s what we’re going to do so however there are two cases case one is that the linear functional we’re looking at is the zero functional what’s the zero function that’s a function sense any cubic space element to zero in the complex numbers if that is the case then take lambda to be the zero in age and you’re done because obviously the inner product of lambda being zero comma something sends everything to zero so that’s the first case you’re done by choosing lambda to be the zero vector however if that is not the case if you have a linear bounded functional that not the zero functional what can you do then well then you know that the kernel of this L you know what the kernel of a linear map is okay for me yes that that’s that’s the M or a rather it’s the set of all the agents that are sent to zero by the map right so then the kernel of the map being defined like this is of

course always a subset of the he’ll be of the domain but in that case it’s certainly a proper subset because there are some sigh that are not being sent to zero because otherwise all the side would be sent to zero and then this would be the zero map okay now the kernel is of course a linear subspace of H okay and in fact by its definition it’s easy to see it’s also a closed linear subspace of H it’s a sub Hilbert space so then the kernel is not the entire Hilbert space but from that it follows that the entire Hilbert space can be written as the kernel of L plus the kernel of L the orthogonal complement why well that’s the projection theorem we had earlier this is closed closed subset closed sub vector space so it’s a sub Hilbert space these two together that’s the decomposition we had before as the whole space one sometimes writes in a direct sum here Hilbert space can be decomposed like this but because this is not the entire Hilbert space we conclude that the orthogonal complement of the kernel is not just the zero it’s not just zero vector in H which it would be if it was the entire hilbert space because the dual of the entire bird space is just a set with the element zero okay so if the horrible complement of the kernel is not only the zero then we can pick some sign from the orthogonal complement of the kernel and some sigh that is not to zero in h why can we do this well because the set doesn’t only contain this one well I’m good maybe be very clear here and say it’s more than this right so now if you can pick a vector that’s a non zero vector then without loss of generality W log wanna be ranking the item in height without loss of generality can choose excite you need vector because if it isn’t u divided by its norm and it becomes one so you see we have existence of such a vector and if we’re now going to use it to construct something we better show independence of the choice later on so there is whatever we’re going to do there’s some well definition issue hanging over our heads but we’ll clarify that in the end we just proceed at this stage I mean question okay so without loss of generality we can choose side to be a unit vector and then take remember we want to take one to provide some lambda that does the trick well in this case in the second case we take lambda to be the application of the abstract map that we’re given now we’re given some L in H star we apply it to the X I what’s the result well it lies in the complex numbers right then we can take the complex conjugate of it bar and we can multiply it by the vector X I and what do we get well obviously we get a vector in the Hilbert space right I mean more precisely we get a vector in the dual sorry in the orthogonal complement of the Connells which is of course part of the Hilbert space well we can do this and why would we do this well obviously because that’s our candidate for the lamp that we need to prescribe in order to get our L for then let’s consider L with this such chosen lambda apply to

some alpha which could be any alpha in the hilbert space and compare the result to if you took the abstract map you were handed okay so if we can show that this difference 0 we showed that the a lambda map is the L map which is wanna show right okay so let’s write this down what is this a lambda on alpha this is by definition lambda comma alpha and the lambda is this one – its L of side bar sy comma alpha – well L of alpha now comes the tricky trick its L of alpha times 1 they want this ah yes I do yes yes okay good sorry got myself confused for a second now this is just a complex number right we can let it jump out and then we let it jump in again but to the second slot so it’s Sai L of Sai Baba nothing anymore alpha minus L of alpha and we let this one jump into the second slot Sai L of alpha X I agree so it changes the power because here it’s anti linear in the first but linear in the second slot so but now by linearity in the second slot additivity in this case only Sai have l sy alpha minus L of alpha X I so again I’m to emphasize that X I is an element of the affordable complement of the kernel right that’s where we chose it from now if we could show so I first claim it and then I show it if we could show that this is an element of the kernel of L then I had an inner product between an element of the kernel and an element of the orthogonal complement so what’s the result if that is true zero so that’s what we want to show so it remains to be shown that this is true so how do you check whether this is an element of the kernel well you apply the map and see whether apply it to this object if yield zero well but because it’s a linear map you can pull this out and you get L of sy l alpha minus L alpha L of sy all complex numbers and that is zero hence this element of the kernel hence this is zero hence we found the lambda well we didn’t find the lambda we found a lambda because our choice of psy which we use to construct the lambda well it was just some choice in the set so we need to finally answer the question where this is a unique lambda we found we certainly found one but do we find a unique lambda in age but did we find this way a unique lambda in age mind you you could choose to side if rent ly but it will always use the same lambda by the above construction so let lambda 1 lambda 2 be two such constructed vectors so that in particular L lambda 1 equals L but the L

