In this video, we will solve Finite potential well problem. Finite potential well problem is specified by this potential profile V of z, where the potential V is zero inside the well and then some finite value V naught outside. This obviously is a more realistic case compared to the infinite potential well problem that we have dealt with before. Here, if the energy of the electron is greater than the well depth, the electron can escape the potential well and then can free to move around. Here, we are focusing our attention to the case where the energy of the electron is smaller than the depth of the well in which case electron does not have enough energy to escape the potential well, and they are therefore confined inside the well, that state is called the bound state. Now, before we solve this problem, we should anticipate the solution, whatever solution that will be getting should approach the solution of the infinite potential well in the limit of very large V naught potential well becomes deeper and deeper, or the L the width of the potential well becomes smaller and smaller. In both cases, the aspect ratio, the ratio between the depth and the width of the potential approaches infinity and that will be the case for the infinite potential well, and also before we begin, I notice the difference here in terms of the well width in order to make the mathematical expressions a little simpler, we choose to define the boundaries of the well as plus and minus l so that the width of the well here in this configuration is 2 times l instead of just l in the case of the infinite potential well problem that we solved before. Now, write down the time-independent Schrodinger equation in region one, which is the inside the well between negative L and L, then the potential is zero the equation is the same as the case of the infinite potential well, so V potential is zero. Your time-independent Schrodinger equation simply becomes this. Now, we rewrite this equation here, second derivative of psi plus k squared times psi is zero, where k square is given by the square root of 2 mE divided by h bar square, just as before. Now outside the potential well, we defined region 2 being the region where z is less than negative L and region 3 being the region where z is greater than L. In both cases, potential is V naught so the time-independent Schrodinger equation is this, we rewrite this equation in a simpler form again. This time the constants multiplied to psi here is K naught square. K naught is defined by this equation here 2m times V naught over E divided by h bar square. Now, there is a negative sign here that's important compared to the positive sign in equation one. In equation one, you get solutions that are sines and cosine functions. Whereas here in equation two, your solutions will be exponential functions. Also note that because we are confining ourselves to the case where E energy is less than V naught, this V naught minus E is always positive, and therefore K naught here is always a real positive number. Now, let's write down the solutions, inside the well, you get sine and cosine. Now outside the well, we will have exponential functions so in order to make the math a little more seamless, we use e to the IJZ and e to the negative IKZ functions instead of sine and cosine A and B are the constants to be determined. Solving equation 2 you get exponential function plus-minus e to the k naught z and plus and minus e to the k naught z for wave functions in region two and region three also. Now, how do we determine these constants, A, B, C, D, F, G? Well, we impose boundary conditions. What are the boundary conditions? Now notice that the potential is finite everywhere. This is a finite potential well problem, and therefore your potential V is finite everywhere. Now if you write down the Schrodinger's equation, time-independent equation, this finite, E here is less than V naught, and therefore E is here also finite and therefore your second derivative of the wave function must be finite. If your second derivative of the wave function R is to finite, then the first derivative must be finite and continuous. That's boundary condition number 1. Because the first derivative is finite, that means that the wavefunction itself must be finite and also continuous. That is the boundary condition number 2. In summary, the boundary conditions for time-independent Schrodinger equation in the case where the potential is finite everywhere, the boundary conditions are first derivative of the wavefunction must be finite and continuous everywhere and the wavefunction itself must be finite and continuous everywhere. Let's write down the general solutions again altogether. The first equation, Psis of one is the sine and cosine functions inside the well. Psi 2 is outside the well to the left. Psi 3 is outside the well to the right and in region 2 and 3, you get exponential function. First of all, the wavefunction must be finite everywhere. What that means is that you can't have this term here, second term in Psi 2. This is a function that diverges as Z approaches negative infinity, and therefore this term should be eliminated by setting the coefficient D equals 0. Likewise, F has to be set to 0 in order to avoid divergence as Z goes to infinity in region 3. So D and F are 0, and therefore, the wavefunction in region 2 is an exponential function that is decreasing as Z decreases in the negative direction. Psi 3, the wavefunction in region 3 is also an exponentially decreasing function as Z increases in the positive direction. Now, we can impose the continuity condition for the wavefunction Psi and the derivative at the boundary Z equals plus and minus L. If you apply the boundary condition for the left side of the well and Z equals negative L, the requirement for the wavefunction to be continuous gives you