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Chapter 7: Problem 12

Consider a quantum system with just three linearly independent states. Suppose the Hamiltonian, in matrix form, is $$\mathrm{H}=V_{0}\left(\begin{array}{ccc} (1-\epsilon) & 0 & 0 \\ 0 & 1 & \epsilon \\ 0 & \epsilon & 2 \end{array}\right)$$ where \(V_{0}\) is a constant, and \(\epsilon\) is some small number \((\epsilon \ll 1)\) (a) Write down the eigenvectors and eigenvalues of the unperturbed Hamiltonian \((\epsilon=0)\) (b) Solve for the exact eigenvalues of \(H\) Expand each of them as a power series in \(\epsilon,\) up to second order. (c) Use first- and second-order non-degenerate perturbation theory to find the approximate eigenvalue for the state that grows out of the nondegenerate eigenvector of \(\mathrm{H}^{0}\). Compare the exact result, from (b). (d) Use degenerate perturbation theory to find the first-order correction to the two initially degenerate eigenvalues. Compare the exact results.

Short Answer

Expert verified
Eigenvalues are \( V_0, V_0(1+\epsilon), 2V_0 - \epsilon^2V_0 \). Perturbation approximations are consistent.

Step by step solution

01

Unperturbed Hamiltonian

When \( \epsilon = 0 \), the Hamiltonian \( \mathrm{H}^0 \) simplifies to \( V_0 \begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 2 \end{pmatrix} \). The eigenvalues are simply the diagonal elements \( \lambda_1 = V_0, \lambda_2 = V_0, \lambda_3 = 2V_0 \). The corresponding unperturbed eigenvectors are \( \begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix}, \begin{pmatrix} 0 \ 1 \ 0 \end{pmatrix}, \begin{pmatrix} 0 \ 0 \ 1 \end{pmatrix} \) respectively.
02

Exact Eigenvalues (Characteristic Polynomial)

To find the exact eigenvalues, solve \( \text{det}(H - \lambda I) = 0 \) where \( H = V_0 \begin{pmatrix} (1-\epsilon) & 0 & 0 \ 0 & 1 & \epsilon \ 0 & \epsilon & 2 \end{pmatrix} \). This gives the polynomial: \( \left| V_0 \begin{pmatrix} (1-\epsilon) - \lambda/V_0 & 0 & 0 \ 0 & 1 - \lambda/V_0 & \epsilon \ 0 & \epsilon & 2 - \lambda/V_0 \end{pmatrix} \right| = 0 \). Solving for \( \lambda \), we get the eigenvalues up to the second order of \( \epsilon \):- \( \lambda_1 = \lambda_2 = V_0(1 + \epsilon) \), and \( \lambda_3 = (2 - \epsilon^2)V_0 \).
03

Perturbation Theory - Non-degenerate

For the non-degenerate eigenvector of \( H^0 \) corresponding to \( 2V_0 \), the first-order correction for \( \lambda_3 \) using non-degenerate perturbation theory is zero, as it is orthogonal to perturbation. Second order gives \( \langle \psi_{(0)}| H' |\psi_{(0)} \rangle = 0 \), resulting in \( \lambda_3 = 2V_0 - \epsilon^2 V_0 \), matching the exact value calculated.
04

Perturbation Theory - Degenerate

For the initially degenerate states (\( V_0 \) eigenvalue), apply degenerate perturbation theory. Construct the matrix:\[ W = V_0 \begin{pmatrix} 0 & 0 \ 0 & \epsilon \end{pmatrix} \] The eigenvalues of this matrix are \( 0, \epsilon V_0 \). Thus, the first-order corrections are \( V_0 \) and \( V_0 (1 + \epsilon) \). The second value matches the exact expansion in Step 2.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Eigenvalues and Eigenvectors
In quantum systems, eigenvalues and eigenvectors play crucial roles in defining the behavior of a system. The eigenvalue problem often arises when dealing with matrices like the Hamiltonian in quantum mechanics. The Hamiltonian represents the total energy of the system.

For a given Hermitian matrix representing a Hamiltonian, finding eigenvalues involves solving the determinant equation \( \det(H - \lambda I) = 0 \), where \( \lambda \) is an eigenvalue and \( I \) is the identity matrix. The solutions, or eigenvalues, represent the energy levels of the system. The corresponding eigenvectors provide the state vectors associated with these energies.

In the context of the exercise, the unperturbed Hamiltonian \( H^0 \) has simple diagonal form, leading directly to its eigenvalues being \( V_0, V_0, \) and \( 2V_0 \), with eigenvectors corresponding to the basis vectors of the matrix. These energies form the foundation upon which perturbations are analyzed.
Degenerate Perturbation Theory
Degenerate perturbation theory is applied when two or more states of a quantum mechanical system share the same energy 鈥� making them degenerate. In such cases, introducing a small perturbative term can cause the degenerate states to split into distinct energy levels.

The primary goal of this theory is to still find the new energies and states after the perturbation. We work with a subspace of the degenerate states and construct a smaller matrix known as the "perturbation matrix." This matrix is derived from the interaction of the perturbation with the degenerate states. By solving the eigenvalue problem for this smaller matrix, corrections to the energy levels are found.

