51

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 third-...


39

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 ...


36

Coming from a math perspective, I would define a dimension as "any property which is orthogonal to all other properties." "Orthogonal" here means you cannot get to one property by applying scalar operations on another. For example, the x-axis dimension can never become a y-axis value, and similarly for time vs. spatial dimensions. For that matter, it'...


32

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 ...


28

In this context, I usually explain it (non-mathematically) by saying that the number of dimensions is the number of values you need to specify where an event occurs. For most people this involves space and time (but for particle physicists it might involve more values ;). Anyway, certainly even people before Einstein would need to specify the time as well ...


26

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 $\mathbb K=$ $\mathbb R$ ...


24

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 ...


24

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 ...


22

To take the partial trace you need to build the sum over the matrix elements w.r.t. the same input and output basis, as you probably already used to calculate the partial traces you gave. In Dirac notation this is often written as: $$ tr_A(L_{AB}) =\sum_i \langle i|_A L_{AB} |i\rangle_A=\langle0|0\rangle\langle 0|0\rangle (|1\rangle\langle0|)_B+\langle1|0\...


21

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 ...


17

As previous answers have correctly noted gamma matrices do not forma a basis of $M(4,\mathbb{C})$. Nevertheless you can construct one from them in the following way 1 the identity matrix $\mathbb{1}$ 4 matrices $\gamma^\mu$ 6 matrices $\sigma^{\mu\nu}=\gamma^{[\mu}\gamma^{\nu]}$ 4 matrices $\sigma^{\mu\nu\rho}=\gamma^{[\mu}\gamma^{\nu}\gamma^{\rho]}$ 1 ...


17

Let $\{|i\rangle\}$ be an orthonormal basis for the Hilbert space of the system. Then the trace of an operator $O$ is given by (See the Addendum below) \begin{align} \mathrm {tr}(O) = \sum_i \langle i|O|i\rangle \end{align} For a given state $|\psi\rangle$, we define an operator $P_\psi$ by \begin{align} P_\psi|\phi\rangle = \langle\psi|\phi\rangle|\psi\...


16

You are starting from the incorrect point. The argument follows by linearity of the equation. Suppose $\Psi_k(x,t)$ is solution of the time dependent Schr$\ddot{\hbox{o}}$dinger equation: $$ i\hbar \frac{\partial }{\partial t}\Psi_k(x,t)=-\frac{\hbar^2}{2m}\frac{\partial^2\Psi_k(x,t)}{\partial x^2}+U(x)\Psi_k(x,t)\, . $$ Then: $$ \Phi(x,t)=a_1\Psi_1(x,t)+...


16

Your top-line question can be answered at many levels. Setting aside issues of forms and covariant/contravariant, the answer is: The dot product is the product of the magnitudes of the two vectors, times the cosine of the angle between them. No matter what basis you compute that in, you have to get the same answer because it's a physical quantity. The ...


15

Dot products, or inner products are defined axiomatically, or abstractly. An inner product on a vector space $V$ over $R$ is a pairing $V\times V\to R$, denoted by $ \langle u,v\rangle$, with properties $\langle u,v\rangle=\langle v,u\rangle$, $\langle u+cw,v\rangle= \langle u,v\rangle+c\langle w,v\rangle$, and $ \langle u,u\rangle\gt0$ if $u\ne0$. In ...


14

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: ...


14

If the Hilbert space of the system in question is finite-dimensional, then in a given basis for the Hilbert space, the Hamiltonian (and every other observable for that matter), will be represented by a matrix. If the Hilbert space is infinite-dimensional, the situation is a bit different. In Quantum Mechanics, we typically assume that the Hilbert spaces we ...


14

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 ...


14

“Scalar”, “vector”, and “tensor” have no meaning without specifying the group of transformations. In physics we focus on groups such as rotations, Galilean transformations, Lorentz transformations, Poincaire transformations, and gauge transformations because these are symmetries of various physical theories, built in to reflect symmetries of the natural ...


13

Ron Maimon is entirely correct when he says that GA is precisely Clifford algebra from a mathematical perspective, as any book or paper using the phrase "Geometric Algebra" is sure to say. But I think he misses both the point of the question and the point of "GA" — which is different from Clifford algebra from a pedagogical perspective. The question I'll ...


13

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 ...


13

There are two concepts of duality for vector spaces. One is the algebraic dual that is the set of all linear maps. Precisely, given a vector space $V$ over a field $\mathbb{K}$, the algebraic dual $V_{alg}^*$ is the set of all linear functions $\phi:V\to \mathbb{K}$. This is a subset of $\mathbb{K}^V$, the set of all functions from $V$ to $\mathbb{K}$. The ...


13

Does this mean that the operator $\hat O$ (an observable) is special in some way? I believe it means there is no such $\hat O$. If $\hat O$ corresponds to an observable, we require the eigenvalues to be real. Let $|o\rangle$ be an eigenket of $\hat O$ with real eigenvalue $o$: $$\hat O |o\rangle = o |o\rangle$$ Now consider the following $$\hat O \...


13

Let $H_A \otimes H_B$ be your Hilbert space, and $O$ be an operator acting on this composite space. Then $O$ can be written has $$ O = \sum_{i,j} c_{ij} M_i \otimes N_j$$ where the $M_i$'s and $N_j$'s act on $H_A$ and $H_B$ respectively. Then the partial trace over $H_A$ defined as $$tr_{H_A}(O) = \sum_{i,j} c_{ij} tr(M_i) N_j ,$$ and similarly for $H_B$.


12

To complement V. Moretti's excellent answer, it's worth emphasizing that the dimension of the four-by-four complex matrices $\mathbb C^{4\times 4}$, when seen as a vector space over $\mathbb C$, is $4\!\times\!4=16$. As such, a set of four matrices (i.e. vectors in $\mathbb C^{4\times 4}$) can never be a basis for it. It's also worth saying that the general ...


12

Call $u_1, u_2, u_3, u_4$ the eigenvectors described by you, respectively. Your claims are all right, but realize that both $u_1$ and $u_2$ share the same eigenvalue, that is $1$, i.e., $Pu_1=u_1$ and $Pu_2=u_2$. Hence, any linear combination of $u_1$ and $u_2$ will also be eigenvectors with the same eigenvalue $1$. Try to find eigenvectors of $H$ of the ...


12

In a vector space over the field of complex numbers the notion of complex conjugation is basis dependent. You might say a vector is "real" if its components in some basis are real numbers, but if you change to another basis and the matrix expressing the new basis in terms of the old has complex entries then the "real" vector will have complex components in ...


11

1) Let us replace the momentum indices ${\bf k}$ and $-{\bf k}$ with abstract indices $1$ and $2$, and ignore the momentum summation. The Hamiltonian then reads $$\tag{1}H ~=~ \begin{pmatrix}a_1^{\dagger} & a_2\end{pmatrix} M\begin{pmatrix}a_1 \\ a_2^{\dagger}\end{pmatrix}, $$ where $$\tag{2} M~:=\begin{pmatrix}A & B\\B^{*} & A \end{pmatrix}~...


11

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 ...


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