# Tag Info

## New answers tagged linear-algebra

2

$D_{\mu\nu} = A g_{\mu\nu}+B k_{\mu} k _{\nu}$ with A and B two unknown functions of the scalar k^2. The two tensor after A and B are the only possible Lorentz invariant tensors . Simply plugin and calculate the unknown functions.

5

I) More generally, Let $V$ be a (say, finite dimensional) vector space over a field $\mathbb{F}$. Let $(e_i)_{i\in I}$ be a basis for $V$. Let $A\in {\rm End}(V)$ be an endomorphism in $V$, i.e. a $\mathbb{F}$-linear map $A:V\to V$. Let the matrix $(M^i{}_j)_{i,j\in I}$ be the unique $\mathbb{F}$-valued matrix that represents the linear map $A$ in the ...

0

The only crucial point is the degeneracy of eigenspaces. Consider the finite dimensional Hilbert space $\cal H$ (the extension to the infinite dimensional case is more difficult also because a part of continuous spectrum may appear) and a pair of commuting Hermitian operators $A$ and $B$ on that space such that the following requirement is satisfied. R.: ...

2

Consider self-adjoint operators $A$ and $B$, on a Hilbert space $\mathscr{H}$. Roughly speaking (forgetting about domains of definition), it is possible only if your Hilbert space can be written as $\mathscr{H}=\mathscr{H}_1\otimes \mathscr{H}_2$; where $\mathscr{H}_1$ and $\mathscr{H}_2$ are Hilbert spaces such that: $A=A_1\otimes 1$ and $B=1\otimes B_2$, ...

2

I can give you a mathematical reason why you have to use components to define contraction. The reason is that contraction doesn't work for infinite dimensional vector spaces. You would have to sum over infinitely many components, and this sum might not converge. So there is no possibility of contraction for infinite dimensional tensors.* But if we had a ...

1

If you have 3 eigenvectors of the stress tensor/matrix $T$, you can choose them as your new basis and it will be diagonal there - and no off-diagonal elements mean no shear stress, since the shear stress on the plane in the $ij$-direction ($i \neq j$) is given by $T_{ij}$, which, for $i \neq j$, will be zero in this basis.

1

The density matrix for a pure state $|\psi\rangle$ is $\rho=|\psi\rangle\langle\psi|$. Note that this is a matrix in the sense that it takes some vector $|\phi\rangle$ to the vector $\rho|\phi\rangle=|\psi\rangle\langle\psi|\phi\rangle$. It has components \rho_{ij}=\langle e_i|\rho|e_j\rangle=\langle e_i|\psi\rangle\langle\psi|e_j\rangle=\psi_i ...

4

This might not be what Nakahara has in mind, but one can make sense of this using the idea of projective Hilbert spaces. Let $\mathcal{P}(\mathcal{H})$ denote the projective space associated to the "normal" space $\mathcal{H}$. The subset of separable states is not a subvectorspace in the proper sense, as Holographer notes. Yet it can be understood as a ...

3

Note that the space of separable states is not a vector space, and in particular not a subspace of the total Hilbert space: the sum of two separable states is unlikely to be separable. So dimension here means something more general than vector space dimension. Having said that, I would disagree with the author on his dimension! I would say that the space of ...

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