# Adjoint of momentum operator

In position basis, we have,

$$\langle x \mid \hat p \mid \Psi(t) \rangle = -\imath \hbar \frac{\partial{\langle x \mid \Psi(t) \rangle}}{\partial{x}}$$

Now I know $\hat{p}$ is a Hermitian operator which should be self adjoint.

The self adjoint operators are said to satisfy :

$$\langle A \psi \mid \phi \rangle = \langle \psi \mid A \phi \rangle$$

But I failed to workout the following :

$$\langle x \mid \hat{p}^\dagger \mid \Psi(t) \rangle$$

For ladder operator $\hat{a}$ I found $\hat{a}^\dagger$ by conjugating in position basis. And clearly $\hat{a}$ is not Hermitian because $$\hat{a}^\dagger \neq \hat{a}$$ in position basis. And thus $\hat{a}$ does not correspond to any observable.

But $\hat{p}$ is Hermitian. But it seems, at position basis, complex conjugating $\hat{p}$ gives a different object.

Where am I making a mistake here?

EDIT:

After posting the question I found some inconsistencies in my argument.

$$\hat a = \frac {\hat x}{\sqrt {\frac {2 \hbar}{m \omega}}} + \frac {i \hat p}{\sqrt {2 \hbar m \omega}}$$

is not a representation in any basis. It's just an operator relation with some scaling factor.

whereas $$-i \hbar \frac {\partial}{\partial x}$$ is a representation of $\hat p$ in position $x$ basis.

So they should not be comparable.

But i still want to know what $\hat{p}^\dagger$ is in position $x$ basis.

I just falsely took a wrong example.

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Where and why do the ladder operators come into the game? – Fabian Dec 15 '12 at 17:10
see my edit. I was comparing falsely. I just asked how should i get a adjoint momentum operator in position basis. – Aftnix Dec 15 '12 at 18:18

$\hat{p}$ is Hermitian and Hermitian operators $O$ satisfy, by definition,
$$\hat{O} = \hat{O}^\dagger$$
Adjoint is not synonym for complex. $\hat{p} = -i\hbar \nabla \rightarrow +i\hbar \nabla^\dagger \rightarrow -i\hbar \nabla =\hat{p}^\dagger$, but $\hat{p} \neq \hat{p}^*$.
@Aftnix $\hat{p}$ can be represented as a matrix, just an infinite-dimensional one. You might find my answer to this question useful. – Mark Mitchison Dec 15 '12 at 18:27
The adjoin of the differential operator $\nabla^\dagger = -\nabla$ can be proved by integration by part. – xslittlegrass Apr 7 '14 at 23:53