# Very elementary notation question

1. Jul 18, 2009

### romistrub

Shankar p68-69 gives a mathematical "derivation" of the action of the X (position) operator, the summary of which is as follows:
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:
Similarly, he writes, of the action of X in the K basis
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?

Last edited: Jul 18, 2009
2. Jul 18, 2009

### romistrub

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

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

and not

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

Last edited: Jul 18, 2009
3. Jul 19, 2009

### 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 $$X|f\rangle$$.

Last edited: Jul 19, 2009
4. Jul 19, 2009

### humanino

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

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