Recent content by julian

The particular derivation I gave requires knowledge of complex analysis and is quite lengthy. I therefore thought that it illustrated why deriving the result is not straightforward. However, I had forgotten that there is a substantially more elementary proof using Fourier series, which are...

Extracting relevant bits from other thread: We can write \begin{align*} \sum_{n=0}^\infty \dfrac{1}{(2n+1)^2} & = \sum_{n=1}^\infty \dfrac{1}{n^2} - \sum_{n=1}^\infty \dfrac{1}{(2n)^2} \nonumber \\ & = (1 - 2^{-2}) \sum_{n=1}^\infty \dfrac{1}{n^2} \nonumber \\ & = \frac34 \sum_{n=1}^\infty...

In another thread I provide a derivation of the general closed formula for ##\zeta(2k)## in terms of Bernoulli numbers: \begin{align*} \zeta(2k) = \sum_{n=1}^\infty \frac{1}{n^{2k}} = (-1)^{k+1} (2 \pi)^{2k} \dfrac{B_{2k}}{2 (2k)!} \end{align*} From which you can obatin your series as...

Write \begin{align*} n^A &= \begin{pmatrix} n^0\\ n^1 \end{pmatrix}, & k^A &= \begin{pmatrix} k^0\\ k^1 \end{pmatrix}. \end{align*} If \begin{align*} m^{AB} = n^A k^B, \end{align*} then the matrix representing ##m^{AB}##, with row index ##A## and column index##B##, is \begin{align*}...

We wish to maximise ##\langle L_x \rangle^2 + \langle L_y \rangle^2 + \langle L_z \rangle^2##. Write \begin{align*} | \psi_0 \rangle = \sum_{m=-l}^l a_m |l,m\rangle \end{align*} I get \begin{align*} & | \langle L_+\rangle |^2 + \langle L_z\rangle^2 = \nonumber \\ & = \hbar^2\, \left|...

To show that ##\sin (\frac{n^2}{n+a} x)## has partially bounded sums for ##x \in [1,2]##: Write \begin{align*} \frac{n^2}{n+a} = n - a + \frac{a^2}{n+a} \end{align*} Then \begin{align*} \sin (\frac{n^2}{n+a} x) = \sin (n - a + r_n) x , \quad where \quad r_n = \frac{a^2}{n+a} \end{align*}...

To show that ##\psi_a^0## and ##\psi_b^0## span the space, we must prove they are linearly independent. Suppose \begin{align*} \kappa \psi_a^0 + \zeta \psi_b^0 = 0. \end{align*} Applying ##A## gives \begin{align*} \kappa \mu \psi_a^0 + \zeta \nu \psi_b^0 = 0. \end{align*} Subtracting ##\mu...

If ##\gamma = \mu##, then you don’t necessarily have ##\langle \psi_a^0 , \psi_\gamma (0) \rangle = 0##.

I think this is the fourth edition: https://archive.org/details/landau-l.-d.-lifshitz-e.-m.-course-of-theoretical-vol-2/page/n3/mode/2up

I believe these notes cover Proposition 4.4.2 - in the author's notes it appears as Proposition 4.3.7. I spoke with the author about his notes, but that was likely over 10 years ago. Things are a bit hectic at the moment, so I won’t be able to revisit this right now.

The "stars and bars" thing: Consider the example where ##T_{ijk}## is symmetric under the interchange of any pair of indices, and the indices take the values ##1,2,3,4##. For example, all these are the same: \begin{align*} T_{112} = T_{121} = T_{211} . \end{align*} These can all be represented...

Say ##i=1,2,\cdots n##, and consider the object ##T_{i_i i_2 \cdots i_k}## that is symmetric under the interchange of any pair of indices. The number of independent components, ##N##, is equal to the number degree-##k## monomials in ##n## variables ##x_1, x_2, \dots, x_n##, i.e., expressions of...

The approach is thorough, though perhaps a bit more detailed than necessary. (1) This symmetry applies to ##i## and ##j##. For each fixed ##k##, ##T_{ijk}## is a symmetric 3×3 matrix in indices ##i,j##. The number of independent components in a symmetric 3×3 matrix is 6. Since ##k## can...

Just to clarify — the expression you gave for the integrating factor wasn’t correct, so I was pointing that out.

It should be ##\int \ln x dx = x \ln x -x +C##, so ##e^{\int \ln x dx} = K x^x e^{-x}##.