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The other answer are all correct, but I want to give another outlook, which I believe is the most self-consistent one. Clearly, we have to start somewhere. My starting point is to define momentum as the quantity that is conserved as a consequence of translational symmetry.

In quantum mechanics the translation operator for a single spin-less particle acts on the states as $$ (T_a \psi)(x)=\psi(x-a) $$ and its generator is$^1$ $$ G=i\hbar\frac{d}{dx} \quad\iff\quad T_a=e^{iaG/\hbar}. $$

If the system is translationally symmetric$^2$ the generator $G$ is conserved, as follows from general arguments in quantum mechanics. We conventionally consider its opposite as the momentum operator, that is clearly conserved as well.

Now, the eigenstates of the momentum operator are the states $\lvert p\rangle$ whose wavefunctions are$^3$ $$ \langle x\vert p\rangle=e^{ipx/\hbar} $$ and therefore the probability of a generic state $\vert\psi\rangle$ to have momentum $p$ is $$ \langle p\vert\psi\rangle=\int\!dx\,\langle p\vert x\rangle\langle x\vert\psi\rangle =\int\!dx\,e^{-ipx/\hbar}\psi(x). $$ This result says that the probability density of the momentum is the Fourier transform of the probability density of the position$^4$. This is how position and momentum are related through the Fourier transform.

useless nitpicking:

$^1$a nice argument for this is \begin{align} (e^{iaG/\hbar}\psi)(x)&=\left(e^{-a\frac{d}{dx}}\psi\right)(x) \\&= \left[\left(1-a\frac{d}{dx}+a^2\frac{d^2}{dx^2}+\dots\right)\psi\right](x) \\&= \psi(x)-\frac{d\psi}{dx}(x)\,a+\frac{d^2\psi}{dx^2}(x)\,a^2+\dots \\&= \psi(x-a) \end{align} (whose downside is that it's valid only for analytic functions)

$^2$this does not mean that the states do not change under translations, but the hamiltonian doesn't (a point that sometimes has confused me in other contexts), as in classical mechanics, or that probabilities amplitude are preserved, as in Wigner theorem

$^3$this is because in position representation the eigenvalue equation for the momentum reads $$ -i\hbar\frac{d\phi_p}{dx}(x)=p\phi_p(x) $$ that has the (unique) aforementioned solution (note that they are not normalizable so they are actually generalized eigenstates)

$^4$a bit loosely speaking, since the probabilities are their $|\cdot|^2$

The other answer are all correct, but I want to give another outlook, which I believe is the most self-consistent one. Clearly, we have to start somewhere. My starting point is to define momentum as the quantity that is conserved as a consequence of translational symmetry.

In quantum mechanics the translation operator for a single spin-less particle acts on the states as $$ (T_a \psi)(x)=\psi(x-a) $$ and its generator is$^1$ $$ G=i\hbar\frac{d}{dx} \quad\iff\quad T_a=e^{iaG/\hbar}. $$

If the system is translationally symmetric the generator $G$ is conserved, as follows from general arguments in quantum mechanics. We conventionally consider its opposite as the momentum operator, that is clearly conserved as well.

Now, the eigenstates of the momentum operator are the states $\lvert p\rangle$ whose wavefunctions are$^2$ $$ \langle x\vert p\rangle=e^{ipx/\hbar} $$ and therefore the probability of a generic state $\vert\psi\rangle$ to have momentum $p$ is $$ \langle p\vert\psi\rangle=\int\!dx\,\langle p\vert x\rangle\langle x\vert\psi\rangle =\int\!dx\,e^{-ipx/\hbar}\psi(x). $$ This result says that the probability density of the momentum is the Fourier transform of the probability density of the position$^3$. This is how position and momentum are related through the Fourier transform.

useless nitpicking:

$^1$a way to see this is to compare the expansions \begin{align} (T_a \psi)(x) &= \psi(x-a) = \psi(x)-a\frac{d\psi}{dx}(x) + o(a^2)\\ &= (e^{iaG/\hbar}\psi)(x) = \psi(x) + \frac{ia}{\hbar}(G\psi)(x) + o(a^2) \end{align}

$^2$this is because in position representation the eigenvalue equation for the momentum reads $$ -i\hbar\frac{d\phi_p}{dx}(x)=p\phi_p(x) $$ that has the (unique) aforementioned solution (note that they are not normalizable so they are actually generalized eigenstates)

$^3$a bit loosely speaking, since the probabilities are their $|\cdot|^2$

The other answer are all correct, but I want to give another outlook, which I believe is the most self-consistent one. Clearly, we have to start somewhere. My starting point is to define momentum as the quantity that is conserved as a consequence of translational symmetry.

