Lecture 06. Quantum Mechanical Tunneling


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Lecture 06. Quantum Mechanical Tunneling
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Lecture 06. Quantum Principles: Quantum Mechanical Tunneling
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Shaka, Athan J.
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UCI Chem 131A Quantum Principles (Winter 2014) Instructor: A.J. Shaka, Ph.D Description: This course provides an introduction to quantum mechanics and principles of quantum chemistry with applications to nuclear motions and the electronic structure of the hydrogen atom. It also examines the Schrödinger equation and study how it describes the behavior of very light particles, the quantum description of rotating and vibrating molecules is compared to the classical description, and the quantum description of the electronic structure of atoms is studied. Index of Topics: 0:00:42 Tunneling 0:29:42 Barrier Penetration 0:44:34 Interpretation
welcome back again the chemistry 131 last time we talked about the particle in a box we had an idealized 1 dimensional quantum mechanical systems but we analyzed quite thoroughly and today I want to talk about a different phenomenon I want talk about quantum-mechanical tunneling which makes may seem like a very esoteric I think the study but in fact it's very important and 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 run into another run 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 supposed heroes to some value but it was still finite the question is why 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 it is is less than the potential energy of of the edge than the particle is bound and it must remain in the well forever you can never escape with no possibility of that quantum mechanically however the particle has some probability of spreading out it gets very small but it's not on 0 and therefore their some probability that 1 we measure the particle might be outside the the barrier it's as if we know that we have a high jumper the high jumper can jump 6 feet but no higher and we set the bar at 8 feet then
we expect them if you've got to go over the bar then you're never going to get there because you can only jump 6 feet but if there is a way to sneak around the bar Cheney not actually go over the top but somehow the all-rounder through and 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 Tuesday cheap 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 now an arrangement like this which have shown in the bottom of the slide where we have an area that's trapped and we have to barriers on either side that's turns out to be quite a long-winded and challenging problem to solve I don't want to do that in the lecture but we may do that as homework but I don't want to do that in the lecture so I'll leave that for another timer for homework for you to do on your own but I just remarked that in this kind of a set up where you have some sort of a well and finite barriers on either side of the actual solution depends a lot on the actual mass the particle the size of the barriers the length of of the of the well and so forth and so on and so against fairly complicated for example if the barrier is too small compared to the length and massive the particle then what we may find with the time independent sharing a equation is that there are no energy Oregon State's at all With them within the well off and on what that means is that the no bound state for example if we had a molecule
consisting of a couple of Adams and we found that the potential well was to use too shallow and not deep enough and it then basically the way we interpret that is just the uncertainty in 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 chemistry 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 diatonic a try atomic and therefore they wanna compare theory and experiment and do the calculations extremely let's start tho With a simpler system much simpler let's start with a free particle which is coming from the left I have written size if it's coming in it has an energy we know the energy E and it collides with the barrier which is a step function which is non-physical but that is easy mathematically again we have a sudden change nothing can change infinitely faster nature like that there has to be some transition region and sometimes I can be very important when you come to a conclusion about tunneling behavior but for the time being let's just take this 1st step function in and the potential and then it stays high forever which is
what the doctor meant to me all the way to insanity so that's equal 0 suddenly goes up and then it goes on for insanity an hour questionnaires what will the particle do not classically the particle comes up images bounces against the wall and it has to go back and it certainly can't go again 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 but 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 yeah To simplify our notation what I'm going to introduce here is something called the way vector K K is equal to the upon H bar and is a vector in 3 D that it had it as a vector with 3 components and just like momentum as a vector points in the in the direction of the wave propagation is just easier to use the over-age bar because it appears so often that if we introduce notation it just makes it much much easier to do it and hear what we have because we're doing a one-dimensional problem is we just have plus minus exits the direction of propagation if if caves positive we're going to the right of caves negative for going back toward minus infinity nothing more complicated the and you might say well but 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 wealth to jump over or exactly what happened in other words you should do a proper time-dependent calculations 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 this time independent but just indicates sort of overall time or at any time let's if we measure what is the chance that we find the particle somewhere and that means that all we have to do is all the time independent shorting