En la Educación Primaria

Multiple-Choice Problems

016 qmult 00700 1 1 3 easy memory: equivalent postulates 1. If two postulates are said to be equivalent, then

a) one can be derived from the other, but not the other from the one. b) the other can be derived from the one, but not the one from the other.

c) each one can be derived from the other. d) neither can be true.

e) both must be true.

016 qmult 00800 1 4 5 easy deducto-memory: variational principle

2. “Let’s play Jeopardy! For $100, the answer is: Usually the demand that an action (or action integral) be stationary with respect to arbitrary variation in a function appearing somehow in the integrand.”

a) What is a Hermitian conjugate, Alex? b) What is an unperturbation principle, Alex?

c) What is a perturbation principle, Alex? d) What is an invariation principle, Alex?

e) What is a variational principle, Alex?

016 qmult 00900 1 1 3 easy memory: quantum mechanics action

3. In non-relativistic quantum mechanics the action of the usual variation principle is: a) the integral of angular momentum.

b) the derivative of angular momentum. c) the expectation value of the Hamiltonian. d) the time independent Schr¨odinger equation.

e) the Dirac equation.

016 qmult 01000 1 1 1 easy memory: stationary action

4. An exact solution |φi to the time-independent Schr¨odinger equation is the one that by the variational principle in quantum mechanics makes the action

E(φ) = hφ|H|φi hφ|φi be stationary with respect to:

a) arbitrary variations of the state |φi (i.e., δE(φ) = 0). b) some variations of the state |φi.

c) no variations of the state |φi.

d) reasonable variations of the state |φi. e) unreasonable variations of the state |φi.

Chapt. 16 Variational Principle and Variational Methods 109 016 qmult 01100 1 1 5 easy memory: simple variational method

5. In the simple variational method one takes a parameterized trial wave function and finds the parameters that make the expectation value of the Hamiltonian:

a) a maximum. b) 1.

c) negative. d) positive.

e) a minimum.

016 qmult 01200 1 4 3 easy deducto-memory: linear variation method

6. “Let’s play Jeopardy! For $100, the answer is: The justification for the linear variational method (or Rayleigh-Ritz method or truncated Hamiltonian matrix eigen-problem).”

a) What is Hermitian conjugation, Alex? b) What is bra/ket notation, Alex?

c) What is the quantum mechanics variational principle, Alex? d) What is the Dirac principle, Alex?

e) What is the cosmological principle, Alex?

016 qmult 01500 1 4 1 easy deducto-memory: repulsion of the energy levels

7. Any perturbation applied to a two-level system that is initially degenerate causes: a) a repulsion of the energy levels.

b) an attraction of the energy levels.

c) a warm and affectionate relationship between the energy levels. d) a wonderful, meaningful togetherness of the energy levels.

e) an eternal soul-bliss of the energy levels.

Full-Answer Problems

016 qfull 00010 1 5 0 easy thinking: equivalent results

1. If two different looking theorems or postulates were said to be equivalent what would that mean?

016 qfull 00020 2 5 0 moderate thinking: variational principle and method

2. Are the variational principle and the variational method the same thing? Explain please. 016 qfull 00030 1 5 0 easy thinking: what is a stationary point?

3. What does it mean to say a function is stationary at a point? 016 qfull 00040 2 3 0 moderate math: differentiation for stationarity

4. Take the derivative of

E(α) = 5 4 h −2 mα2 + 1 14mω 2_α2

and determine the stationary point. Just by imagining the function’s behavior in the large and small α limits determine whether the stationary point is a minimum. Give the analytic expression for E(α) at the stationary point.

016 qfull 00050 2 5 0 moderate thinking: Snell’s law and var. princ.

5. Can Snell’s law be derived using the variational principle (or a variational principle “as you prefer”)? Please explain.

110 Chapt. 16 Variational Principle and Variational Methods

016 qfull 00060 2 5 0 moderate thinking: Schr”od. and var. princ.

6. Can the time-independent Schr¨odinger’s equation be derived using the variational principle? Please explain.

016 qfull 00070 2 5 0 moderate thinking: convert to matrix eigenproblem

7. Convert the braket eigenproblem H|Ψi = E|Ψi to the discrete {|uji} orthonormal basis

representation by expanding |Ψi in terms of the |uji kets and then operating on the equation

with the bra hui|. Find the matrix representation of the eigenproblem.

016 qfull 00080 1 5 0 easy thinking: solving infinite matrix problem

8. Can one literally solve in a numerical procedure an infinite matrix problem: i.e. a problem with an infinite number of terms to number crunch? Why so or why not?

016 qfull 00090 1 5 0 easy thinking: diagonalization defined 9. What is meant by diagonalization in quantum mechanics? 016 qfull 00200 2 5 0 moderate thinking: simple variational method

Extra keywords: simple variational method for excited states

10. The simple variational method can in principle be applied to excited states.

a) Say an unnormalized trial wave function |ψi is orthogonal to all energy eigenstates |φii

of quantum number less than n, where the eigen-energies increase monotonically with quantum number as usual. Show that Etrial≥ En where Etrial is the expectation value of

the Hamiltonian for |ψi. When will the equality hold? Remember there is such a thing as degeneracy.

b) Using the simple variational method for finding excited eigenstate energies isn’t really of general interest since constructing trial functions with the right orthogonality properties is often harder than using the other approaches. However, if the eigenstates have definite parity, definite parity trial wave functions can be used to determine the lowest eigen-energies for wave functions of each kind of parity.

