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Chapter 5: Problem 11

Evaluate the matrix elements (or expectation values) given below. If any vanishes, explain why it vanishes using simple symmetry (or other) arguments. a. \(\langle n=2, l=1, m=0|x| n=2, l=0, m=0\rangle\). b. \(\left\langle n=2, l=1, m=0\left|p_{z}\right| n=2, l=0, m=0\right\rangle\). [In (a) and (b), \(|n l m\rangle\) stands for the energy eigenket of a nonrelativistic hydrogen atom with spin ignored.] c. \(\left\langle L_{z}\right\rangle\) for an electron in a central field with \(j=\frac{9}{2}, m=\frac{7}{2}, l=4\). d. \(\left\langle\right.\) singlet, \(m_{s}=0\left|S_{z}^{(e-)}-S_{z}^{(e+)}\right|\) triplet, \(\left.m_{s}=0\right\rangle\) for an \(s\)-state positronium. e. \(\left\langle\mathbf{S}^{(1)} \cdot \mathbf{S}^{(2)}\right\rangle\) for the ground state of a hydrogen molecule.

Short Answer

Expert verified
a, b: Vanish due to orthogonality; c: \( \frac{7}{2} \hbar \); d: 0 by symmetry; e: \(-\frac{3}{4} \hbar^2\).

Step by step solution

01

Analyze matrix element for (a)

Consider the matrix element \( \langle n=2, l=1, m=0|x| n=2, l=0, m=0\rangle. \) The key is to determine whether the integral vanishes or not. In spherical coordinates, the operator \( x \) does not change the principal quantum number \( n \) but changes the azimuthal quantum number \( l \) according to angular momentum selection rules. The change \( \Delta l = \pm 1 \) is allowed, but since \( \langle l=0 | l'=1 \rangle = 0 \) due to orthogonality, this element vanishes.
02

Analyze matrix element for (b)

Consider the matrix element \( \left\langle n=2, l=1, m=0 \left| p_{z} \right| n=2, l=0, m=0 \right\rangle. \) The operator \( p_z \) is the momentum operator in the \( z \) direction and changes the azimuthal quantum number \( l \) by \( \pm 1 \). Similarly to the \( x \) operator, the state \( | n=2, l=1, m=0 \rangle \) is orthogonal to \( | n=2, l=0, m=0 \rangle \), making the matrix element equal to zero.
03

Compute \( \langle L_z \rangle \) for electron in central field

For an electron in a central field, consider \( j = \frac{9}{2}, m = \frac{7}{2}, l = 4 \). The angular momentum projection \( \langle L_z \rangle \) for a state is \( m \hbar \), where \( m \) is the magnetic quantum number. Therefore, \( \langle L_z \rangle = 7/2 \cdot \hbar = \frac{7}{2} \hbar \).
04

Analyze spin difference for positronium states

For a singlet state \( |S, m_s = 0\rangle \) and triplet state \( |T, m_s = 0\rangle \), the operator \( S_z^{(e-)} - S_z^{(e+)} \) measures the difference in z-component spins of the electron and positron. In a singlet state, spins are oppositely aligned resulting in zero: \( \langle S_z^{(e-)} - S_z^{(e+)} \rangle = 0 \). This expectation value is zero because the singlet state has zero total spin in any direction due to symmetry.
05

Analyze spin product for hydrogen molecule

In the ground state of a hydrogen molecule (H2), the two electrons are in a spin singlet state, which means their spins are paired/opposite. The expectation value of the dot product \( \langle \mathbf{S}^{(1)} \cdot \mathbf{S}^{(2)} \rangle \) is \( -\frac{3}{4} \hbar^2 \). This negative value arises because for a singlet state, total spin \( S(S+1) = 0 \), and hence the dot product of paired spins is \( -\frac{3}{4} \hbar^2 \).

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Key Concepts

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

Hydrogen atom energy levels
Understanding the energy levels of a hydrogen atom is crucial in quantum mechanics. The hydrogen atom is the simplest atomic system, consisting of a single electron orbiting a single proton. The energy levels are quantized, meaning they can only have specific values. These values are determined by the principal quantum number, denoted by \( n \). The energy corresponding to each level can be calculated using the formula: \[ E_n = - \frac{13.6 \, \text{eV}}{n^2} \]- **Principal Quantum Number (\( n \))**: It indicates the size of the orbit and the energy level of the electron.- **Azimuthal Quantum Number (\( l \))**: Relates to the shape of the orbitals. For any given \( n \), \( l \) ranges from 0 to \( n-1 \).- **Magnetic Quantum Number (\( m \))**: Specifies the orientation of the orbital in space.For example, when \( n = 2 \), possible \( l \) values are 0 and 1, leading to different sub-levels or types: 2s (\( l = 0 \)) and 2p (\( l = 1 \)). These specific energy levels lead to various electronic transitions and spectral emissions in hydrogen that are observed in experiments.
Angular momentum selection rules
Selection rules dictate the allowed transitions between states in terms of angular momentum. They are important for predicting which quantum state transitions are possible and therefore observable.Angular Momentum in Quantum Mechanics:The orbital angular momentum \( L \) is quantized and related to \( l \), where the magnitude is \( \sqrt{l(l+1)} \hbar \). The component \( L_z \) (orbital angular momentum along the \( z \)-axis) can take values of \( m \hbar \), where \( m \) is the magnetic quantum number.
  • For a transition to be allowed, \( \Delta l = \pm 1 \).
  • The change in charge, such as\( \Delta m = 0, \pm 1 \), determines the change in the system's magnetic properties.
In exercises where states such as \( |n=2, l=1, m=0\rangle \) and \( |n=2, l=0, m=0\rangle \) are considered, the orthogonality condition \( \langle l=0|l'=1 \rangle = 0 \) implies these states do not overlap. Thus, transitions between them via operators that demand a change in \( l \), such as position \( x \) or \( p_z \), result in vanishing matrix elements, adhering to the selection rules.
Spin states in quantum systems
Spin is an intrinsic form of angular momentum carried by elementary particles, such as electrons and positrons. It's crucial in understanding magnetic and structural properties of materials. Quantum Spin Concepts:- Spin quantum number (\( s \)) for electrons is \( \frac{1}{2} \), signifying two possible states: \( +\frac{1}{2} \) and \( -\frac{1}{2} \).
  • **Singlet and Triplet States**: In systems like positronium, a bound state of electron and positron, singlet states (opposite spins) have total spin \( S = 0 \), while triplet states (same spin direction) have \( S = 1 \).
  • **Spin Symmetry**: In singlet states, spins are opposite, making them symmetric under spin exchanges. This results in zero total spin in any given direction. Hence, expectation operators measuring spin differences, such as \( S_z^{(e-)} - S_z^{(e+)} \), yield zero.
In a hydrogen molecule, the electrons typically form a singlet state in their ground state. This state results in a negative expectation value for their spin product, reflecting their opposite spin alignment: \( \langle \mathbf{S}^{(1)} \cdot \mathbf{S}^{(2)} \rangle = -\frac{3}{4} \hbar^2 \). This is because total spin \( S(S+1) \) equals 0 for such paired opposites.

