# Why does the integration over momentum has normalization constant of volume?

If I Fourier transform a wave function in position space, integration carries no normalization constant:

$$\displaystyle{\phi(k) \equiv \langle k|\psi\rangle = \sum\limits_x \langle k|x\rangle\langle x|\psi\rangle \equiv \int \limits_{-\infty}^{\infty}dx e^{-i kx}\psi(x) }.$$

$$\psi(x)$$ has units of $$1/\sqrt{L}$$, and $$\phi(k)$$ has units of $$\sqrt{L}$$, as expected. If I transform back into position space, it has a normalization constant of $$2\pi$$:

$$\displaystyle{\psi(x) \equiv \langle x|\psi\rangle = \sum\limits_k \langle x|k\rangle\langle k|\psi\rangle \equiv \frac{1}{2\pi}\int \limits_{-\infty}^{\infty}dk e^{i kx}\phi(k) }.$$

But, provided that the system has a finite volume (length), the summation can carry a normalization constant of length in continuum limit. This type of momentum summation is observed in condensed matter physics frequently:

$$\displaystyle{\sum \limits_k} \rightarrow \frac{L}{2\pi}\int \limits_{-\infty}^{\infty}dk.$$

But then units do not match? What is wrong here?

$$\displaystyle{\psi(x) \equiv \langle x|\psi\rangle = \sum\limits_k \langle x|k\rangle\langle k|\psi\rangle \equiv? \frac{L}{2\pi}\int \limits_{-\infty}^{\infty}dk e^{i kx}\phi(k) }.$$

• The mistake is in the very first line. The value of the expression $\langle k | \psi \rangle$ must be dimensionless. Therefore, we need proper dimensioned constants in front of that integral to make it dimensionless, whereas it presently has dimensions of $\sqrt{\text{length}}$ (because $\psi(x)$ has dimensions of $1/\sqrt{\text{length}}$). – DanielSank Jan 10 at 6:37
• Ah, I see... $\langle x|k\rangle = e^{ikx}/\sqrt{L}$? – gurluk Jan 10 at 6:50