The correct statement(s) about spherical harmonics $(Y_l^m)$ is(are)

Question

An electron is confined to move in a one-dimensional box of length 1˚ A. Its energy in the first excited state is approximately

Answer 1

Spherical harmonics $Y_l^m$ are the standard angular functions used for anything with a central potential, from the rigid rotor to the hydrogen atom. Each statement here tests a core property of these functions.

All complex (A): The $\phi$-dependence of $Y_l^m$ sits in the factor $e^{im\phi}$. For $m=0$ this factor is just $1$, and the harmonic collapses to a real Legendre-polynomial function of $\cos\theta$ alone, for instance $Y_0^0$ and $Y_1^0$ are both real. So the word "all" makes this statement false.
Eigenfunctions of $\hat L^2$ (B): The defining equation of the spherical harmonics is $\hat L^2 Y_l^m = l(l+1)\hbar^2 Y_l^m$; this is how $l$ is defined as a quantum number in the first place. True.
Eigenfunctions of $\hat L_z$ (C): They also satisfy $\hat L_z Y_l^m = m\hbar Y_l^m$, which is how the second quantum number $m$ is defined. $\hat L^2$ and $\hat L_z$ commute, so sharing simultaneous eigenfunctions is expected. True.
Degeneracy of $Y_1^1$ and $Y_1^{-1}$ (D): For any spherically symmetric potential, the energy does not depend on $m$ at all, only on $l$. $Y_1^1$ and $Y_1^{-1}$ both have $l=1$, differing only in $m$, so they carry the same energy unless an external field like a magnetic field breaks the spherical symmetry. True.

Three of the four statements survive the check, with only the "all" in option A being too strong a claim.

Let's summarize:

$Y_l^m$ is real only when $m=0$; otherwise it is complex because of the $e^{im\phi}$ factor.
$\hat L^2$ and $\hat L_z$ both act on $Y_l^m$ as eigenoperators, giving eigenvalues $l(l+1)\hbar^2$ and $m\hbar$.
Energy in a central potential depends only on $l$, so different $m$ states at the same $l$, like $Y_1^1$ and $Y_1^{-1}$, are degenerate.

The correct statements are B, C and D.

Answer 2

Spherical harmonics $Y_l^m$ are the standard angular functions used for anything with a central potential, from the rigid rotor to the hydrogen atom. Each statement here tests a core property of these functions.

All complex (A): The $\phi$-dependence of $Y_l^m$ sits in the factor $e^{im\phi}$. For $m=0$ this factor is just $1$, and the harmonic collapses to a real Legendre-polynomial function of $\cos\theta$ alone, for instance $Y_0^0$ and $Y_1^0$ are both real. So the word "all" makes this statement false.
Eigenfunctions of $\hat L^2$ (B): The defining equation of the spherical harmonics is $\hat L^2 Y_l^m = l(l+1)\hbar^2 Y_l^m$; this is how $l$ is defined as a quantum number in the first place. True.
Eigenfunctions of $\hat L_z$ (C): They also satisfy $\hat L_z Y_l^m = m\hbar Y_l^m$, which is how the second quantum number $m$ is defined. $\hat L^2$ and $\hat L_z$ commute, so sharing simultaneous eigenfunctions is expected. True.
Degeneracy of $Y_1^1$ and $Y_1^{-1}$ (D): For any spherically symmetric potential, the energy does not depend on $m$ at all, only on $l$. $Y_1^1$ and $Y_1^{-1}$ both have $l=1$, differing only in $m$, so they carry the same energy unless an external field like a magnetic field breaks the spherical symmetry. True.

Three of the four statements survive the check, with only the "all" in option A being too strong a claim.

Let's summarize:

$Y_l^m$ is real only when $m=0$; otherwise it is complex because of the $e^{im\phi}$ factor.
$\hat L^2$ and $\hat L_z$ both act on $Y_l^m$ as eigenoperators, giving eigenvalues $l(l+1)\hbar^2$ and $m\hbar$.
Energy in a central potential depends only on $l$, so different $m$ states at the same $l$, like $Y_1^1$ and $Y_1^{-1}$, are degenerate.

The correct statements are B, C and D.

The correct statement(s) about spherical harmonics \((Y_l^m)\) is(are)

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The Correct Option is B, C, D

Solution and Explanation

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