Chem 131A. Lec 06. Quantum Principles: Quantum Mechanical Tunneling

welcome back again to chemistry 131a last time we talked about the particle in a box we had an idealized one dimensional quantum mechanical system that we analyzed quite thoroughly and today I want to talk about a different phenomenon I want to talk about quantum mechanical tunneling which may may seem like a very esoteric thing to study but in fact it's very important in practical applications and we're going to see that in the beginning of the next lecture tunneling is the phenomenon that because the particle behaves like a wave it can actually slosh around a little bit in cases where the potential doesn't become infinite where the potential becomes finite the wave function can penetrate slightly in in much the same way that sound waves can go around the corner of a room into another room the question we want to ask then is what would happen if the potential energy did not rise to infinity at the edges of the box suppose it rose to some high value but it was still finite the question is what would we expect could we solve that problem and what could we figure out by solving that problem classically what we know is that if the energy of the particle is less than the potential energy of the edge then the particle is bound and it must remain in the well forever it can never escape no possibility of that quantum mechanically however the particle has some probability of spreading out it gets very small but it's non zero and therefore there's some probability that when we measure the particle might be outside the barrier it's as if we know that we have a high jumper the high jumper can jump six feet but no higher and we set the bar at eight feet then we expect if you've got to go over the bar then you're never going to get there because you can only jump six feet but if there's a way to sneak around the bar cheat not actually go over the top but somehow go around or through and under then you could get to the other side and you would appear on the other side as if you had jumped over but in fact you would have cheated and in fact that's what these particles do is they cheat we call that tunneling because we imagine the barrier like a mountain and rather than going up over the top where we wouldn't have enough energy to do so we tunnel right through and we just appear with some probability on the other side almost magically now an arrangement like this which I've shown in the bottom of this slide where we have an area that's trapped and we have two barriers on either side that turns out to be quite a long winded and challenging problem to solve I don't want to do that in the lecture we may do that as homework but I don't want to do that in the lecture so I'll leave that for another time or for homework for you to do on your own but I'll just remark that in this kind of a setup where you have some sort of a well and finite barriers on either side the actual solution depends a lot on the actual mass of the particle the size of the barriers the length of the of the well and so forth and so on and so it gets fairly complicated for example if the barrier is too small compared to the length and mass of the particle then what we may find with the time-independent schrodinger equation is that there are no energy eigenstates at all within within the well and what that means is that there's no bound state for example if we had a mall molecule consisting of a couple of atoms and we found that the potential well was 22 shallow and not deep enough it then basically the way we interpret that is just the uncertainty and energy from the uncertainty principle is enough to rattle the thing out of the out of the well and so as a result we predict no bound States at all if the wells deeper there may be some bound States and it depends how many physical chemists are extremely interested in that when it comes to studying bonds because they like to know exactly how the bonds are being made and there's a lot of theory about how many bound States there should be for this kind of a diatomic or try atomic and therefore they want to compare theory and experiment and do the calculations extremely accurately let's start though with a simpler system much simpler let's start with a free particle which is coming from the left I've written sigh as if it's coming in it has an energy we know the energy E and it collides with a barrier which is a step function which is non-physical but is easy mathematically again we have a sudden change nothing can change infinitely fast in nature like that there has to be some transition region and sometimes that can be very important when you come to a conclusion about tunneling behavior but for the time being let's just take this step function in the potential and then it stays high forever which is what the dots are meant to mean all the way to infinity so at x equals 0 it suddenly goes up and then it goes on for infinity and our question is what will the particle do now classically the particle comes up and it just bounces against a wall and it has to go back and it certainly can't go in because it doesn't have enough energy to go in its energy is less than the potential energy in the region and quantum mechanically what we want to ask is what is the chance that the particle can tunnel into the barrier and how far can it get inside the barrier and that's purely quantum mechanical behavior of course to simplify our notation what I'm going to introduce here is something called the wave vector K k is equal to P upon h-bar and K is a vector in 3d it has a vector with three components and just like momentum is a vector K points in the in the direction of the wave propagation it's just easier to use P over H bar because it appears so often the if we introduce a notation that just makes it much much easier to do it and here what we have because we're doing a one-dimensional problem is we just have plus or minus axes the