In this video we will discuss the approximation of using finite basis set. The use of finite basis set is quite common but it's rarely discussed in most common mechanics textbooks. The quantum mechanical problems can be reduced to linear algebra, where we express the operators as matrices and functions by vectors. The practical problem of solving these eigenvalue problems or solving these quantum mechanical problems of finding like energy eigenvalues and eigenstates. Then essentially boils down to problem of finding the eigenvectors of a matrix. And usually the exact analytical solution is just not available. And we have to resort to numerical technique to solve these matrix eigenvalue equation. Which means that for practical purpose we have to restrict the matrix to being a finite size matrix. We can even consider analytically of finite matrix and solve for a simple problem exactly. Then we would have an approximate analytical solution. In practice there is no prescribed formula for choosing an appropriate finite basis set. It is something of an art. It requires a lot of experience and good physical intuition. Usually, people choose the energy eigenfunctions of the unperturbed problem if there is one available as their basis set. If that is not available those of a simpler problem, so you simplify the problem that resembles the problem that you want to solve and use the eigenfunctions of those simpler problems as the basis set. So let's consider an example here. And this is an example that we consider before an infinite potential well, with a constant electric field. So this is the Hamiltonian and this part here is the potential energy arising from the applied constant electric field which is linearly proportional to the position. And we chose the potential energy 0 to occur at the center of the potential well. Then we use the unperturbed solution, the solution of the infinite potential well without any electric field as our basis set. And of course those solutions are shown here. And then using that, we construct the matrix for Hamiltonian operator and the matrix element ij. Matrix element is given here. Now if we include all of these infinitely many eigenfunctions, unperturbed eigenfunction side, i then we construct the infinite dimensional matrix for H. And if you can somehow solve that then you will obtain the exact solution. There will be no approximation. But of course solving an infinite dimensional eigenvalue equation is just not practical. And you have to truncate the basis set at some point to obtain a finite size matrix. So this is the approximation of using finite basis set. So as an example, we would just use the first 3 wavefunctions corresponding to the 3 lowest energy eigenstates of the unperturbed problem. And then we obtain 3 x 3 matrix for the Hamiltonian. And for numerical calculation, if you use these parameters, the width of the potential, well is one nanometer in the applied electric field is 10 to the ninth vote per meter. Then you can evaluate the integral given in the previous slide to obtain all of these matrix elements. And express it in terms of E sub 1 which is the ground state energy of the unperturbed problem infinite potential well with no electric field which is given here. So this is the three by three matrix of of the Hamiltonian. We just need to simply diagonalize it and find the energy eigenvalues which you can do by using your favorite software. Then these are the 3 eigenvalues that you would get. So the values are fairly close to the unperturbed values which are E sub 1 ground state energy 4E sub 1 that's the first eigenstate energy in 9E sub 1 is the second eigenstate energy. So these values that we obtain are similar, but the energy shifts lower for the ground state, whereas the second and third energy levels is actually shifted higher. And let us compare these results with the exact solutions using the area functions as the solution. If you recall this, the Schrodinger's equation for these infinite potential, well, with a constant electric field does have an exact solution. And the solution of those differential equation is given by the eigenfunctions. And you can evaluate the energies as shown here. So you can see that these finite basis set approximation using the unperturbed wavefunction as the basis set. That gives a very pretty close eigenvalues to the exact solutions. We can now proceed to calculate the eigenvectors for these 3 eigenvalues. And they are shown here. Explicitly we can write these ground state wavefunction shown in this vector here, in terms of the unperturbed eigenfunctions as shown here. And you can see that the co options for the ground state wavefunction is by far the largest. So it makes the dominant contribution, whereas the second and third wavefunction makes a smaller contribution. So if you plot those wavefunction and compare with the unperturbed wavefunction and the exact solution. You can see that here the red curve shows the unperturbed wave function the sine function and the pink line is the exact solution given by the area functions. And these blue markers shows the approximate solution we obtained using the finite basis set, the 3 functions, 3 lowest energy eigenfunctions of the unperturbed system. They're remarkably accurate. They follow the exact solution very very closely. Now, once again consider the ground state wave function given by this finite basis set approximation. You can interpret this result as being that the applied electric field mixes these different eigenstates. The mixes the ground state and the first eigenstates and mixes the ground state and mixes these second eigenstate as well. And if you recall this is how the perturbation theory works as well. So the perturbation equation, the first order correction to the wavefunction is given by this. So the first order correction actually includes the contribution by all the other unperturbed eigenstates. And as you move away from the energy level, you're interested in energy level of interest as E sub k becomes more and more different from E sub 1. This denominator becomes large and large. And so the co-option becomes smaller, making the mixing or the contribution from that eigenstate smaller and smaller. So for the ground state energy level the ground state wavefunction makes the dominant contribution. First eigenstate smaller and the second eigenstate even smaller in that order. That's how the perturbation approach works as well. And if you write down the perturbation solutions, using the same parameters, you will get this expressions here. So in this solution, perturbation solution, you have size of four, the 4th energy level, eigenfunction instead of side 3, which we used in the finite basis approximation. And the reason that the perturbation solution does not include any contribution from sigh 3 is because there is no mixing between sigh 1and sigh 3 of the symmetry, because of the parity. And this is actually true in the finite basis approximation 2. If you go back and check out the Hamiltonian matrix that we constructed using the finite basis set of sigh 1, sigh 2 and sigh 3. The 13 component and 31 components, which is the matrix element between sigh 1 and sigh 3. Those elements are actually zero. And there is no mixing the contribution from sigh 3 in the finite basis approximation actually arises from the mixing up sigh 2 to sigh 3. Resulting in non zero matrix element of H23 and H32. So this is included in the finite basis approximation, whereas in the part of eigen approximation only these mixing with sigh 1 to various other eigenstates are considered. So in this part of eigen approach, the mixing between 2 and 3, sigh 2 and sigh 3 is simply not included from the construction of the theory. So here are the graphical representation. Once again, the blue curve is the unperturbed state and the exact solutions and finite basis solution and the perturbation solutions are all shown here and they agree pretty well. And all of them predicts that the ground state wavefunction is shifted to the left and that's the region where the potential energy is lower. And therefore the ground state energy will be shifted lower compared to the unperturbed state.