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I understand that in quantum mechanics every vector from the state space (e.g. $|\Psi(t) \rangle$) can be projected on to observables such as the momentum space basis vectors $|p\rangle$ or position space basis vectors $|x\rangle$ as subspaces of the phase space. But it still isn't really clear to me, why exactly the momentum vectors are the Fourier transforms of the position space vectors. Or does one have to think always in the context of De-Broglie waves so that one has to deal with such tranformations? It would really help, if someone coud give an intuitive example why the Fourier-transform relates those two bases.

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How do you want to define momentum to begin with? Perhaps the best way is to define the momentum operator $\hat p$ to be the operator that satisfies the commutation relation $$[\hat x,\hat p]=i\hbar$$ with the position operator $\hat x$. You might like this definition because it has the same structure as the canonical Poisson bracket in classical mechanics $\{x,p\}=1$.

Then suppose we are working in the position basis so that $\hat x |x\rangle=x |x\rangle$. Now let's find the expression for $\hat p$ in this basis. We can do that by looking for an expression that satisfies the commutation relation. You can convince yourself that this is $-i\hbar(d/dx)$.

But this is exactly the relationship that the Fourier transform would give you! Recall that taking the derivative of a function is the same thing as multiplying the Fourier transform of that function by the conjugate variable (up to some factors):

For a function $\psi(x)$ and its Fourier transform $\psi(p)$ we have $$-i\frac{d}{dx}\psi(x) = -i\frac{d}{dx}\int_{-\infty}^\infty e^{ipx} \psi(p)\, dp = \int_{-\infty}^\infty e^{ipx} p\,\psi(p)\, dp.$$ So we see that the Fourier transform $\psi(p)$ is an eigenbasis of the operator $\hat p$.

(Footnote: in the end I set $\hbar=1$ to not clutter the usual form of the Fourier transform.)

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Consider a particle in some state $|\Psi\rangle$, and let's look at how its position and momentum space wavefunctions $\psi(x) = \langle x | \Psi \rangle$ and $\widetilde{\psi}(p) = \langle p | \Psi \rangle$ are related. The first step is to add insert the identity in between either of the equations, using the fact that:

$$\mathbb{I} = \int_{-\infty}^\infty \text{d}p \, |p\rangle\langle p|,$$

$$\psi(x) = \int_{-\infty}^\infty \text{d}p\, \langle x |p\rangle \langle p|\Psi\rangle = \int_{-\infty}^\infty \text{d}p\, \langle x |p\rangle \widetilde{\psi}(p).$$

What remains is to show that $\langle x| p \rangle = e^{ipx/\hbar}$. To do this, we look at the relation that defines $|p\rangle$: $$\hat{p} |p\rangle = p |p\rangle.$$ In the position basis, given the form of the momentum operator, this translates into $$\frac{\hbar}{i}\frac{\text{d}}{\text{d}x} \langle x| p\rangle = p \langle x| p\rangle,$$

which is a simple differential equation that can be solved to show that: $$\langle x | p\rangle = e^{ipx/\hbar},$$ modulo some normalisation factor. Plugging this back into the equation above, $$\psi(x) = \int_{-\infty}^\infty \text{d}p\, e^{ipx/\hbar}\widetilde{\psi}(p),$$

showing that the functions $\psi(x)$ and $\widetilde{\psi}(p)$ are indeed Fourier Transforms of each other.

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    $\begingroup$ Thank you very much, I think this demonstrates what I asked very good. $\endgroup$ – Görgün Dec 27 '20 at 19:15

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