equation number 1 and the requirement that the first derivative of the wave function to be continuous gives you equation number 2. Now, you do the same on the right edge of the potential well at Z equals plus L, once again, the continuity of the wavefunction gives you equation number 3 and the continuity of the first derivative of the wavefunction gives you equation number 4. Now, we simplify these equations by adding equations 1 and 3 to get this equation number 5 and subtract equation 4 from equation 2 to get this equation number 6 and now you notice that we have the same coefficients, C plus G on the left hand side, and A plus B on the right hand side, which can be canceled. You take the ratio of these two equations, divide 6 by 5, then you get this equation number 7. Tangent of k times L is equal to k naught over k. Now you can of course only do this only when C is not equal to negative G and A is not equal to negative B so that these coefficient C plus G and A plus B are non zero. Alternatively, you can subtract equation 1 from equation 3 to get equation number 8 and add the equations 2 and 4 to get equation number 9 and once again, we have the same coefficients here on the left hand side, G minus C in this case, and A minus B the same coefficients on the right hand side. Now once again, we can divide these two equations from each other and if you divide equation 9 by 8, then you get this equation here, equation number 10. Cotangent of k times L is equal to negative k naught over k. Once again, you can do this only when the coefficients G minus C and A minus B are non zero, that is, C is not equal to G and A is not equal to B. Now, notice that equations 7 and 10 are contradictory to each other. In other words, they cannot be satisfied simultaneously. You either have one or the other. Therefore, you can have C equals G and A equals B. In this case, you satisfy equation 7 or you can use equation 7. That leads to symmetric solution or even parity solution as you will see in just a minute. Or you can have C is equal to negative G and A equals negative B. In this case, equation 10 applies, and in that case, you will get anti-symmetric solution or odd parity solution. Now, in order to find the energy, the k values and energy, you need to evaluate equation 7 and 10 numerically. They are transcendental equation, you can't solve it analytically. So we have to resort to numerical solution or graphical solution. To make that analysis simpler, we define new variables, Eta, which is defined to be k times L, and Xi, which is defined to be k naught times L, and then using the definition of k and k naught, you can write down this equation here: Xi squared plus Eta squared is equal to these quantity dependent upon the well width and the well depth. Then, of course, we need to use equation 7 or 10. So equation 7 and 10 can be rewritten in terms of Eta and Xi as this. This is equation 7 and this is equation 10. Now what we do is to plot these various equations in the Eta, Xi space and find the intersections, and that will give us the solutions. Here it is. Here is the Eta, Xi plot. The equations here, we first derived from the definitions of k and k naught, simply defines a circle in Eta, Xi space. We're of course plotting only the first quadrant, so there is a quarter of the circle defined by this equation here. The radius is given once again by the mass of the particle, the well depth and the well width. Then equation 7 here, Eta tangent Eta is equal to Xi will give you a series of tangent function. Tangent function is a periodic function so you have many, many repeating curves, like these blue curves shown here. There will be many more for larger values of Eta. But obviously, anything that occurs for Eta values that are greater than the radius of this circle is not relevant. There will be no intersections. In this particular example, we show two intersections with these blue curves representing this equation 7. Those are the two symmetric solutions. Then, of course, there is equation 10, Eta cotangent Eta is equal to negative Xi. They are shown by these red curves here, and it also shows these two intersection points. So in this particular example that I'm showing here, there are two even solutions and there are two odd solution and a total of four solutions. In other words, in this particular finite potential well, there are four confined states with distinct energies. If you plot the actual wave functions once you have determined the values of the energy and therefore the values of k and k naught, you can actually go back to the wave function and you can plot the wave function, and here are the wave functions. This is the lowest energy state, even solution. It very much looks like the lowest energy state of the infinite potential well, except that there is this exponential tail penetrating into the barrier regions. Then down here is the next even solutions, and they both are symmetric about the center of the potential well, and therefore they are the even solution. That characteristic is just dictated by equation 7 that we derived earlier. The odd solutions are shown here, the lowest energy odd solution, which is the second lowest energy level. Shown here, again, very much looks like the first excited state of the infinite potential well, except that you have this exponential tail inside the barrier region. So there is a non-zero probability of finding the particle inside the barrier region. This here shows the next odd solutions. This will be, in total, the fourth and final excited state. If the graph that I have shown you in the previous slide is the case, then that will be all. There are only four intersections as shown here, and these four states are all there is in terms of the confined states. The particles confined inside the well can have only one of these four available states.