# I Why is momentum the Fourier transform of position?

1. Apr 16, 2017

### entropy1

Apart from the fact that it is, what is the physical significance of the fact that you can get the momentum distribution of a particle by taking the Fourier transform of its position distribution?

2. Apr 16, 2017

### Staff: Mentor

1. Momentum is directly related to a wave property, the wavenumber $k$, via $p = \hbar k$.
2. Wavefunctions have the property of superposition. If $\psi_1 = e^{i k_1 x}$ and $\psi_2 = e^{i k_2 x}$ are solutions of the Schrödinger equation for a certain potential function V(x), then $A \psi_1 + B \psi_2$ is also a solution.

3. Apr 16, 2017

### entropy1

Why is that?
Do you mean an energy potential?

4. Apr 16, 2017

### Staff: Mentor

It comes from solving Schrodinger's equation. The momentum operator is $\hat{p}=i\hbar\frac{\partial}{\partial{x}}$ and that leads to the momentum appearing in the solutions as jtbell described. So your question may come down to asking why the momentum operator is defined the way it is.

5. Apr 16, 2017

### entropy1

I understand it can be found by differentiating the position expectation value with respect to time: $\frac{\mathrm{d} \langle X \rangle}{\mathrm{d} t}$, correct?

6. Apr 16, 2017

### Mentz114

This an 'explanation' I found somewhere.

If you want to know which momentum contributions are present in a given wave function, you have to expand that function in terms of the solutions of definite momentum. Fortunately, these solutions form an orthogonal basis of the Hilbert space so that the expansion becomes a simple inner product, namely $c(\mathbf{k},t) = \int_{-\infty}^\infty \exp(i(\mathbf{k}.\mathbf{r}-\omega t))^* \psi(\mathbf{r})d\mathbf{r}$. We can now apply the complex conjugation and move the time dependency out of the integral:$c(\mathbf{k},t)=\exp(i\omega t)\int_{-\infty}^{\infty}\exp(-i\mathbf{k}.\mathbf{r})\psi(\mathbf{r})d\mathbf{r}$. We have not assumed any time dependency of $\psi$, and so we can just assume that $\psi$ refers to a state at t=0 so that we get rid of the leading factor and have $c(\mathbf{k})=\int_{-\infty}^{\infty}\exp(-i\mathbf{k}.\mathbf{r})\psi(\mathbf{r})d\mathbf{r}$, which is the Fourier transform of $\psi(\mathbf{r})$.

7. Apr 16, 2017

### SergioPL

$\frac{\mathrm{d} \langle X \rangle}{\mathrm{d} t}$ would give the momentum expected value but not the momentum itself, the momentum depend on
$\frac{\mathrm{d} ψ }{\mathrm{d} x}$ and in order to be analyzed must be transformed to the momentum dominion through the Fourier Transform, because of the uncertainty principle, a particles momentum cannot be completely defined, except if the position is completely undefined as is the case of plane waves.

8. Apr 16, 2017

### atyy

Momentum and position being Fourier transform pairs are an expression of commutation relations, which give rise to various uncertainty relations.

9. Apr 17, 2017

### entropy1

I ment the momentum operator. (#5) $\frac{\mathrm{d} \langle X \rangle}{\mathrm{d} t}$ leads to the speed EV <V> which is the momentum EV <P> short for a mass constant m. <P> is $\langle \psi | \hat{p} | \psi \rangle$, and you can read $\hat{p}$ off the equation to be $\hat{p}=i\hbar\frac{\partial}{\partial{x}}$. (And you need the schroedinger equation)

How does that lead to the momentum being the Fourier transform of position?

Last edited: Apr 17, 2017
10. Apr 17, 2017

### vanhees71

It's, obviously!, nonsense to say that position is a Fourier transformation of momentum. What's true is that the position representation of Hilbert-space vectors ("wave function") is the Fourier transformation of the momentum representation (momentum-space wave function).

It can easily be proven that momenta, as in classical mechanics, are the generators of spatial translations, from which one gets the momentum eigendistribution as
$$\langle x|p \rangle=\frac{1}{\sqrt{2 \pi}} \exp(\mathrm{i} p x),$$
where for simplicity I consider only the one-dimensional case and use natural units with $\hbar=1$.

