29 - Quantum Physics - Particle in a box

let's now go over a classic problem known as the particle in a box or the infinite potential well same idea the idea is that we have a particle here located between x equals 0 and X equal L that is trapped in that region because in this region it has no potential energy and the regions to the left and to the right have an infinite amount of potential energy and in other words you can never provide this particle with enough energy for it to jump out of this region so it stays trapped there we're going to do first is write the Schrodinger equation for each region and then we're going to move on to the wave functions now for regions 1 & 3 we're going to have the same Schrodinger equation because these regions are similar we're going to write that minus H bar squared over 2m D 2 sy DX squared plus u times sy is equal to e total x sy now this for region 1 & 3 region 2 is going to be different because U is equal to 0 so there's one less term in the Schrodinger equation D 2 sy d x squared is equal to e total times sine this is for region 2 now as it happens because you want and you three are infinite this equation is going to boil down to something very simple because this term here is basically going to dwarf all these terms because U is so big now that means that we're left with u 1 sy 1 is equal to 0 which can only happen of course if sy 1 is equal to 0 now that makes sense if the particle can't leave this region we don't expect to have a wave function here and we don't expect to have one here which we don't because of course you three times psy 3 is equal to 0 for the same reason this term is so big that it Dwarfs all the other terms in the equation and therefore psy 3 is equal to 0 now for region 2 we'll lash with this region to this equation we don't really know how to solve it and nobody expects you to and what will probably happen is that someone will give you the solution it's going to say situ of X is going to be a sine KX plus B cosine KX and if you're given that then you can do things because you can plug this back into here and figure out a condition on K now let's try that let's try plugging sy 2 back into this we really have to take the double derivative with respect to X so first we have minus h-bar squared over 2m times the double derivative now sine is going to become cosine if I take the derivative again cosine becomes minus sine and for every derivative the chain rule pops out okay so it's going to be minus a K squared sine KX and the same thing happens here cosine becomes minus sign minus sine becomes minus cosine two K's pop out and we're it we end up with B K squared cosine KX and that is equal to e total times sigh I'm just going to write sigh and you'll see why because in fact if we get the K squares out of here and this minus that cancels these two minus signs we end up with H bar squared K squared divided by 2m times a sine KX plus B cosine KX oh wait that's sine in fact we accurate should have sight to here and that is sy to X is equal to e total say to that X which means that this term here has to be equal to this term in other words that K squared is equal to 2m e total divided by H bar squared which means that K is equal to square root of 2m e total divided by H bar squared so that gives you K you know if you have sy to plug it in you get a condition on K that relates it to the total energy and we're then going to see how to relate the total energy to an integer n in the next part we've just derived that K is equal to square root of 2m e total over H bar squared and we've also found that psy 1 is equal to C 3 is equal to 0 and we have sy 2 of X is equal to a sine KX plus B cosine KX so let's try to figure out what a and B are if we can the first thing we're going to do is say that overall the wave function even though it's a piecewise function is a continuous function therefore at zero sigh 1 and sight to have to match and an lc3 inside to have to match in other words at x equals 0 it must be that psy 1 of 0 is equal to side 2 of 0 now of course Taiwan is always 0 so it's still 0 at x equals 0 and then we have 0 is equal to Phi 1 at x equals 0 sine of 0 is 0 that goes away cosine of 0 is 1 so that gives you B therefore B must be equal to 0 and that comes from continuity now if we do the same thing at L situ L is equal to sy 3 L well sy 3 is always 0 so it's still going to be 0 at X equal L and if I plug in X equal L to here knowing that B is 0 I end up with a sine KL is equal to 0 now be very careful if we conclude that therefore a is 0 then a and B are both 0 sy 2 is 0 and we're not really doing much in terms of quantum mechanics so a cannot be 0 otherwise we get a trivial solution what has to be true is that sine KL is 0 that means that KL has to be equal to some integer n times pi with n is equal to 0 plus or minus 1 plus or minus 2 etc but we know that K is equal to that so this means that the energy and is quantized because KL is quantized which incidentally is beautiful but be that as it may we express K as this and we square everything so if we squared K we get 2m e total divided by h-bar squared and then we squared this piece so that gives us n squared pi squared divided by l squared and if we rearrange this just a little bit we get e total and that is N squared PI squared H bar squared divided by 2m l squared and we found 'total now we still have not found a but we're going to get to that eventually what we're going