Recent content by eys_physics

The Legendre polynomials are usually normalized such that ##P_n(1) = 1##.

Your basis is orthogonal but not orthonormal. You need to compute the normalization for your third basis function, i.e. $$\int_{-1}^1 dx (x^2 - 1/3)^2$$.

From the explicit expressions for the spherical harmonics it is straightforward to derive that $$x = \rho \sqrt{\frac{2\pi}{3}}(Y_{1,-1}(\theta, \phi) - Y_{1,1}(\theta, \phi)),$$ $$y = i\rho \sqrt{\frac{2\pi}{3}}(Y_{1,-1}(\theta, \phi) + Y_{1,1}(\theta, \phi)),$$ $$z = \rho...

Have you tried to use the addition formula of ##\tan^{-1}##? That is, $$\tan^{-1}(u) + \tan^{-1}(v) =\tan^{-1}\left(\frac{u + v}{1 - uv}\right)\quad (mod\: \pi)$$

So, in your example you have one tensor ##S## of angular momentum (rank) ##j_1=1/2## and another one ##D## with ##j_2 = 1##. You can form the new tensors (using Clebsch-Jordan symbols) $$T_{J=1/2, M} = \sum_{m_1 m_2} (1/2 m_1 1 m_2 | 1/2 M) S_{1/2, m_1}D_{1, m_2}$$, and $$T_{J=3/2, M} =...

The products of one or more ##\gamma## and ##\gamma_5##, forms a basis for all matrices of dimension 4. ##\Gamma## can thus be written in a basis of them. You need then also to take into account e.g. Lorentz invariance in order to arrive at the expression for ##\Gamma##.

Because, for the eigenvectors are (or can be constructed) orthonormal , i.e. $$<v_1|v_2>=<v_2|v_1> =0, $$ and $$<v_1|v_1>=<v_2|v_2> =1 $$.

Sorry, it was a mistake in the first sentence. It should read "|v_1>" and "|v_2>".

Yes, the formula for ##|x>## is wrong. The plus sign should be a minus sign. That is, ##|x> = \frac{1}{\sqrt{2}}(|v_1> - |v_2>)##

Yes, you either write ##|x>## and ##|y>## in terms of ##|v_1>## and ##|v_2|##, or the other way around. They are connected by a linear transformation (mapping). Similarly, the matrix ##A## can be expressed in the ##\{v_1, v_2\}## basis. In that basis ##A## is diagonal.

You have two states ##|x>## and ##|y>##, which forms an orthonormal basis. Your matrix ##A## is written in this basis. You are solving for the eigenvectors, ##|v_1>## and ##|v_2>##, in this basis. That is, $$|v_1> = a |x> + b|y>$$ $$|v_2> = -a |x> + b|y>,$$ with ##a^2+b^2=1##. The notation...

This is not correct. You have that $$|v_1> = \frac{1}{\sqrt{2}} |x> + \frac{1}{\sqrt{2}}|y>$$ and $$|v_2> = -\frac{1}{\sqrt{2}} |x> + \frac{1}{\sqrt{2}}|y>$$ . What is then ##|x>## and #|y># in terms of ##|v_1>## and ##|v_2>##? Yes, if ##A## is the Hamiltonian? But, it is not obvious from...

Thanks @fresh_42, for a nice article.

Alternatively, one can do it with array sections: real :: A(2), B(2), C(2, 2) A = (/ 1.0, 2.0 /) B = (/ 3.0, 4.0 /) C(1, :) = A C(2, :) = B

Thanks for sharing.