The question asks to identify the set of operators for which the 1s state of the hydrogen atom is a simultaneous eigenfunction in the non-relativistic formalism.
Identifying Key Quantum Numbers
The 1s state corresponds to the following quantum numbers:
- Principal quantum number: $n=1$
- Azimuthal (orbital) quantum number: $l=0$
- Magnetic quantum number: $m_l=0$
Eigenfunctions and Commuting Operators
A quantum state can be a simultaneous eigenfunction of multiple operators only if those operators commute. The operators relevant here are the Hamiltonian ($H$), the square of the orbital angular momentum ($L^2$), the z-component of the orbital angular momentum ($L_z$), and the orbital angular momentum vector ($\vec{L}$).
- Hamiltonian ($H$): The 1s state is the ground state and thus an eigenfunction of $H$ with the lowest energy eigenvalue $E_1$.
- Angular Momentum Squared ($L^2$): The eigenvalue of $L^2$ is given by $l(l+1)\hbar^2$. For the 1s state, $l=0$, so the eigenvalue is $0(0+1)\hbar^2 = 0$. Therefore, the 1s state is an eigenfunction of $L^2$.
- Angular Momentum z-component ($L_z$): The eigenvalue of $L_z$ is given by $m_l\hbar$. For the 1s state, $m_l=0$, so the eigenvalue is $0\hbar = 0$. Therefore, the 1s state is an eigenfunction of $L_z$.
In the context of the hydrogen atom's central potential, $H$, $L^2$, and $L_z$ all commute with each other ($[H, L^2] = 0$, $[H, L_z] = 0$, $[L^2, L_z] = 0$). This allows for simultaneous eigenfunctions.
Operator $\vec{L}$ Consideration
The vector operator $\vec{L} = (L_x, L_y, L_z)$ does not commute with $L_z$ (e.g., $[L_x, L_z] = i\hbar L_y \neq 0$). Therefore, a state cannot be a simultaneous eigenfunction of $\vec{L}$ and $L_z$. While the 1s state has $l=0$, meaning $\vec{L}$ acting on it results in the zero vector ($0$), the operator $\vec{L}$ itself cannot be included in a set of operators that are simultaneously diagonal with $L_z$. The standard basis states are typically defined by eigenvalues of $H$, $L^2$, and $L_z$.
Conclusion
Based on the quantum numbers ($n=1, l=0, m_l=0$) and the commutation relations, the 1s state of the hydrogen atom is a simultaneous eigenfunction of $H$, $L^2$, and $L_z$. This corresponds to Option 1.
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
- 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?
A particle has wavefunction
$\psi(x,y,z) = N ze^{-\alpha(x^2+y^2+z^2)}$,
where $N$ is a normalization constant and $\alpha$ is a positive constant. In this state, which one of the following options represents the eigenvalues of $L^2$ and $L_z$ respectively?
Some values of $Y_l^m$ are:
$Y_0^0 = \sqrt{\frac{1}{4\pi}}$, $Y_1^0 = \sqrt{\frac{3}{4\pi}} \cos\theta$, $Y_1^{\pm 1} = \mp \sqrt{\frac{3}{8\pi}} \sin\theta e^{\pm i\phi}$
