In this video, we will revisit Harmonic oscillator problem. Simple harmonic oscillator, if you recall, the Hamiltonian, is given by this equation here. The kinetic energy term and the potential energy term for a simple oscillator, ideal spring case. We define the annihilation and creation operators. Operator a is defined as this and a-dagger defined like this. Here, notice that x and p are both operators, and p and x are all operators here. I just dropped the hat sign for simplicity, but it's implied that both x and p are position and momentum operators respectively here. Now, with this definition, it follows straightforwardly that the commutator between a and a-dagger, you just substitute this to this commutator and you work out the algebra that you get one. They don't commute. A-dagger times a, again, if you just multiply these two, then after some algebra, you will find that it's H Hamiltonian divided by h-bar omega, minus 1/2. We can define an operator, a-dagger a, as capital N, and we call that a number operator. You can write your Hamiltonian now in terms of a number operator in a very simple form, h-bar omega times N operator plus 1/2. It's obvious that the Hamiltonian and the number operator commute because Hamiltonian is number operator times a constant plus a constant, so they commute and they share the same set of eigenvectors or in the language of linear algebra, they can be diagonalized simultaneously. Now, the eigenkets of a number operator, let's denote them as ket n here, and the eigenvalue is little n as shown here. Then it immediately follows that the ket n is an eigenvector, eigenket of a Hamiltonian as well with an eigenvalue of h bar omega times n plus 1/2. Now, this eigenstate of operator N represented by this ket n is called the number state. These numbers states are eigenstates of Hamiltonian too because Hamiltonian and the number operator commute with each other and the energy eigenvalue for this number state n is h-bar omega times n plus 1/2. Now, let's consider the commutation relation between the number operator and a and a-dagger operators. Because N number operator is defined as a-dagger times a, you can just write it. I'll spell out this commutation relation and use the commutation relation between a-dagger and a, which is just one a and a. They obviously commute with itself. You get these negative a for this commutation between N and a. Similarly, commutation of N and a-dagger gives you a-dagger. Now, we consider N and a-dagger n the eigenket of the number state. Now we represent the N a-dagger in terms of this commutator relation that we found just above. Then you will find that this N a-dagger operating on a number state n, gives you n plus one times a-dagger n. Similarly, N a n gives you n minus one times a operator acting on n. What does this mean? That means that a dagger n here is an eigenstate of number operator N with an eigenvalue n plus 1. Similarly, aN is an eigenstate of number operator N with an eigenvalue n minus 1. If you recall that this number state n is an eigenstate of the number operator with eigenvalue N. If you apply a dagger to that number state n, you create an eigenstate number state with an eigenvalue n plus 1. You're creating one quantum, hence the name creation operator. If you operate a to a number state n, you're creating an eigenstate number state with an eigenvalue n minus 1. You're decreasing, you're annihilating one quantum, hence the name annihilation operator. Because a annihilation operator operating on number state n creates an eigenstate with an eigenvalue n minus 1, this state should be a constant multiple of number state n minus 1. That constant c can be found by the normalization condition. Those states should be normalized. From the normalization condition, we define your c square absolute value square through this inner product with itself, but the number operator N is defined once again as a dagger times a, so this operator here is just a number operator, and therefore, c square should be absolute value square of c should be just equal to n, or c is square root of n. We are free to choose any phase factor once again, so we're just making it real. Therefore, operating an annihilation operator to number state n creates a number state n minus 1, and the constant that should be multiplied in front to make everything consistent and normalized is square root of n. This is an equation. Now, you can do the same thing for the creation operator and then you will find that the creation operator acting on the number state n creates a number state n plus 1 with a constant square root of n plus 1 multiplied to it. Now, since the c absolute value of square is n, n must be non-negative. It's positive definite. Also, if we start with an arbitrary number state n and keep applying annihilation operator and you're reducing n by one every time you apply this annihilation operator to this initial state n. First time, you get n minus 1, you get n minus 2, and then n minus 3, and so on. Now, if n is an integer, then this process stops when n becomes zero, because when n becomes zero, then you get a null state, you multiply zero to this state, and then you get a null state, and whatever operator you may apply to a null state, you will still get a null state, you don't do anything. But if n is not an integer, then you never hit zero. You're skipping over zero, and then you end up going into a negative number, which is forbidden by this condition here that n is equal to c absolute value square. From this consideration, we come to a conclusion that n, the number state n, the number that specifies a number state, or the eigenvalue of the number state should be a non-negative integer. It's zero or positive integer. Now, since the minimum is n equals 0, then we can start with zero, the n equals 0 ket number state and we simply apply a creation operator to produce n equals 1 state. If you keep doing it, you will end up creating all number state. General number state n is simply given by this equation here. Now, we have produced all eigenstates of number operator, and therefore all eigenstates of Hamiltonian operator. All you need to do now is to know this zero, n equals 0 state. To find n equals 0 state, we know that if you apply annihilation operator to the lowest state, you should get null ket zero. From the definition of your annihilation operator a, we get this equation. Here we use position representation, so we choose a specific representation. In position representation, your ket is represented by a wave function, function of position and the momentum operator is given by this differential operator. The equation then becomes this x_naught here is given by this constant here. This is a simple differential equation you can immediately solve and get a Gaussian function as your solution. This constant here is determined by the normalization condition. Now we have found the ground state, lowest state, and then we successively apply the creation operator to this function, and then you generate all wave functions of harmonic oscillator problem, which in previous lecture we obtained by solving the Hermite differential equations. Finally, we look at the position and momentum operator in terms of the creation and annihilation operator. Then we calculate the matrix element of position and momentum operator with respect to the number states. From simple algebra, you will be able to derive these equations. They are non-diagonal for one. You immediately find that for the ground state, n equals 0 state, the expectation value of x and p are both zero. From this equation, you can immediately see this. For x square operator, once again, you can express x square operator in terms of a and a dagger operators. The expectation value of x squared with respect to the ground state n equals 0 state. You will find that it will give you x_naught square over 2. Similarly, for p square operator, you get h bar m Omega divided by 2. Now we define a variance operators, Delta x square and Delta p squared as the difference between x square expectation value and x expectation value squared. Because the x expectation value is zero, it's simply is equal to the x square expectation value, similarly for p. Now we obtain the uncertainty relationship, the product of the variance of x and variance of p ends up giving you h bar square over 2. This is consistent with the momentum position uncertainty relationship. For any excited states n, you end up getting this. Ground-state give you the lowest uncertainty, excited state give you higher uncertainty.