For example, let us consider the simple harmonic oscillator problem in one dimension. We know that the eigenstates are non-degenerate and have definite parity. It is given that the ground state has even parity and the first excited state has odd parity. We can use an odd trial wave function and the variational method to approximately determine the energy of the first excited state. The simple harmonic oscillator eigenproblem in scaled dimensionless variables is  − d 2 dx2 + x 2_{ψ = Eψ ,} where x =r mω h

− xphy and E = Ephy

h

−ω/2 = 2n + 1 .

The n is the SHO energy quantum number (n runs 0, 1, 2, 3, . . .) and the “phy” stands for physical. Consider the odd trial wave function

ψ = x(x

2_{− c}2_{), |x| ≤ c;}

0. |x| > c,

where c is a variational parameter. Normalize this trial wave function, evaluate its expectation energy, and minimize the expectation energy by varying c. How does this variational method energy compare to the exact result which in scaled variables is 3. HINT: There are no wonderful tricks in the integrations: grind them out carefully.

Chapt. 16 Variational Principle and Variational Methods 111

016 qfull 00300 3 5 0 tough thinking: variational hydrogen Extra keywords: (Ha-327:4.1)

11. We know, of course, the ground state for the hydrogenic atom sans perturbations: ψnℓm=

1 √

4π(2a

−3/2_)e−r/a _,

where a = a0/[(m/me)Z] is the radial scale parameter: a0= h− 2

/(mee2) = λCompton/(2πα) =

0.529 ˚A is the Bohr radius, m is the reduced mass, and Z is the nuclear charge (Gr-128, 141). But as a tedious illustration of the simple variational method, let us try find an approximate ground state wave function and energy starting with the trial Gaussian wave function

ψ = Ae−βr2

/a2

.

a) Can we obtain the exact solution with a trial wave function of this form? temitemb) The varied energy is given by

Ev= hψ|H|ψi hψ|ψi = R∞ 0 [ψ(r) ∗_{Hψ(r)] (4πr}2_{) dr} R∞ 0 [ψ(r)∗ψ(r)] (4πr2) dr ,

where H is the Hamiltonian for ℓ = 0 (i.e., the zero angular momentum case) given by

H = −−h 2 2m 1 r2 ∂ ∂rr 2 ∂ ∂r− Ze2 r .

Note the varied energy form does not require a Lagrange undetermined multiplier since we are building the constraint of normalization into the variation. We, of course, need to evaluate A later to normalize the minimized wave function. Convert the varied energy expression into a dimensionless form in terms of the coordinate x = r/a and reduced varied energy ǫv = Ev/[Ze2/(2a)] =

Z−2_(m/m

e)Ev/ERyd≈ Z−2(m/me)Ev/(13.606 eV). HINT: A further integration transformation

can make the analytic form even simpler.

temitemc) Find the explicit analytic expression for ǫv. Sketch a plot of ǫv as a function of β.

HINT: Use an integral table.

d) Now find the minimizing β value and the minimum ǫv. Compare ǫv to exact ground state

value which is −1 in fact.

016 qfull 01000 3 5 0 tough thinking: non-orthogonal linear variation Extra keywords: method for a two level system.

12. You are given two basis states |1i and |2i and want to solve a two-dimensional system with Hamiltonian H in terms of this basis. The basis is not orthogonal although the basis states are normalized of course. Recall in this case that the non-orthogonal linear variational method eigenproblem is

H~c = ES~c ,

where ~c is an unknown eigenvector, E and unknown eigen-energy, and S is the overlap matrix. Let

H = ε1 V V ε2

 .

We have assumed that h1|H|2i = h2|H|1i and designated these elements by V : i.e., the eigenstates are pure real. This assumption is generality that probably pointless for the cases

112 Chapt. 16 Variational Principle and Variational Methods

where this problem is probably of most interest: i.e., in LCAO method (i.e., linear combination of atomic orbitals method) for molecular orbitals. We will also assume V < 0 which is also appropriate for LCAO, and so avoids needles generality. As a fiducial choice assume ε2 ≥ ε1

although all the formulae will not depend this choice in fact. For the overlap matrix let S = h1|1i h1|2i h2|1i h2|2i  = 1 s s 1  .

Your mission Mr. Phelps—if you choose to accept it—is solve for the eigen-energies and eigenvectors. These quantities tend to come out in clumsy forms. So you should try to find nice forms. You may subsume large clumpy expressions into single symbols, but show some restraint. One trick is to re-origin all the energies: i.e., define

¯

ε =ε1+ ε2

2 , −ε

′_{= ε}

1− ¯ε , ε′= ε2− ¯ε , V′= V − ¯εs , and E′= E − ¯ε .

Note with our fiducial assumptions ε′_{≥ 0, but all the formula should work for ε}′_{< 0 too. Note}

also that V′ _{< 0, V}′_{> 0, or V}′ _{= 0 are all possible now. Now subtract}

¯ ε 1 s

s 1 

~c

from both sides of the eigenproblem and solve for the primed eigen-energies and the eigen- vectors in terms of the primed quantities. Having found the solutions, you should examine the special limiting cases: i.e., ε′_{→ 0, and s → 0.}

The State Department confesses that it does not know the ideal forms for the solutions and in any case will disavow all knowledge of your activities.