12. Approx. Methods for Time Ind. Systems 12A. The Variational Principle A Great Way to Find the Ground State Given an arbitrary time-independent Hamiltonian H, we want to approximate The eigenstates |i The eigen-energies Ei Idea behind the variational principle: The ground state is the state with the lowest energy The first excited state is the lowest energy state orthogonal to the ground state Etc. The method: Choose a large number of state vectors Measure their energy Pick the one with the lowest energy

The Ground State Has the Lowest Energy Imagine we knew the eigenstates and energies of the Hamiltonian These states are assumed to be complete and orthonormal Assume the energies are ordered: For an arbitrary state |, we use completeness to write: Let’s find the expectation value of the Hamiltonian for |: It follows that Consider the inner product: We therefore have:

The Variational Principle Select a wide range of trial vectors | Calculate the expectation values at right Choose the one with the lowest ratio at right Use this ratio as an estimate of the ground state energy Use the corresponding state vector | (normalized) as an estimate of ground state vector |1 The variational part: We want as many state vectors as possible, ideally infinitely many Best way to do this is to make | have one or more variational parameters: Calculate the energy as a function of these parameters Find the value of min that minimizes E() Then the estimate of the energy and wave function is:

Theory vs. Practice For homework problems: Pick a set of trial vectors described by a small number ( 2) of parameters Find E() analytically Find the minimum using derivatives Substitute back in min For research problems: Pick a set of vectors described by a large number (100’s) of parameters Find E() numerically Find the minimum using multi-dimensional search algorithms (simplex method, for example)

Good Trial Wave Functions How do you pick a good trial wave function? Discontinuous functions will have infinite derivative Can show this yields infinite P2 and hence infinite H Don’t use discontinuous functions Non-smooth functions are okay, but can be tricky to evaluate Discontinuous first derivatives have infinite second derivative at a point This will contribute non-trivially to Unless  vanishes at the non-smooth point When in doubt, you can avoid this problem with: No! Danger! Fine!

Sample Problem (1) A particle of mass m in 1D lies in potential V = m2x2/2. Estimate the ground-state energy. The potential rises suddenly Try a function that disappears suddenly No need to normalize in this formalism We now need to calculate Work out the pieces, one at a time

Sample Problem (2) A particle of mass m in 1D lies in potential V = m2x2/2. Estimate the ground-state energy. Put the pieces together: Minimize with respect to a Substitute back in to get E(a), an estimate of the energy:

Some comments on how we did We picked a terrible trial wave function We still did pretty well (20% error) The error in the energy is caused by the square of the amount of bad wave functions in the wave function Small things squared are very small The state vector has first order errors Not as reliable We got an overestimate of the energy This will always happen With more parameters, you can do much better This method is powerful; most realistic problems are solved with (advanced) variational approaches

Sample Problem – Hydrogen Estimate (1) A particle of mass m in 3D lies in potential V = - kee 2/r. Estimate the ground-state energy. Do this in class Trial wave functions: First function tried: This is not normalizable, | =  Second function tried: This looks very promising Things we need to find:

Sample Problem – Hydrogen Estimate (2) A particle of mass m in 3D lies in potential V = - kee 2/r. Estimate the ground-state energy. We now calculate the energy function: Find the minimum: Substitute in to get the minimum energy: We got it exactly right! The wave function is Also exactly right

Can We Get Beyond the Ground State? Is there a way to get states beyond the ground state? Yes, if we pick states orthogonal to the ground state Removing the approximate ground state: Assume you have an estimate of the true normalized ground state found by variational method Create a set of states that you estimate are close to the next excited state Remove the portion of these states in the approximate ground state This state is, by construction, orthogonal to |1 Calculate the energy for this state: Minimize this energy and find min Then we approximate the first excited state energy as

Comments on Excited State(s) Is the resulting energy guaranteed to be an overestimate? In general no: The state is guaranteed to contain none of the estimated ground state |1  But it could contain a small mixture of the actual ground state |1  Does this work as well for excited states as it did for the ground state? Error from previous step gets compounded with this step Over many steps, errors accumulate An exception where it does work Suppose the problem has some symmetry Then all eigenstates can be classified by their eigenvalues under this symmetry Choose trial state vectors that have this symmetry eigenvalue The energies you find will be true overestimates of the energy of the lowest state with each symmetry eigenvalues

