 Dwy'n golygu, amser i ysgrifnegadau. Diolch i fynd o'r problema o'r partygol addysg gennym yn y llaw yma. Ie ddweud ddweud, mae'n ddweud ddweud eich rhai oherwydd na ddweud, rydych chi'n gofynu eich bod ddweud oherwydd a'r hollwch ar gyfer cyflawni, ddweud yw'r hollwch yn amlwg i'n gwybod i'ch gweithio, oherwydd eich cyfrifwyr o hynny yn unigol yn gweithio eu mechani a'r gyfrifiadau yn ei wneud o'i cael ei wneud o'r hollwch i'r hollwch ar gweithio'i cyfrifiadau. Ond y gyrfaethau sy'n rhaid i'r cyfle cyfwyr sy'n gyfrannu cymryd yma. mae'r cyfeirio rydw i cynnwys gyda'r cyfle, ac rhaiwn i wneud mewn i'r cyfan, oherwydd y cyfle yw'r lluniau a'r lluniau a'r lluniau mae'r cyfrannu ym mhag sydd yn gydigio'r cyflym yn ei mod. Fel y rhaid i'r cyfle o'r cyflog ymdano i'r cyflog ymif targeddau cyflog yn eu cyfrannu. The energies in that square well are proportional to the square. They go like n squared of some integer, whereas the energies in the harmonic oscillator potential well go like n. That's to say when we took this simple case of the infinitely deep potential, the energy levels go like n squared, not like n. So that gives you a very different behaviour from the simple harmonic oscillator, and I think it's interesting to investigate that to try and recover classical results and see how that dependence on the energy levels on n squared rather than on n manifests itself in the bottom line. We won't take time to do that. The lecture course is short. We're going to move on to this problem, which is more fun. It will allow us to understand how an ammonia maser works, which is a timekeeping device and was the first amazing lasing device. First of all, we're going to solve an abstract problem and then I'll explain why this is relevant for ammonia. We imagine we've got two potential wells, which are divided by some barrier. Ideally, these potential wells would only be a finite depth, so the walls here would cruise off like this with v0, but the computations are made much easier if we let these walls go off to infinity. So the potential goes off to infinity if you go to left of minus x is minus b or right of x is plus b. The potential is zero here. The potential is v0, some number there. So these are basically two of these square wells put adjacent to each other with a finite barrier between the two. Nothing essentially has changed as the computations are made a bit easier by making these go off to infinity left and right. So what are we going to do? We want to find again the stationary states, the states of well-defined energy for this system because they will have an interesting property, which will lead to the ammonia maser. So again, here is the origin, x is zero, this potential well is symmetrical on reflection around the origin. So it's an even function, the potential is an even function of x. That guarantees that the parity operator commutes with the potential energy operator, which guarantees that we can look for states of well-defined parity. So the stationary states can be eigenfunctions of p, so they have well-defined parity. The usual argument that commuting operators mean we can find a complete set of mutual eigenstates. And we're going to assume, so there will be states which have even parity, there will be states that have odd parity, and as yesterday the equation we have to solve is minus h bar squared over 2m d2u by dx squared plus v of x u is equal to e u. So we have to solve this equation and we're looking for even functions and we're looking for odd solutions. We are going to assume that we're going to look for states which are low enough in energy so that they are, as it were, bound by either this well or by this well, as I say, they're classically forbidden in this region. So we're going to look for states which have less energy with e being less than this barrier height, so that classically the particle will be stuck in one well or stuck in the other well and couldn't go between the two, so we'll see quantum mechanically it will go between the two. So that being so, the solutions, so what does that mean? That means that we've got d2u by dx squared, well, d2u by dx squared is going to be 2m over h bar squared v0 minus e times u and this quantity is going to be positive when we're in the middle between x is, when mod x is less than a, so this quantity here is going to be greater than zero and that means that in this middle zone, the solutions to this equation are going to be things like e to the plus big kx, so this means for mod x less than a, we're going to have u goes like e to the plus or minus big kx, this is what we wrote down yesterday, where big k is the square root of 2m v0 minus e over h bar squared, okay? Or, because we're looking for solutions of well-defined parity and these objects don't have well-defined parity, we can take linear combinations of those and we can say, or looking for well-defined parity, we can say u goes like cos kx or shine big kx. These are the solutions of well-defined parity to that very boring differential equation given that this quantity here is positive. So those are the solutions, we know that our solutions will have that form in that middle section, as drawn up there, this is x equals a, this is x equals minus a. And then in the allowed regions, b will be somewhere over here, in the allowed regions either side of the central barrier, we know that we will have sinusoidal behaviour, there we'll be trying to solve the equation. So this is for mod x less than a. If mod x is bigger than a and less than b, we will be trying to solve d2u by the x squared plus nothing times, plus v being nothing in that zone times u equals 2m over h bar squared e times u. So we're going to have solutions like sin kx plus some phase constant where k is going to be the square root of 2me on h bar squared. So if I put in a phase, an unknown phase, this is an unknown phase here, then this sinusoid will represent any linear combination of sine and cosine and so will be a general solution to