lambda 2 reproduces the L 2 then you have L lambda 1 minus L lambda 2 applied to some alpha in the Hubert space well you know it’s 0 because the maps both reproduce the same abstract l but then you have lambda 1 alpha minus lambda 2 alpha is lambda 1 minus lambda true alpha but this is true for all alpha and H this is 0 and because it’s true for all alpha and age you can conclude lambda 1 it’s the same as lambda 2 thus the whole constructions unique and there is lemma is proven remark the proof provides for a bijection from H to H star and because of the construction here it provides for a linear by ejection from 8 to H star okay so this is why I can say the the H and the H star they actually the same space right there the isomorphic to each other’s vector spaces and so that’s this duality in hence very often in quantum mechanics when does the following thing one says why would we bother about this complicated space H star if all it is its obtain its elements are obtained by taking element of H a unique element of age and writing down a bracket where the first slot is filled with this element from age and the second slot is left open so then we have a linear map so it’s a way to think about the H dual of Hilbert spaces that seems to avoid having to understand the H star properly like we did will did a long time ago right so I’m Dirac bra ket notation so the whole thing is is pretty clear since for any element L in the dual space there exists a unique lambda in H due to the Reece lemma as we just showed such that L of alpha is the same as lambda comma alpha or what we could also say is L open tslot so the map L as such is L is lambda comma open slots so the L acting on something you leave this slot open here to you can plug in any element from H right so this is then a map from age to C because it can still eat something Dirac suggested the short notation l equals bracket lambda and because some people write the inner product not with a comma but with a line here like this line and that is it and British humor has it that because this is a bracket well actually it’s brackets but well let’s say it’s a bracket then most obviously they say the first half of it must be the bra and whatever comes behind is the ket and they left out the C okay

and of course bra has other meanings too but you don’t want to think too much about this notation because I don’t think it’s politically correct nowadays anyway so so it’s it’s the it’s the it’s the bracket and and you’re supposed to to write this like that there’s absolutely nothing wrong with it because of the Rhys lemma there’s absolutely nothing wrong with it okay hmm so far and he further suggests it so far so good I’m with him there so he further suggested to then write and now it becomes a little funny to take a lambda in the hilbert space and instead of writing the lambda you already dress it with a ket so whenever other we other people that’s us so whenever we write lambda as an element of age which is perfect notation we know exactly what it means to be an element of a set he suggests that you dressed the lambda already with a ket why well because he says the whole life purpose of such an element here in such an element there’s that they meet right then gray if we have an L and we apply it to a side we can write the L like a lambda say so and we can apply it to the side but if you don’t write brackets and just apply it then the side looks like this according to the rule so that’s the notation and then because quite obviously for some people double lines must become single lines it looks like this I mean is this not so wrong because it says that if you have an inner product you can always consider this inner product as the application of The Associated functional on the side in here yes you can again Reese tells you this is one-to-one this really fits together okay now obviously one could say well it’s a little funny that you use half a bracket whatever that means okay I mean obviously it’s not half the inner product it’s just it’s just a notation it’s a little funny to use it here because this has nothing to do with it in abroad it makes more sense because really finding this duel you need the inner product to associate a dual uniquely with this lambda okay so you see there is it okay now you might wonder why on earth would you want to do this while earth would you not would you exchange a beautiful clear thing a map that takes an element from a hilbert space and maps it to a complex number only because you can during due to the res lemma why would you interpret this thing in a sense always like compulsively as as this as something like this well there is an advantage to it and that advantage is mainly one that you can memorize formulae more easily okay but that’s just for the little people right I mean you don’t want you don’t want to memorize formula easily right you want to memorize them right for the questions whether this easy memorizing comes at a cost when it comes at a terrible cost okay and so that’s why so the easy memorization thing and it’s what people would like to get when they use the Dirac braket notation well let me write down a few things because we’re not going to use it anyway I just make a few remarks of course you must be able to recognize what people mean and sometimes you will have to decide they write down this or that well is it this or is it that and sometimes it’s not so clear so um this all is perfectly fine so there’s nothing wrong with it at this point the gain is that some expressions are quite easily memorized using Dirac

