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Chapter 7: Problem 39

Consider a three-level system with the unperturbed Hamiltonian $$\mathrm{H}^{0}=\left(\begin{array}{ccc} \epsilon_{a} & 0 & 0 \\ 0 & \epsilon_{a} & 0 \\ 0 & 0 & \epsilon_{c} \end{array}\right)$$ \(\left(\epsilon_{a} > \epsilon_{c}\right)\) and the perturbation $$\mathrm{H}^{\prime}=\left(\begin{array}{ccc} 0 & 0 & V \\ 0 & 0 & V \\ V^{*} & V^{*} & 0 \end{array}\right)$$ since the \((2 \times 2)\) matrix \(W\) is diagonal (and in fact identically 0 ) in the basis of states (1,0,0) and \((0,1,0),\) you might assume they are the good states, but they're not. To see this: (a) Obtain the exact eigenvalues for the perturbed Hamiltonian \(\mathrm{H}=\mathrm{H}^{0}+\mathrm{H}^{\prime}\) (b) Expand your results from part (a) as a power series in \(|V|\) up to second order. (c) What do you obtain by applying nondegenerate perturbation theory to find the energies of all three states (up to second order)? This would work if the assumption about the good states above were correct. Moral: If any of the eigenvalues of \(\mathrm{W}\) are equal, the states that diagonalize \(\mathrm{W}\) are not unique, and diagonalizing \(W\) does not determine the "good" states. When this happens (and it's not uncommon), you need to use second-order degenerate perturbation theory (see Problem 7.40 ).

Short Answer

Expert verified
The exact eigenvalues are \(\epsilon_a, \epsilon_a, \epsilon_c + \frac{2|V|^2}{\epsilon_a-\epsilon_c}\).

Step by step solution

01

Write the Perturbed Hamiltonian

We are given the unperturbed Hamiltonian \( \mathrm{H}^{0} \) and the perturbation \( \mathrm{H}^{\prime} \). The total Hamiltonian \( \mathrm{H} \) is the sum of these two matrices. Substituting in the given matrices, we have:\[\mathrm{H} = \begin{pmatrix}\epsilon_{a} & 0 & V \0 & \epsilon_{a} & V \V^{*} & V^{*} & \epsilon_{c}\end{pmatrix}\]
02

Find the Characteristic Equation

To find the eigenvalues of \( \mathrm{H} \), we require the determinant of \( \mathrm{H} - \lambda \mathbb{I} \) to be zero, where \( \lambda \) represents the eigenvalues. Thus, we compute:\[\mathrm{det}(\mathrm{H} - \lambda \mathbb{I}) = \begin{vmatrix}\epsilon_{a} - \lambda & 0 & V \0 & \epsilon_{a} - \lambda & V \V^{*} & V^{*} & \epsilon_{c} - \lambda\end{vmatrix}\]
03

Solve the Determinant Equation

Expanding the determinant:\[(\epsilon_{a} - \lambda)^2(\epsilon_{c} - \lambda) - 2|V|^2(\epsilon_{a} - \lambda) = 0\]This simplifies to the cubic equation:\[(\epsilon_{a} - \lambda)^2( (\epsilon_{c} - \lambda) - 2|V|^2) = 0\]
04

Analyze the Eigenvalues

From the factorized form of the cubic equation, we get:1. \(\lambda_1 = \epsilon_{a}\) 2. \(\lambda_2 = \epsilon_{a}\) 3. \(\lambda_3 = \epsilon_{c} + 2|V|^2/(\epsilon_{a} - \epsilon_{c})\)These are the exact eigenvalues for the perturbed Hamiltonian.
05

Expand Eigenvalues as Power Series

For small \( |V| \), expand \( \lambda_3 \) up to second order:\[\lambda_3 \approx \epsilon_{c} + \frac{2|V|^2}{\epsilon_a-\epsilon_c} - \frac{2|V|^4}{(\epsilon_a-\epsilon_c)^3}\]Both \( \lambda_1 \) and \( \lambda_2 \) remain equal to \( \epsilon_{a} \) in this approximation.
06

Apply Nondegenerate Perturbation Theory

Applying first and second order non-degenerate perturbation theory for eigenvalues:- For \( |1 \rangle = (1,0,0) \) and \( |2 \rangle = (0,1,0) \), no first or second-order corrections because \( W \) is zero.- For \( |3 \rangle = (0,0,1) \), the first-order correction is zero and the second-order correction gives \( \frac{2|V|^2}{\epsilon_a - \epsilon_c} \).Thus, we would find \( E_3 = \epsilon_c + \frac{2|V|^2}{\epsilon_a - \epsilon_c} \).
07

Conclusion

The eigenvalues found using perturbation theory do not match those found by diagonalizing \( \mathrm{H} \) because the states used were not correctly identified as good states due to degeneracy issues in \( W \).