this. Oops, sorry, that's correct. I want e to be negative, right? Sorry, no, no, no, no, e is positive, e is positive, e is positive. Yes, there's a minus sign here coming from the minus h bar squared over 2m, so that's why there's a minus sign here and that's why we have this sinusoidal behaviour. Right. So we know what the solutions look like in the middle, we know what the solutions look like on the sides and all we have to do now to fix everything up is make sure the boundary conditions are satisfied where they join. So we have to, so we say that at x equals a, we have to make sure as yesterday that our wave function is continuous and has a continuous derivative. So continuity of the wave function, let's specialise on the even parity case. Continuity of the wave function is going to say that cos times ka must equal b, that's my constant that multiplies the sine, times sine ka plus phi, that's the continuity of u and then I have to deal with continuity of the gradient of u as yesterday, so we have that k sin ka is equal to kb cos ka plus phi. So we are not really very interested in b, it's the same sort of setup as yesterday, we're not very interested in the constant b, yesterday we had a similar constant a, we're very interested in the value that k takes and this thing here, big k, as yesterday can be expressed as a function of little k so the name of the game is to get rid of b so we divide this equation by this equation so 1 over 2 leads us to the conclusion that the hyperbolic cough of ka is equal to k over k times the, sorry, sorry, sorry, sorry, I'm doing which division, sorry I've got that the wrong way up, so that becomes big k over little k, because I am taking this equation and dividing it by this equation so this k was on the bottom as it were and I brought it up to the top of the other side times the tangent of ka plus phi and now we haven't quite yet because there's one thing we haven't fixed which is we need to find out about this phi thing or what do we have to do with the phi? Phi has to be chosen so that the wave function in that right hand zone vanishes at x equals b remember when we had the infinitely steep sides we concluded that the wave function vanished adjacent to the rise to infinite potential energy that was the point we finished on yesterday so what we can say is that at x equals b we require, because the potential is about to go infinite that u equals 0 so that means that sin kb plus phi is 0 and kb bus fly is so what does that imply? it implies that kb plus phi is some integer number of pi's so that tells us what phi is that tells us that phi obviously is our pi minus b so we put this information into here and we have our equation but while we're doing it it's probably a good idea to express big k in terms of little k so we have expressions for big k and little k on the board and let's work out exactly what we have actually so big k is the square root of 2m v0 minus e on h bar squared which we write as 2m v0 a squared over a squared h bar squared minus 2m e over h bar squared is in fact little k squared and yesterday we identified this object as a dimensionless constant w so this thing is the square root of w over a squared minus k squared so when I take this equation here and that result there in the right hand side and this result here to get rid of those big k's we discover that the hyperbolic cotangent of I want big k a which is going to be the square root of w minus k a squared is equal to on the right hand side big k over little k which is going to be the square root of k sorry, square root of what am I talking about of big w over k a squared minus 1 times the tangent of r pi minus k b minus a now we have what this is so w is the dimensionless constant that characterises the potential wells on each side as yesterday k is the thing of interest to us because it controls the energy it determines the energy of the stationary state and is the only unknown in this equation everything else is known so what we have here is a pretty ghastly equation whose roots determine the values of k which in turn determine the energy and as yesterday the way to solve this is graphically to plot the left hand side of the equation and the right hand side of the equation and separate curves and see where they intersect and I live in hope that my computer has this here so, the computer may have it here but it didn't come up the system has perhaps gone to sleep is anything showing? yeah oh but meanwhile my computer has gone and buried it's here we go right so that should show it's very faint from here it's warming up right but that should show the two sides of the equation the right hand side of the equation sorry no concern that this is not sorry that was that problem we need to go further down it's not the right solution so here is the double potential well and each of these curves is the right hand side of the equation it's coming down like this and there's a different one because we have this tangent so the right hand side of the equation contains a tangent which goes to zero we can forget about the r pi as a matter of fact because the tangent of r pi plus an angle is the same as the tangent of the angle so if I could write this I probably should write this there's w over ka squared minus one times the tangent of this angle here k b minus a and there's a minus sign out front here which is that minus sign so you can add an r pi to a tangent and you don't affect its value so this tangent keeps having zeros as k increases this tangent hits pi 2 pi 3 pi etc the argument of this tangent hits pi 2 pi 3 pi etc and the tangent vanishes and those give you the points where these vertical lines crash down through the origin these points these points that's why there are many branches the other thing that's happening here so the dotted line there is sorry there are two focus for the moment only on the top curve because these things so there are two curves where is this I think this thing is needs a new battery