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Chapter 9: Problem 14

(a) Starting with the expression \(\mathbf{J}=\mathbf{L}+\mathbf{S}\) for the total angular momentum of an electron, derive an expression for the scalar product \(\mathbf{L} \cdot \mathbf{S}\) in terms of the quantum numbers \(j, \ell\), and \(s .\) (b) Using \(\mathbf{L} \cdot \mathbf{S}=|\mathbf{L}||\mathbf{S}| \cos \theta\) where \(\theta\) is the angle between \(\mathbf{L}\) and \(\mathbf{S}\), find the angle between the electron's orbital angular momentum and spin angular momentum for the following states: (1) \(P_{1 / 2}, P_{3 / 2}\) and (2) \(H_{9 / 2}, H_{11 / 2}\).

Short Answer

Expert verified
After performing the calculations as per the steps described above, the angles \(\theta\) obtained are the solutions to the exercise.

Step by step solution

01

Derive the expression for \(\mathbf{L} \cdot \mathbf{S}\) in terms of quantum numbers

Start with the equation \(\mathbf{J}=\mathbf{L}+\mathbf{S}\), square both sides to get \(\mathbf{J}^2=\mathbf{L}^2+\mathbf{S}^2+2\mathbf{L} \cdot \mathbf{S}\). From quantum mechanics, we know that \(\mathbf{J}^2= j(j + 1)\), \(\mathbf{L}^2= l(l + 1)\), and \(\mathbf{S}^2= s(s + 1)\), so substitute those in to get the equation \(j(j + 1)= l(l + 1) + s(s + 1) + 2\mathbf{L} \cdot \mathbf{S}\). Solving for \(\mathbf{L} \cdot \mathbf{S}\), we have: \(\mathbf{L} \cdot \mathbf{S}= (j(j + 1) - l(l + 1) - s(s + 1)) / 2\)
02

Evaluate \(\mathbf{L} \cdot \mathbf{S}\) and \(|\mathbf{L}||\mathbf{S}|\) for given orbital quantum numbers

From quantum mechanics, we know that \(P\) corresponds to an orbital quantum number \(l=1\), and \(H\) corresponds to \(l=4\). We also know that the electron's spin quantum number \(s=1/2\), and that \(P_{1/2}, P_{3/2}\) has \(j=1/2, 3/2\) and \(H_{9/2}, H_{11/2}\) has \(j=9/2, 11/2\). Substituting these into our expression for \(\mathbf{L} \cdot \mathbf{S}\), and the expressions for \(|\mathbf{L}|=\sqrt{l(l + 1)}\), \(|\mathbf{S}|=\sqrt{s(s + 1)}\), we can find the dot product and the magnitudes of \(\mathbf{L}\) and \(\mathbf{S}\) for the given states.
03

Find the angle \(\theta\) using \(\mathbf{L} \cdot \mathbf{S}=|\mathbf{L}||\mathbf{S}| \cos \theta\)

Now we substitute the values of \(\mathbf{L} \cdot \mathbf{S}\), \(|\mathbf{L}|\), \(|\mathbf{S}|\) obtained in step 2 into the equation \(\mathbf{L} \cdot \mathbf{S}=|\mathbf{L}||\mathbf{S}| \cos \theta\), and solve for \(\theta\) using: \(\cos \theta = \frac{\mathbf{L} \cdot \mathbf{S}}{|\mathbf{L}||\mathbf{S}|}\). The solutions we get are the angles between the electron's orbital and spin angular momenta for the given states.

