Complex conjugate of momentum operator

Question

Consider momentum operator representation in position space. $$\hat{p}=-i\frac{\partial}{\partial x} \,\ \text{and its eigen functions are } e^{ipx} \,\text{and} \,\ e^{-ipx}.$$ $$\hat{p}e^{ipx}=pe^{ipx}$$ Taking the complex conjugate of the equation, $${\hat{p}}^* e^{-ipx}=p^* e^{-ipx}\tag{1}$$ Because momentum eigenvalue is real, $p^* =p$. Thus $${\hat{p}}^* e^{-ipx}=p e^{-ipx}$$ Now consider, $$-\hat{p}e^{-ipx}=pe^{-ipx}$$ From these two equations we see that ${\hat{p}}^* =-\hat{p}$.\

Now consider the matrix representation of the momentum operator. In the basis of the momentum eigenstates, the momentum operator matrix (infinite dimensional) is diagonal and the diagonal elements represent the eigenvalues of the momentum operator just as in the case of other finite dimensional operators. This means that the complex conjugate of the $p-$matrix is $p-$matrix itself. However we saw from the above logic that the complex conjugate should be negative of the $p-$matrix.\

I can't see where the problem is!!

Answer 1

The short story is, you can't distribute the * onto the operator like that. You need to keep it as $\left(\hat{p}e^{ipx}\right)^* = {p}\,e^{-ipx}$ because $\hat{p}$ is an operator on a complex vector space which looks like a derivative in the $x$ basis.

$\hat{p} =-i{\partial \over \partial x} $ is the representation of the momentum operator in the $x$ basis, which is to say $\langle x \left| \hat p \right|\psi\rangle = -i{\partial \over \partial x}\langle x\left|\psi\right\rangle$. Complex conjugation is an operation we know how to do on complex numbers, so let's make sure that the objects we're working with are complex numbers first. The complex conjugate of an inner product is $\langle x | \psi \rangle^* = \langle\psi|x\rangle$. The operator $\hat{p}$ changes the vector $|\psi\rangle$ into some other vector $\hat{p}|\psi\rangle = |p\psi\rangle$. Now taking the complex conjugate looks like:

$$ \langle x \left| \hat p \right|\psi\rangle^* = \langle x |p\psi\rangle^* = \langle p\psi|x\rangle $$

The question now is what is $\langle p\psi|$? In order to have $\langle\psi|\psi\rangle = |\psi|^2$ then $\langle \psi|$ must be hermitian adjoint of $|\psi\rangle$. If you think of kets as column vectors, bras as row vectors, and operators as matrices, then this operation is taking the transpose of the operator matrix in addition to complex conjugation.

Answer 2

The complex conjugation of an operator depends on the basis you are choosing. In other words, there is no base independent definition of complex conjugation.

Answer 3

It's easy to find the mistake if you realize that an operator is a function on a vector space with argument the function on its right. For example, $\hat{p} e^{ipx} = p e^{ipx}$ is (apparently confusing) notation for $\hat{p}(e^{ipx}) = pe^{ipx}$, therefore $\hat{p}^*(e^{ipx}) = p e^{-ipx}$. See also George G's answer in Dirac notation.

Answer 4

You are assuming that the Hermitian conjugate of the derivative operator $d/dx$ is the derivative again. This is not the case. The hermitian conjugate of $d/dx$ is $-d/dx$. That's why the $i$ is there.