direction of propagation if if K is positive we're going to the right if K is negative we're going back toward minus infinity nothing more complicated than that and you might say well if you want to understand how far you're going to get you should with the high jumper analogy you should take a video of them you should take a video of them and you should see if they have enough oomph to jump over or exactly what happened in other words you should do a proper time-dependent calculation and we might do that at some later point but for now what we're going to do is we're going to get around that by a very clever dodge what we're going to do is just make up a wave function that's time independent that just indicates sort of overall time or at any time let's say if we measure what is the chance that we find the particle somewhere and that means that all we have to do is solve the time-independent schrodinger equation and we have to make sure we have a wave function that follows our rules it has to be well-behaved we have to be able to normalize it and so forth and so on but it makes it a lot lot easier than actually doing a time-dependent problem where we see something crash in like waves crashing on a shore on the beach and then things happening we can do that but we won't do that at this point with that in mind then let's assume that the barrier begins at the point x equals zero and it continues on to x equals plus infinity and we know on the left-hand side we know exactly what the acceptable solutions are for the wave function if it has an energy it has to be a linear superposition of two momentum eigenstates which we can write in this formula that I've written here if X is less than zero the wave function is some constant a times e to the ikx which represents a wave going to the right crashing into the barrier and a reflected wave if you like + b e to the minus ikx which represents the particle hitting the barrier and going back now keep in mind that this is a time independent calculation but nevertheless when we look at the terms we can interpret them as if there were time and things were actually moving but these are just static things that are going to give a probability distribution that we're going to interpret after the fact is kind of a still shot of how many times the high jumper ended up in the sand pit and how many times they didn't make it and we're going to take that as the probability of the event these free particle solutions we we derive them previously we know we just take the second derivative we said the potential equal to 0 and we get we's waves we know the particle is coming from the left so we know a is not equal to 0 and we know it probably is going to be reflected so we certainly can't say B is equal to 0 we might be tempted to say well if we know the particle started on the left and is coming toward the barrier what that means is that a is something and B is nothing but in fact because this is a snapshot of all time or no time if you like then we have all the failures as well so we have all the reflected waves going back and we have everything just sitting there and so we cannot assume b is equal to 0 that's important because if you make a bad assumption like that then of course the rest of the calculation is completely wrong now for the tunneling part we write the time-independent schrodinger equation minus h-bar squared over 2m the second derivative of sigh of X plus V so I is equal to e psi and now the difference is V the potential energy on once where x is greater than 0 is greater than e the mathematically that's easy for us to do this is just another differential equation to solve no big deal but we may have to see what the solutions mean in this case where we have V is bigger than e but there's there's nothing that jumps out at us and says well there's no solution or or something's haywire with this and it's interesting that when there's nothing haywire it usually actually occurs in nature so if the math says well that's okay then we find it eventually when we look for it hard enough by doing an experiment i'm going to rewrite this slightly now knowing that v is bigger than e i'm going to write just the second derivative of sci of x is equal to 2m times the quantity V minus e over H bar squared so the minus h-bar squared got rid to the sign and that's a positive number so I have the second derivative of size equal to some positive real number x sigh which i'm going to write suggestively as kappa by analogy with k the wave vector for the free particle case i'm going to write that as kappa squared times sigh of x and so Kappa is the square root of two M times V minus e over H bar this then means that once again we have the derivative of something is equal to something times to something we know from previous lectures that what we're going to get is an exponential function the difference now is that because this is a strictly positive number we aren't going to get any imaginary Exponential's and the imaginary Exponential's by oil errs formula were the things that gave us cos theta + I sine theta and those waves those corkscrews indicate motion moving forward these real Exponential's are going to indicate kind of a dying gasp of the wave function as it dies out it starts to try to get through and then runs out of gas but it does so in a very particularly prescribed way here's the solution then which you can verify by putting the solution back into the original differential equation sigh of X is equal to some constant C times e to the e to the Kappa X plus d e to the minus Kappa X and this applies when x is greater than 0 the positive exponential see e to the Kappa X is really counterintuitive what that seems to indicate is that as you go into the barrier part of the wave function decides to get bigger and bigger and bigger and bigger like the population of Earth as you go through even though the potential is much bigger than the energy and that doesn't make