there are two curves running horizontally almost horizontally a dotted one and a dashed one the dotted one is the one we should look at the one that curves upwards at the end that is a plot of this cough alright so when if w is fairly large and ka is fairly small then we're looking at the cough of a large number and that's always one so this thing runs along at one and then eventually when ka becomes on the order of w we're looking at the cough of something smaller than a large number and that becomes that then goes up towards infinity that then goes up towards so that's what's happening there and the magic values of k are given by the intersection of the dotted line and the full lines coming down and you can see you get a series of energies there's only a finite number of them because when ka becomes bigger than w when ka becomes bigger than w this square root goes imaginary and we no longer have any solutions to the equation bad things happen here too so there's only a finite number of energy levels if the well has any finite value of w if we would repeat all this so that's what we have we have a finite number of energy levels this is for the even parity case if we were looking for the odd parity solutions so for odd parity what would happen is that only the only thing that would happen is that this would become a shine and that will become a cosh everything else would stay the same so that the only thing that would happen as we followed through this logic would be that this cough would become a fan a hyperbolic tangent so this is even parity you can follow through the algebra afterwards but believe me I think it's very plausible that all that happens is because you're swapping over a cosh and a shine this becomes for odd we get fan of the square root equals minus the square root etc so the vertically crashing down lines remain the relevant lines the right hand side hasn't changed but the left hand side has changed so it starts off as the tangent the hyperbolic tangent of a large number i.e. one and then it becomes the tangent of a smaller number as ka becomes comparable to w so it turns downwards not upwards because so the point now the crucial point that has physical consequence is that when ka is small in other words when and k is a measure of the energy the energy is proportional to k squared because h bar k is the momentum of the particle as it rattles around inside this well so the energy of the low energy solutions are associated with the oh these things come in pairs the energies come in pairs agreed because we have the even and odd parity things intersect at low k almost exactly in the same place on that diagram you can't see the distinction as you get to higher values of k and the particle is only marginally bound has an energy which is comparable to v0 the the two curves for the right and the left sides diverge and you can see you get different values of k so so once the key thing is the solutions come in pairs because we had two potential wells we get two adjacent solutions in fact if we had three potential wells we would get three if we have an infinite number we would have an infinite number and this is in solid state physics this is crucial because in a crystal you have a 10 to the 24 you have some vast number of potential wells one for each atom and you get a vast number of solutions all crowded together but anyway we just have two wells, we have a pair of solutions they come in pairs of similar energy very similar energy similar E so one is even parity and this actually will be the lower energy solution it has the smaller value of k you can check from the diagram up there and one for the odd parity and this is the higher energy so for this system for a particle trap like that in a well a lowest state and just above it which is of even parity so the ground state here we are, we've got a picture here I think the ground state so the top diagram there shows these wave functions we have a sinusoid on the right a sinusoid on the left the ground state is the two upper full curves it's two sinusoids with a depression in the middle where the barrier is and the particles are classically forbidden and then at an energy just a tiny bit higher you have a very similar pair of sinusoids they are in fact subtly different values of k so they're very subtly different but essentially to a very good approximation the odd parity state is the same as the even parity state except reflected around the origin so it becomes negative on this side so we have two states we have so the wave functions are going to be what we'll call them u plus of x which is the even parity case and u minus of x sorry, sorry u e of x and u odd of x which is the odd parity case if we take linear combinations of these two states we get two other states appsi plus which we'll say is 1 over root 2 of x which is u even of x plus u odd of x and an appsi minus of x so if we take, so appsi plus is the sum of those two wave functions up there so it essentially it vanishes on the left and has a non-zero value on the right so this is the state of being on the right and this is the state of being on the left because if you take them away they add up on the left and they subtract on the right if we put a particle if we actually put the particle into the right-hand well we will set our system up in this state appsi plus this state appsi plus is not a stationary state it is not an energy eigenstate it is not an eigenstate of different energy so if we drop it on the right what our initial condition at t equals 0 this is what our wave function looks like what does our wave function look like generally appsi of x and t is going to be 1 over root 2 times u even of x e to the minus i e even of t over h bar the usual boring time evolution of a stationary state plus u odd of x e to the minus i e odd t over h bar which we can write more conveniently as e to the minus i e even t of h bar over root 2 u even of x plus u odd of x e to the minus i e odd minus e even