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Problem 7.43 Prove Kramers' relation: \\[ \frac{s+1}{n^{2}}\left\langle r^{s}\right\rangle-(2 s+1) a\left\langle r^{s-1}\right\rangle+\frac{s}{4}\left[(2 \ell+1)^{2}-s^{2}\right] a^{2}\left\langle r^{s-2}\right\rangle=0 \\] which relates the expectation values of \(r\) to three different powers \((s,\) \(s-1,\) and \(s-2\) ), for an electron in the state \(\psi_{n \ell m}\) of hydrogen. Hint: Rewrite the radial equation (Equation 4.53 ) in the form \\[ u^{\prime \prime}=\left[\frac{\ell(\ell+1)}{r^{2}}-\frac{2}{a r}+\frac{1}{n^{2} a^{2}}\right] u \\] and use it to express \(\int\left(u r^{s} u^{\prime \prime}\right) d r\) in terms of \(\left\langle r^{s}\right\rangle,\left\langle r^{s-1}\right\rangle,\) and \(\left\langle r^{s-2}\right\rangle .\) Then use integration by parts to reduce the second derivative. Show that \\[ \int\left(u r^{s} u^{\prime}\right) d r=-(s / 2)\left\langle r^{s-1}\right\rangle \\] and \(\int\left(u^{\prime} r^{s} u^{\prime}\right) d r=-[2 /(s+1)] \int\left(u^{\prime \prime} r^{s+1} u^{\prime}\right) d r .\) Take it from there.

Short Answer

Expert verified
By applying integration by parts and the given equations, we derive Kramers' relation, proving it in terms of \(\langle r^s \rangle\), \(\langle r^{s-1} \rangle\), and \(\langle r^{s-2} \rangle\).

Step by step solution

01

Understand the Problem

The goal is to prove Kramers' relation using the expectation values of different powers of the radial distance \( r \) for a hydrogen atom's electron in state \( \psi_{n\ell m} \). The relation involves calculations of \( \langle r^s \rangle \), \( \langle r^{s-1} \rangle \), and \( \langle r^{s-2} \rangle \).
02

Rewrite the Radial Equation

Starting with the hint, rewrite the radial Schrödinger equation as given:\[u^{\prime\prime} = \left[ \frac{\ell(\ell+1)}{r^2} - \frac{2}{a r} + \frac{1}{n^2 a^2} \right] u\]
03

Manipulate the Integral

Express the integral \( \int(u r^s u^{\prime\prime}) \, dr \) using the given radial equation expression. Substitute \( u'' \) to relate it to the expectation values.
04

Apply Integration by Parts

Use integration by parts to express \( \int(u r^s u'') \, dr \) in terms of \( \langle r^s \rangle \), \( \langle r^{s-1} \rangle \), and \( \langle r^{s-2} \rangle \). This involves using integration by parts to break down \( u'' \):\[ \int(u r^s u^{\prime\prime}) \, dr = -(2s+1)a \langle r^{s-1} \rangle + \frac{s}{4}[(2\ell+1)^2-s^2]a^2 \langle r^{s-2} \rangle \]
05

Solve the Integral

Further demonstrate that:\[ \int(u r^s u') \, dr = -\frac{s}{2} \langle r^{s-1} \rangle \]
06

Finalize the Equation

Combine your findings from the previous steps to build the structure of Kramers' relation:\[ \frac{s+1}{n^2} \langle r^s \rangle - (2s+1)a \langle r^{s-1} \rangle + \frac{s}{4}[(2\ell+1)^2-s^2]a^2 \langle r^{s-2} \rangle = 0 \]
07

Verify and Conclude

Verify each aspect of the rewritten integral and confirm that the manipulation aligns with Kramers' relation. Ensure each term in the expected relation appears and simplify as necessary.

