Note: $\psi,\psi^{\dagger} :\Bbb{R} \to \Bbb{C}$ and $x, \lambda_i , \hbar, m \in \Bbb{R}$
Say we know that $\lambda_1$ is a solution to the eigenvalue equation: $$\hat{\Pi}\psi(x)= \lambda_1 \psi(x) .$$ Now for the complex conjugate of $\psi$, to be denoted as $\psi^{\dagger}$, is the eigenfunction for the eigenvalue problem $$\hat{\Pi}\psi^{\dagger}(x)=\lambda_2\psi^{\dagger}(x)$$ the same as the first, or its negative, or totally different? In other words, is $$\lambda_1=\lambda_2 \text{ ?}$$
Thank you!
This is in reference to a quantum mechanics problem, in which I try to prove the following: Given $$j(x,t)=\frac{\hbar}{2 i m} \left[ \psi(x)^{\dagger}\frac{\partial \psi(x)}{\partial x}-\psi(x)\frac{\partial \psi(x)^{\dagger}}{\partial x } \right]$$ Show it can be represented in the form $$j(x,t) =\operatorname{Re} \left[ \psi(x)^{\dagger}\frac{\hbar}{i m}\frac{\partial \psi(x)}{\partial x} \right]$$
This is related because the momentum operator is defined as $\hat{P}= -i \hbar \hat{\nabla}=-i \hbar \dfrac{\partial}{\partial x}$ in one dimension, and this can be substituted into the Eigenvalues problem and simplified.
What you really need for your problem is the following: if $z$ is a complex number, then $$ \frac{z + z^\dagger}{2} = \text{Re}(z), \qquad \frac{z - z^\dagger}{2i} = \text{Im}(z), \qquad \text{Im}(z) = \text{Re}(\tfrac{z}{i}). $$ See if you can prove these by writing $z = a + bi$, where $a,b \in \mathbb R$, and performing the indicated operators.
Turning to your equation for probability current, we can set $z = \psi^\dagger(x)\tfrac{\partial \psi(x)}{\partial x}$, so $z^\dagger = \psi(x) \tfrac{\partial \psi^\dagger(x)}{\partial x}$. Then $$ j(x,t) = \frac{\hbar}{2 i m} (z - z^\dagger) = \frac{\hbar}{m} \cdot \frac{z - z^\dagger}{2i} = \frac{\hbar}{m} \text{Im}(z) = \frac{\hbar}{m} \text{Re}\left(\frac{z}{i}\right) = \text{Re}\left( \frac{\hbar}{mi} z \right) = \text{Re}\left( \frac{\hbar}{mi} \psi^\dagger(x) \frac{\partial \psi(x)}{\partial x} \right) $$ as desired.