In this video, we will discuss wave functions in position and momentum basis. So far, we implicitly assume that eigenvalues are discrete and can be indexed by an integer. A standard equation that we've been writing down was like this, and the eigenvalue b_n and the eigenkets b_n are both indexed by an integer n. However, we already know that eigenvalues are not always discrete and they can be continuously varying. For example, the energy of electron in a potential barrier problem, continuously varying. Momentum of a particle in a free space, also continuously varying. We now need to be able to extend all the results that we have obtained with the discrete case to a continuous case. The general eigenvalue equation can be written like this. Xi here with a hat is an operator, and then the psi here is a scalar eigenvalue, and then the xi ket are the eigenkets eigenvectors. The understanding here is that the eigenvalue xi can be continuously varying. Now, the extension of this discrete case into a continuous case can be done by simply changing the summation into an integral. Here in this page, we summarize the changes that we need to make in order to extend the discrete case into a continuous case. Now, with this understanding, we want to look at the position eigenfunction. Formally, you can write the eigenvalue equation for a position operator like this. x hat here is the position operator, which is just equal to x in the position basis. We already discussed this before. x prime ket here are the eigenkets and x prime here is the eigenvalue, it's a number. Now, we notice that the ket here, eigenket x prime, is actually a delta function, delta x minus x prime. Once again, remember, x here is a variable, x prime is a number. If you multiply x, the left-hand side of this eigenvalue equation, if you multiply x to the delta function here, as shown here, you can simply change x with an x prime. Why? Because the delta function is zero everywhere except x equals x prime. Therefore, when x here is not equal to x prime, the delta function is zero, so it doesn't make a difference and they are the same. You can see that the delta function satisfies this eigenvalue equation. Therefore, this delta function is a position operator with an eigenvalue x prime. They are also orthogonal and actually they are orthonormal. You can see that by considering this inner product from the definition of the inner product, you can write this inner product like this and this delta function product to simply give you delta x prime minus x double-prime, which gives you the orthonormality condition for this position basis function. Now, we consider the eigenfunction of a momentum operator. We already know the momentum operator p_x can be expressed by this differential operator and the eigenfunction for this piece of x operator is given by this exponential function, exponential ikx times a constant C, to be determined by the normalization condition. The eigenvalue is h bar times k. Now, constant C is determined by the orthonormality condition or the normalization condition shown here. This is the condition for the continuously varying case. The inner product in the left-hand side is given here by this integral. If you plug in this exponential function for this psi k double prime and psi k prime, you get this equation. This integral simply gives you a delta function. From this, you can determine the value of c to be just 1 over square root of 2 pi. Therefore, the complete eigenfunction of the momentum operator is given here, fully normalized. This here is the eigenfunction of momentum operator in the position basis set. Now, we formally recognize that the usual wave function that we've been using, we typically say that, oh, my wave function is Psi of x. What do we mean by that? That wave function really is the representation of ket Alpha in the position basis set. In the continuous case, your vector representation is given by this integral here, and if you formally plugin for this eigenfunction Psi position eigenfunction, which is the Delta function, and this inner product are the matrix representation in the Psi basis, and that will be this function here and that is the expression for this function Psi. Now, analogously, we use the same expression but this time, as eigenket, we use momentum eigenfunctions. If we do that, we get this equation here. Now we define an inverse transformation of this integral equation and we can write like this. Here, the momentum function here is taken to the left-hand side, and we have taken the complex conjugate of this exponential to make an inverse transformation here. Now we define our unitary transformation, as this integral transformation, and we can see that this unitary transformation U transforms the wave function in position basis into wave function in momentum basis. Now as an example, let's consider the position eigenstate in the position basis. In the position basis, your position eigenfunction was a Delta function. X_0 here is the eigenvalue. Now let's apply this transformation U integral transformation to this Delta function. That's here. Let's see the result here is the exponential function e^ikx_0. This is a function in the momentum basis set, so the variable here is k, not x. It's a function of k, and this is the position eigenfunction in the momentum basis. Recall, the momentum eigenfunction in the position basis was this exponential function. If you consider position eigenfunction, in the momentum basis, you get similar exponential function. Analogously, let's consider momentum eigenfunction in the position representation. In the position representation or using the position basis set, your momentum eigenfunction was this exponential function. Now let's apply this transformation here once again to get the function in the momentum basis or momentum representation, then this simply led to Delta function. Just as the position eigenfunction in position representation was the Delta function, momentum eigenfunction in the momentum representation is the Delta function. Finally, we consider the transformation of operators. Let's consider this position operator in the momentum representation. Then this is equal to the position operator in position representation transformed by the similarity transformation. U and U dagger are these two integral representation, and the position operator in position basis or position representation is just x. If you perform this double integral, you will find that using this mathematical identity, you are left with this, which simply gives you this simple differential operator. The position operator in momentum representation is i times d dk, and using p momentum is equal to h bar times k, you can rewrite like this. Compare this with the momentum operator in position representation like this, it's very similar except for this negative sign out in front.