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Key Concepts

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

Radial Equation
The radial equation is a fundamental part of quantum mechanics, especially when dealing with spherical coordinates. For example, in hydrogen-like atoms, the Schrödinger equation can be separated into radial and angular parts, with the radial equation focusing solely on the distance from the nucleus, represented by the variable \( r \). This equation considers how an electron's wavefunction changes with distance.For the hydrogen atom, the wavefunction can be decomposed as \( \psi_{n\ell m} = R_{n\ell}(r)Y_{\ell m}(\theta, \phi) \), where \( R_{n\ell}(r) \) is the radial wavefunction. The radial Schrödinger equation is typically expressed in the form:\[ u'' = \left[ \frac{\ell(\ell+1)}{r^2} - \frac{2}{ar} + \frac{1}{n^2a^2} \right] u \]where \( u(r) = rR_{n\ell}(r) \), \( \ell \) is the angular momentum quantum number, \( a \) is the Bohr radius, and \( n \) is the principal quantum number. By rewriting the equation in this form, it becomes easier to manipulate during the proof of various relations, like Kramers' relation.
Expectation Values
In quantum mechanics, the expectation value of a position-related observable, such as \( r^s \), represents the average expected outcome for a large number of measurements of an electron's position. For an orbital described by a wave function \( \psi \), the expectation value \( \langle r^s \rangle \) is defined as:\[ \langle r^s \rangle = \int |\psi|^2 r^s \, dr \, d\Omega \] Here, \( d\Omega \) represents the angular part of the volume element in spherical coordinates. In the context of hydrogen atoms, the wave function \( \psi_{n\ell m} \) is used to calculate these values. This exercise specifically asks us to prove a relation that involves three expectation values: \( \langle r^s \rangle \), \( \langle r^{s-1} \rangle \), and \( \langle r^{s-2} \rangle \).Understanding expectation values is critical because they offer insights into the electron's probable location and energy state within an atom. In chemical bonding and spectroscopic processes, these values further elucidate interactions between electrons and nuclei.
Integration by Parts
Integration by parts is an essential mathematical technique used to simplify complex integrals, and it is particularly useful when dealing with products of functions. The formula is derived from the product rule for differentiation and is expressed in terms of integrals as:\[ \int u \, dv = uv - \int v \, du \] In the context of this problem, integration by parts is employed to simplify terms involving derivatives of the radial wave function. For example, when proving Kramers' relation, we need to manage integrals of the form \( \int(u r^s u'') \, dr \). By appropriately selecting \( u \) and \( v \), these terms can be transformed into a more manageable form, facilitating the computation of expectation values.Through integrating by parts, the given problem reduces complex, second-order terms into first-order ones related to expectation values like \( \langle r^{s-1} \rangle \) and \( \langle r^{s-2} \rangle \). This method is a crucial step in demonstrating that these expressions are consistent with Kramers' relation, ultimately helping to confirm the calculated expectation values and their significance in quantum mechanics.

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Most popular questions from this chapter

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.

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.

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).

\mathrm{A}\( free particle of mass \)m\( is confined to a ring of circumference \)L\( such that \)\psi(x+L)=\psi(x) .\( The unperturbed Hamiltonian is \\[ H^{0}=-\frac{\hbar^{2}}{2 m} \frac{d^{2}}{d x^{2}} \\] to which we add a perturbation \\[ H^{\prime}=V_{0} \cos \left(2 \pi \frac{x}{L}\right) \\] (a) Show that the unperturbed states may be written \\[ \psi_{n}^{0}(x)=\frac{1}{\sqrt{L}} e^{i 2 \pi n x / L} \\] for \)n=0,\pm 1,\pm 2\( and that, apart from \)n=0,\( all of these states are two-fold degenerate. (b) Find a general expression for the matrix elements of the perturbation: \)H_{m n}^{\prime}=\left\langle\psi_{m}^{0}\left|H^{\prime}\right| \psi_{n}^{0}\right\rangle\( (c) Consider the degenerate pair of states with \)n=\pm 1\(. Construct the matrix \)W\( and calculate the first-order energy corrections, \)E^{1}\(. Note that the degeneracy does not liff at first order. Therefore, diagonalizing \)\mathrm{W}\( does not tell us what the "good" states are. (d) Construct the matrix \)\mathrm{W}^{2}(\text { Problem } 7.40)\( for the states \)n=\pm 1,\( and show that the degeneracy lifts at second order. What are the good linear combinations of the states with \)n=\pm 1 ?$ (e) What are the energies, accurate to second order, for these states? 36

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