# Tag Info

8

Flip back a page; Dirac uses real to mean Hermitian when talking about linear operators. So you can see that even if $A$ and $B$ are Hermitian, $AB$ won't be Hermitian unless they commute, whereas those linear combinations will be.

3

The two particles $m_s$ and $m_I$ live in different vector spaces, so you are actually not picking the same basis vectors (because the basis vectors of the different particles belong to two separate vector spaces). Secondly, the tensor product between the basis vectors of the two different vector spaces will form the basis vectors of a new $3 \times 3 = 9$ ...

1

Well here is an attempt at the answer based on your comments and question. Consider your state $\left | \psi \right>$ be $$\left | \psi \right> = \frac{1}{\sqrt2}(\left | 0 \right> +\left | -1 \right>)$$ Now int the $\mathbb{C}^3$ representation, this becomes $$[\left | \psi \right> ] = \frac{1}{\sqrt2} \left[ {\begin{array}{c} 0 ... 1 The rotation operator normalization you have chosen,  R_{\boldsymbol n}(\theta)=e^{-i\theta \boldsymbol \sigma\cdot \boldsymbol n/2} , means that for a rotation by 2\pi about the z axis,$$ R_{\hat z}(2\pi)=e^{-i\pi \sigma_z} =(-1)^{\sigma_z}=-1,$$because (-1)^1=(-1)^{-1}=-1. Thus, your rotation operator has indeed rotated you back to the same point ... 1 You may know already that "symmetry" is not always important when it comes to non-Euclidean spaces. For instance, in quantum mechanics, a symmetric operator is seldom important, but one that is equal to its Hermitian conjugate--one that is Hermitian or "self-adjoint"--is incredibly important, for those operators have real eigenvalues and thus correspond to ... 2 Indeed NowIGetToLearnWhatAHeadIs's comment answers your question: "Simply because I hadn't encountered one that was not." A rotation is a lorentz transformation which is not symmetric. Indeed the transpose of a rotation matrix is its inverse, and only trivial rotations or rotations through half a turn are involutary (self inverse). To see this in ... 1 The first major step would be calculus. Really just becoming familiar with integration and differentiation on all types of functions. From there a little knowledge on differential equations can go a long way. Knowing just this can get you solving some basic problems. "Early Transcendentals" by Thomas is a good calculus book. Then there are some nice ... 6 Because direction cosines are, unlike sines and tans, even functions of the angle which makes the sign of the angle irrelevant and that's a good thing. More importantly, the direction cosines of a unit vector \vec v end up being the coordinates v_x,v_y,v_z, respectively, so the direction cosines obey$$\cos^2 a+\cos^2 b+\cos^2 c = 1$$which is nice. ... 2 Linear combinations of Pauli matrices play particularly nicely with diagonalization. The reason for this is that a linear combination of the form$$ \vec v\cdot\vec\sigma=\sum_j v_j\sigma_j =\begin{pmatrix}v_z&v_x-iv_y\\ v_x+iv_y&-v_z\end{pmatrix}\tag1  represent the density matrix $\rho=\tfrac12(1+\vec v\cdot\vec\sigma)$ of a state at the point ...

2

As the Pauli matrices are hermitian, every linear-combination (with real coefficients) of them is hermitian as well, in particular $(\sigma_x\pm\sigma_y)$. And because every hermitian operator can be diagonalized, the answer to your question is yes. Just write the corresponding 2x2-matrix and try to diagonalize it.

2

The matrix you consider is Hermitian because real linear combination of Hermitian matrices, thus it can be diagonalized (i.e. $S_3$ does exist) for a known general theorem. If a matrix admits eigenvalues it may be non-diagonalizable. Consider the matrix $A$ with the form: 0 1 0 0 The eigenvalues are the complex solutions of $\det(A-\lambda I)=0$. There ...

7

There are at least three notions of basis depending on the mathematical structure you are considering. I will quickly discuss three cases relevant in physics (topological vector spaces are relevant too, but I will not consider them for the shake of brevity). (1) Pure algebraic structure (i.e. vector space structure over the field $\mathbb K=$ $\mathbb R$ ...

0

I believe this is just a matter of vocabulary. Here's how it goes in mathematics: A basis is a linearly independent spanning set of the vector space, ie, a set of vectors such that any vector in the space can be expressed uniquely as a finite linear combination. In an infinite dimensional Hilbert space, such bases aren't so convenient: due to the Baire ...

9

Your doubt is not ridiculous, it is probably simply due to the confused way often mathematics is taught in physics. (I am a physicist too and, during my career, I had to bear ridiculous misconceptions, wasting lot of time in tackling non-existent pseudo-mathematical problems instead of focusing on genuine physical issues). There are sensible mathematical ...

1

Complete set is a well defined expression. The reason why people sometimes differentiate between complete orthonormal set (COS) and a basis, is that any vector can be written as a finite linear combination of elements of the basis (if you use basis in the linear algebra sense). While for the COS you need an infinite linear combination. If you use the ...

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