Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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, $${\hat{p}}^* e^{-ipx}=p^* e^{-ipx}$$ 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!!

share|cite|improve this question

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.

share|cite|improve this answer
Working on the space of infinitely differentiable functions, the distribution of the hermitian conjugation is correct in the way depicted in the question. The only problem would arrise when conjugating the multiplication of operators, in which case the only correction needed is to inverse the order of the operators in the product, like $(\hat{A}\hat{B})^*=\hat{B}^*\hat{A}^*$ – Ajayu Apr 25 '14 at 21:58
@Ajayu Physicists denote complex conjugation by $\hat{A}^*$ and hermitian conjugation by $\hat{A}^\dagger$. – auxsvr Apr 26 '14 at 4:40
$\hat{A}^*$ is coherent with the notation used in the question, which uses $\hat{p}^*$ as the hermitian conjugate of the impulsion operator. Althought not standard among physicist, it's the one chosen to formulate the question. – Ajayu Apr 26 '14 at 18:31
I don't think the OP was aware of the difference between hermitian conjugation and complex conjugation, so it is better to maintain a distinction. – George G Apr 26 '14 at 21:32

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.

share|cite|improve this answer
The hermitian conjugate is defined without reference to a basis. – Robin Ekman Apr 25 '14 at 19:44
@RobinEkman: But the complex conjugate is not the same as the hermitian conjugate. But the fact that the hermitian conjugate is not basis dependent can be used to show that complex conjugation indeed is basis dependent: Hermitian conjugate is the combination of complex conjugate and transpose. Therefore complex conjugate is the same as hermitian conjugate followed by transpose. Now I think it is quite obvious that the transpose is basis dependent. – celtschk Apr 26 '14 at 9:07

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.

share|cite|improve this answer
That has not been assumed anywhere!!! – user35122 Apr 30 '14 at 10:22

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.