We’ll start by splitting up the terms as follows. Now let us return to the hydrogen atom and ask ourselves what is the orbital angular momentum $$\textbf{l}$$ of the electron. where the values of $${\lambda _{\,n}}$$ are given above. The principal eigenvalue of the system and the corresponding eigenfunction are investigated both analytically and numerically. However there really was a reason for it. We started off this section looking at this BVP and we already know one eigenvalue ($$\lambda = 4$$) and we know one value of $$\lambda$$ that is not an eigenvalue ($$\lambda = 3$$). I hope the reader will not perpetuate such a degradation of the English language, and will always refer to "the hamiltonian operator". That is, $$\mathsf{AB}\psi = \mathsf{BA}\psi$$. The total energy of such a particle is the sum of its kinetic and potential energies, which, in nonrelativistic terms, is given by, $E = \frac{p^2}{2m} + V . In those two examples we solved homogeneous (and that’s important!) Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. We will work quite a few examples illustrating how to find eigenvalues and eigenfunctions. Try lx and ly. As with the previous example we again know that $$2\pi \sqrt { - \lambda } \ne 0$$ and so $$\sinh \left( {2\pi \sqrt { - \lambda } } \right) \ne 0$$. This will take a little time, but if you do this, you will obtain two rather complicated expressions for the first two, but the third one turns out to be very simple: \[\mathsf{l}_\mathsf{z} \equiv - i \hbar \frac{\partial}{\partial \phi} . Recall that Schrödinger's Equation is equation 7.8.5, and, for hydrogenlike atoms we use the Equation 7.9.1 for the potential energy. and the eigenfunctions that correspond to these eigenvalues are. So, eigenvalues for this case will occur where the two curves intersect. Doing so gives the following set of eigenvalues and eigenfunctions. $$\underline {\lambda > 0}$$ So, another way to write the solution to a second order differential equation whose characteristic polynomial has two real, distinct roots in the form $${r_1} = \alpha ,\,\,{r_2} = - \,\alpha$$ is. Schrodinger's equation is an example of an eigenvalue equation. Solving for $$\lambda$$ and we see that we get exactly the same positive eigenvalues for this BVP that we got in the previous example. We will be using both of these facts in some of our work so we shouldn’t forget them. Once again, we’ve got an example with no negative eigenvalues. It’s important to recall here that in order for $$\lambda$$ to be an eigenvalue then we had to be able to find nonzero solutions to the equation. $$\underline {1 - \lambda > 0,\,\,\lambda < 1}$$ The three cases that we will need to look at are : $$\lambda > 0$$, $$\lambda = 0$$, and $$\lambda < 0$$. The solution for a given eigenvalue is. They involve only the one quantum number (often called the "principal" quantum number) $$n$$, which can have any nonnegative integral value. But we still don't deny that it is exciting so far.). This will often not happen, but when it does we’ll take advantage of it. The $$z$$-component of angular momentum can have, for a given value of $$l$$, the $$2l+1$$ integral values from $$−l$$ to $$+l$$. Is this by any chance an eigenfunction for the operator $$\ref{7.10.10}$$? Have questions or comments? So, let’s get started on the cases. We’ll need to go through all three cases just as the previous example so let’s get started on that. conditions to see if we’ll get non-trivial solutions or not. If μ A (λ i) = 1, then λ i is said to be a simple eigenvalue. As an equation, this condition can be written as In Example 8 we used $$\lambda = 3$$ and the only solution was the trivial solution (i.e. and note that this will trivially satisfy the second boundary condition just as we saw in the second example above. Recall that we are assuming that $$\lambda > 0$$ here and so this will only be zero if $${c_2} = 0$$. The solution will depend on whether or not the roots are real distinct, double or complex and these cases will depend upon the sign/value of $$1 - \lambda$$. If the wavefunction that describes a system is an eigenfunction of an operator, then the value of the associated observable is extracted from the eigenfunction by operating on the eigenfunction with the appropriate operator. Do any two of these commute? The value of the observable for the system is the eigenvalue, and the system is said to be in an eigenstate. eigenfunction (see the example concerning d2 dx2 above). In the special case where D is defined on a function space, the eigenvectors are referred to as eigenfunctions. Some speakers also treat the word as if it were a noun, talking about "the hamiltonian". Also, we can again combine the last two into one set of eigenvalues and eigenfunctions. You appear to be on a device with a "narrow" screen width (. So, if we let $${c_2} = 0$$ we’ll get the trivial solution and so in order to satisfy this boundary condition we’ll need to require instead that. A walk through of this problem would help a … Having the solution in this form for some (actually most) of the problems we’ll be looking will make our life a lot easier. This means that we can only have. Therefore, for this BVP (and that’s important), if we have $$\lambda = 0$$ the only solution is the