t on h bar the crucial thing is that if we wait long enough if we wait such a time that the argument of this exponential becomes equal to pi we will be looking at e to the i pi which is minus 1 here and our state will have evolved into some phase factor who cares times u even minus u odd which is the state of being on the left so after a time the time required for this to become pi which is t equals pi h bar over e odd minus e even the particles on the left and you can see that this will go on forever you put the particle in on the right it's not a stationary state it moves to the left and then after twice this time it will have come back to the right it's going to oscillate between these two between these two wells forever and ever according to this theory and crucially the time scale is long it takes a long time to get from the right to the left if this energy difference is small and we've seen that this energy difference is small right energy difference being small means that this time scale is long or when is this energy difference small it's small when the barrier between the two wells is high so what we say is that the particle takes a certain time to tunnel through that barrier so we say that the particle whoops tunnels through the barrier in a time which grows it grows very rapidly as a fact with v0 with w with the height of the barrier so a high barrier or a wide barrier means it takes a long time to get through but according to this analysis it will eventually get through ok so that's just a toy problem now let's see what the hell that's got to do with the real world by talking about ammonia somewhere here we have a picture of ammonia so ammonia is a we don't have a picture of ammonia sorry I've somehow lifted up ok so what does ammonia consist of it consists of three hydrogen atoms so these are H's and it consists of a nitrogen atom and the three hydrogens form some kind of a triangle roughly speaking and the nitrogen atom sits well this is the classical picture we'll see quantum mechanical picture isn't like this but this is the classical picture we think of the hydrogen atom as sitting either above the triangle formed by the hydrogen atoms or below the triangle formed by the hydrogen atoms now this is a complicated system it's got four nuclei and ten electrons so it's a really complicated dynamical problem but next term in the next course I'll see a thing called the adiabatic approximation which is what chemists use in order to understand the dynamics of complicated systems like this it allows you to treat all these bonds between the nuclei which are provided by the electrons as springs so what we can do what chemists routinely do do is they calculate the energy of this molecule in what's called a clamped nuclear nuclei approximation so they put they say let the nuclei be let them be here and let's calculate the energy and you can do that for the nitrogen atom for the nitrogen nucleus being at every point along a line that passes through the centroid of the triangle so imagine doing that that leads to a potential energy curve if you like for the nitrogen nucleus which is going to be something it turns out to be something like this so there are two wells the lowest potential energy when it's an appropriate distance either above the plane or below the plane there are two wells and a barrier in the middle because it isn't comfortable being in the middle of the triangle formed by the hydrogens so we have this potential energy curve and now we know what the energy levels of this molecule are going to look like because we know that there are going to be even parities the stationary states are going to divide it's obvious on geometrical grounds this is symmetrical that this well looks just like this well because there's no fundamental difference above the triangle and below the triangle so we know the solutions are going to come in pairs there's going to be a of even parity solutions and odd parity solutions just like in our square well so we also know that if we start if we start with a nitrogen in this well so we start with a nitrogen above the triangle it's going on some characteristic time to move over here and then it's going to move back there and it's going to oscillate between the above the triangle and below the triangle so if we start it in either above the triangle or below the triangle it's going to oscillate between the two those who have done chemistry will know also that this nitrogen is going to be carrying a negative charge and these hydrogens some amount of positive charge so what are we going to have we're going to have an oscillating dipole this molecule is going to have a dipole moment because of the electronegativity of the nitrogen and it's going to be an oscillating dipole if we start it in this in this state of being above the triangle it's going to oscillate to and fro and what does an oscillating dipole electric dipole do? it radiates so this thing is going to radiate and you will be able to if you know what this potential surface looks like you'll be able to calculate the frequency which is going to radiate because the frequency which is going to radiate is going to be given basically by this formula here for the half period to go from there to there so twice this time it's going to be the period of a complete oscillation and that's going to be the the period of the radiation ok so so the molecule if started it's a dipole so started on top if you can achieve that it's going to be an oscillating dipole and the experiments lead to the conclusion that the frequency new is about 150 MHz you can measure it so it's a source of microwave radiation from this 150 MHz you can also work out what the change in the energy what this energy difference is so EOD minus EE turns out to be about 10 to the minus 4 electron volts so that's an energy difference which is small compared to the thermal energies of molecules just here in the room the thermal energies of molecules so K the Boltzmann constant