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I'm starting a more 'formal' course on quantum mechanics, and we're going through momentum and position space, which I'm kind of confused on. I would just like clarification on this since I'm not understanding my teacher or my textbook very well. I'm just going to list what I think is going on and would love it if those who understand more could help me out!

  1. There is a 'big, general' state vector $|s(t)\rangle$ that contains all the information of the system. If I want to find the position wavefunction, that is the description of the system in terms of position, $x$ I compute $\langle x|s(t)\rangle$, where $|x\rangle$ is an eigenfunction of the position operator $\hat{x}$ with eigenvalue $x$. How is this exactly computed? I know the eigenfunction must be of the form $\delta(y-x)$ so do I calculate: $$\int_{-\infty}^{\infty}dy\ \delta(y-x)s(t)$$ but this doesn't seem right.
  2. Similarly, if I wanted to find a description of the system in terms of momentum, $p$ I compute $\langle p|s(t)\rangle$. My professor does the following: He sets up the eigenvalue problem $$P|p\rangle=p|p\rangle$$ and then has $$\langle x|P|p\rangle=p\langle x|p\rangle$$ and then claims $$\langle x|P|p\rangle=-i\hbar\partial_x \langle x|p\rangle$$ Did he just pull the momentum operator out of the bra-ket? Why is he allowed to do this? From what I understand, he seems to be claiming $\langle x|P|p\rangle=P \langle x|p\rangle$? He then goes on to show that $$ \langle x|p\rangle=\dfrac{1}{\sqrt{2\pi\hbar}}e^{(ip/\hbar)x}$$ which I follow, but I'm lost on why he is computing $ \langle x|p\rangle$ in the first place. Why couldn't he just compute $|p\rangle$ instead?
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Did he just pull the momentum operator out of the bra-ket? Why is he allowed to do this? From what I understand, he seems to be claiming $\langle x|P|p\rangle=P \langle x|p\rangle$?

No, that's not what he did. He's using the correct way to interface between the abstract Hilbert-space momentum operator $\hat P$ and its position-representation action as a derivative; this may well be the first time you're exposed to that.

It might help to first hold off from using Dirac notation, and just to consider an arbitrary wavefunction $\psi\in\mathcal H$ in Hilbert space. The momentum operator $\hat P:\mathcal H \to \mathcal H$ takes our wavefunction and gives us a new one, $\phi = \hat P\psi$, whose value is given by the derivative of the old one: $$ \left(\hat P\psi\right)(x) = \phi(x) = -i\hbar \psi'(x). \tag 1 $$

If you now put the Dirac notation back in, the way to connect the abstract state $|\psi⟩$ to its position wavefunction is to query it with the position functional $⟨x|$, so that e.g. $⟨x|\psi⟩=\psi(x)$. With that in mind, then, $(1)$ tells us that the position wavefunction of $\hat P|\psi⟩$ is just the spatial derivative of $⟨x|\psi⟩$, i.e. $$ ⟨x|\hat P|\psi⟩ = -i\hbar\frac{\mathrm d}{\mathrm dx}⟨x|\psi⟩. \tag 2 $$

And now I'm going to do something crazier still. Consider two the following procedures: (i) take a state $|\psi⟩$ act on it with the abstract $\hat P$, and then take its position state wavefunction at $x$; and (ii) take a state $|\psi⟩$, get its position-state wavefunction, and get ($-i\hbar$ times) the derivative of that at $x$. Both of those describe linear functionals $\mathcal H \to \mathbb C$ because they start with states and give us complex numbers, and $(2)$ tells us that they're equal on all their inputs in $\mathcal H$, which means that they're equal as functions: $$ ⟨x|\hat P = -i\hbar\frac{\mathrm d}{\mathrm dx}⟨x|. \tag 3 $$ In other words, you can essentially "cancel out the $|\psi⟩$" from $(2)$ in a rigorous way.

In the specific example of your professor's calculation, he's using the general property $⟨x|\hat P|\psi⟩ = -i\hbar\frac{\mathrm d}{\mathrm dx}⟨x|\psi⟩$ of the momentum operator for the specific case where $|\psi⟩=|p⟩$ is an eigenstate of momentum, $\hat P|p⟩=p|p⟩$, which gets turned by the general property into a differential equation for the position-space wavefunction of $|p⟩$.

I hope that clears up the handling, at least partially.

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  • $\begingroup$ Alright, I think I'm getting a picture of what's going on. Can I put it in this more layman's way: we can't really express $\hat{P}|\psi\rangle$ but we understand how this works in the position basis, $\langle x|\hat{P}|\psi\rangle$, which is just the positional derivative of the position wavefunction times a factor. $\endgroup$ – Ayumu Kasugano Nov 5 '17 at 1:55
  • $\begingroup$ Yeah, that is correct. $\endgroup$ – Emilio Pisanty Nov 5 '17 at 8:46
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This is because $\langle x\vert p\rangle$ is “known”: it is the spatial wave function for a particle with definite momentum, and this must be proportional to the plane wave solution $e^{i p x/\hbar}$ since plane wave have well-defined $k$-vector and thus well-defined $p$. The factor $\sqrt{2\pi\hbar}$ comes from Fourier analysis.

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