a equation and we have to make sure we have a wave function that follows the rules it has to be well-behaved we have to be able to normalize it and so forth and so on but it makes a lot lot easier than actually doing time deep tendon problem where we see something crash like waves crashing on the shore on the beach and then things happening we can do that but we won't do that at this point without a mind then let's assume that the barrier begins at the point taxable 0 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 2 momentum states which we can write in this formula that I've written here effects is less than 0 the wave function is some constant times EDI kayaks which represents a way of going to the right crashing into the barrier and a reflected wave if you like the Class B e to the minus side kayaks which represents the particle hitting the barrier and going back now keep in mind that this is a time independent calculation but nevertheless will we look at the terms we can interpret them as serious and 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 as kind of a still shot of how many times a high-jumper ended up in the centered on how many times they didn't make it and we're going to take that as as the probability of the event these free particle solutions we we derive than previously we know we just take a 2nd derivative we set the potential equal to 0 and we get these we waste we know the particle is coming from the left so we know it is not equal to 0 and we know that 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 part started on the left and is coming toward the barrier what that means is that there is something and be is nothing but in fact because this is a snapshot of all kinds and 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 the is equal to 0 that's important because if you make a bad assumptions like that and of course the rest of the calculation is completely wrong now 4 the tunneling we write the time independent shorting equation minus H bars square over 2 and the 2nd derivative of sigh of X plus B side is equal to the side and now the difference is the potential energy once where X is greater than 0 is greater than b 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 the is bigger than the but there's there's nothing that jumped out at us and says well that the there's no solution an hour or something so haywire with this and it's interesting that when there's nothing haywire usually actually occurs in nature so if the masses well that's OK and then we find it eventually and when we look for hard enough by doing an experiment under the rewrite this slightly now knowing that the ears bigger than the I'm gonna write just the 2nd derivative of cited X is equal to 2 m times the quantity of the minus the Over bars square to the minors aged bars square and got real change the size and that's a positive numbers so I have the 2nd derivative of size equal some positive real number times sigh which I'm going to write suggestively as cap by analogy with Kay the way vector for the free particle case I'm going to write that Capus square times sigh of acts and so capital is the square root of 2 m times the B-minus Over Ridge by this stand means that once again we have the derivative of something Siegel to something times is 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 are going to get any imaginary exponential an imaginary exponential spy Oilers formula where the things again gave us coast there and I signed it and those who waives those corkscrews indicate motion moving forward these real exponential sir going to indicate kind of a dying gasp of the wave function as it dies out if it starts to try to get through and then runs out against but it does so in a in a very particularly prescribed wh and here's the solution then which you can verify by putting the solution back into the original differential equations sigh of X is equal to some constancy time speed to take you to the cap X plus the heat of the miner's cabin acts and this applies when X is greater than the positive exponential seat the cabin eggs it 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 but the potential is much bigger is bigger than the energy and that doesn't make any sense and there's a reason the NYC has to be 0 and the reason why years since the barrier goes to infinity if this thing keeps rising then it doesn't satisfy the condition that the wave function the goal is 0 and insanity any any function it is Infiniti has to go down to 0 and because otherwise when we integrated we get an infinite area and we can normalize the wave function if we can't normalize the wave function we can interpret the wave function in terms of the probability of finding something somewhere because probabilities to and so that's not allowed and what that means is we were allowed here we were allowed said B is equal to 0 but we are allowed to set sees equal to 0 and therefore on the right-hand side of the barrier which is set side of X is equal to some constant D times to the miners X where capitalism as defined on the previous slide now
what we have to do this we have to paste together Our
solutions sort of like a ransom note so that they match up and
that everything is OK and we need and that's good because pasting this thing together so that it looks like it was drawn by 1 hand means that we get conditions on what date b and d I can possibly be and we need those conditions otherwise we can figure out what's going on so we have a couple of conditions here 1st the wave function has to be continuous it it can have a step function at discontinuity if it has a discontinuity we can't take the derivative and 2nd it has to have a continuous derivatives as well because if it doesn't we can take the 2nd Europe and if we can't take the 2nd derivative then how are we going to calculate the energy with the short equation and those 2 conditions give us 2 equations that the function be continuous and that means that the real part of 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 the continuous and matchup means that a plus B it has to be equal to do and the reason why is