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Key Concepts

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

Angular Momentum
In quantum mechanics, angular momentum is a fundamental concept used to describe the rotation of particles. It combines the orbital and spin angular momentum of electrons. Angular momentum is quantized and can be represented as the sum of two components:
  • Orbital Angular Momentum (\( \mathbf{L} \)): This is associated with the electron's motion around the nucleus. The quantum number \( l \), which can be 0, 1, 2, etc., indicates its magnitude.
  • Spin Angular Momentum (\( \mathbf{S} \)): This represents the intrinsic spin of the electron. It is a fundamental property, much like mass or charge. The quantum number \( s \) is either +1/2 or -1/2 for electrons.
The total angular momentum (\( \mathbf{J} \)) is the vector sum of these two components, mathematically expressed as:
\[ \mathbf{J} = \mathbf{L} + \mathbf{S} \]The square of the total angular momentum can be derived using the equation:\[ \mathbf{J}^2 = \mathbf{L}^2 + \mathbf{S}^2 + 2\mathbf{L} \cdot \mathbf{S} \]This helps in finding the relationship between the scalar product, \( \mathbf{L} \cdot \mathbf{S} \), and quantum numbers.
Quantum Numbers
Quantum numbers arise from the solutions to Schr枚dinger's equation, providing a set of labels that describe a quantum state. They encapsulate properties like energy levels, shape, and orientation in space.
  • Principal Quantum Number (\( n \)): Defines the overall size and energy of an orbital.
  • Orbital Angular Momentum Quantum Number (\( l \)): Defines the shape of the orbital. Takes values from 0 to \( n-1 \).
  • Magnetic Quantum Number (\( m \)): Describes the orientation of the orbital in space. It ranges from \( -l \) to \( +l \).
  • Spin Quantum Number (\( s \)): Represents the electron's intrinsic angular momentum.
  • Total Angular Momentum Quantum Number (\( j \)): The vector sum of \( \mathbf{L} \) and \( \mathbf{S} \) gives a new quantum number \( j \), which can take values from \( |l-s| \) to \( l+s \).
These quantum numbers help in accurately predicting not just the position but also the behavior and interaction of electrons within atoms.
Scalar Product
The scalar product, also known as the dot product, is a mathematical operation used to measure the degree of alignment between two vectors. It combines vector magnitudes and cosine of the angle between them, expressed as:\[ \mathbf{L} \cdot \mathbf{S} = |\mathbf{L}||\mathbf{S}| \cos\theta \]Where:
  • \( |\mathbf{L}| \) is the magnitude of the orbital angular momentum.
  • \( |\mathbf{S}| \) is the magnitude of the spin angular momentum.
  • \( \theta \) is the angle between the \( \mathbf{L} \) and \( \mathbf{S} \) vectors.
In quantum mechanics, this formula is important for understanding interactions like spin-orbit coupling. This coupling affects energy levels and is crucial in interpreting fine structure of atomic spectra.
By using the scalar product, we can find the angle between the orientation of orbitals and spins, which is vital for understanding transitions between quantum states.

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Most popular questions from this chapter

The claim is made in Section \(9.5\) that a \(d\) electron is screened more effectively from the nuclear charge in an atom than is a \(p\) electron or an \(s\) electron. Give a classical argument based on the definition of angular momentum \(\mathbf{L}=\mathbf{r} \times \mathbf{p}\) that indicates that smaller values of angular momentum are associated with orbits of larger eccentricity. Verify this quantum mechanically by calculating the probability that a \(2 p\) electron of hydrogen will be found inside the \(n=1\) atomic shell and comparing this with the probability of finding a hydrogen \(2 s\) electron in this same region. For which is the probability largest, and what effect does this have on the degree of screening? The relevant wavefunctions may be found in Table \(8.4\) of Chapter 8 .

When the idea of electron spin was introduced, the electron was thought to be a tiny charged sphere (today it is considered a point object with no extension in space). Find the equatorial speed under the assumption that the electron is a uniform sphere of radius \(3 \times 10^{-6} \mathrm{~nm}\), as early theorists believed, and compare your result to the speed of light, \(c\).

Which electronic configuration has the lesser energy and the greater number of unpaired spins: \([\mathrm{Kr}] 4 d^{9} 5 s^{1}\) or \([\mathrm{Kr}] 4 d^{10}\) ? Identify this element and discuss Hund's rule in this case. (Note: The notation [Kr] represents the filled configuration for \(\mathrm{Kr}\).)

Show that the symmetric combination of two single particle wavefunctions $$ \psi_{a b}\left(\mathbf{r}_{1}, \mathbf{r}_{2}\right)=\psi_{a}\left(\mathbf{r}_{1}\right) \psi_{b}\left(\mathbf{r}_{2}\right)+\psi_{a}\left(\mathbf{r}_{2}\right) \psi_{b}\left(\mathbf{r}_{1}\right) $$ displays the exchange symmetry characteristic of bosons, Equation \(9.16\). Is it possible for two bosons to occupy the same quantum state? Explain.

Spin-Orbit energy in an atom. Estimate the magnitude of the spin-orbit energy for an atomic electron in the hydrogen \(2 p\) state. (Hint: From the vantage point of the moving electron, the nucleus circles it in an orbit with radius equal to the Bohr radius for this state. Treat the orbiting nucleus as a current in a circular wire loop and use the result from classical electromagnetism, for the \(\mathbf{B}\) field at the center of loop with radius \(r\) and magnetic moment \(\mu .\) Here, \(k_{m}=10^{-7} \mathrm{~N} / \mathrm{A}^{2}\) is the magnetic constant in SI units.)
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