any sense and there's a reason then y si has to be zero and the reason why is since the barrier goes to infinity if this thing keeps rising then it doesn't satisfy the condition that the wave function go to zero at infinity any any function that at infinity has to go down to zero because otherwise when we integrate it we get an infinite area and we can't normalize the wave function if we can't normalize the wave function we can't interpret the wave function in terms of the probability of finding something somewhere because it probability is too big and so that's not allowed and what that means then is we were allowed here we weren't allowed to set B is equal to zero but we are allowed to set C is equal to 0 and therefore on the right-hand side of the barrier we just have died of X is equal to some constant D times e to the minus Kappa X where Kappas as I defined it on the previous slide now what we have to do is we have to paste together our solutions sort of like a ransom note so that they match up and that everything is okay and we need and that's good because pasting this thing together so that it looks like it was drawn by one hand means that we get conditions on what a B and D can possibly be and we need those conditions otherwise we can't figure out what's going on so we have a couple of conditions here first the wave function has to be continuous it can't have a step function discontinuity if it has a discontinuity we can't take the derivative and second it has to have a continuous derivative as well because if it doesn't we can't take the second derivative and if we can't take the second derivative then how are we going to calculate the energy with the Schrodinger equation and those two conditions give us two equations that the function be continuous and that means that the real part is continuous and the imaginary part is continuous in other words the whole complex thing is continuous because a B and D can be complex numbers the condition that the wave function be continuous and match up means that a plus B has to be equal to D and the reason why is because when x is 0 all those Exponential's go away whether it's e to the I x 0 or e to the minus something else x 0 it's still one and luckily all that stuff goes away the second condition that the derivative be that the derivative match up at the barrier condition means that I k times a minus ik x b is equal to minus cap at times d now we've got three unknowns a B and D and we've got two equations and that means we can't solve it uniquely for them the third equation that we would get to if we actually wanted to solve uniquely would arise from the normalization of the wave function we'd have to say if we integrate the wave function everywhere that it comes to one when we take size star psy and that third equation then would give us a relationship between a B and D which we could use with the other two to solve it we don't really need to do that in this case though because we can just get ratios so we can leave one of them in determinant and still figure out what the others regardless of the particular values of a B and D we can see right away that the wave function cannot be a hundred percent reflected right at the barrier because we have a be a plus B is dee dee is not zero right so we we can see right away between the two equations that there's there's no no possibility that we can have a hundred percent reflection from the barrier it is going to penetrate then to a to an extent that's going to depend on Kappa which depends on things like the mass and and the difference in the potential and the total energy of the particle we can rearrange the equations then like this so we can take the two equations and rearrange them and say look I k the second equation I k times a minus B is equal to minus Kappa times D but from the first equation kappa or sorry d is a plus B and so we get I k times a minus B is equal to minus Kappa times a plus B and we can rearrange that to get a on one side and we get ik plus Kappa times a is equal to ik minus Kappa times B and then if we solve that for the ratio be over a we get I kappa + oh excuse me we get ik plus Kappa in the numerator and we get ik minus Kappa in the denominator if you haven't done equations with complex numbers and you start getting I this plus that it's very easy to get mixed up because it seems very unfamiliar and foreign compared to simultaneously equations with real variables is basically the same thing but there are a couple of tricks if we look at this ratio of these numbers bna as ik plus Kappa in the numerator and I came minus Kappa we might not know how big that is but we should always think of complex numbers as vectors and the size of the complex number is the length of the vector it's always the hypotenuse of the triangle so let's imagine K is positive and Kappas positive if we have ik that's something sticking up like this if we have ik plus Kappa that's something then that's moved over here and the size of that is this length of this triangle whatever it is if we have ik minus Kappa goes over here you can see by symmetry that these two have exactly the same length and therefore this thing whatever it is I k plus kappa / ik minus Kappa has to have length 1 because these two when I take a ratio of them they had the same length they have to have length 1 and if they have length 1 that means that no matter what it is as a complex number i can write it as e to the I theta because e to the I theta when theta 0 it's one when theta is 90 it's I and so forth and it just goes around the unit circle as I crank theta around and therefore i can write it as 1 term e to the I theta and then that cleans it up a lot for us for this ratio be upon a yeah on this slide then what I've worked out is this ratio and I've worked out the length in detail the length is be over a absolute squared which means you take whatever that number is ik plus ik plus cap / ik minus Kappa you take a