Why is the momentum state of a particle in quantum mechanics given by the Fourier transform of its position state? For instance, in one dimension given by

$$\varphi(p)=\frac{1}{\sqrt{2\pi\hbar}}\int \mathrm dx \, e^{-i p x/\hbar} \psi(x).$$

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    $\begingroup$ Well, what else would it be? If you want to go from position space to momentum space (or equivalently time to frequency) then you perform a Fourier transform. $\endgroup$ – JamalS Dec 25 '16 at 22:12
  • $\begingroup$ Aren't there uncoutable many alternative possibilities of linear transformations to other representations? My question is, why do we choose the specific form above? $\endgroup$ – user56224 Dec 26 '16 at 16:05

Let's start from scratch. Take the positions eigenvectors, $\left|x\right>$. They are such that $X\left|x\right> = x\left|x\right>$. Now, take a general ket for a wavefunction, $\left|\psi\right>$. If we want to know $\psi(x)$, that is, the wavefunction in the position representation, then we take the following scalar product : $\left<x\right|\left|\psi\right> = \psi(x)$. Indeed, this is true since the position representation of $\left|x\right>$ is $\delta(x)$ (I can show this if need be). From this is also follows that $\int\left|x\right>\left<x\right|dx = I$ where I is the identity (called the completeness relation).

So, let's get back to the question. Analogously, we have that $\psi(p) = \left<p\right|\left|\psi\right> =\int \left<p\right|\left|x\right>\left<x\right|\left|\psi\right>dx$ using the completeness relation. All we have to do now, is determine $\left<p\right|\left|x\right>$. This is done by the defining equation of $\left|p\right>$ which simply is $P\left|p\right> = p\left|p\right>$.

Taking the scalar product with $\left<x\right|$ and using the positiong representation of $P = -i\hbar\nabla$ we get the following equation :

$$ -i\hbar\frac{d p(x)}{d x} = pp(x)$$

Where $p(x) = \left<x\right|\left|p\right>$

Solving this equation you find $p(x) = Ae^{ip/\hbar x}$

Finally, using the hermiticity properties of the scalar product and plugging back in our initial integral we get :

$$\psi(p) = \int Ae^{-ip/\hbar x}\psi(x)$$

The constant $A$ is taken to be $\frac{1}{\sqrt{2\pi\hbar}}$ arbitrarily to get the usual form of the Fourier transform. This is because since the position representation of the $p$ eigenvectors cannot be normalised, this constant $A$ is arbitrary.

  • $\begingroup$ As far as I understand, you solve the momentum eigenvalue equation in position space and then write down the inverse of the general solution, while the prefactor of the transformation is obtained by choosing a normalized basis. This scheme seems promising to me. So, what if the state in position space has compact support? For instance, a particle in a infinite well. Would you still apply the Fourier transformation with continuous momentum spectrum or better the Fourier series with a discrete momentum spectrum? What is the reason to prefer either of them? $\endgroup$ – user56224 Dec 26 '16 at 16:53
  • $\begingroup$ My guess would be that yes, you still apply the same fourier transform. Indeed, in the derivation above, I make no use of the Hamiltonian of the particle. So the results obtained above should hold regardless of wether the $\psi$ is a free particle solution or something else. $\endgroup$ – Frotaur Dec 26 '16 at 17:28

In general, a Fourier transform takes functions on a group $G$, or a space $X$ on which $G$ acts, and decomposes them in terms of characters of the group, such as $\chi : G \to S^1$, and the coefficients of the decomposition are encoded in the transformed function in the Pontryagin dual $\hat G$ of $G$.

Now for Euclidean space, $\mathbb R^n$, we can identify the Pontryagin dual $\hat{ \mathbb{R}}^n$ with itself. In particular, it is a locally compact group with $\mathbb R^n$, identifying $\xi \in \mathbb R^n$ as the frequency, for which $x \to \xi \cdot x.$ The Pontryagin dual in general is the group of all characters of $G$.

In general, the Fourier transform for $f \in L^1(G)$ is given by,

$$\hat f (\chi) = \int_G f(x)\bar{\chi(x)} d \mu(x)$$

where $d\mu$ is the Haar measure. Specializing now to the aforementioned case, one has,

$$\hat{f}(\xi) = \int_{\mathbb R^n} f(x)e^{-2\pi i \xi \cdot x} dx.$$

If we interpret the domain of $f$ as time, then the corresponding domain of the transform is in frequency space. For position, one has momentum space. That we can take $\psi(x)$ to $\hat \psi(p)$ is not exclusive to the wavefunction but can be performed on any suitable function.

Another check: in the exponential one has $e^{-i\omega t}$ and so one can deduce $\omega$ has dimensions of frequency for the argument to be dimensionless. Now, for position, you'd get something like $[L]^{-1}$ which is actually the wave vector $k$, but $p = \hbar k$ and so we can express the Fourier transform either in terms of the wave vector or the momentum.

We also normally work in natural units wherein $\hbar = 1$ and so we use either interchangeably.

  • $\begingroup$ Of course, Fourier transformations can be defined between various spaces. That is why I ask for the specific choice mentioned above. Actually, a check of dimensions can not be sufficient to justify the point. $\endgroup$ – user56224 Dec 26 '16 at 16:14

First, we need some definitions of the concepts we will deal with: the position and momentum representations. The position and momentum operators satisfy the commutation relation: \begin{equation} [X,P]=i\hbar \end{equation}

A representation of this algebra over the Hilbert space $L^2\left(\mathbb{R}\right)$, the position representation, is given by $(Xf)(x)=x\,f(x)$ and $Pf=-i\hbar f'$. The momentum representation is similar, just swapping $X$ and $P$ (and changing a sign), to obtain $(Pf)(p)=p\,f(p)$ and $Xf=i\hbar f'$.

Why do we use these definitions? Well, it makes sense to call position representation the one for which the position operator acts as multiplication by the argument of the function, because then that argument is interpreted as position. The same thing happens for the momentum representation. In that case, the argument of the functions can be interpreted as the momentum, as $P$ acts as multiplication by it.

Now we can address the question. The statement is that the Fourier transform takes functions in the position representation to the momentum representation. To prove it, let $f$ be a function in the position representation and observe that \begin{align} (X\hat{f})(p) &= \frac{1}{\sqrt{2\pi\hbar}}\int dx \,x\,e^{-ipx/\hbar}f(x) = \frac{i\hbar}{\sqrt{2\pi\hbar}}\frac{d}{dp}\int dx\,e^{-i px/\hbar}f(x) =i\hbar\,\hat{f}'(p) \\ (P\hat{f})(p) &= \frac{-i\hbar}{\sqrt{2\pi\hbar}}\int dx e^{-ipx/\hbar}f'(x) \overset{\text{by parts}}= \frac{p}{\sqrt{2\pi\hbar}}\int dx e^{-ipx/\hbar}f(x) =p\,\hat{f}(p), \end{align} where $\hat{f}$ is the Fourier transform.

  • $\begingroup$ In your explanation you already applied what I am asking for. I didn't ask for the momentum representation of the position operator etc. $\endgroup$ – user56224 Dec 26 '16 at 16:22
  • $\begingroup$ I defined the position and momentum representations and then showed that the Fourier transform takes one into the other. I thought that was what you were asking. I mean, I can't tell you why the Fourier transform gives you the momentum space from the position one if we don't have a definition for them to begin with. I'll edit to make this clearer $\endgroup$ – coconut Dec 26 '16 at 16:29
  • $\begingroup$ I have added a paragraph explaining why these definitions make sense. Maybe that helps you $\endgroup$ – coconut Dec 26 '16 at 16:43

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