Sample Problem (1) A particle of mass m in 1D lies in potential V = m2x2/2. Estimate the first excited state energy. Is it an overestimate? The potential is symmetric under parity The ground state, previously discussed, has even parity Let’s find the lowest energy odd parity state Try an odd wave function Work out the pieces we need, one at a time

Sample Problem (2) A particle of mass m in 1D lies in potential V = m2x2/2. Estimate the first excited state energy. Is it an overestimate? Minimize the energy: Substitute it back in: Since our state is odd, the lowest odd energy state must be lower than this value Actual coefficient is 1.5 (15% off)

12B. The WKB Approximation A Method For High-Energy States The variational method is good for ground state and other low energy states The WKB method is good for highly excited states Idea behind the WKB approximation If the energy is large, the wave function will be quickly oscillating The potential will then look like it’s slowly varying Start from Schrödinger’s equation in 1D: Define: Then Schrödinger’s equation becomes: If we think of k(x) as a constant, this solution would be like eikx This suggests breaking  into a magnitude and a phase A(x) and (x) are real functions

Breaking it into two equations Substitute in (denote derivatives with primes): Match real and imaginary parts: Multiply second equation by A Anything with a zero derivative is a constant Rename  as W, then

0th Order WKB Approximation Expand out that second derivative Substitute it back in If energy is large, so is k(x) To zeroth order, assume that k2 dominates the other two terms

Bound Problems with Steep Boundaries Suppose we want to consider bound states For now, assume the potential goes to infinity suddenly at x = a and x = b For bound states, we generally prefer real wave functions: Sines, cosines, or some linear combination We have two additional constraints: Wave function must vanish at x = a Wave function must vanish at x = b Vanishing at x = a requires that Then vanishing at x = b requires that Where n = 0, 1, 2, … Recall So we have

Bound Problems with Steep Boundaries Suppose we want to consider bound states For now, assume the potential goes to infinity suddenly at x = a and x = b For bound states, we generally prefer real wave functions: Sines, cosines, or some linear combination We have two additional constraints: Wave function must vanish at x = a Wave function must vanish at x = b Vanishing at x = a requires that Then vanishing at x = b requires that Where n = 0, 1, 2, … Recall So we have

Near a Soft Boundary: Airy Functions Consider now the case where the potential rises smoothly across the boundary This is the typical case We assumed k2 is large, but this is not true near x = a, b By definition, k2 vanishes there Close to x = a, Taylor expand k2(x) We are trying to solve Compare to the Airy equation General solution to Airy equation is Ai and Bi are called Airy functions The general solution for  will be Ai Bi

Phase Shift Near One Soft Boundary Ai Bi Bi diverges as x   We want only Ai, so The Ai function as x  – behaves like This implies as x goes above a, we would have Compare this with the WKB wave function in the allowed region We therefore conclude:

Quantization Condition For Soft Boundaries When we had steep boundaries, we demanded wave function vanishes at x = a and x = b This implied The softening of the boundary at x = a, changed the boundary condition there to Similarly, the softening of the boundary at x = b changes the boundary condition there to Recall the definition of k(x): We therefore have:

Three Different Cases We can get quantization conditions for three types of problems: Hard boundaries: Soft boundaries: One hard, one soft:

How to use these formulas Given a potential V(x), can we estimate the energy eigenstates? Pick the energy E Find the classical turning points a and b, the points where the particle would stop classically Write the relevant integral, usually Do the integral Solve for En as a function of n

Sample Problem Estimate the eigenenergies of a particle of mass m in the 1D Harmonic oscillator with V(x) = m2x2/2 We pick an energy E We solve for the turning points: Substitute into the integral: Make a trigonometric substitution: By coincidence, we got it exactly right

