Complex conjugate of a real wave function [closed]

Question

In the context of applying operators to find expectation values; is the 'complex conjugate' of a wave function, $\psi^*$, where $\psi$ has no complex numbers, just simply itself? For example, given the wave function:

$$ \psi(x)=\sqrt{\frac2L}\sin\biggl(\frac{2\pi}Lx\biggr) $$

Is this set up to find the expectation value of the momentum, $p_x$, an accurate starting point:

$$ \begin{align} \langle p_x \rangle &= \int^{+\infty}_{-\infty} \psi^*[p_x]\psi\ \ dx \\ & = -i\hbar\int^{+\infty}_{-\infty} \psi^*\frac{\partial}{\partial x}\psi\ \ dx \\ & = -i\hbar\int^{+\infty}_{-\infty} \Biggl(\sqrt{\frac2L}\sin\biggl(\frac{2\pi}Lx\biggr)\Biggr)^*\frac{\partial}{\partial x}\Biggl(\sqrt{\frac2L}\sin\biggl(\frac{2\pi}Lx\biggr)\Biggr)\ \ dx \end{align} $$

Answer 1

Yes, it is the correct setup and the complex conjugate of a real function is the function itself.

But notice, the form of the wavefunction and its normalization suggest that you are looking at a particle in a box. So the integral should extend only to the box size. In alternative, you can integrate over the entire x axis, but then the wavefunction would be zero outside of the box.

Answer 2

Yes, the complex conjugate of a real number is the same real number. Complex conjugation of functions is defined pointwise and hence the same property applies.