star and you multiply that by the same thing I k plus cap / ik minus Kappa and whenever you have a star you grab every I that you see and you just mechanically change it to minus I that's all you got to do and if you do that and you work it out what you find out is you get K squared plus Kappa squared in the numerator and K squared plus Kappa squared in the denominator so this is just making mathematically more formal what we already knew from looking at this geometrical picture and it's good to practice looking at things both ways be able to work out something with an equation just knock it out but then also try to understand what it means if you can get a picture for it you can oftentimes work much more quickly because you can just see what something is going to be whereas the calculation may take a long time to do and if you can see what something's going to be you can oftentimes make very fast progress the ratio be over a has unit length and so as I said we can write it as e to the I alpha where alpha is a strictly real number and that's just again using the Euler formula which I went over before we can also get the ratio of D over a and it's important to realize that when we don't think we can get everything one way to do it is to try to get the ratio that's why I got be over a and D over a because I knew from experience that that's going to be the way to tackle this problem if you don't know that and you don't try to solve for these ratios you will go round and round in circles for quite some time before you come to any any satisfactory solution if we take d over a what we find using the two equations is that 2a is equal to D minus cap / ik x d and that means that i can figure out d over a in terms of isolating d and figuring d over a and what I find is that the ratio of D over a is to I k / ik minus Kappa again and i can write that and in a very tricky way which you wouldn't normally think of doing unless you had already done problems like this which is why of course you you're taking courses like this to get exposed to this kind of thing I write it as ik plus kappa / ik minus kappa plus ik minus kappa / ik minus kappa and the top of course is just to ik written in a way in which the Kappas go away but when I write it this way I see that the first one is my same thing e to the I alpha and I see that the second thing as long as I k minus cap is not zero in which case I might be in trouble the second thing is 1 and therefore the ratio of D over a works out to e to the I alpha plus 1 and the ratio of be over a is just e to the I alpha so now I've got these two things in hand now I can get what I want to get we substitute these two ratios in for for the in the wave function we get an explicit form for the solution up to normalization constant big a which we're going to ignore because we don't really care how big the wave is we could normalize that separately we get to a times e to the I alpha over 2 times the cosine of KX minus alpha over 2 for x less than 0 for x greater than 0 we get to AE to the i alpha over 2 times cosine alpha over two times e to the minus Kappa X clearly shows that on the on the negative side we have a wave that's just going on forever consists of two counter-rotating corkscrews like this one coming in one going back and inside the barrier we have something that comes in and then stops oscillating it stops oscillating because the potential is higher than the energy so there's no nothing left to oscillate but it dies out last gasp e to the minus Kappa X and we can make an explicit plot which I've done in this slide for what this might look like here the energy is half the energy of the barrier and in comes this black wave sigh comes in it hits the barrier and then it starts dropping down and then it tails off as it goes into the barrier and dies away but we can see that there are some kind of penetration there and what that means in which we're going to see in a second is that if the barrier doesn't go on forever then even though this thing's dying out if the barrier finally stops there's a slim but nonzero chance that there's some probability that leaks out and once it leaks out it'll keep oscillating and go over to the right-hand side and that explicitly will will represent the chance that we had a particle we we thought we had it in jail and suddenly it's outside it doesn't happen very often but it does happen so let's expand our vision now to imagine a free particle which is coming from the left and it collides with a barrier and the barrier has high V again but now it has with l and l is finite and therefore it's not it's not this infinite situation what's the chance that the particle can tunnel through the barrier if we come in with a certain energy we hit the barrier what's the chance that we're going to see the particle on the other side once again we are not going to do this as a time-dependent problem we're just going to do this as a time-independent problem we're going to start with a wave function on the left we're going to get the solutions we're going to get the solutions in the barrier which are similar to the solutions we already got and then we're going to get the solutions on the right which are going to be similar to the solutions we already got and then we only have to worry about the two places where the barrier starts which is that x equals zero and where the barrier ends which is at x equals L and we need to then get equations to make sure our wave function is continuous at those two places therefore if we assume the barrier begins at x equals 0 and X and ends at X equal L we know what the acceptable solutions are going to be on the left and right of the barrier on the left it's a e to the ikx plus b e to the minus ikx depending what the energy is that's for x less than 0 on the right hand side it's e e to the minus e to the ikx plus f e to the minus ikx that's if x is greater than L there is these are just the free particle solutions