An electron in the Coulomb field of a proton is in the following state of coherent superposition of orthonormal states $\psi_{nlm}$
$\Psi = \frac{1}{3}\psi_{100} + \frac{1}{\sqrt{3}}\psi_{210} - \frac{\sqrt{5}}{3}\psi_{320}$
Let $E_1, E_2$, and $E_3$ represent the first three energy levels of the system. A sequence of measurements is done on the same system at different times. Energy is measured first at time $t_1$ and the outcome is $E_2$. Then total angular momentum is measured at time $t_2 > t_1$ and finally energy is measured again at $t_3 > t_2$. The probability of finding the system in a state with energy $E_2$ after the final measurement is $P/9$. The value of $P$ is ______________ (in integer).
The initial quantum state of the electron is given by the superposition:
$ \Psi = \frac{1}{3}\psi_{100} + \frac{1}{\sqrt{3}}\psi_{210} - \frac{\sqrt{5}}{3}\psi_{320} $
Here, $E_1, E_2, E_3$ represent the energy levels corresponding to principal quantum numbers $n=1, 2, 3$. The states $\psi_{nlm}$ are orthonormal energy eigenstates.
- $\psi_{100}$ corresponds to $n=1$ (Energy $E_1$).
- $\psi_{210}$ corresponds to $n=2$ (Energy $E_2$).
- $\psi_{320}$ corresponds to $n=3$ (Energy $E_3$).
Energy Measurement Outcome E2 at t1
At time $t_1$, energy is measured, yielding $E_2$. In quantum mechanics, this measurement causes the wave function to collapse into the eigenstate(s) associated with the measured eigenvalue. Here, the measured energy is $E_2$, which corresponds to the state $\psi_{210}$ ($n=2$).
After this measurement, the system's state collapses to:
$ \psi'_{1} = \psi_{210} $
Angular Momentum Measurement on Collapsed State at t2
At time $t_2 > t_1$, total angular momentum ($L^2$) is measured. The system is in the state $\psi_{210}$, which has quantum numbers $n=2, l=1, m=0$.
The operator for total angular momentum squared is $L^2$, with eigenvalues $l(l+1)\hbar^2$. For $l=1$, the eigenvalue is $1(1+1)\hbar^2 = 2\hbar^2$.
Since $\psi_{210}$ is an eigenstate of $L^2$ (with $l=1$), the measurement of $L^2$ will yield $2\hbar^2$ with certainty. This measurement does not alter the state of the system.
Therefore, the state remains:
$ \psi'_{2} = \psi_{210} $
Energy Measurement Probability at t3
At time $t_3 > t_2$, energy is measured again. The system is in state $\psi'_{2} = \psi_{210}$. This state corresponds to $n=2$, meaning its energy is precisely $E_2$.
Consequently, measuring the energy of the system in the state $\psi_{210}$ is guaranteed to yield $E_2$. The probability of finding the system with energy $E_2$ is 1.
Probability Calculation and P Value
The problem states this final probability is $P/9$. We have determined the probability to be 1.
Equating these values:
$ 1 = \frac{P}{9} $
Solving for $P$ yields:
$ P = 9 $
Important Questions from Angular Momentum Operators Eigenvalues Clebsch Gordan
- $H$ 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?
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:
$Y_0^0 = \sqrt{\frac{1}{4\pi}}$, $Y_1^0 = \sqrt{\frac{3}{4\pi}} \cos\theta$, $Y_1^{\pm 1} = \mp \sqrt{\frac{3}{8\pi}} \sin\theta e^{\pm i\phi}$- The spin $ \vec{S}$ and orbital angular momentum $ \vec{L}$ of an atom precess about $ \vec{J}$, the total angular momentum. $ \vec{J}$ precesses about an axis fixed by a magnetic field $ \vec{B}_1 = 2B_0 \hat{z}$, where $B_0$ is a constant. Now the magnetic field is changed to $ \vec{B}_2 = B_0( \hat{x} + \sqrt{2} \hat{y} + \hat{z})$. Given the orbital angular momentum quantum number $l = 2$ and spin quantum number $s = 1/2$, $ \theta$ is the angle between $ \vec{B}_1$ and $ \vec{J}$ for the largest possible values of total angular quantum number $j$ and its $z$-component $j_z$. The value of $ \theta$ (in degree, rounded off to the nearest integer) is ________
