Adjoint of position operator

In summary: 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##Is this
  • #1
Kashmir
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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?
 
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  • #2
Kashmir said:
##\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.
 
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  • #3
PeroK said:
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}##?
 
  • #4
Kashmir said:
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:
$$| \alpha \rangle = \int dx \ |x \rangle \langle x| \alpha \rangle = \int dx \ \alpha(x) |x \rangle$$
 
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  • #5
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!
 
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  • #6
PeroK said:
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!

PeroK said:
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?
 
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  • #7
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##.
 

1. What is the adjoint of position operator?

The adjoint of position operator is the Hermitian conjugate of the position operator, denoted as x̂†. It is a mathematical operation that involves taking the complex conjugate of the operator and then transposing it.

2. What is the significance of the adjoint of position operator in quantum mechanics?

In quantum mechanics, the adjoint of position operator is used to calculate the expectation value of a particle's position in a given state. It is also used in the Heisenberg uncertainty principle and in the formulation of quantum mechanical operators.

3. How is the adjoint of position operator related to the momentum operator?

The adjoint of position operator and the momentum operator are related through the canonical commutation relation, which states that [x̂†, p̂] = iħ. This relation is essential in quantum mechanics as it describes the uncertainty in the measurement of position and momentum of a particle.

4. Can the adjoint of position operator be used to find the position of a particle at a specific time?

No, the adjoint of position operator does not give the position of a particle at a specific time. It is used to calculate the expectation value of the position operator in a given state. To find the position of a particle at a specific time, the time-dependent Schrödinger equation must be solved.

5. How is the adjoint of position operator related to the position eigenstates?

The adjoint of position operator is closely related to the position eigenstates. The position eigenstates are the eigenfunctions of the position operator, and the adjoint of position operator acts on these eigenstates to give the corresponding eigenvalues. This relationship is fundamental in understanding the position measurement in quantum mechanics.

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