that we found and once again we know a is not equal to 0 because we're assuming that the particle is coming from the left and moving right and therefore that term can't be 0 but we can't say that B is equal to 0 because the particle could be reflected and again we're looking at the thing in totality like freeze-frame averaged over everything on the other hand when it comes to the other side once we've gone through there is nothing physically reflecting the particle back it's just free space after that and so so if the particle were moving to the right and got through the barrier it would continue moving to the right even quantum mechanics isn't that strange and therefore we wouldn't have any incident waves coming from plus infinity coming backwards because there's physically nothing there that can be reflecting the particle in that manner and what that means mathematically which is important is that we can safely set F equals to zero and therefore our starting point for the solutions on the left is a eat ooh the ikx plus b the minus ikx on the right it's just e capital e times e to the ikx for X greater than L what are the acceptable solutions within the barrier well we we saw from before we go back to the time-independent schrodinger equation to find out and we got the solution in terms of Kappa and now to make an acceptable solution we have to find that the derivative must be continuous and the function must be continuous and now it must be so on both sides there is a difference here though before we got rid of the positive exponential with the term e to the plus Kappa X and the argument is that as X goes to infinity we can't have that term because it gets too big but now we can have it and the reason why we can have it is that that solution stops at x equals L and that's finite eita the something e to the kappa l is some finite number that doesn't violate anything about the a function getting too large to normalize and so we must keep both solutions while we do that this is just a picture of what you cannot have you can't have either sigh on the left I've shown be discontinuous where it's got a kink or you can't have the slope be discontinuous where it comes in and then suddenly changes slope and of course we don't have to worry about either desire the derivative being discontinuous except at the boundaries because we have these nice functions we know that they have an infinite number of derivatives these Exponential's so there's no problem except at the actual discontinuity of the potential the four conditions then at the start and at the end of the barrier gives us four equations to relate to the coefficients that we've got and as usual we have five coefficients so it's more more tedious at x equals zero the continuity of zai which I've written here as sigh from the minus side has to equal to sigh from the plus side that means the real part and the imaginary part gives us a plus B is equal to C plus D and the continuity of the derivative deep sigh DX means that a times ik plus B times minus ik is equal to C times kappa which we can have that term we have to keep plus d x minus kappa those are the two equations that hold on the left-hand side of the barrier where x is equal to 0 the same conditions hold on the other side of the barrier but now the Exponential's have a term with L and E we don't know that that's one we did know that e to the 0 when x is 0 is 1 but now we've got an extra thing to keep track of and at x equals L we have the continuity of zai gives us the equation SI e to the kappa l plus d e to the minus kappa l that's what SIA is inside the barrier is equal to e there is no f capital e times e to the I KL that's the wave function on the right and the continuity of the derivative deep sigh DX means that see e to the kappa l times kappa + d e to the minus kappa l x minus kappa is equal to capital e x ik times e to the ik l these four equations then are what you have to battle with in order to figure out what the chance of getting through the barrier is clearly the probability of observing the particle on the right hands on the right hand side depends on the square of the coefficient e because the e to the plus ikx when x is bigger than l all we need to know is the amount of that thing and we know what the chances that it got through the probability that the particle was on the left and moving to the right depends on the square modulus of capital A or capital a squared and therefore we want to have a ratio of the probability that it was on the left moving to the right versus the probability that's on the right and still moving to the right that's what we need to figure out and the ratio of the square of these coefficients is called the transmission coefficient t it's basically the probability that the particle tunneled through the barrier and emerged on the other side of the barrier still going forward and that transmission coefficient is the ratio than of the square of capital e divided by the square of capital a now we have to get a relationship between them with all these equations and I can tell you that when you first try to do this it's not very easy so you want to get in a quiet room and you want to get a big piece of paper and you want to keep a clear head because it's very easy when you have these equations unfortunately to go around in a circle like a dog chasing its tail you just keep substituting in this for that and you go round and round and it's probably good to do that a couple of times because if you do do that then you'll learn the proper way to do it mathematically where you don't actually waste time like that because you know exactly what you're doing and that's worth the price of admission right there again we have four equations five unknowns we can't solve uniquely but we can certainly solve for the ratio because I showed you before we could solve for the ratio of things and that's what we did what we need to do we can get the ratio 40 and it