QM1 - Tutorial 2 Schrodinger Equation, Hamiltonian and Free Particle

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1 QM - Tutorial Schrodinger Equation, Hamiltonian and Free Particle Yaakov Yuin November 07 Contents Hamiltonian. Denition Example: Hamiltonian of a Free Particle Wave Function 3 Schrodinger Equation 3. Statement of the Equation Physical Interpretation: Conservation of Energy Time Independent Schrodinger Equation General Solution of the Schrodinger Equation Free Particle 5 4. General Solution of the Schrodinger Equation Time Evolution of a Wave Packet Hamiltonian. Denition Conservation of Energy In any closed system the energy is conserved. We say: the energy is a constant of motion. No matter what the particle does (what its motion is the total energy will always be the same. Kinetic and Potential Energy the potential energy V. The total energy E can be broken down into two parts. The kinetic energy T and E = T + V In many cases T depends only on the velocity and V depends only on the position. E.g. for a mass m tied to a spring k (harmonic oscillator we have T = mv, V = kx, E = mv + kx The conservation of energy implies that as the position x increases the velocity v decreases and vice versa. Hamiltonian Using the connection p = mv the energy can be ressed in terms of momentum p instead of velocity v. For the harmonic oscillator this is written E = p + kx When the energy is ressed in terms of p and x we call it the Hamiltonian H. H (p, x = T (p + V (x

2 . Example: Hamiltonian of a Free Particle Statement of the Problem Find the Hamiltonian of a free particle. Free means that no force acts on it and hence V (x = 0. Solution The total energy E = T + V consists in this case of just the kinetic part T. We plug in v = p/m to obtain Wave Function E = T = mv H = p The wave function of a system contains the information about the state it is in. If the wave function of a particle is ψ(x, t the probability to nd the particle at the point x (or rather in the interval [x, x + dx] and at time t is We also introduce the probability density P (x, t = ψ(x, t dx = ψ (x, tψ(x, tdx ρ (x, t = ψ(x, t Note that, because it is a density, ρ (x, t has units (length while the probability P (x, t = ρ (x, t dx is dimensionless. The probability of nding the particle between the point x = a and the point x = b is P (a < x < b, t = b a ψ(x, t dx If we look for the particle in all of space [, ] we will denitely nd it, hence, an integral over all of space must give. ψ(x, t dx = This is known as the normalization condition. It is true for any instance in time, so for all values of t. 3 Schrodinger Equation 3. Statement of the Equation The equation that governs the evolution in time of the wave function ψ(x, t is the so called Schrodinger equation. It is i ψ(x, t = t ψ(x, t + V (xψ(x, t In the coming lessons we will solve this equation for dierent potentials V (x and analyze the solutions. For now we will take the simplest example, namely V (x = Physical Interpretation: Conservation of Energy Lets give the Schrodinger equation a physical interpretation. Consider a particle of mass m. Its momentum is given by p = k and its energy is E = ω. Satement of the Problem Assume the particle's wave function is ψ(x, t = A [ikx iωt] Plug the wave function into Schroedinger's equation. What did you nd? Notice that ψ (x, t is the comlex form of a plane wave.

3 Solution The left hand side is i ψ(x, t = ωψ(x, t = Eψ(x, t t and on the right hand side we get Putting it together we get This is just the total energy of the particle. ψ(x, t + V (xψ(x, t = p {}}{ E = p + V (x 3.3 Time Independent Schrodinger Equation k ψ(x, t + V (xψ(x, t Seperation of the Space Variable x from the Time Variable t For almost all cases we will consider in this course the right hand side of the Schroedinger will be time independent. (The potential is constant in time. In this case the wave function can be written as the product of two functions. Plugging this into the Schrodinger equation gives T (t i φ(x = t ψ(x, t = φ(xt (t We divide this by ψ(x, t = φ(xt (t on both sides to obtain φ(x x T (t + V (xφ(xt (t T i T = φ φ + V (x Both sides must be equal to a constant which we call E (the energy. We thus get two equations. One for T (t and one for φ (x. T (t i = ET (t t φ (x + V (xφ(x = Eφ(x The equation for φ(x is called the time independent Schroedinger equation. Solving the Time Dependent Equation The equation for T (t is easy to solve i T t = ET dt T = ie dt T (t = A ( i E t The Hamiltonian Operator Lets take a closer look at the time independent Schrodinger equation. Note that it may be written as [ φ (x + V (xφ(x = Eφ(x ] x + V (x φ(x = Eφ(x The ression in the square brackets is an operator. It tells us to do some action on the function following it ( φ (x. In this case the action is taking its second derivative and multiplying it by some function V (x. For reasons that will be made clear shortly we will call this operator the Hamiltonian operator H op = x + V (x 3

