Hermitian Operator Condition

An operator $\hat{A}$ is Hermitian if its adjoint $\hat{A}^\dagger$ equals itself. This condition is formally expressed using the inner product: $ \int \psi^* (\hat{A} \phi) \, dV = \int (\hat{A} \psi)^* \phi \, dV $ for all suitable wavefunctions $\psi$ and $\phi$. In cylindrical coordinates $(s, \varphi, z)$, the volume element is $dV = s \, ds \, d\varphi \, dz$. We will evaluate each option using this definition.

Operator Analysis in Cylindrical Coordinates

Option 1: $\frac{1}{i} \frac{\partial}{\partial s}$

For $\hat{A} = \frac{1}{i} \frac{\partial}{\partial s}$, direct calculation using the Hermitian condition with $dV = s \, ds \, d\varphi \, dz$ shows that $\int \psi^* (\hat{A} \phi) \, dV \neq \int (\hat{A} \psi)^* \phi \, dV$. Therefore, this operator is not Hermitian.

Option 2: $\frac{1}{i} (\frac{\partial}{\partial s} + \frac{1}{s})$

For $\hat{A} = \frac{1}{i} (\frac{\partial}{\partial s} + \frac{1}{s})$, applying the definition of the Hermitian operator and performing integration by parts reveals that the condition $\int \psi^* (\hat{A} \phi) \, dV = \int (\hat{A} \psi)^* \phi \, dV$ is not satisfied. This operator is not Hermitian.

Option 3: $\frac{1}{i} (\frac{\partial}{\partial s} + \frac{1}{2s})$

Let's verify the Hermitian condition for $\hat{A} = \frac{1}{i} (\frac{\partial}{\partial s} + \frac{1}{2s})$.

Left-Hand Side (LHS) Calculation

$ \text{LHS} = \int \psi^* \hat{A} \phi \, dV = \int \psi^* \frac{1}{i} \left(\frac{\partial \phi}{\partial s} + \frac{1}{2s} \phi\right) s \, ds \, d\varphi \, dz $

$ \text{LHS} = \frac{1}{i} \int \psi^* \frac{\partial \phi}{\partial s} s \, ds \, d\varphi \, dz + \frac{1}{2i} \int \psi^* \phi \, ds \, d\varphi \, dz $

Using integration by parts on the first term $\int \psi^* \frac{\partial \phi}{\partial s} s \, ds$: $ \int \psi^* \frac{\partial \phi}{\partial s} s \, ds = [\psi^* \phi s]_{s_{min}}^{s_{max}} - \int \left(\frac{\partial \psi^*}{\partial s} s + \psi^*\right) \phi \, ds $

Assuming boundary terms vanish, the LHS becomes: $ \text{LHS} = \frac{1}{i} \left[-\int \left(\frac{\partial \psi^*}{\partial s} s + \psi^*\right) \phi \, ds \, d\varphi \, dz\right] + \frac{1}{2i} \int \psi^* \phi \, ds \, d\varphi \, dz $

$ \text{LHS} = \frac{1}{i} \int \left[-s \frac{\partial \psi^*}{\partial s} - \psi^* + \frac{1}{2} \psi^*\right] \phi \, ds \, d\varphi \, dz $

$ \text{LHS} = \frac{1}{i} \int \left[-s \frac{\partial \psi^*}{\partial s} - \frac{1}{2} \psi^*\right] \phi \, ds \, d\varphi \, dz $

Right-Hand Side (RHS) Calculation

$ \text{RHS} = \int (\hat{A} \psi)^* \phi \, dV = \int \left(\frac{1}{i} \left(\frac{\partial \psi}{\partial s} + \frac{1}{2s} \psi\right)\right)^* \phi \, s \, ds \, d\varphi \, dz $

Since $(\frac{1}{i})^* = -\frac{1}{i}$ and assuming $s$ is real, $(\frac{1}{2s})^* = \frac{1}{2s}$: $ \text{RHS} = \int -\frac{1}{i} \left(\frac{\partial \psi^*}{\partial s} + \frac{1}{2s} \psi^*\right) \phi \, s \, ds \, d\varphi \, dz $

$ \text{RHS} = \frac{1}{i} \int \left(-\frac{\partial \psi^*}{\partial s} s - \frac{1}{2} \psi^*\right) \phi \, ds \, d\varphi \, dz $

$ \text{RHS} = \frac{1}{i} \int \left[-s \frac{\partial \psi^*}{\partial s} - \frac{1}{2} \psi^*\right] \phi \, ds \, d\varphi \, dz $

Since LHS = RHS, this operator is Hermitian.

Option 4: $\frac{\partial}{\partial s} + \frac{1}{s}$

For $\hat{A} = \frac{\partial}{\partial s} + \frac{1}{s}$, the application of the Hermitian condition shows that $\int \psi^* (\hat{A} \phi) \, dV \neq \int (\hat{A} \psi)^* \phi \, dV$. This operator is not Hermitian.

Conclusion

Based on the analysis, the operator $\frac{1}{i} (\frac{\partial}{\partial s} + \frac{1}{2s})$ satisfies the definition of a Hermitian operator in cylindrical coordinates.