- #1

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Is there a fairly simple explanation for this ? What leads us to this relation ?

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- Thread starter DoobleD
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- #1

- 259

- 20

Is there a fairly simple explanation for this ? What leads us to this relation ?

- #2

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$$[\hat{x},\hat{p}]=\mathrm{i},$$

where I set ##\hbar=1## for convenience. From that you can derive that in the position representation

$$\hat{p} \psi(x)=\langle x|\hat{p} \psi \rangle=-\mathrm{i} \partial_x \psi(x)=-\mathrm{i} \langle x|\psi \rangle.$$

The generalized momentum eigenstates are definied by

$$\hat{p} u_{p}(x)=p u_p(x).$$

Solving this equations for the eigenstates leads to

$$u_{p}(x)=N \exp(\mathrm{i} p x).$$

The normalization is most conveniently chosen such that

$$\langle p|p' \rangle=\int_{\mathbb{R}} \mathrm{d} x u_p^*(x) u_{p'}(x) = |N|^2 \int_{\mathbb{R}} \mathrm{d} p \exp[\mathrm{i}(p'-p)x] = (2 \pi)^3 |N|^2 \delta(p-p') \stackrel{!}{=} \delta(p-p') \; \Rightarrow \; N=\frac{1}{\sqrt{2 \pi}}.$$

Thus we have

$$u_p(x)=\langle x|p \rangle=\frac{1}{\sqrt{2 \pi}} \exp(\mathrm{i} p x).$$

From this you finally get for the wave function in momentum representation

$$\tilde{\psi}(p)=\langle p|\psi \rangle = \int_{\mathbb{R}} \mathrm{d} x \langle p|x \rangle \langle x |\psi \rangle = \int_{\mathbb{R}} \mathrm{d} x u_p^*(x) \psi(x) = \int_{\mathbb{R}} \mathrm{d} x \frac{1}{\sqrt{2 \pi}} \exp(-\mathrm{i} p x) \psi(x),$$

which is nothing else than the Fourier transformation of the wave function in position representation. In the same way you immediately get the inverse transformation

$$\psi(x)=\int_{\mathbb{R}} \mathrm{d} p u_p(x) \tilde{\psi}(p) = \int_{\mathbb{R}} \mathrm{d} p \frac{1}{\sqrt{2 \pi}} \exp(\mathrm{i} p x) \tilde{\psi}(p).$$

- #3

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This provided some insight. However my knowledge of QM and the associated maths is very incomplete and quite shaky. I think I'll come back to your answer later, it'll probably be much clearer then.

There's just one part I wonder about. This :

Here is what I read in plain english : "There exist a momentum operator and an associated eigenfunction. The eigenvalue is the momentum." If that's a correct reading, then why do we know there must exist a momentum operator, and its eigenfunction ? Hope the question makes sense. I wonder why someone would come with that idea.The generalized momentum eigenstates are definied by

glish ^pup(x)=pup(x).

Also in the mean time I have read the first chapter of Griffiths's textbook "Introduction to Quantum Mechanics (2nd Edition)". To introduce momentum, he takes the time derivative of the expected value of position, and using Schrödinger equation, he ends up with an expression for the expected value of momentum. Which does look like a Fourier transform. He seems to say this isn't really good "theoretical foundation" though, I suppose it serves more as an intro.

- #4

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- #5

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Thank you for the recommandations, I'll have a look at those !

- #6

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The underlying physical reason for this is the properties of the vacuum (of spacetime, specifically) and the complementarity of conservation of momentum and continuous symmetry of physics over position under Noether's Theorem. Position and momentum are intimately related by the nature of spacetime itself, and this leads to complementary Heisenberg uncertainty of position and momentum as well as the fact about the Fourier transforms of their wavefunctions that you are trying to understand.

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