In quantum mechanics the translation operator for a single spin-less particle acts on the states as $$ (T_a \psi)(x)=\psi(x-a) $$ and its generator is$^1$ $$ G=i\hbar\frac{d}{dx} \quad\iff\quad T_a=e^{iaG/\hbar}. $$

If the system is translationally symmetric$^2$ the generator $G$ is conserved, as follows from general arguments in quantum mechanics. We conventionally consider its opposite as the momentum operator, that is clearly conserved as well.

Now, the eigenstates of the momentum operator are the states $\lvert p\rangle$ whose wavefunctions are$^3$ $$ \langle x\vert p\rangle=e^{ipx/\hbar} $$ and therefore the probability of a generic state $\vert\psi\rangle$ to have momentum $p$ is $$ \langle p\vert\psi\rangle=\int\!dx\,\langle p\vert x\rangle\langle x\vert\psi\rangle =\int\!dx\,e^{-ipx/\hbar}\psi(x). $$ This result says that the probability density of the momentum is the Fourier transform of the probability density of the position$^4$. This is how position and momentum are related through the Fourier transform.

useless nitpicking:

$^1$a nice argument for this is \begin{align} (e^{iaG/\hbar}\psi)(x)&=\left(e^{-a\frac{d}{dx}}\psi\right)(x) \\&= \left[\left(1-a\frac{d}{dx}+a^2\frac{d^2}{dx^2}+\dots\right)\psi\right](x) \\&= \psi(x)-\frac{d\psi}{dx}(x)\,a+\frac{d^2\psi}{dx^2}(x)\,a^2+\dots \\&= \psi(x-a) \end{align} (whose downside is that it's valid only for analytic functions)

$^2$this does not mean that the states do not change under translations, it is the hamiltonian that doesn't (a point that sometimes has confused me in other contexts), as in classical mechanics

$^3$this is because in position representation the eigenvalue equation for the momentum reads $$ -i\hbar\frac{d\phi_p}{dx}(x)=p\phi_p(x) $$ that has the (unique) aforementioned solution (note that they are not normalizable so they are actually generalized eigenstates)

$^4$a bit loosely speaking, since the probabilities are their $|\cdot|^2$

The other answer are all correct, but I want to give another outlook, which I believe is the most self-consistent one. Clearly, we have to start somewhere. My starting point is to define momentum as the quantity that is conserved as a consequence of translational symmetry.

In quantum mechanics the translation operator for a single spin-less particle acts on the states as $$ (T_a \psi)(x)=\psi(x-a) $$ and its generator is$^1$ $$ G=i\hbar\frac{d}{dx} \quad\iff\quad T_a=e^{iaG/\hbar}. $$

If the system is translationally symmetric$^2$ the generator $G$ is conserved, as follows from general arguments in quantum mechanics. We conventionally consider its opposite as the momentum operator, that is clearly conserved as well.

Now, the eigenstates of the momentum operator are the states $\lvert p\rangle$ whose wavefunctions are$^3$ $$ \langle x\vert p\rangle=e^{ipx/\hbar} $$ and therefore the probability of a generic state $\vert\psi\rangle$ to have momentum $p$ is $$ \langle p\vert\psi\rangle=\int\!dx\,\langle p\vert x\rangle\langle x\vert\psi\rangle =\int\!dx\,e^{-ipx/\hbar}\psi(x). $$ This result says that the probability density of the momentum is the Fourier transform of the probability density of the position$^4$. This is how position and momentum are related through the Fourier transform.

useless nitpicking:

$^1$a nice argument for this is \begin{align} (e^{iaG/\hbar}\psi)(x)&=\left(e^{-a\frac{d}{dx}}\psi\right)(x) \\&= \left[\left(1-a\frac{d}{dx}+a^2\frac{d^2}{dx^2}+\dots\right)\psi\right](x) \\&= \psi(x)-\frac{d\psi}{dx}(x)\,a+\frac{d^2\psi}{dx^2}(x)\,a^2+\dots \\&= \psi(x-a) \end{align} (whose downside is that it's valid only for analytic functions)

$^2$this does not mean that the states do not change under translations, but the hamiltonian doesn't (a point that sometimes has confused me in other contexts), as in classical mechanics, or that probabilities amplitude are preserved, as in Wigner theorem

$^3$this is because in position representation the eigenvalue equation for the momentum reads $$ -i\hbar\frac{d\phi_p}{dx}(x)=p\phi_p(x) $$ that has the (unique) aforementioned solution (note that they are not normalizable so they are actually generalized eigenstates)

$^4$a bit loosely speaking, since the probabilities are their $|\cdot|^2$

The other answer are all correct, but I want to give another outlook, which I believe is the most self-consistent one. Clearly, we have to start somewhere. My starting point is to define momentum as the quantity that is conserved as a consequence of translational symmetry.

