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## Main Question or Discussion Point

I would just like some clarification and some assertion that I've got the right idea. Please correct everything I say!

For any observable [itex]A[/itex] over a finite-dimensional vector space with orthonormal basis kets [itex]\{|a_i\rangle\}_{i=1}^n[/itex] we can write

[tex]A = IAI = \left(\sum_{i=1}^n |a_i\rangle\langle a_i|\right) A \left(\sum_{j=1}^n |a_j\rangle\langle a_j|\right) = \sum_{i=1}^n |a_i\rangle\langle a_i| \sum_{j=1}^n A|a_j\rangle\langle a_j| = \sum_{i=1}^n \sum_{j=1}^n |a_i\rangle\langle a_i|A|a_j\rangle\langle a_j| = \sum_{i,j} |a_i\rangle\langle a_j| \: \langle a_i|A|a_j\rangle \; \; (1)[/tex]

and we say that [itex]\langle a_i|A|a_j\rangle[/itex] is the

My professor commonly talks about the coordinate representation of the momentum and position operators. I know these live in an infinite-dimensional vector space. Is this a Hilbert space? A rigged Hilbert space?

We know that [itex]\hat{X}[/itex] acts on vectors [itex]|x\rangle[/itex] such that [itex]\hat{X}|x\rangle = x|x\rangle[/itex]. Why can we write the following? What does it mean?

[tex]\int_{-\infty}^{\infty} dx \; x \;|x\rangle\langle x| \; \; (2)[/tex]

For an arbitrary operator [itex]A[/itex] in this infinite-dimensional space, (suppressing the bounds of integration and writing x' to denote another variable and not the derivative of x),

[tex]A = IAI = \left(\int dx' \; |x'\rangle\langle x'|\right) A \left(\int dx \; |x\rangle\langle x|\right) = \int\int dx' dx \; |x'\rangle\langle x'|A|x\rangle\langle x| = \int\int dx' dx \; |x'\rangle\langle x| \; \langle x'|A|x\rangle \; \; (3)[/tex]

This is analogous to the finite-dimensional case in the sense that the coordinate representation of [itex]A[/itex] is given by [itex]\langle x'|A|x\rangle[/itex].

Now if [itex]A = \hat{X}[/itex] then [itex]\langle x'|\hat{X}|x\rangle = \langle x'|x|x\rangle = x \langle x'|x\rangle = x \delta(x' - x)[/itex], and with this, (3) simplifies into (2) (by the sifting property of the Dirac delta function).

I don't know how to formally proceed if [itex]A = \hat{P}[/itex] because it seems a little bit more tricky. Can anybody help there? Also, why do most textbooks avoid this integration business and simply state that [itex]\hat{X} = x[/itex] and [itex]\hat{P} = -i\hbar \frac{\partial}{\partial x}[/itex]?

Thanks!

For any observable [itex]A[/itex] over a finite-dimensional vector space with orthonormal basis kets [itex]\{|a_i\rangle\}_{i=1}^n[/itex] we can write

[tex]A = IAI = \left(\sum_{i=1}^n |a_i\rangle\langle a_i|\right) A \left(\sum_{j=1}^n |a_j\rangle\langle a_j|\right) = \sum_{i=1}^n |a_i\rangle\langle a_i| \sum_{j=1}^n A|a_j\rangle\langle a_j| = \sum_{i=1}^n \sum_{j=1}^n |a_i\rangle\langle a_i|A|a_j\rangle\langle a_j| = \sum_{i,j} |a_i\rangle\langle a_j| \: \langle a_i|A|a_j\rangle \; \; (1)[/tex]

and we say that [itex]\langle a_i|A|a_j\rangle[/itex] is the

*coordinate representation*of A.My professor commonly talks about the coordinate representation of the momentum and position operators. I know these live in an infinite-dimensional vector space. Is this a Hilbert space? A rigged Hilbert space?

We know that [itex]\hat{X}[/itex] acts on vectors [itex]|x\rangle[/itex] such that [itex]\hat{X}|x\rangle = x|x\rangle[/itex]. Why can we write the following? What does it mean?

[tex]\int_{-\infty}^{\infty} dx \; x \;|x\rangle\langle x| \; \; (2)[/tex]

For an arbitrary operator [itex]A[/itex] in this infinite-dimensional space, (suppressing the bounds of integration and writing x' to denote another variable and not the derivative of x),

[tex]A = IAI = \left(\int dx' \; |x'\rangle\langle x'|\right) A \left(\int dx \; |x\rangle\langle x|\right) = \int\int dx' dx \; |x'\rangle\langle x'|A|x\rangle\langle x| = \int\int dx' dx \; |x'\rangle\langle x| \; \langle x'|A|x\rangle \; \; (3)[/tex]

This is analogous to the finite-dimensional case in the sense that the coordinate representation of [itex]A[/itex] is given by [itex]\langle x'|A|x\rangle[/itex].

Now if [itex]A = \hat{X}[/itex] then [itex]\langle x'|\hat{X}|x\rangle = \langle x'|x|x\rangle = x \langle x'|x\rangle = x \delta(x' - x)[/itex], and with this, (3) simplifies into (2) (by the sifting property of the Dirac delta function).

I don't know how to formally proceed if [itex]A = \hat{P}[/itex] because it seems a little bit more tricky. Can anybody help there? Also, why do most textbooks avoid this integration business and simply state that [itex]\hat{X} = x[/itex] and [itex]\hat{P} = -i\hbar \frac{\partial}{\partial x}[/itex]?

Thanks!