I Position operator explicit form

[cianfa72]

TL;DR

About the position explicit form definition

I've a doubt about the following definition from PSE thread.
The first answer says that the position representation of the position operator $\hat{x}$ is:
$$\bra{x}\hat{x} = \bra{x}x$$ I believe there is a typo, it should actually be $$\bra{x}\hat{x} = x \bra{x}$$
Does it make sense ? Thanks.

 

[PeroK]

TL;DR Summary: About the position explicit form definition

I've a doubt about the following definition from PSE thread.
The first answer says that the position representation of the position operator $\hat{x}$ is:
$$\bra{x}\hat{x} = \bra{x}x$$ I believe there is a typo, it should actually be $$\bra{x}\hat{x} = x \bra{x}$$
Does it make sense ? Thanks.

In standard linear algebra we tend to write the scalar before a vector. In Dirac notation, however, it's important to write the scalar after a bra. That allows all the typographical wizardry of Dirac's notation.

 

[cianfa72]

In standard linear algebra we tend to write the scalar before a vector. In Dirac notation, however, it's important to write the scalar after a bra. That allows all the typographical wizardry of Dirac's notation.

Ok, however applying the operator $\hat{x}$ to the state/vector $\ket{\psi}$, one gets $$\bra{x}\hat{x} \ket{\psi} = \bra{x}x\ket{\psi}$$ and, by defining the wavefunction $\psi(x):=\braket{x|\psi}$, one doesn't get the expression $$\bra{x}\hat{x} \ket{\psi} = x\psi(x)$$ as expected...

 

[PeroK]

Ok, however applying the operator $\hat{x}$ to the state/vector $\ket{\psi}$, one gets $$\bra{x}\hat{x} \ket{\psi} = \bra{x}x\ket{\psi}$$ and, by defining the wavefunction $\psi(x):=\braket{x|\psi}$, one doesn't get the expression $$\bra{x}\hat{x} \ket{\psi} = x\psi(x)$$ as expected...

Well, yes you do. Note that in general:
$$\langle \phi |\hat A |\psi \rangle = \langle \psi |\hat A^{\dagger}|\phi \rangle^*$$

 

[PeroK]

Actually, you don't need that in this case:
$$\langle x|x|\psi\rangle = x \langle x|\psi \rangle$$As the $x$ that comes out is a number/eigenvalue.

 

[cianfa72]

Actually, you don't need that in this case:
$$\langle x|x|\psi\rangle = x \langle x|\psi \rangle$$As the $x$ that comes out is a number/eigenvalue.

Ah ok...so the $x$ that comes out isn't a function, it is just a number (actually the eigenvalue) let's say associated with the specific position eigenstate $\ket {x}$.

Therefore, since it is just "a real number", it comes out from the inner product.

 

[PeroK]

Ah ok...so the $x$ that comes out isn't a function

It's a function value. The inner product itself is a functional, I.e. a function/mapping into the complex numbers.

We can define a complex valued function on the real numbers as follows. For each real number $x$ we have:
$$f(x) = \langle x |\hat x|\psi\rangle = \langle x|x|\psi\rangle = x \langle x|\psi \rangle = x\psi(x)$$

 

[PeroK]

PS note the subtlety that $\hat x$ is the position operator and is not dependent on $x$, which represents a specific real number and eigenstate associated with that real number.

 

[Demystifier]

TL;DR Summary: About the position explicit form definition

I've a doubt about the following definition from PSE thread.
The first answer says that the position representation of the position operator $\hat{x}$ is:
$$\bra{x}\hat{x} = \bra{x}x$$ I believe there is a typo, it should actually be $$\bra{x}\hat{x} = x \bra{x}$$
Does it make sense ? Thanks.

Multiplication of a vector by a scalar is commutative, i.e. $x\bra{x}=\bra{x}x$.

 

[cianfa72]

PS note the subtlety that $\hat x$ is the position operator and is not dependent on $x$, which represents a specific real number and eigenstate associated with that real number.

In other words, for a given $x$ value (say $x=2$) the relevant specific real number (eigenvalue) and eigenstate are 2 and the bra $\bra{2}$ respectively -- the latter is the dual vector associated via musical isomorphism with the ket $\ket{2}$.