1. Use a gaussian trial function (Equation 7.2) to obtain the lowest
upper bound you can on the ground state energy of (a) the linear
potential: V(x) ... (b) the quartic potential: .... ... Get solution
2. Find the best bound on Egs for the one-dimensional harmonic oscillator using a trial wave function of the form...where A is determined by normalization and b is an adjustable parameter. Get solution
3. Find the best bound on ...for the delta-function potential V(x) = ..., using a triangular trial function (Equation 7.10, only centered at the origin). This time a is an adjustable parameter. ... Get solution
4. (a) Prove the following corollary to the variational principle: If 〈ψ | ψgs〉 = 0, then 〈H〉 ≥ Efe, where Efe is the energy of the first excited state.Thus, if we can find a trial function that is orthogonal to the exact ground state, we can get an upper bound on the first excited state. In general, it's difficult to be sure that is orthogonal to ψgs, since (presumably) we don't know the latter. However, if the potential V(x) is an even function of x, then the ground state is likewise even, and hence any odd trial function will automatically meet the condition for the corollary.(b) Find the best bound on the first excited state of the one-dimensional harmonic oscillator using the trial functionΨ (x)= Axe–bx. Get solution
5. (a) Use the variational principle to prove that first-order non-degenerate perturbation theory always overestimates (or at any rate never underestimates) the ground state energy. (b) In view of (a), you would expect that the second-order correction to the ground state is always negative. Confirm that this is indeed the case, by examining Equation 6.15. ... Get solution
6. Using Egs = – 79.0 eV for the ground state energy of helium, calculate the ionization energy (the energy required to remove just one electron). Hint: First calculate the ground state energy of the helium ion, He+, with a single electron orbiting the nucleus; then subtract the two energies. Get solution
7. Apply the techniques of this Section to the H− and Li+ ions (each has two electrons, like helium, but nuclear charges Z = 1 and Z = 3, respectively). Find the effective (partially shielded) nuclear charge, and determine the best upper bound on Egs, for each case. Comment: In the case of H− you should find that 〈H〉>–13.6 eV, which would appear to indicate that there is no bound state at all, since it would be energetically favorable for one electron to fly off, leaving behind a neutral hydrogen atom. This is not entirely surprising, since the electrons are less strongly attracted to the nucleus than they are in helium, and the electron repulsion tends to break the atom apart. However, it turns out to be incorrect. With a more sophisticated trial wave function (see Problem 7.18) it can be shown that Egs–13.6 eV, and hence that a bound state does exist. It's only barely bound, however, and there are no excited bound states,11 so H− has no discrete spectrum (all transitions are to and from the continuum). As a result, it is difficult to study in the laboratory, although it exists in great abundance on the surface of the sun12 Get solution
8. Evaluate D and X (Equations 7.45 and 7.46). Check your answers against Equations 7.47 and 7.48. Get solution
9. Suppose we used a minus sign in our trial wave function (Equation 7.37) ... Without doing any new integrals, find F(x) (the analog to Equation 7.51) for this case, and construct the graph. Show that there is no evidence of bonding.13 (Since the variational principle only gives an upper bound, this doesn't prove that bonding cannot occur for such a state, but it certainly doesn't look promising). Comment: Actually, any function of the form ... has the desired property that the electron is equally likely to be associated with either proton. However, since the Hamiltonian (Equation 7.35) is invariant under the interchange ... its eigenfunctions can be chosen to be 'simultaneously eigenfunctions of P. The plus sign (Equation 7.37) goes with the eigenvalue +1, and the minus sign (Equation 7.52) with the eigenvalue -1; nothing is to be gained by considering the ostensibly more general case (Equation 7.53), though you're welcome to try it, if you're interested. ... ... ... Get solution
10. The second derivative of F(x), at the equilibrium point, can be used to estimate the natural frequency of vibration (ω) of the two protons in the hydrogen molecule ion (see Section 2.3). If the ground state energy (hω/2) of this oscillator exceeds the binding energy of the system, it will fly apart. Show that in fact the oscillator energy is small enough that this will not happen, and estimate how many bound vibrational levels there are. Note: You're not going to be able to obtain the position of the minimum—still less the second derivative at that point—analytically. Do it numerically, on a computer. Get solution
11. (a) Use a trial wave function of the form ... ... Example 7.3 Find an upper bound on the ground state energy of the one-dimensional infinite square well (Equation 2.19), using the "triangular" trial wave function (Figure 7.1).3 ... Get solution
