Recent content by ccesare

There is a difference between the transformation that you give and the infinitesimal transformation generated by the third Pauli matrix, namely that your transformation is not unitary, while the transformation generated by the third Pauli matrix is. Recall that the transformations need to form a...

To first order in what? dt? That is precisely what changes their lengths. None of the Pauli matrices have determinant zero, since they are elements of SU(2) with a physically irrelevant phase factored out. Representations of elements of SU(2) must be special, i.e. determinant one, and unitary...

Also, choosing the generators to be Hermitian does make some sense, since their expectation values can then be interpreted in a useful way in quantum mechanics.

The two vectors you list have different lengths. Thus, the transformation does not preserve the length of vectors. Since you already agreed that your transformation is not even the representation of an SU(2) element, I am failing to see your point.

The determinant of the matrix you gave above is (1+2*i*dt-dt^2). Additionally, that linear transformation does not preserve the lengths of vectors. The infinitesimal dt has no direction associated to it; it is just a small (real) number. I don't understand your statement about its orthogonality...

You can treat these, formally, as postulates. However, they are physically motivated postulates. For example the first two are related to causality (with the fields all taken at the same time), and the last one is analogous to QM.

The matrix you have given is neither special nor unitary, so it certainly is not a representation of an element of SU(2).

It appears in relation to the quantum dimension of the (nontrivial) anyon in the Fibonacci anyon model. Whether or not you consider this a physical example is another matter...

If a system is in the state \left|\psi\right>, the expectation value of an Hermitian operator A is given by \left<\psi\right|A\left|\psi\right> This has nothing to do with an operator "being anywhere," but rather with the average value you would see if you performed an experiment many...

I was under the impression that large extra dimensions could be ruled out. For instance, if you consider the one-dimensional infinite potential well and add just one compactified extra dimension, the first observable change to the original problem occurs at energy scales that are much higher...

I've found Griffiths great. The problems are sometimes more challenging than what he covers in the chapter, but there are also many solid problems that will give you an understanding. He covers both the variational principle and perturbation theory. Since you seem to be coming from a...

\hbar should be 1.05*10^{-34} Js

Correct, you would set the radius to \Delta x. If you solved the uncertainty principle for the minimum momentum in this case, you could easily find the minimum kinetic energy.

The reason for using the m's and n's is really to make the derivation more general. If, for instance, you have two different eigenfunctions so that you have \psi_{m} and \psi_{n}, then taking their inner product would give you: \int \psi^{*}_{m} \psi_{n} dx Now, because the set of all the...

It should definitely read greater than 0. The wavefunction for a three-dimensional particle in a box is just three one-dimensional wavefunctions multiplied together. Thus, a zero for any of the three quantum numbers reduces the total wavefunction to zero. Therefore, the lowest energy state is...