Very elementary notation question

[romistrub]

Shankar p68-69 gives a mathematical "derivation" of the action of the X (position) operator, the summary of which is as follows:

[tex]\left\langle x \left| \textbf{X} \right| f \right\rangle = \dots = xf(x)[/tex]

I followed the logic without a problem, since it only involves using the matrix elements of X in the basis of eigenfunctions of X. However, the next paragraph reads:

We can summarize the action of X in Hilbert space as
[tex]\textbf{X} \left| f(x) \right\rangle = \left|xf(x)\right\rangle[/tex].

Similarly, he writes, of the action of X in the K basis

[tex]\textbf{X} \left| g(k) \right\rangle = \left|i\frac{dg(k)}{dk}\right\rangle[/tex]

Now, to me

[tex]f(x) = \left\langle x | f \right\rangle[/tex]

and

[tex]g(k) = \left\langle k | g \right\rangle[/tex]

are scalars. Hence I cannot comprehend what is intended by Shankar's notation. Any insight?

 

[romistrub]

I am wondering if the next computation gives some insight: Shankar then computes the matrix elements of X in the K basis as:

[tex]\left\langle k \left| \textbf{X} \right| k' \right\rangle = \frac{1}{2\pi}\int^{\infty}_{-\infty}e^{-ikx}xe^{ikx}dx[/tex]

where, again, to me

[tex]\left\langle x|k\right\rangle \propto e^{ikx}[/tex]

and not

[tex]\left|k\right\rangle \propto e^{ikx}[/tex]

 

[kanato]

Inside the integral you have

[tex]e^{-ikx} x e^{ik'x}[/tex]

[tex]= e^{-ikx} \frac{1}i \frac{d}{dk'} e^{ik'x}[/tex]

which is what leads him to write that [tex]\textbf{X} \left| g(k) \right\rangle = \left|i\frac{dg(k)}{dk}\right\rangle[/tex].The notation [tex]|f(x)\rangle[/tex] as the ket corresponding to f(x) (which he says near the top of pg. 69) is sloppy, but I don't think there is anything wrong with it. He's not saying that [tex]|k\rangle \propto e^{ikx}[/tex], he's just using it as a notation to express the ket that comes out of the operation [tex]X|f\rangle[/tex].

 

[humanino]

[tex]\left\langle x|k\right\rangle \propto e^{ikx}[/tex]

and not

[tex]\left|k\right\rangle \propto e^{ikx}[/tex]

This is correct. To go from left to right, simply insert
[tex]1=\int_{-\infty}^{\infty}\text{d}x\left|x\right\rangle\left\langle x\right|[/tex]