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Chapter 7: Problem 45

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.

Short Answer

Expert verified
(a) No shift in ground state energy. (b) \(n=2\) splits into two levels. (c) Good wave functions exhibit a permanent dipole moment.

Step by step solution

01

Understanding the Problem

The problem involves analyzing the Stark effect for hydrogen atoms placed in an external electric field. In particular, we look at how energy levels are shifted for states with quantum numbers \(n=1\) and \(n=2\). We are tasked with showing that certain energy levels are not affected by the perturbation in first order, determining first-order corrections for degenerate states, and finding the expectation value of the electric dipole moment for specific wave functions.
02

Perturbation Theory Overview

In perturbation theory, we treat the external electric field as a small perturbation to the hydrogen atom's Hamiltonian. The perturbation Hamiltonian given is \(H_S' = e E_{\mathrm{ext}} z = e E_{\mathrm{ext}} r \cos \theta\). We use first-order perturbation theory to find the changes in energy levels.
03

Ground State Energy Shift Analysis

For the \(n=1\) ground state of hydrogen, the wave function is spherically symmetric, \(\psi_{100} = R_{10}(r) Y_{00}(\theta, \phi)\). The perturbation \(H_S' = e E_{\mathrm{ext}} z\) is odd in \(z\). The integral \(\langle \psi_{100} | H_S' | \psi_{100} \rangle\) involves an odd function over a symmetric range, ideal for the selection rule that odd integrals over symmetric ranges are zero. Therefore, the first-order energy shift is zero.
04

Degenerate Perturbation Theory for First Excited State

The first excited state \(n=2\) is fourfold degenerate, with spherical harmonics \(\psi_{200}, \psi_{211}, \psi_{210}, \psi_{21-1}\). We must find the matrix elements \(W_{ij} = \langle \psi_{2im} | H_S' | \psi_{2jn} \rangle\). Use selection rules: only terms where \(\Delta l = \pm 1\) contribute. Non-zero contributions are \(W_{13} = W_{31} = -3 e a E_{\mathrm{ext}}\), splitting the energy levels.
05

Finding "Good" Wave Functions

To find the "good" wave functions, diagonalize the perturbation matrix. From symmetry, some states do not mix, while others form new linear combinations, representing the actual observed wave functions. These 'good' states correspond to the new eigenstates of the perturbation matrix.
06

Expectation Value of Dipole Moment

The expectation value of the dipole moment \(\langle \mathbf{p}_e \rangle = -e \langle \mathbf{r} \rangle\) is calculated for the 'good' states. Since \(\mathbf{p}_e\) aligns with the new potential minima, only specific combinations yield non-zero moments, resulting in a permanent dipole moment independent of the external field direction, confirming the hint's implication.

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Key Concepts

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

Perturbation Theory
The Stark effect in hydrogen atoms is a perfect application of perturbation theory, which is a powerful tool in quantum mechanics. This theory helps us understand how small changes in the system (like an external electric field) affect its properties. Here, the small change or "perturbation" is due to the electric field applied to the hydrogen atom. The perturbed Hamiltonian is represented as \[ H_{S}^{\prime}=e E_{\mathrm{ext}} r \cos \theta \]By considering this additional potential energy, we can calculate the effect on energy levels.
  • Perturbation theory assumes that the perturbation is small compared to the unperturbed system.
  • In first-order perturbation theory, we calculate how energy levels shift due to the perturbation.
  • We use the original, unperturbed wave functions to calculate these shifts.
The main advantage of this method is that it allows us to approximate the behavior of a complicated system by adding small corrections to a simpler, well-understood system. In our problem, we're assuming the external field is a weak perturbation to the hydrogen atom's Hamiltonian.
Degenerate Perturbation Theory
When dealing with systems where states share the same energy, like the first excited state of hydrogen, we apply degenerate perturbation theory. The first excited state (\(n=2\)) in hydrogen has four degenerate states: - \(\psi_{200}\)- \(\psi_{211}\)- \(\psi_{210}\)- \(\psi_{21-1}\)In this scenario, perturbation theory must account for interactions between these states, as they are all initially at the same energy level. A matrix of the perturbation operator with regards to these states is constructed. Only terms satisfying the selection rule \(\Delta l = \pm 1\) affect the matrix elements:
  • Non-zero contributions result in specific energy level shifts.
  • Diagonalization of this matrix yields new energy states and their respective energies.
Degenerate perturbation theory ensures that all potential shifts and mixings of the degenerate states are accounted for. This method is critical since it provides insights into systems when simple, non-degenerate perturbation theory is insufficient.
Electric Dipole Moment
An electric dipole moment measures the separation of positive and negative charges within a system. When hydrogen atoms are subjected to an external electric field, this can induce or reveal an intrinsic electric dipole moment related to the system's spatial charge distribution.In hydrogen's first excited state, 'good' wave functions are combinations of degenerate states that respond uniquely to the external electric field. For these, the electric dipole moment \(\mathbf{p}_e\) expectation value is calculated using:\[ \langle \mathbf{p}_e \rangle = -e \langle \mathbf{r} \rangle \]Here are some key points about electric dipole moments in this context:
  • The Stark effect can reveal a permanent electric dipole moment even without an external electric field.
  • Only particular linear combinations of states ("good" states) present non-zero dipole moments.
  • This moment results from spatial asymmetry in the charge distribution.
By calculating the expectation value, one can determine the dipole moment associated with each good state. This characteristic illustrates how quantum states can have inherent dipole moments influenced by external factors.

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Most popular questions from this chapter

Estimate the correction to the ground state energy of hydrogen due to the finite size of the nucleus. Treat the proton as a uniformly charged spherical shell of radius \(b,\) so the potential energy of an electron inside the shell is constant: \(-e^{2} /\left(4 \pi \epsilon_{0} b\right) ;\) this isn't very realistic, but it is the simplest model, and it will give us the right order of magnitude. Expand your result in powers of the small parameter ( \(b / a\) ), where \(a\) is the Bohr radius, and keep only the leading term, so your final answer takes the form $$\frac{\Delta E}{E}=A(b / a)^{n}$$ Your business is to determine the constant \(A\) and the power \(n\). Finally, put in \(b \approx 10^{-15} \mathrm{m}\) (roughly the radius of the proton) and work out the actual number. How does it compare with fine structure and hyperfine structure?

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.

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 ).

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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