4 The time independent Schrodinger equation may now be written H op φ(x = Eφ(x The interpretation is simple. When the Hamiltonian operator H op acts on φ (x it extracts the energy E. Since the classical Hamiltonian H (p, x is nothing more than the total energy this makes sense. In addition we dene the momentum operator as the derivative with respect to x and the position operator as the multiplication by x Inserting p op and x op into H op gives p op = i x x op = x H op = p op + V (x op This is looks like the classical Hamiltonian H (p, x but it is not exactly it. We have turned it into a quantum operator. How to Solve the Time Independent Schrodinger Equation will be solving the time independent Schrodinger equation. One of the main subjects of the coming lessons H op φ(x = Eφ(x, H op = x + V (x Generally one nds a set of possible solutions {φ n (x}. It is therefore customary to add a subscript n in the equation. H op φ n (x = E n φ n (x There is a solution for each n. The set of functions {φ n (x} are called eigen-functions or eigen-states while the set of numbers {E n } are called eigen-values or eigen-energies. The equation H op φ n = E n φ n is called the eigen-value equation. Properties of φ n and E n Orthogonality: φ n φ m = φ n(xφ m (xdx = δ nm Completeness: The set {φ n (x} are span a basis. Eigen-values are real: E n R 3.4 General Solution of the Schrodinger Equation Superposition i.e. The most general solution to the Schrodinger equation is a superposition of all possible states φ n (x, ψ(x, t = n= ( C n φ n (x i E n t The coecient C n is the probability to nd the particle in the n-th state. If it is in the n-th state its wave function is φ n (x and its energy is E n. Formula for the Coecient C n the eigen-states φ n (x Next we wish to derive a formula for C n. To this end we use the orthogonality of φ n(xφ m (xdx = δ nm Since the coecients C n are time independent we may nd them at t = 0. This makes our lives a little easier. We take ψ(x, 0 = C n φ n (x n= 4

5 and multiply both sides by φ m(x. Then we integrate over x. We get dxψ(x, 0φ m(x = dx C n φ n (xφ m(x On the rhs the integral can be done before the sum. We recognize the Kronecker delta. After replacing it the sum over n is trivial dxψ(x, 0φ m(x = C n dxφ n (xφ m(x = n= n= C n δ nm n= =C m We thus nd C m = φ m(xψ(x, 0dx 4 Free Particle 4. General Solution of the Schrodinger Equation When we say a particle is free we mean V (x = 0. The Schrodinger equation is then i t ψ(x, t = H opψ(x, t, H op = p op, p op = i x Since H op is time independent the solutions are ( φ k (x i E k t where φ k (x = (ikx, E k = k π The function φ k (x is an eigen-state of H op with the eigen-energy E k. There is a solution for each value of k. Each k corresponds to the particles momentum p = k. The most general wave function for a free particle is thus a superposition of all possible k's, i.e. ψ( r, t = π g(k (ikx iωt where g(k plays the role of the coecients C n above. We have used E = ω. From here we can also get the dispersion relation for a free particle 4. Time Evolution of a Wave Packet Statement of the Problem is given by the wave packet Find its wave function ψ(x, t at time t > 0. ω(k = k A D free particle is initially (more or less localized at x = 0. Its initial wave function ψ(x, 0 = ( x (πσ /4 σ Expressing the Initial Condition as a Superposition of Eigen-States is ψ(x, 0 = g(k (ikx π The most general initial wave function We will use the given initial condition ψ(x, 0 to nd the coecient function g (k. g(k (ikx = ( x π (πσ /4 σ 5

6 As usual we use the orthogonality dx π [i(k k x] = δ(k k We multiply both sides by ( ik x/ π and integrate over x. dx dx ( x π ( ik x π g(k (ikx ( ik x = (πσ /4 σ g(kδ(k k = (πσ /4 πσ ( σ k π σ g(k = ( π σ k The Wave Function at Time t Having found g(k we plug it back into the most general wave function for time t. We also plug in the dispersion relation ω(k. ψ(x, t = g(k (ikx iωt π = σ π π ( σ k (ikx i k t This is the wave function for any time t. The ression can be simplied by computing the integral. Lets do so [ ( σ σ ψ(x, t = π π + i ] t k + ikx We complete the square and use the formula to obtain e ax dx = π a [ σ x ] ψ(x, t = π σ + it σ + it /m m Next it is advisable to write the onent as (a + ib. This way, giving physical interpretation is much easier. We thus multiply the numerator and denominator inside the onent by ( σ it /m. [ σ x ψ(x, t = π σ + it σ + ( m mσ t + i σ 4 x t + m t In addition the coecient can be ressed in polar form, i.e as r (iθ where r and θ are real. σ + it m = σ i m t (σ + ( m t = (σ + ( [ m t The wave function is hence [ σ ψ(x, t = π (σ + ( x m t σ + ( mσ t + i σ 4 i arctan x ] ( ] t mσ t + m t i arctan ( ] mσ t It can be seen that as a function of time the absolute value of the wave function ψ (x, t becomes wider and lower. 6

7 Figure : Graph of ρ(x, t = ψ(x, t for t =,, 3. We see how the probability disperses. Here we chose σ = m = =. 7