A hydrogen atom is in the state $\Psi = \sqrt{\frac{8}{21}} \Psi_{200} - \sqrt{\frac{3}{7}} \Psi_{310} + \sqrt{\frac{4}{21}} \Psi_{321}$, where $n, l, m$ in $\Psi_{nlm}$ denote the principal, orbital and magnetic quantum numbers, respectively. If $\hat{L}$ is the angular momentum operator, the average value of $\hat{L}^2$ is ________ $\hbar^2$.
Hydrogen Atom Average $\hat{L}^2$ Value
The task is to find the average value of the squared angular momentum operator, $\hat{L}^2$, for a hydrogen atom in the given quantum state:
$\Psi = \sqrt{\frac{8}{21}} \Psi_{200} - \sqrt{\frac{3}{7}} \Psi_{310} + \sqrt{\frac{4}{21}} \Psi_{321}$
Here, $\Psi_{nlm}$ represents the stationary states, and $n, l, m$ are the principal, orbital, and magnetic quantum numbers.
The operator $\hat{L}^2$ acts on $\Psi_{nlm}$ as $\hat{L}^2 \Psi_{nlm} = l(l+1)\hbar^2 \Psi_{nlm}$. The average value (expectation value) of $\hat{L}^2$ for a superposition state $\Psi = \sum_i c_i \Psi_i$ is $\langle \hat{L}^2 \rangle = \sum_i |c_i|^2 \lambda_i$, where $\lambda_i$ are the eigenvalues $l(l+1)\hbar^2$.
State Quantum Numbers and Eigenvalues
We examine each component of the superposition state:
- For $\Psi_{200}$: $n=2, l=0, m=0$. The eigenvalue of $\hat{L}^2$ is $l(l+1)\hbar^2 = 0(0+1)\hbar^2 = 0$.
- For $\Psi_{310}$: $n=3, l=1, m=0$. The eigenvalue of $\hat{L}^2$ is $l(l+1)\hbar^2 = 1(1+1)\hbar^2 = 2\hbar^2$.
- For $\Psi_{321}$: $n=3, l=2, m=1$. The eigenvalue of $\hat{L}^2$ is $l(l+1)\hbar^2 = 2(2+1)\hbar^2 = 6\hbar^2$.
Calculating the Average $\hat{L}^2$
The coefficients are $c_{200} = \sqrt{\frac{8}{21}}$, $c_{310} = -\sqrt{\frac{3}{7}}$, $c_{321} = \sqrt{\frac{4}{21}}$. The squares of these coefficients are:
- $|c_{200}|^2 = \frac{8}{21}$
- $|c_{310}|^2 = \frac{3}{7}$
- $|c_{321}|^2 = \frac{4}{21}$
Now, we compute the average value $\langle \hat{L}^2 \rangle$:
$\langle \hat{L}^2 \rangle = |c_{200}|^2 \cdot (0) + |c_{310}|^2 \cdot (2\hbar^2) + |c_{321}|^2 \cdot (6\hbar^2)$
Substitute the squared coefficients and eigenvalues:
$\langle \hat{L}^2 \rangle = \left(\frac{8}{21}\right) \cdot 0 + \left(\frac{3}{7}\right) \cdot 2\hbar^2 + \left(\frac{4}{21}\right) \cdot 6\hbar^2$
Simplify the expression:
$\langle \hat{L}^2 \rangle = 0 + \frac{6}{7}\hbar^2 + \frac{24}{21}\hbar^2$
Combine the terms, noting $\frac{24}{21} = \frac{8}{7}$:
$\langle \hat{L}^2 \rangle = \frac{6}{7}\hbar^2 + \frac{8}{7}\hbar^2$
$\langle \hat{L}^2 \rangle = \frac{6 + 8}{7}\hbar^2$
$\langle \hat{L}^2 \rangle = \frac{14}{7}\hbar^2$
$\langle \hat{L}^2 \rangle = 2\hbar^2$
Conclusion
The calculated average value of $\hat{L}^2$ is $2\hbar^2$. This result falls within the specified range of 1.99 to 2.01.
Important Questions from Angular Momentum Operators Eigenvalues Clebsch Gordan
- Consider two particles with angular momenta $j_1 = 2\hbar$ and $j_2 = \hbar/2$. If the expression
$$|j = 5/2, m = 3/2\rangle = \begin{cases} c_1|j_1 = 2, m_1 = 1\rangle|j_2 = 1/2, m_2 = 1/2\rangle + \\ c_2|j_1 = 2, m_1 = 2\rangle|j_2 = 1/2, m_2 = -1/2\rangle \end{cases}$$
gives an eigenstate of the total angular momentum of the two particles, using standard notation. Which of the following is true?
(Hint: $\hat{J}_{\pm}|j, m\rangle = \sqrt{j(j + 1) - m(m \pm 1)} |j, m \pm 1\rangle$) - A system of three non-identical spin $\frac{1}{2}$ particles has the Hamiltonian $H = \frac{A}{\hbar^2} (\vec{S}_1 + \vec{S}_2) \cdot \vec{S}_3$, where $\vec{S}_1, \vec{S}_2$ and $\vec{S}_3$ are the spin operators of particles labelled $1,2$ and $3$ respectively and $A$ is a constant with appropriate dimensions. The set of possible energy eigenvalues of the system is
- $H$ is the Hamiltonian, $\vec{L}$ the orbital angular momentum and $L_z$ is the $z$-component of $\vec{L}$. The $1s$ state of the hydrogen atom in the non-relativistic formalism is an eigen function of which one of the following sets of operators?
- An atom with non-zero magnetic moment has an angular momentum of magnitude $\sqrt{12}\hbar$. When a beam of such atoms is passed through a Stern-Gerlach apparatus, how many beams does it split into?
- In the vector model of angular momentum applied to atoms, what is the minimum angle in degrees (in integer) made by the orbital angular momentum vector and the positive $z$ axis for a $2p$ electron?
