I Position operator explicit form

cianfa72
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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.
 
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cianfa72 said:
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.
 
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PeroK said:
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...
 
cianfa72 said:
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:
$$\langle \phi |\hat A |\psi \rangle = \langle \psi |\hat A^{\dagger}|\phi \rangle^*$$
 
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.
 
PeroK said:
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.
 
cianfa72 said:
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:
$$f(x) = \langle x |\hat x|\psi\rangle = \langle x|x|\psi\rangle = x \langle x|\psi \rangle = x\psi(x)$$
 
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.
 
cianfa72 said:
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##.
 
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PeroK said:
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}##.
 
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