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I know that mathematically it doesn't matter what sign of $i$ we use to Fourier-transform a wavefunction from real- to momentum-space and vice versa, as long as we consistently change the sign when transforming it back to its original space. But I've seen many text-books (lecture notes) use $-i$, i.e.

$$ \phi(\vec{k}) = N^{-1} \int_{V_r} \psi(\vec{r})\ \ \exp(-i\vec{k}\cdot\vec{r})\ \ d^3r\tag{1}$$

to FT from the real- to momentum-space, and $+i$, i.e.

$$ \psi(\vec{r}) = N^{-1} \int_{V_k} \phi(\vec{k})\ \ \exp(+i\vec{k}\cdot\vec{r})\ \ d^3k\tag{2}$$

to FT from the momentum-space back to real-space. I might have missed something from my quantum mechanics course, but is there some physical reason(s) behind this convention?

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    $\begingroup$ @eranreches OK, it's now absorbed in $N^{-1}$. $\endgroup$
    – rnels12
    Nov 20, 2017 at 11:50
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    $\begingroup$ Then necessarily $N^2=(2\pi)^3$. $\endgroup$
    – Qmechanic
    Nov 26, 2017 at 13:05

2 Answers 2

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A wave that propagates in the $\vec{k}$ direction is given by

$$\left<\vec{r}\Big|\vec{k}\right>=\frac{1}{\left(2\pi\right)^{\frac{3}{2}}}e^{+i\vec{k}\cdot\vec{r}}$$

and thus it is common to decompose a function as

$$\psi\left(\vec{r}\right)=\frac{1}{\left(2\pi\right)^{\frac{3}{2}}}\int\tilde{\psi}\left(\vec{k}\right)e^{+i\vec{k}\cdot\vec{r}}{\rm d}^{3}k$$

If you add a minus sign to the exponent, then $\tilde{\psi}\left(\vec{k}\right)$ is the amplitude of a wave propagating in the $-\vec{k}$ direction.

EDIT 1: As @ZeroTheHero has pointed out, I've implicitly assumed that the time dependent component is

$$T\left(t\right)=e^{-i\omega t}$$

such that a wave traveling in the positive direction is given by

$$\psi\left(\vec{r},t\right)=\frac{1}{\left(2\pi\right)^{\frac{3}{2}}}e^{i\left(\vec{k}\cdot\vec{r}-\omega t\right)}$$

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  • $\begingroup$ But why does an arbitrary (wave)function $\psi(\vec r)$ need to be decomposed by wavefunctions propagating in the $+ \vec k$ direction? After all plane waves in FT are just elements of a complete basis, right? $\endgroup$
    – rnels12
    Nov 20, 2017 at 10:28
  • $\begingroup$ As you said, it is just a convention. The integration domain covers also the "$-\vec{k}$" cases. $\endgroup$
    – eranreches
    Nov 20, 2017 at 10:32
  • $\begingroup$ your argument just makes us going around in circles. $\endgroup$
    – rnels12
    Nov 20, 2017 at 11:56
  • $\begingroup$ @rnels12 I don't see why. The choice of sign merely tells you if $\tilde{\psi}\left(\vec{k}\right)$ is the amplitude of the $\vec{k}$ or $-\vec{k}$ frequency component. $\endgroup$
    – eranreches
    Nov 20, 2017 at 12:01
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    $\begingroup$ Actually this depends on the relative sign of $kx$ and $\omega t$. A wave moving in the positive $x$ direction would equally well be written as $e^{i(\omega t-kx)}$ or $e^{i(kx-\omega t)}$, i.e. the sign of the spatial part is not enough to pin down the direction. $\endgroup$ Nov 20, 2017 at 14:06
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The correct way to pin this down is to look at the full, time-dependent exponential. You can choose to describe a wave moving towards $+x$ either as \begin{align} \Psi(x,t)&=e^{i(kx-\omega t)}\, ,\tag{1} \\ \hbox{ or }\quad \Psi(x,t)&= e^{i(\omega t-kx)}\, .\tag{2} \end{align} Whichever you choose then determines the sign of the Fourier transform and its inverse. Assuming (1) for the purpose of the example, then the spatial part of a plane wave moving in the $+x$ direction would be $$ \psi(x)=\langle x\vert p\rangle=N e^{ipx/\hbar}\, . \tag{3} $$ with $N$ a normalization factor. The rest is basically using the completeness relation. If, for symmetry reason, one sets $$ \hat I=\int dx\,\vert x\rangle \langle x\vert = \int dp\,\vert p\rangle \langle p\vert \tag{4} $$ (which I think is easier to remember) then \begin{align} \langle p\vert \psi\rangle = \psi(p)&= \int dx\,\langle p\vert x\rangle \langle x\vert\psi\rangle\, ,\\ &= \int dx N^* e^{-ipx/\hbar}\psi(x)\, , \tag{5} \end{align} Note that in (5) I've explicitly used $\langle p\vert x\rangle = N^*e^{-ipx/\hbar}$, which will pin down the $N$ through \begin{align} \langle x\vert \psi\rangle = \psi(x)&= \int dp\,\langle x\vert p\rangle \langle p\vert\psi\rangle\, ,\\ &= \int dp N e^{ipx/\hbar}\psi(p)\, ,\\ &= \int dp dx N N^* \psi(x)\, ,\\ \end{align} With this you find $NN^*=1/(2\pi\hbar)$. The choice of $N=1/\sqrt{2\pi \hbar}$ makes the transformation between the direct and inverse transformation very symmetric since $$ \langle x\vert p\rangle = \frac{1}{\sqrt{2\pi\hbar}} e^{ipx/\hbar} =\langle p\vert x\rangle^*\, . $$

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