Question 705028

SCQMEDIUM

A quantum mechanical system is in the angular momentum state l=4,lz=4|l=4, l_z=4\rangle. The uncertainty in LxL_x is

(A)

2\hbar\sqrt{2}

(B)

22\hbar

(C)

0

(D)

\hbar

Detailed Solution

The quantum mechanical system is given to be in the angular momentum state ψ=l=4,m=4|\psi\rangle = |l=4, m=4\rangle. We need to find the uncertainty in the x-component of angular momentum (LxL_x), which is given by the standard formula: ΔLx=Lx2Lx2\Delta L_x = \sqrt{\langle L_x^2 \rangle - \langle L_x \rangle^2}

Step 1: Find the expectation value Lx\langle L_x \rangle For any eigenstate of LzL_z (like the given state l,m|l, m\rangle), the expectation values of the transverse components LxL_x and LyL_y are always zero. Lx=0\langle L_x \rangle = 0

Step 2: Find the expectation value Lx2\langle L_x^2 \rangle We know the relation for the total angular momentum squared operator: L2=Lx2+Ly2+Lz2L^2 = L_x^2 + L_y^2 + L_z^2

Taking the expectation value of this operator for the state l,m|l, m\rangle: L2=Lx2+Ly2+Lz2\langle L^2 \rangle = \langle L_x^2 \rangle + \langle L_y^2 \rangle + \langle L_z^2 \rangle

Due to the symmetry of the LxL_x and LyL_y operators in an LzL_z eigenstate, their expectation values squared are equal: Lx2=Ly2\langle L_x^2 \rangle = \langle L_y^2 \rangle

Substituting this into the expectation value equation: L2=2Lx2+Lz2\langle L^2 \rangle = 2\langle L_x^2 \rangle + \langle L_z^2 \rangle 2Lx2=L2Lz22\langle L_x^2 \rangle = \langle L^2 \rangle - \langle L_z^2 \rangle Lx2=L2Lz22\langle L_x^2 \rangle = \frac{\langle L^2 \rangle - \langle L_z^2 \rangle}{2}

We know the standard eigenvalue results for these operators: L2=l(l+1)2\langle L^2 \rangle = l(l+1)\hbar^2 Lz2=m22\langle L_z^2 \rangle = m^2\hbar^2

For the given state l=4l=4 and m=4m=4: L2=4(4+1)2=202\langle L^2 \rangle = 4(4+1)\hbar^2 = 20\hbar^2 Lz2=(4)22=162\langle L_z^2 \rangle = (4)^2\hbar^2 = 16\hbar^2

Now, substitute these values to calculate Lx2\langle L_x^2 \rangle: Lx2=2021622=422=22\langle L_x^2 \rangle = \frac{20\hbar^2 - 16\hbar^2}{2} = \frac{4\hbar^2}{2} = 2\hbar^2

Step 3: Calculate the Final Uncertainty ΔLx\Delta L_x Substitute the values back into the uncertainty formula: ΔLx=Lx2Lx2\Delta L_x = \sqrt{\langle L_x^2 \rangle - \langle L_x \rangle^2} ΔLx=220\Delta L_x = \sqrt{2\hbar^2 - 0} ΔLx=2\Delta L_x = \sqrt{2}\hbar

Final Answer: The uncertainty in LxL_x is 2\hbar\sqrt{2}. This corresponds to Option (A).

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