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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?

Qmechanic
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Thiago
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2 Answers2

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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:

  • $\left|\psi\right>$ 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.

  • $\left|x_0\right>$ 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.

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

  • $\left<x_0\middle|\psi\right>$, 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 $\left<x\middle|\psi\right>$ for arbitrary $x$, rather than for some specific $x_0$, you can recover the Schroedinger wavefunction.

  • Similarly, $\left<p_0\middle|\psi\right>$ 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, $\tilde\psi(p)$.

  • The representation in position space of a wavefunction with definite momentum $p_0$ is a plane wave with a single frequency, $\left<x\middle|p_0\right> = \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.

rob
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Sorry, but I have not enough rep to reply just as comment.

Those are Dirac Delta Operations. The Bra and Ket vectors.

You first equation, I am not familiar, maybe I have to review again.

Second equation, The left hand side says the wave function as a function of x is equals to operator x operating on "state $\psi$", which basically is the general idea.

Third equation, (also the same), but p stands for momentum, you see, x operator(position operator) and p operator(momentum operator) are just 2 basic operators in Formalism of Quantum mechanics.

Fourth equation is the famous form of heisenberg uncertainty principle. You would need be familiar to idea of commutation to fully understand that.

Anyways, I'm quite sure you have lots of things to study to understand the concept, just to answer your question, you are dealing with "Formalism in Quantum Mechanics" & "Dirac Delta : Bra & Ket"

Anyways, I remember that this series helped me get oriented at those Formalism 6 months ago in my QM class: http://www.youtube.com/playlist?list=PL04722FAFB07E38E1

Goodluck.

arvil
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    I think it's quite incorrect to consider a bra like $\langle x \vert$ as a operator, since the bra maps vectors to scalars. But an operator maps vectors into vectors. The first equation is normalization and I'm unsure the last one is the uncertainty principle. – jinawee Jun 05 '14 at 19:32