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Most popular questions from this chapter

Consider a composite system made up of two spin \(\frac{1}{2}\) objects. For \(t<0\), the Hamiltonian does not depend on spin and can be taken to be zero by suitably adjusting the energy scale. For \(t>0\), the Hamiltonian is given by $$ H=\left(\frac{4 \Delta}{\hbar^{2}}\right) \mathbf{S}_{1} \cdot \mathbf{S}_{2} $$ Suppose the system is in \(|+-\rangle\) for \(t \leq 0\). Find, as a function of time, the probability for being found in each of the following states \(|++\rangle,|+-\rangle,|-+\rangle\), and \(|--\rangle\). a. By solving the problem exactly. b. By solving the problem assuming the validity of first-order time-dependent perturbation theory with \(H\) as a perturbation switched on at \(t=0\). Under what condition does (b) give the correct results?

Estimate the lowest eigenvalue \((\lambda)\) of the differential equation $$ \frac{d^{2} \psi}{d x^{2}}+(\lambda-|x|) \psi=0 $$ where \(\psi \rightarrow 0\) as \(|x| \rightarrow \infty\) using the variational method with $$ \psi=\left\\{\begin{array}{ll} c(\alpha-|x|) & \text { for }|x|<\alpha \\ 0 & \text { for }|x|>\alpha \end{array} \quad(\alpha \text { to be varied })\right. $$ as a trial function. (Caution: \(d \psi / d x\) is discontinuous at \(x=0\).) The exact value of the lowest eigenvalue can be shown to be \(1.019\).

The unperturbed Hamiltonian of a two-state system is represented by $$ H_{0}=\left(\begin{array}{cc} E_{1}^{0} & 0 \\ 0 & E_{2}^{0} \end{array}\right) \text {. } $$ There is, in addition, a time-dependent perturbation $$ V(t)=\left(\begin{array}{cc} 0 & \lambda \cos \omega t \\ \lambda \cos \omega t & 0 \end{array}\right) \quad(\lambda \mathrm{real}) $$ a. At \(t=0\) the system is known to be in the first state, represented by $$ \left(\begin{array}{l} 1 \\ 0 \end{array}\right) $$ Using time-dependent perturbation theory and assuming that \(E_{1}^{0}-E_{2}^{0}\) is not close to \(\pm \hbar \omega\), derive an expression for the probability that the system be found in the second state represented by $$ \left(\begin{array}{l} 0 \\ 1 \end{array}\right) $$ as a function of \(t(t>0)\). b. Why is this procedure not valid when \(E_{1}^{0}-E_{2}^{0}\) is close to \(\pm \hbar \omega\) ?

a. Suppose the Hamiltonian of a rigid rotator in a magnetic field perpendicular to the axis is of the form (Merzbacher 1970, Problem 17-1) $$ A \mathbf{L}^{2}+B L_{z}+C L_{y} $$ if terms quadratic in the field are neglected. Assuming \(B \gg C\), use perturbation theory to lowest nonvanishing order to get approximate energy eigenvalues. b. Consider the matrix elements $$ \begin{gathered} \left\langle n^{\prime} l^{\prime} m_{l}^{\prime} m_{s}^{\prime}\left|\left(3 z^{2}-r^{2}\right)\right| n l m_{l} m_{s}\right\rangle \\ \left\langle n^{\prime} l^{\prime} m_{l}^{\prime} m_{s}^{\prime}|x y| n l m_{l} m_{s}\right\rangle \end{gathered} $$ of a one-electron (for example, alkali) atom. Write the selection rules for \(\Delta l, \Delta m_{l}\), and \(\Delta m_{s}\). Justify your answer. c. Use degenerate perturbation theory to find the first-order energy shifts \(\Delta\). For all eight states, plot \(\Delta / A\) as a function of \(B / A\). See Figure \(5.3\). Explain why the resulting spectrum looks qualitatively different for \(B / A \ll 1\) and \(B / A \gg 1\).

A simple harmonic oscillator (in one dimension) is subjected to a perturbation $$ H_{1}=b x $$ where \(b\) is a real constant. a. Calculate the energy shift of the ground state to lowest nonvanishing order. b. Solve this problem exactly and compare with your result obtained in (a).
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