How to prove the Hermiticity of the angular momentum operator's X-component? [closed]

Question

I understand that the momentum operator is Hermitian (thanks to this proof), as demonstrated by verifying the inner product relation. However, I am unsure how to prove that the angular momentum operator, specifically its $L_x$ component, is Hermitian and thus observable.

The $L_x$ operator can be expressed as:
$$ L_x = y p_z - z p_y, $$
where $p_z = -i\hbar \frac{\partial}{\partial z}$ and $p_y = -i\hbar \frac{\partial}{\partial y}$ are the momentum operators in the $z$- and $y$-directions, respectively.

Alternatively, in spherical coordinates, $L_x$ is expressed as:
$$ L_x = -i\hbar \left( -\sin\phi \frac{\partial}{\partial \theta} - \cos\phi \cot\theta \frac{\partial}{\partial \phi} \right). $$

To prove $L_x$ is Hermitian, I believe we need to check whether it satisfies the Hermitian condition:
$$ \langle \psi_1 | L_x \psi_2 \rangle = \langle L_x \psi_1 | \psi_2 \rangle^*, $$
for arbitrary wavefunctions $\psi_1$ and $\psi_2$ in a suitable Hilbert space.

Attempting the proof:

1. I substitute $L_x$ into the inner product expression:
   $$ \langle \psi_1 | L_x \psi_2 \rangle = \int \psi_1^* \, (y p_z - z p_y) \psi_2 \, dV. $$

2. Compute the $y p_z$ and $z p_y$ terms explicitly:
   $$ \langle \psi_1 | y p_z \psi_2 \rangle = \int \psi_1^* y \left( -i\hbar \frac{\partial}{\partial z} \psi_2 \right) dV, $$
   $$ \langle \psi_1 | -z p_y \psi_2 \rangle = \int \psi_1^* \left( -z \left( -i\hbar \frac{\partial}{\partial y} \psi_2 \right) \right) dV. $$

3. Use integration by parts to see if the boundary terms vanish for well-behaved wavefunctions.

My questions:

• Do both terms satisfy the Hermitian condition?
• Is this enough to confirm that $L_x$ is Hermitian and thus observable?
• Are there specific pitfalls in extending this analysis to spherical coordinates?

Could someone confirm if my approach is correct and point out any errors or assumptions I may have overlooked?

Answer 1

How to Prove the Hermiticity of the Angular Momentum Operator's X-Component?

I understand that the momentum operator is Hermitian...

I assume you also understand that the position operator is Hermitian.

I assume you also understand that the position and momentum obey canonical commutation relations: $$ [r_i, p_j] = i\hbar \delta_{ij}\;, $$ which means, for example, that $$ y p_z = p_z y $$ and $$ z p_y = p_y z\;. $$

I assume you also understand that if $$ C = AB $$ then $$ C^\dagger = B^\dagger A^\dagger $$

I assume you also understand that if $$ D = A + B $$ then $$ D^\dagger = A^\dagger + B^\dagger $$

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The (L_x) operator can be expressed as:
$$ L_x = y p_z - z p_y, $$

Thus, we have: $$ L_x^\dagger = (y p_z)^\dagger - (z p_y)^\dagger $$ $$ = p_z^\dagger y^\dagger - p_y^\dagger z^\dagger $$ $$ =p_z y - p_y z $$ $$ =y p_z - z p_y $$

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Do both terms satisfy the Hermitian condition?

Yes. For example, $(p_z y)^\dagger = p_z y$.