notation so for instance take this statement that any sigh can be written in terms of an orthogonal basis as this decomposition I mean this is for advanced students because you have to remember that the sigh goes to the second slot and then this I hear the I hear and nothing I mean you memorize this formula anyway however if you’re inclined to follow this Dirac notation say aha Hilbert space element it gets dressed it was to make it before it gets a ket now you have I in I now you see not a problem here you have a bracket anyway so let’s write this e I sigh well single lines are omitted first extra rule and you write the e I here also with a well I should actually write it like this VI sigh if you’d write the eat you should rest bei as well so far nothing has happened because this is a vector this is a complex number vectors are multiplied from the left by complex numbers because you have an S multiplication nothing has happened I only only dressed up the whole thing well that is not the the big achievement it goes further you say aha but oh it would be also nice if I wrote it like this ai ai and you put the sigh here or in other words you allow the cats to be multiplied also from the right by a complex number well in high school you do this right vector times two two times the vector or something like this well we can do this so we invent an extra s multiplication that acts from the right with the complex number on a vector and yields exactly what the S multiplication from the left yields okay second extra rule you have to keep in mind you see all these extra rules you have to keep in mind in order to make sense out of this now it gets a little more interesting because what you do then often is you’re right this is e I kept bra AI and you say well quite obviously but I’m not too sure quite obviously you can take this sum out you first execute the sum and then you apply it to the side aha so what do you mean quite obviously because now I first take a sum over half an inner product and here there’s also huh okay so they say no no no we’re not that naive actually there’s a tensor product involved okay let’s do that so you write it like this which is actually which you will see very often I assume this expression you will see very often that this is written like this okay but remember up to this point we already had three extra rules the third extra rule I tell you now and that extra rule comes with a sub extra read to interpret it I’m being mean right but it’s a little like that right okay I wanted to make you sensitive for all the extra implicit rules you have and then the question is if they’re so implicit why do you have them at all if you don’t need them in the first place but okay so um so the whole thing one says is no no no it’s not that simple or not that naive what you actually have here where’s the yellow what you actually do have here is some I’m sorry what you actually do have here is you have a tensor product here they say it’s just the suppressed tensor product symbol okay we can live with that so there was a suppressed tensor product symbol that that acts under on the side okay we suppress this that was the third extra rule now what is what tensor product is this well this guy here lives in H this guy here lives in H star agree because one is a kept and one is a bra so the whole thing and certainly the Sun because that’s the definition of the tensor product the sum this LM this whole element here is an element of that space that doesn’t look too bad because that’s exactly what an endomorphism is right it keeps an element from age and produces an element from age and this is the decomposition however usually if you have a tensor product of two maps F tensor of G then