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Key Concepts

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

Three-Level Quantum System
In quantum mechanics, a three-level quantum system refers to a model that involves three distinct energy states typically characterized by three eigenvalues and their corresponding eigenvectors. This kind of system is fundamental in understanding various phenomena in quantum physics, such as quantum transitions and interactions.
In the exercise, the unperturbed Hamiltonian matrix describes this three-level system, where we have two degenerate states, denoted by \( \epsilon_a \), and a non-degenerate state, represented by \( \epsilon_c \) (where \( \epsilon_a > \epsilon_c \)). These degenerate states are critical in analyzing the effects of perturbations, especially when tackling degeneracy through perturbative methods.
Three-level systems, like the one described here, allow us to explore how differences in energy levels interact under certain conditions, which are often altered by the introduction of a perturbation.
Perturbed Hamiltonian
A perturbed Hamiltonian is a modification of the original Hamiltonian to include an additional potential or interaction. This is essential in exploring how the introduction of external or internal influences can change the energy state of a quantum system.
In this problem, the perturbed Hamiltonian \( \mathrm{H} \) is composed of the unperturbed Hamiltonian \( \mathrm{H}^{0} \) and the perturbation \( \mathrm{H}^{\prime} \). The perturbation term \( \mathrm{H}^{\prime} \) introduces non-diagonal entries, represented by the coupling terms \( V \) and \( V^{*} \), which reflect interactions between the states.
The inclusion of these terms highlights a coupling between the two degenerate states and the third state, thus reflecting a change in the system's energy levels. This provides a rich landscape for studying how systems evolve and respond to different perturbations, a fundamental aspect in quantum mechanics and physical chemistry.
Nondegenerate Perturbation Theory
Nondegenerate perturbation theory is applied in quantum mechanics when dealing with systems where the energy levels are not degenerate or when handling states individually within a degenerate set. Its application allows us to calculate energy corrections due to small perturbations.
In the exercise, although the states \( |1\rangle \) and \( |2\rangle \) are treated via this theory, it results in errors. Here, the erroneous assumption was that \( |1\rangle \) and \( |2\rangle \) would remain unchanged because the matrix \( W \) was diagonally zero. Since these states are degenerate in the absence of perturbation, nondegenerate perturbation theory fails to give the correct energies for these states owing to its assumption of isolated states.
This problem illustrates that in settings where degeneracy exists, nondegenerate methods can be misleading, necessitating the use of degenerate perturbation theory for accurate results.
Quantum Mechanics Problem Solving
Solving problems in quantum mechanics often involves understanding complex interactions within a system. The approach often uses matrices such as Hamiltonians to describe the system's energy.
In this instance, the process involves calculating eigenvalues of a perturbed matrix to predict how energy states will adjust in response to introduced perturbations. Initially, transitioning from complex determinants to simpler eigenvalue equations is crucial, as seen in the expanded determinant approach leading to a cubic equation for eigenvalues.
While solving these problems, a critical aspect is the recognition of degeneracies and the correct methodological approach, whether degenerate or nondegenerate categories, as underscored by this exercise. This understanding helps avoid common traps like incorrect assumptions about "good" states, ultimately leading to reliable and accurate solutions.

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Most popular questions from this chapter

Evaluate the following commutators: (a) \([\mathbf{L} \cdot \mathbf{S}, \mathbf{L}],(\mathbf{b})[\mathbf{L} \cdot \mathbf{S}, \mathbf{S}]\) \((\mathrm{c})[\mathbf{L} \cdot \mathbf{S}, \mathbf{J}],(\mathrm{d})\left[\mathbf{L} \cdot \mathbf{S}, L^{2}\right],(\mathrm{e})\left[\mathbf{L} \cdot \mathbf{S}, S^{2}\right],(\mathrm{f})\left[\mathbf{L} \cdot \mathbf{S}, J^{2}\right] .\) Hint: \(\mathbf{L}\) and \(\mathbf{S}\) satisfy. the fundamental commutation relations for angular momentum (Equations 4.99 and 4.134 ), but they commute with each other.

Consider a charged particle in the one-dimensional harmonic oscillator potential. Suppose we turn on a weak electric field \((E),\) so that the potential energy is shifted by an amount \(H^{\prime}=-q E x\) (a) Show that there is no first-order change in the energy levels, and calculate the second-order correction. Hint: See Problem 3.39 (b) The Schr枚dinger equation can be solved directly in this case, by a change of variables: \(x^{\prime} \equiv x-\left(q E / m \omega^{2}\right) .\) Find the exact energies, and show that they are consistent with the perturbation theory approximation.

In the text I asserted that the first-order corrections to an \(n\) -fold degenerate energy are the eigenvalues of the \(W\) matrix, and I justified this claim as the "natural" generalization of the case \(n=2 .\) Prove it, by reproducing the steps in Section \(7.2 .1,\) starting with $$\psi^{0}=\sum_{j=1}^{n} \alpha_{j} \psi_{j}^{0}$$ (generalizing Equation 7.17 ), and ending by showing that the analog to Equation 7.27 can be interpreted as the eigenvalue equation for the matrix \(\mathrm{W}\)

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.

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).
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