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I read that any observable operator may be represented as:

$$\Omega = \sum_n \omega _n | \omega _n \rangle \langle \omega_n |$$

Where the little omegas are the eigenvectors/eigenvalues of the operator. Now, this definition is completely useless for doing any sort of calculation.

  1. So suppose I want to find the position representation of the momentum operator. I have $$P = \int_{-\infty}^{\infty}dp \ p | p \rangle \langle p|.$$ In position representation, this becomes $$\langle x | P = \int_{-\infty}^{\infty}dp \ p \langle x| p \rangle \langle p|.$$ This is supposed to be equal to $-i \hbar \partial_x$. But how does one get there?

  2. How to find out the overlap $\langle x| p \rangle$?

  3. What additional input does one have to make?

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Comment to the question (v2): Subquestions 2 and 3 is essentially a duplicate of this Phys.SE post. –  Qmechanic Mar 7 at 20:18
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1 Answer 1

One gets there by noting that $\langle x | p \rangle = e^{i p x/\hbar}$ is a plane wave, and you have to throw on the test wave function to talk about the derivative operation. So, you are worried about \begin{equation} \langle x| P | \Psi \rangle = \int dp ~p~ e^{i p x/\hbar} ~\langle p | \Psi \rangle \end{equation} From there you note that you can get the $p$ inside the integral by taking an $x$-derivative and after some algebra and removing the identity operator for $|p \rangle \langle p|$ you find that \begin{equation} \langle x| P | \Psi \rangle = - i \hbar \partial_x \langle x | \Psi \rangle \end{equation}

The point is that you really need to be careful about an operator and its basis-dependent representation. $\hat{P} = - i \hbar \partial_x$ in the x basis. It has other representations in other basis sets, and you have to specify that when you are writing down a concrete representation. It's similar to how the position vector $\vec{r}$ exists, but then the individual components have a basis representation in Cartesian, or cylindrical, or hyperbolic, or …, coordinates.

Edit: just worked through the derivation You want to know the matrix element $\langle x|P|\Psi \rangle$. Then you have, clearly, that \begin{equation} \langle x | P | \Psi \rangle = \int dx' \langle x | P | x' \rangle \langle x' | \Psi \rangle \end{equation} From the commutator relation matrix element, you get that \begin{equation} \langle x'| \left [ X, P \right ]| x \rangle = - i \hbar \langle x' | x \rangle \end{equation} which leads to \begin{equation} - i \hbar \frac{\langle x' | x \rangle}{x' - x} = \langle x' | P | x \rangle \end{equation} so then we get that \begin{equation} \langle x | P | \Psi \rangle = \lim_{x' \rightarrow x} - i \hbar \frac{\langle x' | \Psi \rangle}{x' - x} = - i \hbar \partial_x \langle x | \Psi \rangle \end{equation} Make $|\Psi \rangle = | p \rangle$ and you get the plane wave solution, with the caveat that there may be a sign error or two in here.

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I don't understand where this came from: $\langle x | p \rangle = e^{i p x/\hbar}$ –  DepeHb Mar 7 at 21:10
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@DepeHb: i.imgur.com/IchgG0c.png Momentum is that which is conserved if you translate the system. Turns out the operator $T_d$ which shifts a function $f(x)$ by $d$ can be written as $\exp\left(i\hbar\,d\,\left(\frac{1}{i\hbar}\frac{\partial}{\partial x}\right)\right)$. The generator is the momentum operator $\propto\frac{\partial}{\partial x}$ and so as function of $x$, the eigenstates $\left|p\right\rangle$ of that operator must be $\mathrm{e}^{cpx}$. –  NikolajK Mar 7 at 22:15
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Actually, you can derive it from the commutator relations. I can't remember the derivation off the top of my head, but Sakurai's Modern Quantum Mechanics in Chapter 1 I seem to recall has it spelled out in detail. –  webb Mar 7 at 23:50
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