to say is that now that we have situ as in we know that B is zero and K has to be quantized like this well the normalization condition is going to give us pay now that we found that situ is a sine KX with K satisfying KL is equal to n pi and that this is given to us for later convenience we can find what a must be because all wave functions must satisfy the normalization condition and so there's only one a that's going to work now the normalization condition says that the integral from minus infinity to plus infinity of our wave function overall when I square it don't forget to square it is equal to 1 now our wave function overall has three pieces it has psy 1 psy - psy three but psy 1 and psy 3 are 0 so if you integrate from minus infinity to 0 you get nothing if you integrate from L to plus infinity you get nothing and this integral here in fact boils down to the integral from 0 to L of psy 2 squared xD X so this condition which is normalization condition is in fact the integral from zero to L of situ squared of X DX is equal to 1 which is if we use the expression here for sy 2 is a squared sine squared KX DX and that's why you're given this integral because this problem is math heavy enough that we don't have to expect you to also know how to integrate this though probably at some point in your math career you knew how to do that any of this is given this means that we can compute this and we're back to a condition here that can be written 1 is equal to a squared over 2 and then we evaluate X minus 1 over 2 K sine 2 KX between the limits 0 and L and that gives us a squared over 2 when X is equal to L we get L minus 1 over 2 K sine 2 K L minus 1 X is equal to 0 this is 0 sine of 0 is 0 so minus 0 nothing very interesting there here all we have to do is remember now that KL is equal to n pi and therefore we get a squared over 2 times L minus 1 over 2 K sine of 2 K L is really 2n pi and that's nice because sine 2 and pi is always equal to 0 because if N equals 0 that's a 0 if N equals 1 it's 2 pi you've got around the trig circle once and then you're left with sine of 2pi or zero same thing as zero so this in fact always goes away because KL satisfies and PI and therefore we're left with one is equal to a squared L over two and that means that we can derive what a must be a must be equal to square root of two over L that's what the normalization condition tells you in order for this to be a wave function a has to be equal to root of 2 over L we've now established that sy 2 is root 2 over L sine n PI x over L and all of a sudden we realize that there isn't just one wave function for region 2 there is in fact an infinite number of them because sy 2 is a function of n and n can be equal to 1 2 3 etc now I know we said previously that we could have n is equal to 0 and that's true mathematically however physically that doesn't make much sense because if n is equal to 0 sine of 0 0 and then sy 2 is just 0 so we have nothing there for the smallest value for n that we're going to use is N equals 1 and we might want to graph sy 2 for a few different values of n just to see what it looks like and if we do that we're going to find something like this for n equal 1 something like this for n equal 2 and then something like this for N equals 3 in fact if you remember that really really looks like standing wave patterns for a string attached at both ends having said that let's say we want to find the probability that the particle is in this region here between the left end and let's say L over 4 well the probability that the particle is in that region is given by an integral but we need a probability density function remember that situ if you square it is the probability density function and therefore is going to be equal to the integral from zero to L over four of situ squared X DX in other words it's equal to the integral from zero to L over 4 of 2 over L sine n PI x over L this is squared of course DX but remember we said n is equal to 1 so that makes our life a little bit easier and here is going to go away and if we integrate this using this we find that 2 over L times 1 over 2 that gives us 1 over L times X minus 1 over 2k K in fact is PI over L because n is equal to 1 so it's PI over L the L comes up sine of two PI over L X between 0 and L over 4 now this if we apply the limits I'm going to be 1 over L L over 4 minus L over 2 pi sine of 2 PI over L times L over 4 the elves are going to go away and we're left with sine of PI over 2 and minus 0 because if X is 0 obviously this is 0 and sine of 0 is 0 so minus 0 and sine of PI over 2 is conveniently equal to 1 and therefore simplifying the LS here we get 1 over 4 or minus 1 over 2pi which is approximately point zero nine one which means there's roughly a 9% chance you'll find the particle in this region so that's how it works you have to be careful that the value of n determines which wavefunction you consider and you need to remember of course that if you square your wavefunction you get a probability density function and you can compute the probability of finding the particle in a given region

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Keywords: Introductory physics, Physics, SAT, MCAT, AP Physics, Berkeley, Pre-Med, Pre-Health, Pre-Dental, Physics 8A, Physics 8B, particle in a box, schrödinger equation, wavefunction, normalization condition, probability density

Id: CsVwSWT7s30

Length: 18min 6sec (1086 seconds)

Published: Sun Jul 24 2016