# Tag Info

14

A second-order tensor can be represented by a matrix, just as a first-order tensor can be represented by an array. But there is more to the tensor than just its arrangement of components; we also need to include how the array transforms upon a change of basis. So tensor is an n-dimensional array satisfying a particular transformation law. So, yes, a ...

12

The dual of a tensor you refer to is the Hodge dual, and has nothing to do with the dual of a vector. The word "dual" is used in too many different contexts, and in this case it is even used the same $*$ symbol. One usually specifies "Hodge dual", or "Hodge star operator", to avoid confusion. Both these "duals" are isomorphisms between vector spaces endowed ...

10

Matrices are often first introduced to students to represent linear transformations taking vectors from $\mathbb{R}^n$ and mapping them to vectors in $\mathbb{R}^m$. A given linear transformation may be represented by infinitely many different matrices depending on the basis vectors chosen for $\mathbb{R}^n$ and $\mathbb{R}^m$, and a well-defined ...

9

The wording used in your textbook was sloppy. $A$ acts as $A^*$ on a bra, as $\langle u\rvert A\lvert v\rangle:=\langle u\lvert Av\rangle~$ is the same as $\langle u\rvert A\lvert v\rangle=\langle A^*u\lvert v\rangle~$, by definition of the adjoint. The latter formula also shows that $\langle A^*u\rvert=\langle u\rvert A$. Everything becomes very simple ...

8

The ground state of the harmonic oscillator $|0\rangle$ obeys $$a|0\rangle = 0$$ which means that the action of $a$ can't be undone: once you act with it on a state, you set to zero the coefficient in front of $|0\rangle$ in the decomposition into occupation eigenstates. Any candidate inverse operator $a^{-1}$ acting on zero will give you zero again; you ...

8

The slave particle approach is based on the assumption of spin-charge separation in the strongly correlated electron systems (typically Mott insulators). It was proposed that the electrons can decay into spinons and chargons (holons/doublons). But to preserve the fermion statistics of the electrons, the spinon-chargon bound state must be fermionic, so the ...

7

Quantum mechanics "lives" in a Hilbert space, and Hilbert space is "just" an infinite-dimensional vector space, so that the vectors are actually functions. Then the mathematics of quantum mechanics is pretty much "just" linear operators in the Hilbert space. Quantum mechanics Linear algebra ----------------- -------------- wave function vector ...

7

There is quite a lot of very important information hidden in the term hermitian. For an operator $A$ on a finite-dimensional Hilbert space $\mathcal H$, one can show that there exists an orthonormal basis for the Hilbert space consisting of eigenvectors of the operator $A$. Moreover, one can show that the eigenvalues corresponding to these eigenvectors ...

7

The moment-of-inertia (MOI) tensor is real (no imaginary terms), symmetric, and positive-definite. Linear algebra tells us that for any (3x3) matrix that has those three properties, there's always a set of three perpendicular axes such that the MOI tensor can be expressed as a diagonal tensor in the basis of those axes. These are called the principal axes ...

6

Usually the Clifford group is defined to be the group of unitaries that preserve the Pauli group under conjugation, so no proof is needed. If instead you are asking, how can we prove that a certain unitary (such as the controlled-NOT) is in the Clifford group, the usual straightforward way to do this is just to calculate. Conjugation is a group ...

6

Length and distance are not vector quantities (they are scalar quantities), but position and displacement are vector quantities (at least according to common terminological conventions). Here is how all of these are defined. Note that I am restricting the discussion here to vectors in three-dimensional Euclidean space $\mathbb R^3$. Every point in ...

5

The composition law for quantum systems is always a tensor product. Your problem arises from a confusion over what the tensor product is applied to: you are trying to tensor product the spatial coordinates together, when it is in fact the basis vectors of the Hilbert space you should be tensoring together. More formally, take two quantum systems A and B, ...

4

If I understand correctly, your question basically comes down to identifying a basis for the space of square-integrable functions, $L^2(\mathbb{R})$, since any physical state $|\Psi\rangle$ can be constructed by performing the integral you listed in your question with a function $\Psi_x(x)\in L^2(\mathbb{R})$. $L^2$ is known to be a vector space, so a basis ...

4

There's no wonder you're confused - the author obviously was as well. First, the operations he's talking about are direct sum $U\oplus V$ and tensor product $U\otimes V$ of vector spaces. This has nothing to do with the vector product (an ambiguous term which most often denotes the cross product you probably know from school). Both are two different ways ...

4

Assume the Lorentz transformation $\Lambda$ is not invertible. Then it is in particular not injective and there exists $0\not=u\in\ker\Lambda$. The inner product $g$ is nondegenerate so there's a vector $v$ with $g(u,v)\not=0$ and we end up with the contradiction $$0\not=g(u,v)=g(\Lambda u, \Lambda v)=g(0,\Lambda v)=0$$ where we have used the fact that ...