times T the room temperature is something like 0.03 electron volts so this is much bigger than E0 minus EE what does that mean from your statistical mechanics course you either know now or will soon know that that means that we expect at room temperature there are essentially equal numbers of molecules in the even and the odd states because the energy required to go from the ground state this is the ground state or the energy of the ground state to the first excited state is less than the characteristic thermal energy knocking around in the room so there will be large numbers of molecules in both these even and odd states and the idea behind anamonia masa is that we is to find a way of separating the molecules which are in the excited state the odd state leading them into some kind of a resonant cavity where they will then because they're in the excited state they will be they will decay into the ground state if you leave them if they're left alone they will decay into the ground state emitting a photon with the balance of the energy so they will radiate away at 150 MHz so that's that's the strategy behind a masa so in a masa to get a masa you have to isolate the half the roughly a half of the molecules in the first excited state and the way you do this is you exploit the fact that well let's just imagine putting these molecules into an electric field so we're going to put the molecules and the strength of this electric field I'm going to denote with curly E to distinguish it so that electric field is distinguished from energy ordinary E roman E so we put these molecules into an electric field E and ask ourselves how does that change since these are dipolar molecules these molecules are electric dipoles we're expecting that that changes the energy of the molecules putting them in a field and maybe we can get them separated by exploiting this fact so you put them in an electric field and ask so how does that change we're doing quantum mechanics changing energy means changing Hamiltonian so we have to ask ourselves now how does this change the Hamiltonian of these molecules what term in the Hamiltonian is introduced by this electric field and the way to do that is to think in terms of the Hamiltonian write the Hamiltonian in the basis so we're going to let plus be so it's going to be mathematically 1 over root 2 of the even parity plus the ground state plus the odd parity using a different notation even parity plus odd parity so this is the ground state this is the first excited state this linear combination we're going to call it plus and it'll be the state in which the molecule is definitely above the triangle and similarly minus is going to be 1 over root 2 even parity state is going to be below so this is below triangle this is above this is exactly what we did over here with our psi plus and psi minus we're working there with wave functions here we're working with the underlying kets the thing is the dipole moment if the dipole moment of these two states have opposite signs because this one has the negative charge at positive z and this one has the negative charge at negative z so they have opposite dipole moments and let p now be p be the dipole operator it was the parity operator earlier on but now let it be the dipole operator then these states are going to be eigen states of this dipole operator they have well defined dipoles p plus is in fact going to be minus qs times plus well what am I saying I'm saying this is some charge this is some distance the product of a charge in a distance gives me units of dipole electric dipole moment I've got a minus sign here because because this state I said has a negative charge above at positive z so the dipole moment points in the opposite direction to the location of the nitrogen atom and therefore this one has a negative dipole moment and this eigen value is the dipole moment this is the eigen value of this operator because this thing is a state of well defined dipole moment and this is where the dipole moment appears and basically this is just some number which has the dimensions of a charge times a distance which the charge is going to be on the order some fraction of an electron charge and this distance is going to be this characteristic size of the atom a tenth of a nanometre similarly p minus is going to be plus qs minus so this encapsulates the crucial point that these two states have opposite these signed dipole moments what is the energy of a dipole in an e field well the answer is that the energy is minus this is just classical physics is minus the field times the dipole moment so now we know how much the energy of this state and this state have changed by the electric field it's it's given by this here where we put in those kind of eigen values so now what we want to do is write down the matrix that represents the Hamiltonian in this plus minus basis so we want to write the Hamiltonian as plus h plus matrix this is a complex number right plus h minus minus h plus so these four complex numbers we want to write down that matrix Hamiltonian consists of the Hamiltonian of the undisturbed of the undisturbed nitrogen molecule plus the contribution here this is plus the contribution here from the electric field what do we have for the undisturbed let's calculate this separately so I need to calculate what what plus h plus is for the isolated atom well that is going to be one over two even whoops because these are linear combinations of the eigen states even and odd even plus odd h even plus odd right because is this is this but h on this is an eigen function of this operator if this is the isolated operator so we get h on even produces e even times even even is orthogonal to odd but meets up with this so we get an even and h on odd produces e odd times odd which is orthogonal to this but meets up with this to produce an e odd so at the end of the day we get a half of the even plus e odd which is the average energy let's call it e bar similarly we have that