In the exercise, for the initially degenerate states \( V_0 \), the first-order perturbation theory reveals that the previously degenerate states now have slightly different energies. This is achieved by constructing and solving a matrix, leading to the new energies \( V_0 \) and \( V_0 (1 + \epsilon) \), showcasing how small perturbations affect system degeneracy.
Non-Degenerate Perturbation Theory
When dealing with non-degenerate states, perturbation theory takes a different form. Non-degenerate perturbation theory is applicable when the state's associated energy level is distinct compared to others.

For this process, the primary task is to find corrections to the energy levels and possibly the states. This is done using perturbative expansions, where corrections to the eigenvalues and eigenstates are calculated perturbatively in powers of a small parameter.

In the exercise's context, for the non-degenerate eigenstate corresponding to the eigenvalue \( 2V_0 \), applying perturbation theory reveals no first-order correction, because the perturbation does not mix with other states. Secondly, second-order perturbations provide precise adjustments to match previously calculated exact values, reflecting the accuracy and usefulness of non-degenerate perturbation theory for refining our understanding of quantum systems.

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Most popular questions from this chapter

Let a and b be two constant vectors. Show that \(\int(\mathbf{a} \cdot \hat{r})(\mathbf{b} \cdot \hat{r}) \sin \theta d \theta d \phi=\frac{4 \pi}{3}(\mathbf{a} \cdot \mathbf{b})\) (the integration is over the usual range: \(0 < \theta < \pi, 0 < \phi < 2 \pi\) ). Use this result to demonstrate that \(\left\langle\frac{3\left(\mathbf{S}_{p} \cdot \hat{r}\right)\left(\mathbf{S}_{e} \cdot \hat{r}\right)-\mathbf{S}_{p} \cdot \mathbf{S}_{e}}{r^{3}}\right|=0\) for states with \(\ell=0 .\) Hint: \(\hat{r}=\sin \theta \cos \phi \hat{\imath}+\sin \theta \sin \phi \hat{\jmath}+\cos \theta \hat{k} .\) Do the angular integrals first.

Van der Waals interaction. Consider two atoms a distance \(R\) apart. Because they are electrically neutral you might suppose there would be no force between them, but if they are polarizable there is in fact a weak attraction. To model this system, picture each atom as an electron (mass \(m\), charge \(-e\) ) attached by a spring (spring constant \(k\) ) to the nucleus (charge \(+e),\) as in Figure \(7.13 .\) We'll assume the nuclei are heavy, and essentially motionless. The Hamiltonian for the unperturbed system is $$H^{0}=\frac{1}{2 m} p_{1}^{2}+\frac{1}{2} k x_{1}^{2}+\frac{1}{2 m} p_{2}^{2}+\frac{1}{2} k x_{2}^{2}$$ The Coulomb interaction between the atoms is $$H^{\prime}=\frac{1}{4 \pi \epsilon_{0}}\left(\frac{e^{2}}{R}-\frac{e^{2}}{R-x_{1}}-\frac{e^{2}}{R+x_{2}}+\frac{e^{2}}{R-x_{1}+x_{2}}\right)$$ (a) Explain Equation 7.104 . Assuming that \(\left|x_{1}\right|\) and \(\left|x_{2}\right|\) are both much less than \(R,\) show that $$H^{\prime} \cong-\frac{e^{2} x_{1} x_{2}}{2 \pi \epsilon_{0} R^{3}}$$ (b) Show that the total Hamiltonian ( \(H^{0}\) plus Equation 7.105 ) separates into two harmonic oscillator Hamiltonians: $$H=\left[\frac{1}{2 m} p_{+}^{2}+\frac{1}{2}\left(k-\frac{e^{2}}{2 \pi \epsilon_{0} R^{3}}\right) x_{+}^{2}\right]+\left[\frac{1}{2 m} p_{-}^{2}+\frac{1}{2}\left(k+\frac{e^{2}}{2 \pi \epsilon_{0} R^{3}}\right) x_{-}^{2}\right]$$ under the change variables $$x_{\pm} \equiv \frac{1}{\sqrt{2}}\left(x_{1} \pm x_{2}\right), \quad \text { which entails } \quad p_{\pm}=\frac{1}{\sqrt{2}}\left(p_{1} \pm p_{2}\right)$$ (c) The ground state energy for this Hamiltonian is cvidently $$E=\frac{1}{2} \hbar\left(\omega_{+}+\omega_{-}\right), \quad \text { where } \quad \omega_{\pm}=\sqrt{\frac{k \mp\left(e^{2} / 2 \pi \epsilon_{0} R^{3}\right)}{m}}$$ Without the Coulomb interaction it would have been \(E_{0}=\hbar \omega_{0},\) where \(\omega_{0}=\sqrt{k / m} .\) Assuming that \(k \gg\left(e^{2} / 2 \pi \epsilon_{0} R^{3}\right),\) show that $$\Delta V \equiv E-E_{0} \cong-\frac{\hbar}{8 m^{2} \omega_{0}^{3}}\left(\frac{e^{2}}{2 \pi \epsilon_{0}}\right)^{2} \frac{1}{R^{6}}$$ Conclusion: There is an attractive potential between the atoms, proportional to the inverse sixth power of their separation. This is the van der Waals interaction between two neutral atoms. (d) Now do the same calculation using second-order perturbation theory. Hint: The unperturbed states are of the form \(\psi_{n_{1}}\left(x_{1}\right) \psi_{n_{2}}\left(x_{2}\right),\) where \(\psi_{n}(x)\) is a one-particle oscillator wave function with mass \(m\) and spring constant \(k ; \Delta V\) is the second-order correction to the ground state energy, for the perturbation in Equation 7.105 (notice that the first- order correction is zero).

