Is the Adjoint of the Position Operator Self-Adjoint?

[Kashmir]

I'm trying to find the adjoint of position operator.

I've done this:

The eigenvalue equation of position operator is

$\hat{x}|x\rangle=x|x\rangle$

The adjoint of position operator acts as

$\left\langle x\left|\hat{x}^{\dagger}=x<x\right|\right.$

Then using above equation we've
$\left\langle x\left|x^{\dagger}\right| x\right\rangle=x\langle x \mid x\rangle$

or

$\langle x|( x^{\dagger} |x\rangle)=\langle x|(x| x\rangle)$

Then

$x^{\dagger}|x\rangle=x|x\rangle$

Hence
$x^{\dagger}=x$

Is this correct?

 

[PeroK]

$\hat{x}|x\rangle=x|x\rangle$

The adjoint of position operator acts as

$\left\langle x\left|\hat{x}^{\dagger}=x<x\right|\right.$

Then using above equation we've
$\left\langle x\left|x^{\dagger}\right| x\right\rangle=x\langle x \mid x\rangle$
$\langle x|( x^{\dagger} |x\rangle)=\langle x|(x| x\rangle)$

Then

$x^{\dagger}|x\rangle=x|x\rangle$

Hence
$x^{\dagger}=x$

Is this correct?

I think you've shown that any operator that has a complete spectrum of eigenvectors and real eigenvalues is Hermitian (this is not the same as self-adjoint, but that may be the book you are using).

An important identity for any operator and vectors is:
$$\langle u |Q^{\dagger}|v \rangle = \langle v| Q|u \rangle^*$$To show that $Q$ is Hermitian you need to show that:$$\forall u, v: \ \langle u |Q|v \rangle = \langle v| Q|u \rangle^*$$Your proof is not wrong, but you could add a bit more to it, perhaps.

 

[Kashmir]

I think you've shown that any operator that has a complete spectrum of eigenvectors and real eigenvalues is Hermitian (this is not the same as self-adjoint, but that may be the book you are using).

An important identity for any operator and vectors is:
$$\langle u |Q^{\dagger}|v \rangle = \langle v| Q|u \rangle^*$$To show that $Q$ is Hermitian you need to show that:$$\forall u, v: \ \langle u |Q|v \rangle = \langle v| Q|u \rangle^*$$Your proof is not wrong, but you could add a bit more to it, perhaps.

I'm thinking about it again.

In your opinion how can I find $x^{\dagger}$ then? Also is my result correct about $x^{\dagger}$?

 

[PeroK]

I'm thinking about it again.

In your opinion how can I find $x^{\dagger}$ then? Also is my result correct about $x^{\dagger}$?

In your proof you didn't emphasise that $|x \rangle$ was any eigenvector of $\hat x$ and that every vector can be expressed as an integral over these eigenvectors (which form an uncountable basis). In general:
$$| \alpha \rangle = \int dx \ |x \rangle \langle x| \alpha \rangle = \int dx \ \alpha(x) |x \rangle$$

 

[PeroK]

You already know that the position operator is Hermitian, so the question is whether your proof is valid and complete. There's no question about the conclusion!

 

[Kashmir]

									You already know that the position operator is Hermitian, so the question is whether your proof is valid and complete. There's no question about the conclusion!

You already know that the position operator is Hermitian, so the question is whether your proof is valid and complete. There's no question about the conclusion!

We can write for any kets:

$\langle\varphi|\hat{x}| \psi\rangle=\int x d x\langle\varphi \mid x\rangle\langle x \mid \psi\rangle$

Similarly
$\langle\psi|\hat{x}| \varphi\rangle^{*}=\int x d x(\langle \psi\mid x\rangle\langle x \mid \varphi\rangle)^{*}=\int x d x\langle x \mid \psi\rangle\langle\varphi \mid x\rangle$$\begin{aligned} \therefore \quad &\langle\varphi|\hat{x}| \psi\rangle=\langle\psi|\hat{x}| \varphi\rangle^{*} \\ & \Rightarrow x^{\dagger}=x \end{aligned}$

Is this valid and complete now?

 

[dextercioby]

Discussing operators and their adjoints in bra-ket notation is wrong, or at most confusing. Just stick to regular scalar product notation $\langle a, b\rangle$.