12. (a) Generalize Problem 7.2, using the trial wave function14...for arbitrary n. Partial answer: The best value of b is given by...(b) Find the least upper bound on the first excited state of the harmonic oscillator using a trial function of the form...Partial answer: The best value of b is given by...(c) Notice that the bounds approach the exact energies as n → ∞ Why is that? Hint: Plot the trial wave functions for n = 2,n = 3, and n = 4, and compare them with the true wave functions (Equations 2.59 and 2.62). To do it analytically, start with the identity... Get solution
13. Find the lowest bound on the ground state of hydrogen you can get using a gaussian trial wave function...where A is determined by normalization and b is an adjustable parameter. Answer:–11.5 eV. Get solution
14. If the photon had a nonzero mass (mγ ≠ 0), the Coulomb potential would be replaced by the Yukawa potential,...'Where µ = mγc/h. With a trial wave function of your own devising, estimate the binding energy of a "hydrogen" atom with this potential. Assume µa ≪ 1, and give your answer correct to order (µa)2. Get solution
15. Suppose you’re given a quantum system whose Hamiltonian H0 admits just two eigenstates ψa (with energy Ea), and ψb (with energy Eb). They are orthogonal, normalized, and nondegenerate (assume Ea is the smaller of the two energies). Now we turn on a perturbation H', with the following matrix elements:... [7.55]where h is some specified constant.(a) Find the exact eigenvalues of the perturbed Hamiltonian.(b) Estimate the energies of the perturbed system using second-order perturbation theory.(c) Estimate the ground state energy of the perturbed system using the variational principle, with a trial function of the form [7.56]...where ϕ is an adjustable parameter. Note: Writing the linear combination in this way is just a neat way to guarantee that Ψ is normalized.(d) Compare your answers to (a), (b), and (c). Why is the variational principle so accurate, in this case? Get solution
16. (As an explicit example of the method developed in Problem 7.15, consider an electron at rest in a uniform magnetic field ..., for which the Hamiltonian is (Equation 4.158): ... The eigenspinors, ..., and the corresponding energies..., are given in Equation 4.161. Now we turn on a perturbation, in the form of a uniform field in the x direction: ... (a) Find the matrix elements of H', and confirm that they have the structure of Equation 735. What is h? (b) Using your result in Problem 7.15(b), find the new ground state energy, in second-order perturbation theory. (c) Using your result in Problem 7.15(c), find the variational principle bound on the ground state energy. ... ... ... Get solution
17. Although the Schrodinger equation for helium itself cannot be solved exactly, there exist "helium-like" systems that do admit exact solutions. ... Get solution
18. In Problem 7.7 we found that the trial wave function with shielding (Equation 7.27), which worked well for helium, is inadequate to confirm the existence of a bound state for the negative hydrogen ion. Chandraselchar16 used a trial wave function of the form ... Where ... In effect, he allowed two different shielding factors, suggesting that one electron is relatively close to the nucleus, and the other is farther out. (Because electrons are identical particles, the spatial wave function must be symmetrized with respect to interchange. The spin state—which is irrelevant to the calculation—is evidently ... ... Get solution
19. The fundamental problem in harnessing nuclear fusion is getting the two particles (say, two deuterons) close enough together for the attractive (but short-range) nuclear force to overcome the Coulomb repulsion. The "bulldozer" method is to heat the particles up to fantastic temperatures, and allow the random collisions to bring them together. A more exotic proposal is muon catalysis, in which we construct a "hydrogen molecule ion," only with deuterons in place of protons, and a muon in place of the electron. Predict the equilibrium separation distance between the deuterons in such a structure, and explain why muons are superior to electrons for this purpose. Get solution
20. Quantum dots. Consider a particle constrained to move in two dimensions in the cross-shaped region shown in Figure 7.8. The "arms" of the cross continue out to infinity. The potential is zero within the cross, and infinite in the shaded areas outside. Surprisingly, this configuration admits a positive-energy bound state.(a) Show that the lowest energy that can propagate off to infinity is...any solution with energy less than that has to be a bound state. Hint: Go way out one arm (say, x ≫ a), and solve the Schrödinger equation by separation of variables; if the wave function propagates out to infinity, the dependence on x must take the form exp(ikxx) with kx > 0....FIGURE 7.8: The cross-shaped region for Problem 7.20.(b) Now use the variational principle to show that the ground state has energy less than Ethreshold. Use the following trial wave function (suggested by Krishna Rajagopal):...Normalize it to determine A, and calculate the expectation value of H. Answer:...Now minimize with respect to α, and show that the result is less than Ethreshold. Hint: Take full advantage of the symmetry of the problem—yon only need to integrate over 1/8 of the open region, since the other 7 integrals will be the same. Note however that whereas the trial wave function is continuous, its derivatives are not—there are "roof-lines" at x = 0, y = 0, x = ± a, and y = ± a, where you will need to exploit the technique of Example 7.3. Get solution
