Recent content by George Keeling

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...

The video does a diagonalisation but I think I got that right and it was confirmed by wolfram alpha as per my post #5. The point is that neither I nor wolfram found a unitary diagonalising matrix.... PS thanks for your diligence!

It's not significant. My eigenvector for eigenvalue ##2## is ##(1+i)## time emathhelp's. The other one is ##(−0.5+0.5i)## time emathhelp's. You can always multiply an eigenvector by a scalar to get another eigenvector for the same eigenvalue.

In the first of the questions we had $$T=\left(\begin{matrix}1&1-i\\1+i&0\\\end{matrix}\right)$$Part (d) of the question is "Construct the unitary diagonalizing matrix ##S## , and check explicitly that it diagonalizes ##T##." I got a diagonalising matrix...