So now we proceed with the technical solution of the quantum harmonic oscillator problem. And I should mention that there exist many equivalent ways to solve this problem. But today I am going to present a purely algebraic solution which is based on so-called creation/annihilation operators. I'll introduce them in this video. And as you will see, the harmonic oscillator spectrum and the properties of the wave functions will follow just from an analysis of these creation/annihilation operators and their commutation relations. We're not going to even write down the Schroedinger equation as the differential equation. We're not going to worry too much about any boundary conditions. So these operators and their commutation relations are going to be sufficient for us to determine pretty much everything we want to know. And the problem that we're actually solving is the eigenvalue problem for this operator H with this quadratic potential corresponding to the harmonic oscillator. So again, the eigenvalue problem is H psi is equal to E psi. Now there are only three dimensional quantities that appear in our problem on which our final results, in particular, the energy spectrum, can possibly depend. And these quantities are the particle mass, the frequency of the oscillator, and the Planck constant, which is always there. And it turns out that you can construct just one parameter out of these guys which has the physical dimension of energy, and this parameter is h times omega. So this has the physical dimension of energy. So whatever our spectrum is going to look like, it must scale as h omega. It must be proportional to h omega. So motivated by this, let me write down my Hamiltonian H divided by this h omega. And it's going to be simply I first will write the potential energy term m omega x squared over 2h bar plus the kinetic energy p squared over 2m omega over h bar. So I just switch the order in which these two terms appear in this Hamiltonian. And the next step I'm going to use is also going to be pretty strange from any point of view. So I'm going to describe the first term as the (square root of m omega over 2h bar x) squared plus the second term likewise is going to be (p over 2m omega h bar) squared. And now so the reason I wrote it this way is because I want to use an analogue of the following expression that we know from elementary calculus. Namely, that if we have two variables, let's say A squared minus B squared, we can write it as (A-B)(A+B). Or an equivalent expression to this involving complex numbers, if we have A squared plus B squared, so we can write it as (A-iB)(A+iB). And when we square the imaginary constant, it's going to give rise to the plus sign here. So in this expression I'm going to associate the first term with A and the second term with B. And so what I'm going to write is going to be the following. So I will represent my Hamiltonian divided by h omega, this energy scale in the problem, as, well, this guy square root of m omega over 2h bar x minus i momentum divided by this whole thing, times the same thing but with a plus sign here. So m omega over 2h bar x plus i p over 2m omega h bar. At this stage I have to admit that I have cheated very seriously in this derivation and this equation contains an error. And as a matter of fact, there are some missing terms that I have omitted. And in the next individual quiz, you're suppose to catch me and tell me where exactly I have cheated. So hopefully most of you have figured out where the problem is. And of course, the problem is in that the objects that appear in our quantum mechanical problem are not just some variables. Indeed, these are operators. And for the operators, in particular for operators x and p, it actually matters in which order they appear in an equation. So x times p is not equal to p times x. So in other words, these operations do not commute. And so if we try to re-derive this identity for these operators, we're going to see, of course, that they're going to be cross terms appearing. And so therefore, there is an addition to this guy, so let me write it explicitly. So these terms that I have omitted so it's equal to -i over 2 h bar, the commutator of x and p. And so this commutator of x and p is equal to xp-px. You can verify that this term indeed appears just by expanding this product term by term and making sure that we've reproduced the Hamiltonian in the left-hand side. One other thing that we can verify by looking at this expression is that the first bracket is at Hermitian conjugate of the second bracket. So indeed, x and p are physical operators and as such they are Hermitian operators. So x dagger is equal to x and p dagger is equal to p by definition. Therefore, if I Hermitian conjugate, let's say, the second bracket so x will remain x, p will remain p, all the constants here are real and the only thing that's going to happen is that i will become -i, so we will produce the first bracket in this expression. Based on this fact, let me introduce a new operator, so we'll just call the second bracket an operator a. And the first bracket is going to be Hermitian conjugate too, therefore a dagger. And going a bit ahead of myself, let me mention that these operators, a dagger and a, are called creation and annihilation operators. And the following discussion and derivation will contain a proof that these operators, these guys, indeed deserve the name of creation/annihilation operators. But to explain the reason behind this terminology right away, let me advertise the main result before we actually derive it. And the main result here is the energy spectrum or the harmonic oscillator, which, as we shall see, contains a series of equidistant energy levels, that is, energy levels such that any neighboring pair of levels are separated from one another by the same energy. And this energy happens to be h omega, exactly the energy scale we discussed previously. And the energy of the ground state, the lowest-energy state is h omega over 2. And by the way, this h omega over 2 comes exactly from this additional term that we have in this identity. And so the importance of creation/annihilation operators are in the following. So let's say if we prepare our quantum oscillator in the ground state, so let me symbolically represent it by a dot here. So let's say we have an oscillator in the ground state. And if we apply a creation operator a dagger, it will create essentially a quantum of energy h omega by promoting this oscillator from the ground state to the first excited state. If we apply it again, so then we will go from the first excited state to the second excited state, etc., etc. So if we apply it, let's say, a dagger ten times, we will go from the ground state to the tenth excited state. And you can probably guess that the action of the operator a is opposite to it. So if we, say, have a quantum state with n equals 1, so this is the first excited state, but line a were going to go back to the ground state. Okay, so essentially these operators a and a dagger, they move us between the states in this energy landscape. And in the next video we're going to prove all these statements, and this proof will rely in a very essential way on various commutation relations between the operators involved here, x, p, a, and then a dagger.