Electromagnetic Pulse Momentum Uncertainty
This problem requires calculating the uncertainty in a photon's momentum ($\Delta p$) based on the electromagnetic pulse's width ($\Delta t$), applying quantum mechanical principles.
Heisenberg Uncertainty Principle for Energy-Time
Heisenberg's Uncertainty Principle connects the uncertainty in energy ($\Delta E$) and time ($\Delta t$) for a system:
$ \Delta E \Delta t \ge \frac{\hbar}{2} $
For a photon, energy ($E$) is related to momentum ($p$) and the speed of light ($c$) by $E = pc$. Therefore, the uncertainty in energy is $\Delta E = c \Delta p$. The reduced Planck constant is $\hbar = \frac{h}{2\pi}$.
Substituting these into the principle:
$ (c \Delta p) \Delta t \ge \frac{1}{2} \left( \frac{h}{2\pi} \right) $
$ c \Delta p \Delta t \ge \frac{h}{4\pi} $
Rearranging to solve for the uncertainty in momentum ($\Delta p$):
$ \Delta p \ge \frac{h}{4\pi c \Delta t} $
Calculating Photon Momentum Uncertainty
Given values are:
- Pulse width: $\Delta t = 10^{-3}$ s
- Planck's constant: $h = 6.6 \times 10^{-34}$ J s
- Speed of light: $c = 3 \times 10^8$ m s$^{-1}$
Using the approximation for the order of magnitude:
$ \Delta p \approx \frac{h}{4\pi c \Delta t} $
Plugging in the values:
$ \Delta p \approx \frac{6.6 \times 10^{-34} \text{ J s}}{4\pi \times (3 \times 10^8 \text{ m s}^{-1}) \times (10^{-3} \text{ s})} $
$ \Delta p \approx \frac{6.6 \times 10^{-34}}{12\pi \times 10^5} \text{ kg m s}^{-1} $
Calculating the numerical coefficient (using $\pi \approx 3.14159$):
$ \Delta p \approx \frac{6.6}{37.699} \times 10^{-39} \text{ kg m s}^{-1} $
$ \Delta p \approx 0.175 \times 10^{-39} \text{ kg m s}^{-1} $
$ \Delta p \approx 1.75 \times 10^{-40} \text{ kg m s}^{-1} $
Determining the Exponent N
The uncertainty in momentum is given to be of the order of $10^{-N}$ kg m $s^{-1}$. We set our calculated value equal to this form:
$ 1.75 \times 10^{-40} \approx 10^{-N} $
To find $N$, we take the base-10 logarithm:
$ \log_{10}(1.75 \times 10^{-40}) \approx \log_{10}(10^{-N}) $
$ \log_{10}(1.75) + \log_{10}(10^{-40}) \approx -N $
$ \log_{10}(1.75) - 40 \approx -N $
Using $\log_{10}(1.75) \approx 0.243$:
$ 0.243 - 40 \approx -N $
$ -39.757 \approx -N $
$ N \approx 39.757 $
The calculated value of $N$ is approximately 39.757, which falls within the range of 39 to 40.
Important Questions from Operators Commutators Heisenberg Picture
The wavefunction of a particle in one dimension is given by
$\psi(x) = \begin{cases} M, & -a < x < a \\ 0, & \text{otherwise.} \end{cases}$
Here $M$ and $a$ are positive constants. If $\phi(p)$ is the corresponding momentum space wavefunction, which one of the following plots best represents $|\phi(p)|^2$ ?- From the pairs of operators given below, identify the ones which commute. Here $l$ and $j$ correspond to the orbital angular momentum and the total angular momentum, respectively.
- In cylindrical coordinates $(s, \varphi, z)$, which of the following is a Hermitian operator?
- Let $|m\rangle$ and $|n\rangle$ denote the energy eigenstates of a one-dimensional simple harmonic oscillator. The position and momentum operators are $\hat{X}$ and $\hat{P}$, respectively. The matrix element $\langle m|\hat{P}\hat{X}|n\rangle$ is non-zero when
Let $|\psi_1\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$, $|\psi_2\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ represent two possible states of a two-level quantum system. The state obtained by the incoherent superposition of $|\psi_1\rangle$ and $|\psi_2\rangle$ is given by a density matrix that is defined as $\rho ≡ c_1|\psi_1\rangle\langle\psi_1| + c_2|\psi_2\rangle\langle\psi_2|$. If $c_1 = 0.4$ and $c_2 = 0.6$, the matrix element $\rho_{22}$ (rounded off to one decimal place) is ________