Now for any true Hilbert-space vector $|\psi \rangle$ the position representation is given by
$$\psi(x)=\langle x|\psi \rangle=\int_{\mathbb{R}} \mathrm{d}p \langle x|p \rangle \langle p|\psi \rangle=\int_{\mathbb{R}} \mathrm{d} p \frac{1}{\sqrt{2 \pi}} \exp(\mathrm{i} p x) \tilde{\psi}(p).$$
The inverse is
$$\tilde{\psi}(p)=\langle p|\psi \rangle = \int_{\mathbb{R}} \mathrm{d} x \langle p|x \rangle \langle x|\psi \rangle = \int_{\mathbb{R}} \mathrm{d} x \frac{1}{\sqrt{2 \pi}} \exp(-\mathrm{i} p x) \psi(x).$$

11. Apr 17, 2017

### entropy1

How does one come to this result?

12. Apr 17, 2017

### SergioPL

Yes... you are right.

13. Apr 18, 2017

### vanhees71

You use the commutator relations
$$[\hat{x},\hat{p}]=\mathrm{i}.$$
Now consider the unitary operator
$$\hat{T}(\xi) = \exp(-\mathrm{i} \hat{p} \xi)$$
and the corresponding transformation of the position operator
$$\hat{X}(\xi)=\hat{T}^{\dagger}(\xi) \hat{x} \hat{T}(\xi).$$
Take the derivative wrt. to $\xi$
$$\mathrm{d}_{\xi} \hat{X}(\xi)=-\mathrm{i} \hat{T}^{\dagger}(\xi) [\hat{x},\hat{p}] \hat{T}(\xi) = \hat{T}^{\dagger}(\xi) \hat{T}(\xi)=\hat{1}.$$
From this you get by integration
$$\hat{X}(\xi)=\hat{x}+\xi \hat{1}, \qquad (1)$$
where
Now let $|x \rangle$ be a generalized position eigenvector of $\hat{x}$ with eigenvalue $x$ (which you assume to exist). Then define
$$|\xi \rangle=\hat{T}(x)|x \rangle$$
and check what $\hat{x}$ does to it, using (1)
$$\hat{x} |\xi \rangle=\hat{x} \hat{T}(\xi) |x \rangle=\hat{T}(\xi) \hat{X}(\xi) |x \rangle=\hat{T}(\xi)(\hat{x}+\xi) |0 \rangle=(x+\xi) \hat{T}(\xi) |0 \rangle=(x+\xi) |\xi \rangle,$$
i.e., $|\xi \rangle$ is eigenvector of $\hat{x}$ with eigenvalue $x+\xi$.

This means, if $x$ has a generalized eigenvector with eigenvalue $x$ (which is necessarily real, because $\hat{x}$ is self-adjoint), then $\hat{x}$ as entire $\mathbb{R}$ as a spectrum, and we can write
$$|x \rangle=\hat{T}(x) |0 \rangle.$$
An analogous calculation shows the $\hat{x}$ generates translations in momentum space and thus, if $\hat{p}$ has any generalized eigenstate with a real eigenvalue, then it has entire $\mathbb{R}$ as a spectrum.

Now consider the momentum eigenvectur in position representation:
$$u_p(x)=\langle x|p \rangle=\langle \hat{T}(x) 0|p \rangle=\langle 0|\hat{T}^{\dagger}(x) p \rangle=\exp(\mathrm{i} p x) \langle 0|p \rangle=N_p \exp(\mathrm{i} p x).$$
Now you want to normalize this generalized function to a $\delta$ function in the sense
$$\langle p'|p \rangle=N_{p'}^* N_{p} \int_{\mathbb{R}} \mathrm{d} x \exp[\mathrm{i}(p-p') x]=2 \pi |N_{p}|^2 \delta(p-p') \stackrel{!}{=} \delta(p-p').$$
Thus, up to an irrelevant phase factor, $N_p=1/\sqrt{2 \pi}$, i.e.,
$$u_p(x)=\frac{1}{\sqrt{2 \pi}} \exp(\mathrm{i} x p),$$
which shows that the position and momentum representation of true Hilbert-space vectors is the Fourier transform to each other and one possible realization of the Hilbert space is as the function space $\mathrm{L}^2(\mathbb{R},\mathbb{C})$ in position or momentum representation ("wave mechanics").

The commutator relations of operators that represent observables thus determine the realization of the quantum-theoretical (rigged) Hilbert space. The commutator relations follow from symmetry principles via the Noether theorem. In our case it was the definition of momentum as the generator of spatial translations. Note that we assumed the existence of a position operator.

The full analysis of the Gailei group, which is the logically complete derivation of how the most general Hilbert space for a particle in non-relativistic QT may look, shows that for any physically interpretable unitary ray representation of the Galilei group, a position operator that fulfills the above commutation relation, exists, and mass is introduced as a socalled central charge of the Lie algebra of the Galilei group. For details, see Ballentine's book on QM.

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