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:

$$\left\langle x \left| \textbf{X} \right| f \right\rangle = \dots = xf(x)$$

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
$$\textbf{X} \left| f(x) \right\rangle = \left|xf(x)\right\rangle$$.

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

$$\textbf{X} \left| g(k) \right\rangle = \left|i\frac{dg(k)}{dk}\right\rangle$$

Now, to me

$$f(x) = \left\langle x | f \right\rangle$$

and

$$g(k) = \left\langle k | g \right\rangle$$

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:

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

where, again, to me

$$\left\langle x|k\right\rangle \propto e^{ikx}$$

and not

$$\left|k\right\rangle \propto e^{ikx}$$

 

[kanato]

Inside the integral you have

$$e^{-ikx} x e^{ik'x}$$

$$= e^{-ikx} \frac{1}i \frac{d}{dk'} e^{ik'x}$$

which is what leads him to write that $$\textbf{X} \left| g(k) \right\rangle = \left|i\frac{dg(k)}{dk}\right\rangle$$.The notation $$|f(x)\rangle$$ 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 $$|k\rangle \propto e^{ikx}$$, he's just using it as a notation to express the ket that comes out of the operation $$X|f\rangle$$.

 

[humanino]

$$\left\langle x|k\right\rangle \propto e^{ikx}$$

and not

$$\left|k\right\rangle \propto e^{ikx}$$

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