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Chapter 5: Problem 18

Work out the quadratic Zeeman effect for the ground-state hydrogen atom \([\langle\mathbf{x} | 0\rangle=\) \(\left.(1 / \sqrt{\pi a_{0}^{3}}) e^{-r / a_{0}}\right]\) due to the usually neglected \(e^{2} \mathbf{A}^{2} / 2 m_{e} c^{2}\) -term in the Hamiltonian taken to first order. Write the energy shift as \\[ \Delta=-\frac{1}{2} \chi \mathbf{B}^{2} \\] and obtain an expression for diamagnetic susceptibility, \(\chi .\) The following definite integral may be useful: \\[ \int_{0}^{\infty} e^{-\alpha r} r^{n} d r=\frac{n !}{\alpha^{n+1}} \\]

Short Answer

Expert verified
\( \chi = \frac{e^2}{8 m_e c^2} 3 a_0^2 \) is the diamagnetic susceptibility.

Step by step solution

01

Identify the Relevant Term in the Hamiltonian

To find the quadratic Zeeman effect, we focus on the perturbative term \( e^2 \mathbf{A}^2 / 2m_e c^2 \) in the Hamiltonian. This term is usually neglected but contributes to the energy shift due to a magnetic field \( \mathbf{B} \). Since \( \mathbf{A} = \frac{1}{2} \mathbf{B} \times \mathbf{r} \), we compute \( \mathbf{A}^2 \) to assess its effect on the energy levels.
02

Calculate the Vector Potential

The magnetic vector potential \( \mathbf{A} = \frac{1}{2} \mathbf{B} \times \mathbf{r} \) leads to \( \mathbf{A}^2 = \frac{1}{4} (\mathbf{B} \times \mathbf{r})^2 \). The magnetic field is considered constant for simplicity, and \( \mathbf{r} \) is the position vector. This results in \( (\mathbf{B} \times \mathbf{r})^2 = B^2 x^2 + B^2 y^2 \) when expanded, considering the ground state symmetry.
03

Incorporate the Ground State Wavefunction

The ground state hydrogen atom wavefunction is given by \( \psi_0(\mathbf{r}) = (1/\sqrt{\pi a_0^3}) e^{-r/a_0} \). To find the shift in energy, calculate the expectation value \( \langle \mathbf{A}^2 \rangle \) over this wavefunction. Use spherical coordinates for integration, noting \( r^2 = x^2 + y^2 + z^2 \).
04

Evaluate the Necessary Integrals

To compute \( \langle r^2 \rangle \), integrate over all space: \[ \langle r^2 \rangle = \int_0^\infty \int_0^\pi \int_0^{2\pi} r^2 (\psi_0^2) \,r^2\, \sin\theta \, dr \, d\theta \, d\phi \] Apply the definite integral formula \( \int_0^\infty e^{-\alpha r} r^n \, dr = \frac{n!}{\alpha^{n+1}} \) with \( \alpha = 2/a_0 \) and \( n = 2 \), leading to: \[ \langle r^2 \rangle = \frac{3a_0^2}{2} \].
05

Calculate the Energy Shift

Use the previously determined \( \langle r^2 \rangle \) to find the energy shift:\[ \Delta E = -\frac{1}{2} \frac{e^2 B^2}{2 m_e c^2} \langle r^2 \rangle \] Substitute \( \langle r^2 \rangle = \frac{3 a_0^2}{2} \) to get:\[ \Delta E = -\frac{1}{2} \frac{e^2 B^2}{2 m_e c^2} \left( \frac{3 a_0^2}{2} \right) \].
06

Determine Diamagnetic Susceptibility

Relate the energy shift to the expression: \( \Delta E = -\frac{1}{2} \chi B^2 \). Comparing gives:\[ \chi = \frac{e^2}{4m_e c^2} \frac{3 a_0^2}{2} \]. This is the expression for the diamagnetic susceptibility \( \chi \) in the given context.

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Key Concepts

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

Ground-State Hydrogen Atom
The ground-state hydrogen atom is the simplest atom model, consisting of one proton and one electron. The ground state of the hydrogen atom refers to its lowest energy state, where the electron occupies the closest orbit to the nucleus. This state is primarily used as a starting point to understand more complex atomic systems.

In quantum mechanics, the ground-state wavefunction of a hydrogen atom is an exponential function represented as \( \psi_0(\mathbf{r}) = \frac{1}{\sqrt{\pi a_0^3}} e^{-r/a_0} \). This equation helps us define the probability distribution of an electron around the nucleus.

