# position state as a sum of momentum states

Leonard Susskind, in his lecture (minute 41) about QFT states that when a field acts on a vacuum state it gives a position state- $$\Psi^\dagger(x)=\sum_{k}e^{-ikx}a^\dagger(k)\left|0\right>=\sum_{k}e^{-ikx}\left|k\right>=\left|x\right>$$

But i'm not sure how it works. The position states are $\left|x'\right>=\delta(x-x')$ and together with momentum states they satisfy $\left<x'|k\right>=e^{ikx'}$, so - $$\left<x'|k\right> =\sum_{k}e^{-ikx}\left<x'|k\right>=\sum_{k}e^{-ikx}e^{ikx'}=\sum_{k}e^{-ik(x-x')}$$

does this give $\delta(x-x')$ ?

• 1) Note that $\left|x'\right>=\delta(x-x')$ is a meaningless expression: the l.h.s. is a vector and the r.h.s. is a (generalised) function. 2) do you know how Fourier transforms work? – AccidentalFourierTransform Aug 24 '17 at 12:51
• @AccidentalFourierTransform 1)meaningful or meaningless? im not sure what youre intention. 2) yes. – proton Aug 24 '17 at 12:53
• 1) yes, sorry, I meant meaningless. 2) What happens if you write $\delta(x-x')$ as a Fourier series? what are its Fourier coefficients? – AccidentalFourierTransform Aug 24 '17 at 12:54
• @AccidentalFourierTransform 1) why meaningless? a function is a vector (at least mathematically). this is how you represent state positions to my knowledge. 2)the coefficients of $F[\delta(x-x')]$ would be $\cos(nx')$ or $\sin(nx')$ – proton Aug 24 '17 at 12:59
• proton, $|x'\rangle$ is a ket while $\delta(x - x') = \langle x | x' \rangle$ is a distribution and so $|x'\rangle = \delta(x - x')$ is as (not even) wrong as $|x'\rangle = \langle x | x' \rangle$ is. – Alfred Centauri Aug 24 '17 at 14:52

You've got the right idea, but there are several issues with the equations you've written down.

The complex Fourier series representation of the delta function is given by the identity:

$$\delta(x-x') = {1 \over 2\pi} \sum_{k=-\infty}^{\infty}e^{-ik(x-x')}$$

(See Eq 1.17.20 and 1.17.21 here for how to derive this as a limit of $\sum_{k=-n}^{n}$ as $n\to\infty$

So your conclusion is almost right, just missing a factor of $2\pi$.

The representation of a momentum state in terms of position eigenstates is $$|k\rangle = {1 \over 2\pi} \int_{x=-\infty}^{\infty}e^{ikx}|x\rangle$$

As a couple people complained about in the comments, you write $|x'\rangle = \delta(x-x')$ which is sloppy notation. Yes, $\delta(x-x')$ is the function in the position basis which represents $|x'\rangle$, but the way to write that is not with an equals sign. Or if you really want to use an equals sign, you write it like this: $\langle x|x'\rangle = \delta(x-x')$

I think to make everything work out here, using the normalization Lenny is using in his lecture, you have to assume that:

$$\langle k | k' \rangle = {1 \over 2\pi} \delta_{kk'}$$

Putting all this together, we get:

$$\langle k |x\rangle = \sum_{k'} e^{-ik'x}\langle k|k'\rangle = {1\over 2\pi} e^{ik'x}\delta_{kk'} = {1 \over 2\pi} e^{-ikx}$$

Or if you prefer to go the other direction:

$$\langle x |k\rangle = {1\over 2\pi} \int_{x'=-\infty}^\infty e^{ikx'}\langle x | x'\rangle = {1\over 2\pi} \int_{x' = -\infty}^\infty e^{ikx'}\delta(x-x') = {1 \over 2\pi} e^{ikx}$$

There is a more symmetric notation you can also use, where you have a bunch of factors of $\sqrt{2\pi}$ instead. This is more common to find in texts on non-relativistic quantum mechanics.