takes a fair bit of algebra but it's worth doing at least once and what we get is the following after being pretty careful we get the transmission coefficient is 1 over 1 plus V squared cinch squared kappa l divided by 4 e times V minus e it's going to take a while to figure out that and you may not be familiar even with the hyperbolic sine function but the hyperbolic sine function is not an enemy to be feared it's your friend because it makes things much simpler here's what we can figure out the transmission probability is not going to be 1 because we have 1 over 1 plus some other thing and the other thing can do nothing except add it's got a bunch of squares and it's got D minus E and we know V is bigger than E so these are all positive things over there so the transmission coefficient can't be a hundred percent but and it can't be very big if these things in the denominator get big because that's going to make the probability small those things have kappa they have L and they have the difference between V and E and those are exactly the kinds of things that we expect to enter in to a calculation like this the hyperbolic sine function is defined as cinch theta is equal to e to the theta minus e to the minus theta divided by two it's a purely real function it's exactly the analogous thing of the sign except there's no I the sign is e to the I theta minus e to the minus I theta over 2 I if you work out the transmission coefficient following along the lines that I did for the step function biting the ratios of those things and then substituting them in its that's a good week's work actually to work that out in detail from scratch without looking up anything or trying to find some shortcut but just sitting down and working it out that's a very good problem the take-home message is that linear equations with complex numbers are just like linear equations with real numbers but there can be tricks to simplify certain ratios to see that certain things must be unit length or length 2 or e to the I alpha plus 1 and once you write it like that it's just much much easier to see where to go when you have all these extra terms floating around it you stare at it and you may find it very hard to see which way to go what is the interpretation well if the barrier is sufficiently impermeable because that's going to turn out to be a very important case where we think it's pretty unlikely actually that the particle would get through the barrier classically of course it's it's impossible but still there are levels of impossibility in this game now we think in quantum mechanically it's not likely if the transmission is low the barriers impermeable in the sense that Kappa times L is is much much bigger than 1 so we we know that the exponential has died away to a low low value if we can make that assumption then we can simplify the transmission coefficient further by some some tricks with the Cinch function and we get 16 times a over B times 1 minus e / v times e to the minus 2 kappa l and what that means is that that's clearly now a dying exponential so when the barrier gets big it starts to look a lot like the step function barrier which went on for infinity in other words when it gets big the only thing that can come through is the dying exponential the positive exponential that contributed to the cinch there can't contribute anymore because it would get too large it would keep going up and it can't do that because it's very unlikely that the particle went through and therefore we get this simple formula and what that means then is the transmission will decrease exponentially as the barrier region is made thicker and that's the basis of a very very potent kind of microscopy that we're going to talk about in the next lecture which is called scanning tunneling microscopy in which we actually use the fact that electrons are light and can tunnel to actually design a microscope to look at surfaces with absolutely fantastic resolution so much better than an optical microscope or even an electron microscope that it's amazing and a court resulted in a Nobel Prize the transmission also responds exponentially to the square root of the mass and what that means is that light particles can tunnel easily and the lightest kinds of particles that we usually deal with in chemistry are electrons and electrons therefore we have to be extremely careful with because they can certainly tunnel around and after that we have protons are next they're quite a bit heavier so the tunneling behavior will will not be so good for them and then after that once we get to heavier atoms than protons the tunneling behavior really goes down quite a lot so protons are a special case because the next one is a deuteron and that's mass two and mass two is twice as big as one and so it's quite a lot and of course the electron being so much lighter really can can slosh around a lot and we have to be very careful with the electrons and the conclusion then is light particles tunnel much more easily and that's exactly why we never see you put your fist toward a wall and your fist somehow ends up on the other side of the wall because we're talking about very heavy things and you can maybe try to punch your fist through the wall I don't recommend that but you're never going to have your fist magically appear on the other side of the wall with some probability because we're talking about very massive things there where the probability of that is absolutely miniscule we'll leave it there and we'll pick up next time with an actual application of what seems like this esoteric description of this of quantum mechanical tunneling and we'll see how it made people a ton of money got people a Nobel Prize and has contributed greatly to all kinds of device physics and including inspecting all kinds of circuits and tiny nano structures to actually see exactly what they look like so we'll do that next time


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