let's have a look now at plus h minus the off diagonal element that is going to be one over two even plus odd h even minus odd so things are rather similar except we now get from here here and here e even and from here here and here we get minus e odd so we get a half of e even minus e odd which I think I have been calling we're going to call this minus a so a is so we're going to call this the definition of a and we're going to put this minus sign in here because we know that odd the odd energy is bigger than the even energy so this quantity here is a negative number and putting that minus sign I get this positive number so this matrix up here what's it looking like it's looking like e bar from here sorry a this is a Hermitian matrix so what appears down here has to be the complex conjugate of that real number a in other words a and we will find that this one is also e bar that's for the isolated atom then we have to add on to it the contribution for the from this which but this thing is an eigenfunction of p so the only contributions down the diagonal there are no off diagonal contributions from this additional piece to the operator to the Hamiltonian and what I have is plus what I need is the dipole operator so I have p on plus which produces minus qs times plus so this minus sign and the minus sign that I've just spoken cancel and we get plus q and here we get minus q curly e s so this is what the Hamiltonian looks like in this in this basis I'm running out of time so let me just sketch how the thing goes so what we have got is an explicit expression for the Hamiltonian in the presence of an electric field what we would like to know is what the energy levels are that result from this and how those energy levels vary as a function of the electric field but that's a dead cinch what you've got up there is a 2x2 matrix anybody can find the eigenvalues of that 2x2 matrix and those will be the energies that the molecule has when you stuff it in this electric field so we find the eigenvalues of this matrix and we plot how they depend on curly e so what they actually are is the average energy plus or minus the square root I believe of a2 minus q just make sure I didn't make a mistake that I wouldn't bring yeah so these are the possible energies and if we plot this graphically what do we see here we have the strength of the electric field we find that we have solutions that behave like this here we have ebar plus a here we have ebar minus a and here we have the strength of the electric field this is just a graph of those two things you can easily check that that's the case here of course is ebar so this is the ground state if you switch off the electric field we have two states here and here separated by this small amount which is 2a as you switch on the electric field the energy of the ground state goes down but the energy of the first excited state goes up there's a lot of interest in the way in which this energy behaves here it's behaving quadratically as a function this is a sort of parabola as a function of e and eventually it becomes a straight line as a function of e on both top and bottom and what that's telling you is that in this regime here the even state the ground state which is the even state has no dipole moment intrinsically because the particle is equally likely to be above the nitrogen is equally likely to be above or below the plane so in the ground state there's no dipole moment you switch on the electric field and you make a bias so that the upper field say the upper position gets to have a lower energy in the lower position because the electric field is pushing it that way because the nitrogen starts to spend a bit more time above than below and now the molecule requires a dipole moment because it's not spending the nitrogen is not spending equal time above and below so it requires a dipole moment the magnitude of this dipole moment is proportional to the energy and therefore the energy that you get the change in the energy which is equal as it says up there e is minus electric field times the dipole moment becomes proportional to dipole moment squared so that's why we have a quadratic dependence like this once you're in this regime you have a fairly strong electric field you've made such a strong bias you've made being up so energetically more favourable than being down that the nitrogen spending all its time up the dipole moment is now independent of the strength of the electric field because it's increasingly electric field doesn't cause it to spend any more time up than it or it's already spending all its time up so that the energy becomes proportional to this how do you make your maser work you make your maser work well the fact that the energy of one of these goes down the other one goes up means that as the field increases that if you have an inhomogeneous field if you pass your beam through a region in which the electric field is varying in strength so I have some sort of capacitor plates I have a pointy thing a needle and a cup then the electric field goes like this and we have a high field up here and a low field down there so I have a region of inhomogeneous field the ones in this state are going to be bent sorry they're going to be bent upwards it lowers their energy so they move into the region of high field these ones move into the region of low field so this is the ground and this is the excited and then you can put these ones into your resonant cavity and hear it sing so there's an example of how with a with a very simple minded model you can explore so quite interesting physics physics which is very inherently quantum mechanical this energy level is coming in pairs because we have even an odd parity states this ability of a molecule which you would think was inherently dipolar in the ground state have no dipole moment because the nitrogen can be simultaneously above and below all of these features are very quantum mechanical and with a simple minded model we're able to explore them