When we assess various interactions, such as those involving magnetic fields, we often perform calculations that use this wavefunction to evaluate expectations like the average position of the electron. This average, or expectation value, is critical when determining how external factors, like magnetic fields, affect the atom's energy levels.
Diagnetic Susceptibility
Diamagnetic susceptibility \(\chi\) is a measure of an atom or molecule's response to an external magnetic field. It is important in determining how substances behave in magnetic environments.

Diamagnetic materials create an opposing magnetic field when subjected to an external magnetic field, often leading to a slight repulsion. For the hydrogen atom in a quadratic Zeeman effect setting, we derive the diamagnetic susceptibility through energy shifts.

The resulting expression for \(\chi\) involves fundamental constants and parameters:
  • The charge of an electron \(e\)
  • The mass of an electron \(m_e\)
  • The speed of light \(c\)
  • The Bohr radius \(a_0\)
These constants combine to give us \( \chi = \frac{e^2}{4 m_e c^2} \frac{3 a_0^2}{2} \). This formulation simplifies the calculation of shifts in energy levels due to magnetic fields by relating them back to a material property directly connected to the fundamental characteristics of hydrogen.
Vector Potential
The vector potential \(\mathbf{A}\) plays a significant role in electromagnetic theory. In the context of the quadratic Zeeman effect, it describes how a magnetic field affects the electron within an atom.

For a uniform magnetic field \(\mathbf{B}\), the magnetic vector potential is given by \( \mathbf{A} = \frac{1}{2} \mathbf{B} \times \mathbf{r} \), where \(\mathbf{r}\) represents the position vector of the electron. The vector potential is vital because it allows us to connect the magnetic field with quantum mechanical properties like the wavefunction of a particle.

When calculating the energy shifts due to magnetic fields, we integrate the square of this vector potential over the ground-state wavefunction. This operation reveals the change in energy levels of the electron and is crucial for assessing the influence of external magnetic effects on atomic systems.

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Most popular questions from this chapter

Consider a particle bound to a fixed center by a spherically symmetrical potential \(V(r)\) (a) Prove \\[ |\psi(0)|^{2}=\left(\frac{m}{2 \pi \hbar^{2}}\right)\left\langle\frac{d V}{d r}\right) \\] for all \(s\) -states, ground and excited. (b) Check this relation for the ground state of a three-dimensional isotropic oscillator, the hydrogen atom, and so on. (Note: This relation has actually been found to be useful in guessing the form of the potential between a quark and an antiquark.)

Consider a particle bound in a simple harmonic-oscillator potential. Initially \((t < \) 0 , it is in the ground state. At \(t=0\) a perturbation of the form \\[ H^{\prime}(x, t)=A x^{2} e^{-t / \tau} \\] is switched on. Using time-dependent perturbation theory, calculate the probability that after a sufficiently long time \((t \gg \tau),\) the system will have made a transition to a given excited state. Consider all final states.

Estimate the lowest eigenvalue \((\lambda)\) of the differential equation \\[ \frac{d^{2} \psi}{d x^{2}}+(\lambda-|x|) \psi=0, \quad \psi \rightarrow 0 \quad \text { for }|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 \(x=0 .\) ) Numerical data that may be useful for this problem are \\[ 3^{1 / 3}=1.442, \quad 5^{1 / 3}=1.710, \quad 3^{2 / 3}=2.080, \quad \pi^{2 / 3}=2.145 \\] The exact value of the lowest eigenvalue can be shown to be 1.019

A hydrogen atom in its ground state \([(n, l, m)=(1,0,0)]\) is placed between the plates of a capacitor. A time-dependent but spatially uniform electric field (not potential!) is applied as follows: \(\mathbf{E}=\left\\{\begin{array}{ll}0, & \text { for } t<0 \\ \mathbf{E}_{0} e^{-t / \tau}, & \text { for } t>0\end{array} \quad\left(\mathbf{E}_{0} \text { in the positive } z \text { -direction }\right)\right.\) Using first-order time-dependent perturbation theory, compute the probability for the atom to be found at \(t \gg \tau\) in each of the three \(2 p\) states: \((n, l, m)=(2,1,\pm 1 \text { or } 0)\) Repeat the problem for the \(2 s\) state: \((n, l, m)=(2,0,0) .\) You need not attempt to evaluate radial integrals, but perform all other integrations (with respect to angles and time).

A particle of mass \(m\) constrained to move in one dimension is confined within \(0 < x < L\) by an infinite-wall potential \\[ \begin{array}{ll} V=\infty & \text { for } x < 0, x>L \\ V=0 & \text { for } 0 \leq x \leq L \end{array} \\] Obtain an expression for the density of states (that is, the number of states per unit energy interval) for high energies as a function of \(E .\) (Check your dimension!)
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