you need to feed it with elements lambda mu and the definition is that the F acts on the first entry and the GE acts on the second entry and in between here you have a multiplication okay which if this these are maps that send you to see is the multiplication see and whatnot so actually we need to submit here not only one element that can be eaten by the second one but we also need to submit in the first argument that can be eaten by the first one however there is the sub extra rule that it’s not meant that way and that you keep the first slot open so actually the whole thing is meant I in i ii i.e i tensor tensor III so the whole thing actually keeps an open slot here and then submits this Desai in there aha mm-hmm okay so this is how it is meant if I also make this extra rule so my counting is roughly about four point five extra rules so far all right for what for what extra rules just in order to write that the identity on H is I equals one to infinity e AI e I which looks pretty and you can say oh I just inserted the identity in front here well what hasn’t been totally clarified yet is how does it work with the infinite sum can you actually pull it in and can you pull it out well of course you can if you think about it in the proper formulism you can prove all of this so if these were finite sums then no okay if the infinite sums you need to check anyway well you can check and it kind of works out and so on but it’s a lot of extra baggage it’s a simple notation which looks like you see the appeal the appeal people talk about which I fully understand right of course I understand the appeal the appeal is deceit something from the right and once it added it becomes a number and then you have vectors left okay but please please I mean is this so difficult right and here and you see here the you submit something which is a ket and and then maybe it comes in the bracket and it it’s it’s a mess right I find it’s a total mess and there are many other places if you want to talk about self adjoint operators hermitian operators and so on well once you know everything is fine the notation doesn’t produce inconsistencies anymore so if you know that all the objects you use already have all the properties you want from them in quantum mechanics the notation kind of works but how do you write down the conditions using this notation because what if it doesn’t work then the notation doesn’t work so then we formulate conditions on our objects in terms of a language that only works if the objects already satisfy the condition this is all not so good right it’s all not so good nevertheless it’s heavily used and once you see something like this I think best the best thing to do is to say well this is meant in this way this is meant in that way and that is meant in this way and then it’s very clear in the first place so you translate this into that and you’re done and here we prove all the theorems ok so that’s my argument and here you know the bracket is a linear a pie well is a sesquilinear map from H cross H to C has these properties is continuous you know what to do with the sum sure you know what this object is and nobody on earth knows what this map is sai goes through cat’s-eye what is this right this is not a map this is not something that’s happening that what’s the meaning of this right it’s just notation why would you do that ok so so my point is yes this is heavily used in the physics literature yes you need to know how to deal with this but no it doesn’t clarify things at the end of the day it doesn’t simplify things if you’re really serious about the formulae and what you’re going to derive and prove it’s not making it simpler it’s making it terribly complicated because why on earth do I need a tensor product with all of this in order to write down something as simple as a basis expansion right so this is my little polemics against the Dirac braket notation which at times it’s it’s quite neat to to understand you can insert this but really well if

you can prove all the other stuff you don’t need this really ok good so did you did you use that in the experimental part so far did you see a bra a ket no okay good good good okay so we’re still ahead there’s nothing well I mean there’s the problem because I obviously need to provide for the for the experimental part I need to provide some theoretical bits and but I’m happy so so so far this we’re not behind that’s good that’s good okay good so this actually completes our discussion and the techniques that we need to understand Hilbert spaces as far as we need them what we’re going to start next time on Wednesday is we’re going to look at self adjoint operators and we’ll make very clear what the difference between self adjoint nests and symmetry is symmetry almost looks like it it’s what is usually substituted for self adjoint nasaan finite energy vector space is exactly the same thing or for bounded operators for unbounded operators like the momentum in the position almost any operator in quantum mechanics if you’re not just looking at a spin 1/2 system right but any other system in quantum mechanics that’s of interest in the lectures right harmonic oscillator hydrogen actor molecules whatnot okay particle in the well all of that they heavily employ a unbounded operators and we need to get self adjoint as under control for all these cases and we need to get the spectral theorem we need to understand the generalization of the eigenvalue decomposition that you normally have right for finite dimensional hermitian matrices or real symmetric is the real version of that you can decompose into eigenspaces and you get a diagonalization of an endomorphism on a vector space in a certain basis in the eigenbasis diagonal has the eigenvalues on the on the diagonal right so but this is only in the finite dimensional case and it doesn’t even suffice well it doesn’t suffice in the infinite dimensional case certainly not if your spectrum has a continuous bit because then the eigenvalues can impossibly can’t constitute a continuum because your basis is at most countable so your matrix if you think of matrices well we don’t have any that’s the point but if you thought you had any could go on forever I you say well that’s still accountable right but the point is you provide lambda 1 lambda 2 lambda 3 lambda 4 but how would you provide a continuum of lambdas right you can’t so the mathematics of matrices and projections to also normal bases just thinking of matrices simply hits the roof it’s not sufficient it’s not the right mathematics even though it almost looks like it could be but it’s very simple to show that it cannot be the right mathematics and in fact we already provide the groundwork to really write down the exact definition of self-adjoint operators and they will then be decomposed via the so-called spectral theorem that’s the next big goal we have we’ll have to work a little until we arrive there and that the constructions in there like these new measures this projection valued measures from the axioms of the first lecture and the P sub a’s the projectors so order the family of projectors associated with a self-adjoint map they will actually fall out of the spectral theorem and then we have really all elements together to understand mathematically what precisely these axioms mean and then we’re in a position to tackle all the physics problems okay so that’s sort of the first little part of the of the course is completed at this point now we do self adjoint operators and the spectral theorem then we review the axioms and then we start exploiting all of this for physics okay thank you