because when axes 0 all those exponential go away with the exceeded the eye time 0 Laredo the mine something else times 0 it's still 1 and likely all that stuff goes away the 2nd condition but the derivatives the but the derivative matchup at the barrier conditions means that IKE times a day -minus IK times B is equal to minus capital Wednesday now we've got 3 unknowns and we got to equations and that means we can't solve it uniquely for the the 3rd equation that we would get the if we actually wanted to solve uniquely would arise from the normalization of the wave function we have to say if we integrate the wave function everywhere that it comes to 1 when we take sides stop and that's 3rd equation and would give us a relationship between a BND which we could use with the other to solve it we don't really need to do that in this case there because we can just get ratios so so we can leave 1 of them In determinant and still figure out what the others are but regardless of the particular values of a B and D we can see right away but the wave function cannot be 100 per cent reflected right at the barrier because we have a B Cosby is doing but the is not 0 right so that this week we can see right away between the 2 equations that there's there's no no possibility that we can have 100 per cent reflections from the barrier it is going penetrate them to to an extent that's going to depend on capital which depends on things like the Manson and the difference in the potential and the total energy the particle we can rearrange the equations than like this so we can take the 2 equations and rearrange them and say Look I carried the 2nd equation I times a minus B is equal to minus capital items and but from the 1st equation Kappa Ahtisaari D is a plus B and so we get IKB finds a minor speed is equal to minus capital time today plus B and we can rearrange style To get a on 1 side and we get IK plus cap times 8 is equal to IK -minus capital times indeed and then if we solve that for the ratio b over a we get Icap plus was excuse me we get I plus capital in the numerator and we get minus cap in the denominator if you haven't done equations with complex numbers and you start getting this plus that it's very easy to get mixed up because it seems very unfamiliar and foreign compared to simultaneous equations with real variables is basically the same thing but there are a couple of tricks if we look at this and ratio Of these number the end as I K plus but in the numerator 19 -minus camp we might not know why how big that ideas but we should always think of complex numbers as vectors and the size of a complex number is the length of the vector it's always the hypotenuse of the triangle so let's imagine k is positive and cap positive if we have our I that's something sticking up like this if we have like a plus cap that's something then that's moved over here and the size of that is that like that this triangle whatever evidence if we have I K -minus kept goes over here you can see by cemetery that these 2 have exactly the same length and therefore this thing whatever it is I K +plus Kappa divided by came minus has to have length 1 because these 2 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 a complex number I can write it as to the isolated because EDI Theda when there is 0 it's 1 1 Davis 90 all I and so forth and it just goes around the unit circle as I crank Vader around and therefore I can write it as 1 term needed the I think and then that cleans it up a lot for us for this ratio of the upon it they on this slide than what I've worked out is this ratio and I've worked out the length and detail the lands it is be over a absolute square which means you take whatever that number is like a plot by Kay plus capital Reich A-minus Kappa you take a star and you multiply that by the same thing I K plus capital rightly miner's cabin and when every you have a you grab every eye that you see and you just mechanically change at a minus sign as so I gotta do and if you do that and you work it out we you find out is you get case where caseworkers cap square in the numerator and case squared plus cap square 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 to 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's going to be whereas the calculation may take a long time to do and if you can see what sums gonna be can oftentimes make very fast progress "quotation mark the ratio the over has unit length and so as I said we can write it is easy to the Alpha where Alpha is a strictly real number and that's just again using the boiler formula which I went over before we can also get the ratio of over and it's important to realize that when we don't think we can get everything the 1 way
to do it is to try to get the ratio that's why I don't be over it Andy over 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 solutions if we take deal over what we find using the 2 equations is that to a is equal to a D minus capital over I K times daily and that means that I can figure out d over a in terms of isolating D and figuring D over it and what I find is that the ratio of D over a this too I care divided by I came minus Kappa again and the I can write that 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 is I K plus capital divided by Ike came minus cap I came minus Skepple divided by I T minus camp and the top of course is just too like a written in a way in which the capital's go away but when I arrived it this way I see that the first one is my same thing needed the I Alpha and I see that the 2nd thing as long as I came minus cap not 0 in which case I might be in trouble the 2nd thing is 1 and therefore the ratio deal over a works out to heed to the eye Alpha plus 1 and the ratio of be over it's just 2 the I also so now I've got these 2 things in hand now I can get what I won again but we substitute these 2 ratios-an for a 4 2 in the wave function we get an explicit form for the solution of up to normalization constant Big Eight which were going to ignore because we don't really care how big the way this we could normalize that separately we get to 8 times EDI off over 2 times the co-signer