QM: Coordinate & Momentum Representation w/ Fourier Transform

[LagrangeEuler]

Coordinate representation $\psi=\psi(x)$
Momentum representation $\psi=\psi(p)$
Fourier transform
\psi(p)=\frac{1}{\sqrt{2\pi\hbar}}\int^{\infty}_{-\infty}\psi(x)e^{-\frac{ipx}{\hbar}}dx
I'm confused with this $\hbar$? Why not
\psi(p)=\frac{1}{\sqrt{2\pi}}\frac{1}{\hbar}\int^{\infty}_{-\infty}\psi(x)e^{-\frac{ipx}{\hbar}}dx?

 

[Bill_K]

Because ∫exp(ikx) dk ≡ 2π δ(x). And therefore ∫exp(ipx/ħ) dp = 2π δ(x/ħ) = 2πħ δ(x).

Look at the normalization integral, with the 1/√2πħ factors included:

1 = ∫ψ*(p) ψ(p) dp = 1/(2πħ) ∫∫∫ψ*(x) exp(ipx/ħ) ψ*(x') exp(-ipx'/ħ) dp dx dx'

Do the p integral, using line 1:

= 1/(2πħ) ∫∫ψ*(x) ψ*(x') (2πħ) δ(x - x') dx dx'

Next do the x' integral, using the δ-fn. See we again get the correct normalization:

1 = ∫ψ*(x) ψ*(x) dx

Note that the factors of 2πħ canceled, showing that the 1/√2πħ was necessary!

EDIT: Hold on! I see you asked the exact same question on two consecutive threads!