4

If you think of $\mid u \rangle$ as column vectors and of $\langle u \mid$ as row vectors, then $A$ is just a $n \times n$ matrix (possibly with $n = \infty$). You can then think of $A \mid u \rangle$ as the matrix $A$ acting on a vector $u$. However, since $\langle v \mid$ is a row and not a column vector, you cannot (for a sensible row vector) multiply ...

4

You should be careful not to mix symbolic and index notation. $\text{Tr}(g^{\mu \nu}\delta g_{\mu \nu})$ does not make sense since $g^{\mu \nu}\delta g_{\mu \nu}$ is just a number. The correct symbolic notation would be: $$\delta \text{det}(\mathbf{g})=\text{Tr}(\text{adj}(\mathbf{g})\delta ... 4 In principle, the operator H is not a matrix. However, you can write down a matrix representation. For that, you need a basis, which should be complete (=very large matrix, probably infinitely large) - for many illustrating cases in quantum mechanics, it's not complete. If your basis consists of |a\rangle and |b\rangle, it's not complete, but you can ... 4 The fundamental difficulty here is that If two elements are orthogonal, it means that measuring one component does not give any information about the other. is incorrect. Orthogonality mean exactly that the inner product between the two things is zero. If a \cdot b = 0 then a and b are said to be orthogonal. In a Cartesian space this has the ... 4 It depends on how you define orthogonality, or, as OSE puts it in his comment, "Orthogonality is usually tested using some defined inner product." I'll expand on this a bit. In order to mathematically answer the question Is direction A orthogonal to direction B? we need a definition of the terms "direction" and "orthogonal." The standard ... 3 Any gate of the form diag(1,1,1,\exp(i\phi)) is not in C_n for any n unless \phi = 2\pi k/2^n for some integers k and n. This can be proven by induction using the similar result for single-qubit gates. I'm not sure if this is included in any published paper. We don't have a good characterization of gates in C_n for n > 2, so there is no ... 3 You may order the matrices like this:$$ \Lambda_\rho^\mu g_{\mu\nu} \Lambda^\nu_\sigma = g_{\rho\sigma} $$I suppose all the letters should have been Greek. They're called mu, nu, rho, sigma, good to learn them. In my form, one may view \mu as the summed over index in the first product on the left hand side and \nu as the summed over index in the ... 3 A slow construction would go...$$ \begin{pmatrix}a&b\\c&d\end{pmatrix} = a\begin{pmatrix}1&0\\0&0\end{pmatrix} +b\begin{pmatrix}0&1\\0&0\end{pmatrix} +c\begin{pmatrix}0&0\\1&0\end{pmatrix} +d\begin{pmatrix}0&0\\0&1\end{pmatrix}  \begin{pmatrix}1&0\\0&0\end{pmatrix} =\frac{1}{2} ...

3

Strictly speaking matrices and rank 2 tensors are not quite the same thing, but there is a close correspondence that works for most practical purposes that physicists encounter. A matrix is a two dimensional array of numbers (or values from some field or ring). A 2-rank tensor is a linear map from two vector spaces, over some field such as the real ...

3

The eigenfunctions of a self adjoint operator lie outside the Hilbert space of square integrable functions on the line. One solution is to work with a basis of eigenfunctions of a non-self adjoint operator such as $x+ip$. Of course these are the coherent states. For the coherent states, one has an ovecomplete basis and a partition of unity, thus it is not ...

3

For several years I have been teaching Clifford (geometric) algebra as part of the Vector Analysis Course for undergraduate physics majors in Ateneo de Manila University. I strictly use Cl_{n,0}, even for Special Relativity. 18-year old students do not complain how difficult geometric algebra is. They just learn the math and the geometric interpretations: ...

3

Starting from $$g_{\mu\nu} \Lambda^\mu{}_\rho \Lambda^\nu{}_\sigma = g_{\rho\sigma}$$ we contract with $g^{\sigma\tau}$ $$g_{\mu\nu} \Lambda^\mu{}_\rho \Lambda^\nu{}_\sigma g^{\sigma\tau} = g_{\rho\sigma} g^{\sigma\tau} = \delta_\rho^\tau$$ and reorder the factors  g_{\mu\nu} g^{\sigma\tau} \Lambda^\nu{}_\sigma \cdot \Lambda^\mu{}_\rho = ...

3

Well, learn linear algebra. An advanced text (on linear algebra over "field" number systems) is these lecture notes [pdf] from UC Davis. Once you get that done, you should study differential equations. Or if you want to skip ahead, perhaps Fourier analysis. A free reference would be my notes [pdf]. It's mildly physics-oriented, but connects the ideas back ...

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