of kayaks myself over it 4 x less than 0 4 X greater than 0 we get to to the off over 2 times co-signed also over to times to the miners cap X clearly shows that on the on negative side we have a way this is going on forever consists of 2 counter-rotating corkscrews like this 1 coming in and 1 going back and inside the barrier we have something that comes in and then stops oscillating it stops oscillating because the potentials higher than the energy there's not nothing left to oscillate but it dies out last gasp even minus kept the text and we can make an explicit plot which I have done in this slide for what this might look like here the energies have the energy of the barrier and incomes this black wave side comes in it hits the barrier and then it starts dropping down and then it tails off as a goes into the barrier and dies away but we can see that there is some kind of penetration there and what that means in which we're going to see in the 2nd is that if the barrier doesn't go on forever then even know this things dying out if the barrier finally stops there's a slam but non 0 chance that there's some probability that leaks out and once it leaks out it'll keep isolating 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 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 the barrier and the barrier has Hi V again but now it has with and Bell 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 deep tendon problems were just going to do this is a time independent problem organist start with a wave function on the left were going to get the solutions were going to get the solutions in the barrier witches similar to the solutions we already got and they were going to get the set of solutions on the right which you go berserk similar to the solutions we already got and then we only have to worry about the 2 places where the barrier starts which is at actually will 0 in on the barrier and switches at X equals L and we needed then get equations to make sure our way function is continuous at those 2
places Our therefore if we assume the barrier begins at X equals 0 Annex and NZX equal we know With the acceptable solutions are going to be on the left and right of the barrier on the left it's 82 the IKR explicit BE the minus side kayaks pending with the energy is as for excellence in 0 on the right-hand side it's Eden the miners are the 2 that K explosive the minus side kayaks that's axes greater than there is that these are just the free particles solutions that we found and once again we know based cycle equal to 0 because we're assuming that the particles coming from the left and moving right and therefore that term can be 0 but we can't say that B is equal to 0 because the particle could be reflected in 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 is just a 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 insanity coming backwards because there's physically nothing there that can be reflecting the particle In that and what that means mathematically which is important is that we can safely say equals to 0 and therefore are starting point for the solutions on the left is heeded the eye-care explicitly the mind's eye care acts on the right is just be cut the capital Times needed the eye-care 6 4 greater than now 1 of the acceptable solutions within the barrier while we we saw From before and we go back to the times independent showed equation to find out and we got the solution in terms of capital and now out to make an acceptable solution we have to find a that the derivative must be continuous and the function must be continuous and now must be so on both sides there is a difference here before we got rid of the positive exponential with the term the plus Kappa acts in the argument is that as X goes to infinity we can't have that term because he gets too big but now we can have it and the reason why we can't have it is that that solutions stops at XP equals spell and that's fine I the something even to cap is some finite number that doesn't violate anything about the way 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 side on the left of Shelby discontinuous where it's gonna kid or you can have a sloping discontinuous work comes in and then suddenly changes slope and of course we don't have to worry about he the sire the derivatives being discontinuous except at the boundaries because we have these nice functions we know that they have an infinite number of derivatives these exponential so there's no problem except at the actual discontinuity Of the potential the 4 conditions than at the start and at the end of the barrier give us for equations to relate to the coefficients that we've got and as usual we have 5 coefficients so it's more more tedious at X equals 0 the continuity of sigh which written here sigh from the minus side has to equal to sigh from the plus side that means the real power and the measure this is a plus B a single perceived and the continuity of the derivatives the side means that a times I carried a plus B times minus cited is equal to see times capped which we can have a term we have to keep plus times minus kaput those are the 2 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 have a term with L and the we don't know that that's what we knew did know that he needed the 0 annexes heroes 1 but now we've got an extra things to keep track of and that X equals we have the continuity of science gives us the equation e to the capital "quotation mark St to the minus Kappa L that's what size inside the barrier is equal to there was no answer capital E times that I tale that's the way function on the right and the continuity of the derivative debates IDX means that see the 2 The Capital Times Kabel plus the 2 the minus Kappa L times minus Kemper is equal to capital times times UDI and these 4 equations that are what you have to battle with an order to figure out what the chance of getting through the barrier but clearly the probability of observing the particle on the right hand on the right-hand side depends on the square of the coefficient eh because the EDA the plus I K X 1 X is bigger than now all we need to know is the amount of that thing and we know but the chances that a got through the probability that the particle was on the left and moving to the right depends on the square modulus of capital a capital a squared and therefore we want to have a ratio 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 the coefficients is called the transmission coefficient tea it's it's basically the probability but the particle tunnel 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 but now we have to get a relationship between them with all these equations and I can tell you that when you 1st tried to do this it's not very easy so you wanna get in a quiet room and you want to get a big piece of paper and you wanna 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 the then were you go round and round and it's probably good to do that a couple
