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:
$Y_0^0 = \sqrt{\frac{1}{4\pi}}$, $Y_1^0 = \sqrt{\frac{3}{4\pi}} \cos\theta$, $Y_1^{\pm 1} = \mp \sqrt{\frac{3}{8\pi}} \sin\theta e^{\pm i\phi}$
Quantum Mechanics Wavefunction Analysis
The given wavefunction is $\psi(x,y,z) = N ze^{-\alpha(x^2+y^2+z^2)}$. We need to find the eigenvalues of the angular momentum operators $L^2$ and $L_z$. These operators act on the angular part of the wavefunction in spherical coordinates.
Convert to Spherical Coordinates
First, convert the wavefunction to spherical coordinates $(r, \theta, \phi)$. We know that $z = r \cos\theta$ and $x^2+y^2+z^2 = r^2$. Substituting these, the wavefunction becomes:
$\psi(r, \theta, \phi) = N (r \cos\theta) e^{-\alpha r^2}$
We can separate this into a radial part $R(r)$ and an angular part $Y(\theta, \phi)$: $R(r) = N r e^{-\alpha r^2}$ $Y(\theta, \phi) = \cos\theta$
Identify Angular Momentum Quantum Numbers
The eigenvalues of $L^2$ and $L_z$ depend on the angular momentum quantum numbers $l$ and $m$, respectively, which characterize the angular part of the wavefunction (Spherical Harmonics, $Y_l^m$).
The eigenvalue equation for $L_z$ is $L_z Y_l^m = m\hbar Y_l^m$. The eigenvalue equation for $L^2$ is $L^2 Y_l^m = l(l+1)\hbar^2 Y_l^m$.
We are given some Spherical Harmonics:
- $Y_0^0 = \sqrt{\frac{1}{4\pi}}$
- $Y_1^0 = \sqrt{\frac{3}{4\pi}} \cos\theta$
- $Y_1^{\pm 1} = \mp \sqrt{\frac{3}{8\pi}} \sin\theta e^{\pm i\phi}$
Our angular part is $Y(\theta, \phi) = \cos\theta$. We can express $\cos\theta$ in terms of the given $Y_1^0$:
$\cos\theta = \sqrt{\frac{4\pi}{3}} Y_1^0(\theta, \phi)$
Thus, the angular part of our wavefunction is proportional to $Y_1^0$. This means the quantum numbers are $l=1$ and $m=0$.
Calculate Eigenvalues
Now, we calculate the eigenvalues using $l=1$ and $m=0$:
- $L^2$ Eigenvalue: $l(l+1)\hbar^2 = 1(1+1)\hbar^2 = 1(2)\hbar^2 = 2\hbar^2$.
- $L_z$ Eigenvalue: $m\hbar = 0 \cdot \hbar = 0$.
The eigenvalues for $L^2$ and $L_z$ are $2\hbar^2$ and $0$, respectively.
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?