2. Find the best bound on Egs for the one-dimensional harmonic oscillator using a trial wave function of the form...where A is determined by normalization and b is an adjustable parameter. Get solution
3. Find the best bound on ...for the delta-function potential V(x) = ..., using a triangular trial function (Equation 7.10, only centered at the origin). This time a is an adjustable parameter. ... Get solution
4. (a) Prove the following corollary to the variational principle: If 〈ψ | ψgs〉 = 0, then 〈H〉 ≥ Efe, where Efe is the energy of the first excited state.Thus, if we can find a trial function that is orthogonal to the exact ground state, we can get an upper bound on the first excited state. In general, it's difficult to be sure that is orthogonal to ψgs, since (presumably) we don't know the latter. However, if the potential V(x) is an even function of x, then the ground state is likewise even, and hence any odd trial function will automatically meet the condition for the corollary.(b) Find the best bound on the first excited state of the one-dimensional harmonic oscillator using the trial functionΨ (x)= Axe–bx. Get solution
5. (a) Use the variational principle to prove that first-order non-degenerate perturbation theory always overestimates (or at any rate never underestimates) the ground state energy. (b) In view of (a), you would expect that the second-order correction to the ground state is always negative. Confirm that this is indeed the case, by examining Equation 6.15. ... Get solution
6. Using Egs = – 79.0 eV for the ground state energy of helium, calculate the ionization energy (the energy required to remove just one electron). Hint: First calculate the ground state energy of the helium ion, He+, with a single electron orbiting the nucleus; then subtract the two energies. Get solution
7. Apply the techniques of this Section to the H− and Li+ ions (each has two electrons, like helium, but nuclear charges Z = 1 and Z = 3, respectively). Find the effective (partially shielded) nuclear charge, and determine the best upper bound on Egs, for each case. Comment: In the case of H− you should find that 〈H〉>–13.6 eV, which would appear to indicate that there is no bound state at all, since it would be energetically favorable for one electron to fly off, leaving behind a neutral hydrogen atom. This is not entirely surprising, since the electrons are less strongly attracted to the nucleus than they are in helium, and the electron repulsion tends to break the atom apart. However, it turns out to be incorrect. With a more sophisticated trial wave function (see Problem 7.18) it can be shown that Egs–13.6 eV, and hence that a bound state does exist. It's only barely bound, however, and there are no excited bound states,11 so H− has no discrete spectrum (all transitions are to and from the continuum). As a result, it is difficult to study in the laboratory, although it exists in great abundance on the surface of the sun12 Get solution
8. Evaluate D and X (Equations 7.45 and 7.46). Check your answers against Equations 7.47 and 7.48. Get solution
9. Suppose we used a minus sign in our trial wave function (Equation 7.37) ... Without doing any new integrals, find F(x) (the analog to Equation 7.51) for this case, and construct the graph. Show that there is no evidence of bonding.13 (Since the variational principle only gives an upper bound, this doesn't prove that bonding cannot occur for such a state, but it certainly doesn't look promising). Comment: Actually, any function of the form ... has the desired property that the electron is equally likely to be associated with either proton. However, since the Hamiltonian (Equation 7.35) is invariant under the interchange ... its eigenfunctions can be chosen to be 'simultaneously eigenfunctions of P. The plus sign (Equation 7.37) goes with the eigenvalue +1, and the minus sign (Equation 7.52) with the eigenvalue -1; nothing is to be gained by considering the ostensibly more general case (Equation 7.53), though you're welcome to try it, if you're interested. ... ... ... Get solution
10. The second derivative of F(x), at the equilibrium point, can be used to estimate the natural frequency of vibration (ω) of the two protons in the hydrogen molecule ion (see Section 2.3). If the ground state energy (hω/2) of this oscillator exceeds the binding energy of the system, it will fly apart. Show that in fact the oscillator energy is small enough that this will not happen, and estimate how many bound vibrational levels there are. Note: You're not going to be able to obtain the position of the minimum—still less the second derivative at that point—analytically. Do it numerically, on a computer. Get solution
11. (a) Use a trial wave function of the form ... ... Example 7.3 Find an upper bound on the ground state energy of the one-dimensional infinite square well (Equation 2.19), using the "triangular" trial wave function (Figure 7.1).3 ... Get solution
12. (a) Generalize Problem 7.2, using the trial wave function14...for arbitrary n. Partial answer: The best value of b is given by...(b) Find the least upper bound on the first excited state of the harmonic oscillator using a trial function of the form...Partial answer: The best value of b is given by...(c) Notice that the bounds approach the exact energies as n → ∞ Why is that? Hint: Plot the trial wave functions for n = 2,n = 3, and n = 4, and compare them with the true wave functions (Equations 2.59 and 2.62). To do it analytically, start with the identity... Get solution