of times because if you do that then you 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 I again we have 4 equations 5 unknowns we can't solve uniquely but we can certainly sulfur the ratio because I showed you before we could solve the ratio of things and that's what we did what what we need to do we can get the ratio for fatigue 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 the squared cinch squared cap L divided by 4 80 times the minuses but it's gonna take a while to figure out there and you may not be familiar even with the hyperbolic signed functions but the hyperbolic signed 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 of probability is not going to be 1 because we have won over 1 plus some other thing and the other thing I can do nothing except at it's got a bunch of squares and it's got to be minus B and we know the is bigger than the so these are all positive things over there so the transmission coefficient Gambia 100 per cent but and it can be very big ears these things In the denominator get big because that's going to make the probability is small those things have Kappa they have held and that the difference between V n ee and those are exactly the kinds of things that we expect to enter in to a calculation like this the hyperbolic sign function is defined as cinch the data is equal to the to to the theater -minus to the minors they divided by 2 it's a purely real function it's exactly the analogous thing of the signs except there's no on the sign is needed the I favor minus into the minus sign the over to us if you work out the transmission coefficient following along the lines that I did for the step function but to getting the ratios of those things and and substituting the man but it's a that's a good weeks work actually to work that out in detail from scratch without looking up anything they're trying to find some shortcuts 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 that there can be tricks To simplify certain ratios to see that certain things must be Rick unit length the lengths to worry the I also plus 1 and what 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 but if you stare at it and you may have to find it very hard to see which way to go what is the interpretation well if the barrier it is sufficiently impermeable because that's going to turn out to be a very important case where we think it's pretty unlikely actually but the particle would get through the barrier classically corsets it's impossible but still there levels of impossibility in this game and now we think quantum mechanically it's not likely but if the transmission is low the various impermeable in the sense that cap at times at all this is much much bigger than 1 so 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 synch function and we get 16 times you over times 1 minus the over 3 times the the minus 2 capital but 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 he gets bigger the only thing that can come through as the dying exponential the positive exponential that contributed to the censure there can contribute anymore because it would get too large it would keep going up and it can't do that because it's very unlikely but the particle went through and therefore we get this simple formula and what that means that as the transmission will decrease exponentially as the barrier region has made and that's the basis of a very very potent kind of my cross the beer that we're going to talk about in the next lecture which is called scanning tunneling my cross be in which we actually use the fact that electrons are light and can tunnel to actually designed a microscope to look at surfaces With absolutely fantastic resolution so much better then an optical microscope or even an electron microscope but it's amazing and core and resulted in a Nobel Prize the transmission also responds exponentially to the square root of the matter and what that means is that light particles can call easily and the lightest kinds of particles that we usually deal with in chemistry or electrons and electrons therefore we have to be extremely careful with because they can certainly tell all around and after that we have protons are next there quite a bit heavier so the tunneling behavior will will not be sold good for them and then after that once we get to heavier atoms of protons the tunneling behavior really goes down quite lots of protons are a special case because the next 1 is a Deuteronomy that's mass to land mass to is twice as big as 1 and so it's quite a lot and cost 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 that is light particles tunnel much more easily and that's exactly why we never see you put your fist to a wall in your face somehow winds 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 minuscule will leave it there and will pick up next time with an actual application of what seems like this as a satiric description of this phenomenon of quantum mechanical tunneling and will see how it made people a ton of money got people a Nobel Prize and has contributed greatly to all kinds of devices physics and including inspecting all kinds of circuits and tiny national structures to actually see exactly what they look like so will do that next time


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