trivial solution and so $$\lambda = 0$$ cannot be an eigenvalue for this BVP. Therefore, in this case the only solution is the trivial solution and so, for this BVP we again have no negative eigenvalues. Now, before we start talking about the actual subject of this section let’s recall a topic from Linear Algebra that we briefly discussed previously in these notes. Everyone knows what commuting operators are. In summary then we will have the following eigenvalues/eigenfunctions for this BVP. Therefore. The solution of the Schrödinger equation is tantamount to seeking a function that is an eigenfunction of the operator in parentheses. As a rule, an eigenvalue problem is represented by a homogeneous equation with a parameter. By our assumption on $$\lambda$$ we again have no choice here but to have $${c_1} = 0$$. Also, in the next chapter we will again be restricting ourselves down to some pretty basic and simple problems in order to illustrate one of the more common methods for solving partial differential equations. The general solution to the differential equation is identical to the first few examples and so we have. Seen from the point of view of wave mechanics, however, there is nothing at all mysterious about it, and indeed it is precisely what one would expect. Writing these equations in operator form, we have: \[\mathsf{l}_\mathsf{x} \equiv - i \hbar \left( y \frac{\partial}{\partial z} - z \frac{\partial}{\partial y} \right) \label{7.10.8} \tag{7.10.8}$, and similar expressions for the operators ly and lz. Next let’s take a quick look at the graphs of these functions. Introduction to Quantum Mechanics I. a map between two vector spaces , : such that a) (1+2)=(1)+(2) b) (2)=(2) + ( )+T( )=( + ) ( ) ( ) a linear transformation: Lecture 13: Eigenvalues and eigenfunctions. Nine-Point Finite Difference Approximation $$\underline {1 - \lambda = 0,\,\,\,\lambda = 1}$$ Example 6.3For λ ∈R, solve y00+λy= 0, y(0)−y(π) = 0, y0(0)−y0(π) = 0. Now, applying the first boundary condition gives. We'll leave it to the mathematically inclined to work through the algebraic details, but what we get is the very same expression, Equation 7.4.7, that we got for the energy levels in Section 7.4 when we were dealing with the Bohr model - but this time without the arbitrary Bohr assumptions. This means that we have. This is so easy to see that it is almost a truism. Just carry out that simple operation, and you will immediately find that, $\mathsf{l}_\mathsf{z} | lmn \rangle = m | lmn \rangle . The question is: What is the significance of two operators that commute? I shall not go further into the algebra, which you can either do yourself (it is very straightforward) or refer to books on quantum mechanics, but if you write out in full the operator l2 (and you can work in either rectangular or spherical coordinates) you will soon find that it commutes with lz and hence also with $$\mathsf{H}$$, and hence $$|lmn\rangle$$ is an eigenfunction of it, too. For example, let $$\psi$$ be a function that is simultaneously an eigenfunction of two operators $$\mathsf{A}$$ and $$\mathsf{B}$$, so that $$\mathsf{A}\psi = a\psi$$ and $$\mathsf{B}\psi = b\psi$$. Then we can take !1 = "1 and !2 = "2 +c"1 and we can choose c=!" The four examples that we’ve worked to this point were all fairly simple (with simple being relative of course…), however we don’t want to leave without acknowledging that many eigenvalue/eigenfunctions problems are so easy. We’re working with this other differential equation just to make sure that we don’t get too locked into using one single differential equation. $$\underline {\lambda = 0}$$ FINDING EIGENVALUES AND EIGENVECTORS EXAMPLE 1: Find the eigenvalues and eigenvectors of the matrix A = 1 −3 3 3 −5 3 6 −6 4 . 14. eigenfunction Expansions 1 1. then we called $$\lambda$$ an eigenvalue of $$A$$ and $$\vec x$$ was its corresponding eigenvector. This is an Euler differential equation and so we know that we’ll need to find the roots of the following quadratic. Example 11.1.2 and so in this case we only have the trivial solution and there are no eigenvalues for which $$\lambda < 1$$. We then found the eigenfunction decomposition of the initial temperature $$f(x)=u(x,0)$$ in terms of the eigenfunctions \[f(x)= \sum_{n=1}^{\infty}c_nX_n(x).$ It is easy to show that if is a linear operator with an eigenfunction, then any multiple of is also an eigenfunction of . y'' + ay = 0; y'(0) = 0, y(1) = 0. So, now that all that work is out of the way let’s take a look at the second case. In rectangular coordinates it is easy to write down the components of this vector product: $l_x = yp_z - zp_y , \label{7.10.5} \tag{7.10.5}$, $l_y = z p_x - x p_z , \label{7.10.6} \tag{7.10.6}$, $l_z = xp_y - yp_x . So, in the previous two examples we saw that we generally need to consider different cases for $$\lambda$$ as different values will often lead to different general solutions. So, we’ve worked several eigenvalue/eigenfunctions examples in this section. So, we now know the eigenvalues for this case, but what about the eigenfunctions. with two different nonhomogeneous boundary conditions in the form. Applying the first boundary condition and using the fact that cosine is an even function (i.e.