Probability Density in WKB and Classical Consider the wave function in the classically allowed region in the WKB approximation: The probability density is: WKB works for high energy, k(x) large Rapidly oscillating sine function: So we have k is like momentum, which is like velocity Classically, fraction of time spent in region of size dx is given by

12C. Time Independent Perturbation Theory A Series Expansion Consider a situation where the Hamiltonian can be split into a large, easily solved piece, and a small piece: Replace W by W, where  = 1 The large part is assumed to have known eigenvalues and complete, orthonormal eigenvectors: Let |n be the exact eigenstates of H with energies En In the limit   0, the |n’s will be |n’s and the En’s will go to n’s It makes sense, therefore, to imagine a series expansion in terms of the parameter  for |n and |n

The Idea Behind The Method We write out Schrödinger’s equation as follows: Expand to some order: Since this must be true for all , the coefficients of every power of  must match

A Small Ambiguity Problem Schrödinger’s time-independent equation does not completely determine the eigenstates |n We can always multiply by an arbitrary complex number c: Therefore these states are slightly ambiguous In the limit W = 0, we know that One way to resolve the ambiguity is to simply demand The problem: this means final states |n are not normalized Can always be normalized later Irrelevant for energy computation Only relevant in 2nd or higher order state vector

The Procedure For each order (p): Act on the left with n| Let H0 act on n|: Act on the left with m| for m  n: Reconstruct |n(p) using completeness Iterate order by order When done, reconstruct |n and En using  = 1:

First and Second Order: First order: Second order:

Third Order Energy and Summary: Put it all together:

Sample Problem Find the ground state energy to second order and eigenstate to first order for a particle in 1D with mass m and potential V(x) = m2x2/2 + x4 when  is small. Split Hamiltonian into H0 and perturbation W: Find eigenstates and energies of H0: We now need to find

Sample Problem (2) Find the ground state energy to second order and eigenstate to first order for a particle in 1D with mass m and potential V(x) = m2x2/2 + x4 when  is small. We therefore have:

Validity of Perturbation Theory Perturbation theory can only be trusted if subsequent terms get smaller Let’s compare first and second order energy contribution: Let  be smallest gap between n and m: Then we have Compare to 'n Roughly, perturbation theory works if So we need   W

12D. Degenerate Perturbation Theory The Problem and Its Solution Look at the second order expression for energy or first order for state vector: If two states have the same unperturbed energy, we will get in trouble The problem disappears if we get “lucky” by having If we have such states, then the unperturbed eigenstates are in fact ambiguous We can use this ambiguity to change basis for our unperturbed states Thereby ensuring the problem goes away Then proceed with perturbation theory as normal

Procedure for Degenerate Perturbation Theory Suppose we have a set of degenerate states Define the reduced W matrix: Find g normalized eigenvectors and eigenvalues for this matrix: Change basis to: Then in the new basis, we have: And: So we can use these as our new basis states!

Working in the New Basis States In the new basis, the matrix elements of W vanish for any pair of distinct states with the same energy This gets rid of problem terms Note that the basis states we start with are determined partly by the perturbation Even to leading (0th) order, we need to include perturbation theory The perturbation breaks the degeneracy and determines our states Note that the first correction to the energies are just First order energy easy to find just from eigenvalues of

Sample Problem A particle of mass m in two dimensions has Hamiltonian as given at right, where b is small (a) What are the eigenstates and energies in the limit b = 0? (b) For the first pair of degenerate states, determine the eigenstates to leading order and energies to first order in b. H0 is two harmonic oscillators The x-direction has frequency  The y-direction has frequency 2 The unperturbed states and energies are: The first two states are nondegenerate The second excited states are degenerate Need to use degenerate perturbation theory!

Sample Problem (2) (b) For the first pair of degenerate states, determine the eigenstates to leading order and energies to first order in b. We place our degenerate states in some order: We then calculate the perturbation matrix

Sample Problem (3) (b) For the first pair of degenerate states, determine the eigenstates to leading order and energies to first order in b. We now find eigenstates and eigenvalues of this matrix: The energies and eigenstates are therefore: Note that to find the energies, all we needed was the eigenvalues