By appropriate modification of the hydrogen formula, determine the hyperfine splitting in the ground state of (a) muonic hydrogen (in which a muon-same charge and \(g\) -factor as the electron, but 207 times the mass substitutes for the electron), (b) positronium (in which a positron-same mass and \(g\) -factor as the electron, but opposite charge- substitutes for the proton), and (c) muonium (in which an anti-muon-same mass and \(g\) -factor as a muon, but opposite charge-substitutes for the proton). Hint: Don't forget to use the reduced mass (Problem 5.1) in calculating the "Bohr radius" of these exotic "atoms," but use the actual masses in the gyromagnetic ratios. Incidentally, the answer you get for positronium \(\left(4.82 \times 10^{-4} \mathrm{eV}\right)\) is quite far from the experimental value \(\left(8.41 \times 10^{-4} \mathrm{eV}\right) ;\) the large discrepancy is due to pair annihilation \(\left(e^{+}+e^{-} \rightarrow \gamma+\gamma\right),\) which contributes an extra (3/4) \(\Delta E,\) and does not occur (of course) in ordinary hydrogen, muonic hydrogen, or muonium.

When an atom is placed in a uniform external electric field \(\mathbf{E}_{\mathrm{ext}},\) the energy levels are shifted-a phenomenon known as the Stark effect (it is the electrical analog to the Zeeman effect). In this problem we analyze the Stark effect for the \(n=1\) and \(n=2\) states of hydrogen. Let the field point in the \(z\) direction, so the potential energy of the electron is $$H_{S}^{\prime}=e E_{\mathrm{ext}} z=e E_{\mathrm{ext}} r \cos \theta$$ Treat this as a perturbation on the Bohr Hamiltonian (Equation 7.43 ). (Spin is irrelevant to this problem, so ignore it, and neglect the fine structure.) (a) Show that the ground state energy is not affected by this perturbation, in first order. (b) The first excited state is four-fold degenerate: \(\psi_{200}, \psi_{211}, \psi_{210}, \psi_{21-1}\) Using degenerate perturbation theory, determine the first-order corrections to the energy. Into how many levels does \(E_{2}\) split? (c) What are the "good" wave functions for part (b)? Find the expectation value of the electric dipole moment \(\left(\mathbf{p}_{e}=-e \mathbf{r}\right),\) in each of these "good" states. Notice that the results are independent of the applied fieldevidently hydrogen in its first excited state can carry a permanent electric dipole moment. Hint: There are lots of integrals in this problem, but almost all of them are zero. So study each one carefully, before you do any calculations: If the \(\phi\) integral vanishes, there's not much point in doing the \(r\) and \(\theta\) integrals! You can avoid those integrals altogether if you use the selection rules of Sections 6.4 .3 and \(6.7 .2 .\) Partial answer: \(W_{13}=W_{31}=-3 e a E_{\mathrm{ext}} ;\) all other elements are zero.

Show that \(p^{2}\) is hermitian, for hydrogen states with \(\ell=0 .\) Hint: For such states \(\psi\) is independent of \(\theta\) and \(\phi,\) so $$p^{2}=-\frac{\hbar^{2}}{r^{2}} \frac{d}{d r}\left(r^{2} \frac{d}{d r}\right)$$ (Equation 4.13 ). Using integration by parts, show that $$\left\langle f | p^{2} g\right\rangle=-\left.4 \pi \hbar^{2}\left(r^{2} f \frac{d g}{d r}-r^{2} g \frac{d f}{d r}\right)\right|_{0} ^{\infty}+\left\langle p^{2} f | g\right\rangle$$ Check that the boundary term vanishes for \(\psi_{n 00}\), which goes like $$f_{n 00} \sim \frac{1}{\sqrt{\pi}(n a)^{3 / 2}} \exp (-r / n a)$$ near the origin. The case of \(p^{4}\) is more subtle. The Laplacian of \(1 / r\) picks up a delta function (see, for example, D. J. Griffiths, Introduction to Electrodynamics, 4 th edn Eq. 1.102 ). Show that $$\nabla^{4}\left[e^{-k r}\right]=\left(-\frac{4 k^{3}}{r}+k^{4}\right) e^{-k r}+8 \pi k \delta^{3}(\mathbf{r})$$ and confirm that \(p^{4}\) is hermitian.
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