$$\cos \left( { - x} \right) = \cos \left( x \right)$$) and that sine is an odd function (i.e. Well go back to the previous section and take a look at Example 7 and Example 8. So, taking this into account and applying the second boundary condition we get. If you are still holding on to the idea of a hydrogen atom being a proton surrounded by an electron moving in circular or elliptical orbits around it, you will conclude that the only orbits possible are those that are oriented in such a manner that the $$z$$-component of the angular momentum must be an integral number of times $$\hbar$$, and you will be entirely mystified by this magical picture. a. e-3x b. cos(3x) c. ln(3x)thir d. 3x e. Sooner or later any books on quantum mechanics will bring in these words. As with the previous two examples we still have the standard three cases to look at. Here is that graph and note that the horizontal axis really is values of $$\sqrt \lambda$$ as that will make things a little easier to see and relate to values that we’re familiar with. Often the equations that we need to solve to get the eigenvalues are difficult if not impossible to solve exactly. Orbital angular momentum can take the values $$\sqrt{l(l+1)}\hbar$$, where, for a given $$n$$, $$l$$ can have the $$n$$ integral values from $$0$$ to $$n-1$$. And any operators that commute with the hamiltonian operator will also commute with each other, and all will have equation 7.9.5 as an eigenfunction. Applying the second boundary condition gives. In fact, you may have already seen the reason, at least in part. Why are commutating pairs of operators of special interest? In fact mathematically-minded people have already done that for us, and I have reproduced the result as Equation 7.9.5. In this case since we know that $$\lambda > 0$$ these roots are complex and we can write them instead as. This will often happen, but again we shouldn’t read anything into the fact that we didn’t have negative eigenvalues for either of these two BVP’s. By golly − it is, too! Before working this example let’s note that we will still be working the vast majority of our examples with the one differential equation we’ve been using to this point. and so we must have $${c_2} = 0$$ and once again in this third case we get the trivial solution and so this BVP will have no negative eigenvalues. 1 #"  2 d%" 1 #"  1 d%. The work is pretty much identical to the previous example however so we won’t put in quite as much detail here. We have just found that the function $$|lmn\rangle$$ is an eigenfunction of the operator lz and that the operator has the eigenvalue $$m$$, a number that, for a given $$l$$ can have any of the $$2l+1$$ integral values from $$−l$$ to $$+l$$. ), Let us return briefly to the wavefunction that describes a moving particle discussed at the end of section 7.8, and specifically to the time-dependent equation 7.8.9. FINDING EIGENVALUES • To do this, we ﬁnd the values of λ … So, let’s take a look at one example like this to see what kinds of things can be done to at least get an idea of what the eigenvalues look like in these kinds of cases. Recalling that $$\lambda > 0$$ and we can see that we do need to start the list of possible $$n$$’s at one instead of zero. For a given value of $$n$$ there is a total of $$n^2$$ possible combinations of $$l$$ and $$m$$. Eigenspaces, geometric multiplicity, and the eigenbasis for matrices If two operators $$\mathsf{A}$$ and $$\mathsf{B}$$ commute, then it doesn't matter in which order they are performed - you get the same result either way. There are BVP’s that will have negative eigenvalues. Let us give some examples. If μ A (λ i) equals the geometric multiplicity of λ i, γ A (λ i), defined in the next section, then λ i is said to be a semisimple eigenvalue. Then, \[\mathsf{AB} \psi = \mathsf{A} b \psi = b \mathsf{A} \psi = ba \psi = ab \psi$, \[\mathsf{BA} \psi = \mathsf{B} a \psi = a \mathsf{B} \psi = ab \psi . For example, suppose !1 and !2 both have eigenvalue a with respect to operator A. There is a maximal (negative) discrete eigenvalue, the corresponding eigenfunction u is called the ground state. The general solution here is. 5. So, this homogeneous BVP (recall this also means the boundary conditions are zero) seems to exhibit similar behavior to the behavior in the matrix equation above. Watch the recordings here on Youtube! If we compare this with equation 7.8.9 we see that we can write this in operator form if we replace $$E$$ by the operator $$i \hbar \frac{\partial}{\partial t}$$ and $$\textbf{p}$$ by the operator $$-i \hbar \nabla$$ (or, in one dimension, $$p_x$$ by $$-i \hbar \frac{\partial}{\partial x}$$). Or do not get too locked into the cases only have the standard three cases to at. 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To an origin ( i.e time, unlike the previous example so let ’ s not too bad will... To summarize the results of the observable for the purposes of this stuff! Yourself in the two sets of eigenfunctions, one corresponding to that case we... Solution was the trivial solution and so the “ official ” list of all possible for! T ) is not even possible to get them couple of times gives us the following eigenvalues/eigenfunctions for case!

## eigenvalue and eigenfunction examples

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