In quantum mechanics the translation operator for a single spin-less particle acts on the states as $$ (T_a \psi)(x)=\psi(x-a) $$ and its generator is$^1$ $$ G=i\hbar\frac{d}{dx} \quad\iff\quad T_a=e^{iaG/\hbar}. $$

If the system is translationally symmetric$^2$ the generator $G$ is conserved, as follows from general arguments in quantum mechanics. We conventionally consider its opposite as the momentum operator, that is clearly conserved as well.

Now, the eigenstates of the momentum operator are the states $\lvert p\rangle$ whose wavefunctions are$^3$ $$ \langle x\vert p\rangle=e^{ipx/\hbar} $$ and therefore the probability of a generic state $\vert\psi\rangle$ to have momentum $p$ is $$ \langle p\vert\psi\rangle=\int\!dx\,\langle p\vert x\rangle\langle x\vert\psi\rangle =\int\!dx\,e^{-ipx/\hbar}\psi(x). $$ This result says that the probability density of the momentum is the Fourier transform of the probability density of the position$^4$. This is how position and momentum are related through the Fourier transform.

useless nitpicking:

$^1$a nice argument for this is \begin{align} (e^{iaG/\hbar}\psi)(x)&=\left(e^{-a\frac{d}{dx}}\psi\right)(x) \\&= \left[\left(1-a\frac{d}{dx}+a^2\frac{d^2}{dx^2}+\dots\right)\psi\right](x) \\&= \psi(x)-\frac{d\psi}{dx}(x)\,a+\frac{d^2\psi}{dx^2}(x)\,a^2+\dots \\&= \psi(x-a) \end{align} (whose downside is that it's valid only for analytic functions)

$^2$this does not mean that the states do not change under translations, it is the hamiltonian that doesn't (a point that sometimes has confused me in other contexts), as in classical mechanics

$^3$this is because in position representation the eigenvalue equation for the momentum reads $$ -i\hbar\frac{d\phi_p}{dx}(x)=p\phi_p(x) $$ that has the (unique) aforementioned solution (note that they are not normalizable so they are actually generalized eigenstates)

$^4$a bit loosely speaking, since the probabilities are their $|\cdot|^2$ì

The other answer are all correct, but I want to give another outlook, which I believe is the most self-consistent one. Clearly, we have to start somewhere. My starting point is to define momentum as the quantity that is conserved as a consequence of translational symmetry.

In quantum mechanics the translation operator for a single spin-less particle acts on the states as $$ (T_a \psi)(x)=\psi(x-a) $$ and its generator is$^1$ $$ G=i\hbar\frac{d}{dx} \quad\iff\quad T_a=e^{iaG/\hbar}. $$

If the system is translationally symmetric$^2$ the generator $G$ is conserved, as follows from general arguments in quantum mechanics. We conventionally consider its opposite as the momentum operator, that is clearly conserved as well.

Now, the eigenstates of the momentum operator are the states $\lvert p\rangle$ whose wavefunctions are$^3$ $$ \langle x\vert p\rangle=e^{ipx/\hbar} $$ and therefore the probability of a generic state $\vert\psi\rangle$ to have momentum $p$ is $$ \langle p\vert\psi\rangle=\int\!dx\,\langle p\vert x\rangle\langle x\vert\psi\rangle =\int\!dx\,e^{-ipx/\hbar}\psi(x). $$ This result says that the probability density of the momentum is the Fourier transform of the probability density of the position$^4$. This is how position and momentum are related through the Fourier transform.

useless nitpicking:

$^1$a nice argument for this is \begin{align} (e^{iaG/\hbar}\psi)(x)&=\left(e^{-a\frac{d}{dx}}\psi\right)(x) \\&= \left[\left(1-a\frac{d}{dx}+a^2\frac{d^2}{dx^2}+\dots\right)\psi\right](x) \\&= \psi(x)-\frac{d\psi}{dx}(x)\,a+\frac{d^2\psi}{dx^2}(x)\,a^2+\dots \\&= \psi(x-a) \end{align} (whose downside is that it's valid only for analytic functions)

$^2$this does not mean that the states do not change under translations, it is the hamiltonian that doesn't (a point that sometimes has confused me in other contexts), as in classical mechanics

$^3$this is because in position representation the eigenvalue equation for the momentum reads $$ -i\hbar\frac{d\phi_p}{dx}(x)=p\phi_p(x) $$ that has the (unique) aforementioned solution (note that they are not normalizable so they are actually generalized eigenstates)

$^4$a bit loosely speaking, since the probabilities are their $|\cdot|^2$