13. Find the lowest bound on the ground state of hydrogen you can get using a gaussian trial wave function...where A is determined by normalization and b is an adjustable parameter. Answer:–11.5 eV. Get solution
14. If the photon had a nonzero mass (mγ ≠ 0), the Coulomb potential would be replaced by the Yukawa potential,...'Where µ = mγc/h. With a trial wave function of your own devising, estimate the binding energy of a "hydrogen" atom with this potential. Assume µa ≪ 1, and give your answer correct to order (µa)2. Get solution
15. Suppose you’re given a quantum system whose Hamiltonian H0 admits just two eigenstates ψa (with energy Ea), and ψb (with energy Eb). They are orthogonal, normalized, and nondegenerate (assume Ea is the smaller of the two energies). Now we turn on a perturbation H', with the following matrix elements:... [7.55]where h is some specified constant.(a) Find the exact eigenvalues of the perturbed Hamiltonian.(b) Estimate the energies of the perturbed system using second-order perturbation theory.(c) Estimate the ground state energy of the perturbed system using the variational principle, with a trial function of the form [7.56]...where ϕ is an adjustable parameter. Note: Writing the linear combination in this way is just a neat way to guarantee that Ψ is normalized.(d) Compare your answers to (a), (b), and (c). Why is the variational principle so accurate, in this case? Get solution
16. (As an explicit example of the method developed in Problem 7.15, consider an electron at rest in a uniform magnetic field ..., for which the Hamiltonian is (Equation 4.158): ... The eigenspinors, ..., and the corresponding energies..., are given in Equation 4.161. Now we turn on a perturbation, in the form of a uniform field in the x direction: ... (a) Find the matrix elements of H', and confirm that they have the structure of Equation 735. What is h? (b) Using your result in Problem 7.15(b), find the new ground state energy, in second-order perturbation theory. (c) Using your result in Problem 7.15(c), find the variational principle bound on the ground state energy. ... ... ... Get solution
17. Although the Schrodinger equation for helium itself cannot be solved exactly, there exist "helium-like" systems that do admit exact solutions. ... Get solution
18. In Problem 7.7 we found that the trial wave function with shielding (Equation 7.27), which worked well for helium, is inadequate to confirm the existence of a bound state for the negative hydrogen ion. Chandraselchar16 used a trial wave function of the form ... Where ... In effect, he allowed two different shielding factors, suggesting that one electron is relatively close to the nucleus, and the other is farther out. (Because electrons are identical particles, the spatial wave function must be symmetrized with respect to interchange. The spin state—which is irrelevant to the calculation—is evidently ... ... Get solution
19. The fundamental problem in harnessing nuclear fusion is getting the two particles (say, two deuterons) close enough together for the attractive (but short-range) nuclear force to overcome the Coulomb repulsion. The "bulldozer" method is to heat the particles up to fantastic temperatures, and allow the random collisions to bring them together. A more exotic proposal is muon catalysis, in which we construct a "hydrogen molecule ion," only with deuterons in place of protons, and a muon in place of the electron. Predict the equilibrium separation distance between the deuterons in such a structure, and explain why muons are superior to electrons for this purpose. Get solution
20. Quantum dots. Consider a particle constrained to move in two dimensions in the cross-shaped region shown in Figure 7.8. The "arms" of the cross continue out to infinity. The potential is zero within the cross, and infinite in the shaded areas outside. Surprisingly, this configuration admits a positive-energy bound state.(a) Show that the lowest energy that can propagate off to infinity is...any solution with energy less than that has to be a bound state. Hint: Go way out one arm (say, x ≫ a), and solve the Schrödinger equation by separation of variables; if the wave function propagates out to infinity, the dependence on x must take the form exp(ikxx) with kx > 0....FIGURE 7.8: The cross-shaped region for Problem 7.20.(b) Now use the variational principle to show that the ground state has energy less than Ethreshold. Use the following trial wave function (suggested by Krishna Rajagopal):...Normalize it to determine A, and calculate the expectation value of H. Answer:...Now minimize with respect to α, and show that the result is less than Ethreshold. Hint: Take full advantage of the symmetry of the problem—yon only need to integrate over 1/8 of the open region, since the other 7 integrals will be the same. Note however that whereas the trial wave function is continuous, its derivatives are not—there are "roof-lines" at x = 0, y = 0, x = ± a, and y = ± a, where you will need to exploit the technique of Example 7.3. Get solution
