3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Where and why do the ladder operators come into the game? $\endgroup$
    – Fabian
    Dec 15, 2012 at 17:10
  • $\begingroup$ see my edit. I was comparing falsely. I just asked how should i get a adjoint momentum operator in position basis. $\endgroup$
    – Aftnix
    Dec 15, 2012 at 18:18

1 Answer 1

5
$\begingroup$

$\hat{p}$ is Hermitian and Hermitian operators $O$ satisfy, by definition,

$$\hat{O} = \hat{O}^\dagger$$

Adjoint is not a synonym for complex conjugate. $\hat{p} = -i\hbar \nabla \rightarrow +i\hbar \nabla^\dagger \rightarrow -i\hbar \nabla =\hat{p}^\dagger$, but $\hat{p} \neq \hat{p}^*$.

$\endgroup$
5
  • $\begingroup$ Thanks. Is there any general rule for getting adjoint for non matrix operators? $\endgroup$
    – Aftnix
    Dec 15, 2012 at 18:20
  • $\begingroup$ @Aftnix $\hat{p}$ can be represented as a matrix, just an infinite-dimensional one. You might find my answer to this question useful. $\endgroup$ Dec 15, 2012 at 18:27
  • $\begingroup$ @Aftnix: What do you mean by non matrix operators? $\endgroup$
    – juanrga
    Dec 15, 2012 at 19:01
  • $\begingroup$ after going through linear algebra resources i found exact answer. Thanks for helping. $\endgroup$
    – Aftnix
    Dec 16, 2012 at 18:24
  • 4
    $\begingroup$ The adjoin of the differential operator $\nabla^\dagger = -\nabla$ can be proved by integration by part. $\endgroup$ Apr 7, 2014 at 23:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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