Recent content by George Keeling

Oh yes. I quickly get to $$i\hbar\frac{\partial\psi}{\partial t}=-\frac{\hbar^2}{2m_0}\nabla^2\psi+m_0c^2\psi$$Which is the Schrödinger equation for a free particle in a (rather large) constant potential of ##m_0c^2##. I guess that will do. And adding a constant potential is the gauge equivalent...

I am reading Relativistic Quantum Mechanics Wave Equations, 3rd ed. - W. Greiner and I'm on section 1.3 looking at a solution ##\psi## to the Klein-Gordon equation in the nonrelativistic limit. The solution is split up: $$\psi=\phi\exp{\left[-\frac{i}{\hbar}m_0c^2t\right]}$$and we then show...

I am looking at the Laughlin wave function and it contains the term $$\prod_{j<k}^{N}\left(z_j-z_k\right)^q$$ In Wikipedia the lower index on ##\Pi## is ##1\le i<j\le N## and there is no upper index. I'm not sure what either mean. For example would...

The problem from the book (post #7) asked to show something about $$\frac{d}{dt}\left\langle x p\right\rangle$$so I suppose that means an observable that is two other observables multiplied together. So it looks like Griffiths can do math with observables. After that, observing its rate of...

I think the problem is due to the hats in the expectation value expression. In my book one normally has no hats: ##\left\langle Q\right\rangle## not ##\left\langle\hat{Q}\right\rangle##. ##Q## is an observable and ##\hat{Q}## is its operator. Where I started from was...

That's exactly what I was doing so that makes ##\left\langle\frac{\partial}{\partial t}\left(\hat{x}\hat{p}\right)\right\rangle=0## I have also now found and read Schrödinger's picture. "operators (observables and others) are mostly constant with respect to time (an exception is the Hamiltonian...

To begin with I thought you meant politeness conventions and I had offended you :cry: But on further reflection I think you mean Schrödinger picture conventions which I am having difficulty getting used to. Thanks. I was using that famous product rule convention in calculus.

So I think what you are saying is that the 'interesting term' should be $$-i\hbar\left[\frac{\partial}{\partial t}\left(x\frac{\partial}{\partial x}\right)\right]\Psi$$and so, as the part in square brackets contains no ##t##s, it vanishes when differentiated. I disagree. I would expand the...

Mine is the third edition. We get to ##\left\langle\frac{\partial}{\partial t}\left(\hat{x}\hat{p}\right)\right\rangle=0## after using equation 3.73. Hamiltonian comes into it.

I'm doing Griffiths & Schroeter Problem 3.37 and I would like to prove that the expectation value of a rather peculiar observable vanishes: $$\left\langle\frac{\partial}{\partial t}\left(\hat{x}\hat{p}\right)\right\rangle=0$$My cheat sheet expands the LHS to an integral and continues …...

Thanks! I did miss that. In the 3rd edition it's on page 14, note 15. Slightly new question: is a function square integrable if and only if it is normalisable?

I have the third edition. Did I miss that? Where does Griffiths write it?

Thanks!

I am meeting the momentum space wave function ##\Phi\left(p,t\right)## in chapter 3 of Griffiths & Schroeter. I have an integral which I can integrate by parts: $$\int_{-\infty}^{\infty}{\frac{\partial}{\partial p}\left(e^\frac{ipx}{\hbar}\right)\Phi d...

Gosh thanks! I didn't have to normalise the eigenvectors I don't think. I took the eigenvectors which are in the ##S^{-1}## of my #5 and multiplied each by a constant to get $$S^{-1}=\left(\begin{matrix}2a&b\\\left(1+i\right)a&\left(-1-i\right)b\\\end{matrix}\right)$$##S^{-1}## must be unitary...