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I have read in a couple of places that $\psi(p)$ and $\psi(q)$ are Fourier transforms of one another (e.g. Penrose). But isn't a Fourier transform simply a decomposition of a function into a sum or integral of other functions? Whereas the position and momentum wavefunctions are essentially different but related. They must preserve expectation values like the relationship of classical mechanics, $<p>=m~\frac{d<q>}{dt}$ (where $<p>$ and $<q>$ are now expectation values).

For example, a momentum wave packet that has a positive expectation value constant over time implies a position wave packet that moves over time in some direction. Simply saying there is Fourier transform seems to obscure this important relation.

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I am more or less agree. But that is not a question. –  Kostya Jan 26 '11 at 13:56
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Dear user1602, yes, $\psi(x)$ and $\tilde\psi(p)$ are Fourier transforms of one another. This answers the only real question you have asked. So if one knows the exact wave function as a function of position, one also knows the wave function as a function of momentum, and vice versa.

In particular, there is no "wave function" that would depend both on $x$ and $p$. Indeed, such a "wave function" would violate a basic principle of quantum mechanics, the uncertainty principle.

The wave function - that only depends on $x$ or that only depends on $p$ - remembers everything that a particle can and needs to remember about its position and momentum. For example, a good wave function describing a particle localized around $x_0$ and moving with momentum around $p_0$ is given by $$\psi_{x_0,p_0}(x) = C \exp\left(-K(x-x_0)^2 + ip_0 x/\hbar\right)$$ The constant $K$ determines the width but you see that because of the quadratic term, the wave function is only non-vanishing near $x_0$. On the other hand, the $ipx$ term guarantees that the particle is moving to the right with the right momentum. It's all encoded in the changing phase of the wave function. The more quickly the phase of $\psi(x)$ changes with $x$, the higher is the momentum of the particle. If the phase rotates clockwise or counter-clockwise, the particle is moving to the right or to the left, respectively.

The Fourier transform of the wave function above is something like $$\tilde \psi_{x_0,p_0}(p) = C' \exp\left(-(p-p_0)^2/K' - ip x_0/\hbar\right)$$ Just try it. Schrödinger's equation will guarantee that the wave packet is moving in the right direction - and by the right speed - encoded in $p_0$, and the center-of-mass position of the packet will change accordingly, too. The normalization constants $C,C'$ are physically irrelevant but may be chosen to normalize the state vectors to unity. The parameters $K,K'$ specifying the width are equal, up to a multiplication by a numerical constant and a power of $\hbar$: but it's true that the width in the $x$ representation is inverse to the width in the $p$ representation. That's implied by the uncertainty principle, too.

It is not true that one needs "wave functions" that would depend both on position and momentum. It's the whole point of the uncertainty principle that you may only specify the amplitudes with respect to one of these quantities - the other one doesn't commute with it. If one chooses $\psi(x)$, the position operator is a multiplication by $x$ and the momentum $p$ is simply the operator $-i\hbar\partial/\partial x$. Similarly, for $\tilde\psi(p)$, the momentum operator is the multiplication by $p$ and the position operator $x$ equals $+i\hbar \partial/\partial p$. It's pretty much symmetric with respect to $x,p$.

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I think what I had not understood is that the two functions express a single state in different terms. –  Adrianos Jan 26 '11 at 20:40
    
Yes, @user1602, it's just a different choice of basis to express the same state vector (in this case, both bases are "continuous", so sums over bases are replaced by integrals, but the logic is the same). –  Luboš Motl Feb 2 '11 at 15:37
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If you have a state $|\psi\rangle $, the position wave function is:

\begin{equation} \langle \vec{x}\mid \psi\rangle \end{equation}

and the momentum wave function is:

\begin{equation} \langle \vec{p}\mid \psi\rangle = \int\;d\vec{x}\; \langle\vec{p}\mid\vec{x}\rangle\langle \vec{x}\mid\psi\rangle \end{equation}

The reason why the two expressions are "different but related" is the same why they preserve the expectation values. It is called completeness relation $\int\; d\vec{x}\;\mid\vec{x}\rangle\langle\vec{x}\mid = 1$, $\int\; d\vec{p}\;\mid\vec{p}\rangle\langle\vec{p}\mid = 1$:

\begin{equation} \int\;d\vec{x}\; d\vec{x}^{\prime} \langle \psi\mid\vec{x}\rangle\langle\vec{x}\mid A\mid\vec{x}^{\prime}\rangle\langle\vec{x}^{\prime}\mid\psi\rangle \end{equation}

\begin{equation} =\int\;d\vec{x}\; d\vec{x}^{\prime}\; d\vec{p}\; d\vec{p}^{\prime} \langle \psi\mid\vec{p}\rangle\langle \vec{p}\mid\vec{x}\rangle\langle\vec{x}\mid A\mid\vec{x}^{\prime}\rangle\langle \vec{x}^{\prime}\mid\vec{p}^{\prime}\rangle\langle\vec{p}^{\prime}\mid\psi\rangle \end{equation}

\begin{equation} =\int\; d\vec{p}\; d\vec{p}^{\prime} \langle \psi\mid\vec{p}\rangle\langle\vec{p}\mid A\mid\vec{p}^{\prime}\rangle\langle\vec{p}^{\prime}\mid\psi\rangle \end{equation}

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I think the main thing that is giving you trouble here is what you understand the Fourier transform to be. When you say "But isn't a Fourier transform simply a decomposition of a function into a sum or integral of other functions?" I think you are confusing the 'Fourier transform' and the 'Fourier series'. Fourier was a pretty prolific guy, and as is always the case having many things named after the same person can be confusing. In any case, you now have your answer. If you can't work out how to compute a Fourier transform using the answers above (which are both correct) then have a look online for some examples. It's an incredibly powerful tool :)

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