operators - Representations in quantum mechanics

This might be a very simple question. I just want someone to point me the right direction to understand things like this: $$ \langle x|x'\rangle=\delta(x-x') \\ \psi(x)=\langle x|\psi\rangle \\ \tilde{\psi}(p) = \langle p|\psi\rangle \\ \langle x|p\rangle=\frac{1}{\sqrt{2\pi \hbar}}\exp(ipx/\hbar) $$ I am using Griffiths textbook, but it's too confusing. Where (on the internet) can I find a easy approach to understand these representations?

Answer

Bra-ket notation is so terribly convenient that it gets lots of abuse. There are a lot of common conventions which don't necessarily seem consistent with each other at first glance. Here are some comments which may help you:

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  $\left|\psi\right>$ is a general particle state, perhaps specified elsewhere, perhaps to be specified later. If you have several complicated states to refer to, usually people use other letters from near the end of the Greek alphabet: $\left|\chi\right>$, $\left|\phi\right>$, etc.

  
  


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  $\left|x_0\right>$ represents an eigenstate of the position operator $\hat x$: a particle which is absolutely, definitely located at $x_0$. Similarly, $\left|x_1\right>$ is another position eigenstate, a particle at $x_1$. These two states will have zero overlap unless $x_0$ and $x_1$ are the same, so the overlap $\left< x_1 \middle| x_0\right>$ is a delta function, $\delta(x_0-x_1)$. If there's only one position in the problem, you might call it $x$ instead of $x_0$; sometimes this is too close to the name of the operator, $\hat x$, and I get confused.

  
  


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  Similarly, $\left|p_0\right>$ represents an eigenstate of the momentum operator $\hat p$: a particle that really, definitely has momentum $p_0$.

  
  


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  $\left$, the overlap between a general state and the eigenstate definitely located at $x_0$, gives you the matrix element (whose square is the probability) for finding the particle at $x_0$. In the Schroedinger notation this is written as $\psi(x_0)$, the value of the wavefunction at $x_0$. However if you can do the integral $\left$ for arbitrary $x$, rather than for some specific $x_0$, you can recover the Schroedinger wavefunction.

  
  


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  Similarly, $\left$ gives the probability of finding the particle with momentum $p_0$. If you can do the overlap integral for arbitrary $p$, you will have an expression for the wavefunction in momentum space. This is a terribly useful technique, especially for solid-state physics; it amounts to taking the Fourier transform of the more familiar position-space wavefunction. It's the same state $\left|\psi\right>$ that we used to find $\psi(x)$, but it probably has a completely different functional form, so it gets a different-but-related name, $\tilde\psi(p)$.

  
  


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  The representation in position space of a wavefunction with definite momentum $p_0$ is a plane wave with a single frequency, $\left = \psi_{p=p_0}(x) = e^{ip_0x/\hbar} $ (up to a normalization).

  
  
  

The textbooks will leave off the specific superscripts because there are many interesting ways to generalize these relationships.