In what base should we express the transformation coefficient for a wavefunction?

Question

We have, for example, that $psi(p) = int e^{ixp/hbar} psi(x), dx$, which isn't really hard to derive.
But in what base should we express $e^{ixp/hbar}$? In the $x$ or the $p$ basis?
Also, what are we integrating over? I suppose if we're in the $x$ basis, we go from negative infinity to infinity, but what about the reverse, from the $x$ to the $p$ basis? Do we integrate/sum over all possible momenta?

ZeroTheHero · Accepted Answer

The simplest way to remember this is to realize that the resolution of the identity for discrete bases
$$
hat{1}=sum_n vert nranglelangle nvert
$$
is a sum over all basis states.  The continuous case replaces the discrete sum by a continuous sum so that
begin{align}
hat{1}&=int_{-infty}^infty dx vert xrangle langle xvert, ,
&=int_{-infty}^infty dp vert prangle langle pvert  
end{align}
with $langle xvert prangle = langle pvert xrangle^* = frac{1}{sqrt{2pi hbar}} e^{ipx/hbar}$.  This is just a continous sum (i.e. an integral) over all basis states.
Note there that with these definitions direct and inverse Fourier transforms are symmetric in the placement and powers of the $2pi$ factors.  The extra $hbar$ comes in because the FT is usually defined using $k=p/hbar$ rather than $p$.

Buzz · Answer

The expression
$$tilde{psi}(p)=frac{1}{sqrt{2pihbar}}int_{-infty}^{+infty} dx,exp(-ipx/hbar)psi(x)$$
(using a slightly different notation from in the question*) is just an ordinary integral, and it tells you how to change the basis.  The wave function $psi(x)$ is the wave function in the position representation, $psi(x)=langle x|phirangle$.  What the equation says is that the Fourier transform, $tilde{psi}(p)$, is the wave function in the momentum representation, $psi(p)=tilde{psi}(p)=langle p|psirangle$.
Since the form taken by an eigenstate of position with eigenvalue $x_{0}$ in the momentum representation is $langle p|x_{0}rangle=exp(-ipx_{0}/hbar)$, the Fourier transform formula can actually be seen as an expansion of the momentum-state wave function $tilde{psi}(p)$ in a complete set of position eigenstates.  The function $tilde{psi}(p)$ is a $x$ states (an integral instead of a sum, since the eigenvalues of $p$ form a continuous, rather than discrete, spectrum), with the weight of each eigenstate given by $psi(x)=langle x|psirangle$. Note that this means the whole formula can be rewritten as
$$langle p|psirangle=frac{1}{sqrt{2pihbar}}int_{-infty}^{+infty} dx,langle p|xranglelangle x|psirangle=langle p|left(frac{1}{sqrt{2pihbar}}int_{-infty}^{+infty} dx,|xranglelangle x|right)|psirangle,$$
which makes sense, since
$${bf 1}=frac{1}{sqrt{2pihbar}}int_{-infty}^{+infty} dx,|xranglelangle x|$$
is just the identity operator, expressed as a integral over a complete set of position eigenstates.
The inverse relation is in integral over all $p$, to get $psi(x)$ from $tilde{psi}(p)$,
$$psi(x)=frac{1}{sqrt{2pihbar}}int_{-infty}^{+infty} dp,exp(ipx/hbar)tilde{psi}(p).$$
*I have included tilde to mark the Fourier transform, but it is not necessary.  The limits of integration are also given, and the constant factor ensures that if $psi(x)$ is normalized, then so is $tilde{psi}(p)$.  Most significantly, I have corrected the signs in the exponentials to match the usual conventions of quantum mechanics.

In what base should we express the transformation coefficient for a wavefunction?

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