# Tag Info

23

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 ...

16

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 ...

14

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 ...

12

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 ...

11

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 ...

11

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 ...

11

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 ...

10

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 ...

10

Flip back a page; Dirac uses real to mean Hermitian ($A^{\dagger} = A$) 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.

9

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

9

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 ...

9

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 ...

9

A matrix is a special case of a second rank tensor with 1 index up and 1 index down. It takes vectors to vectors, (by contracting the upper index of the vector with the lower index of the tensor), covectors to covectors (by contracting the lower index of the covector with the upper index of the tensor), and in general, it can take an m upper/n-lower tensor ...

9

Hint: When an eigenvalue for an operator $P$ is degenerate, there are more than one way to chose a set of eigenvectors. If the other commuting operator $H$ lifts that degeneracy, there will be a preferred choice of common eigenvectors. More generally, a set of diagonalizable operators commutes if and only if the set is simultaneously diagonalizable.$^1$ ...

8

The expectation value of energy is something else than the energy in a particular experiment. With your choice of the initial states, the photons emitted (negative difference) or absorbed (positive difference) will have energies either $$E_1-E_0 \text{ or } E_2-E_0 \text{ or } E_1-E_5 \text{ or } E_2-E_5$$ If each of the four transitions were equally ...

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 ...

6

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, ...

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 ...

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. ...

6

It is easy. Assume that $\psi \in \cal H$ is normalized to $1$. In this case, $|\psi\rangle\langle\psi|$ it is nothing but the orthogonal projector $P_\psi$ onto the one-dimensional linear space generated by the vector $\psi$. Putting $|\psi\rangle$ and $\langle\psi|$ together simply means to exploit the tensor product. If $\phi' \in \cal H'$ the ...

6

I suggest to think of it like this: $$\left|\psi\right.\rangle = \left(\begin{array}{c}\psi_1\\\psi_2\\\cdots\\\psi_n\end{array}\right)\;,\qquad \langle\left.\psi\right| = \left(\begin{array}{c}\psi_1^*\,\psi_2^*\,\cdots\,\psi_n^*\end{array}\right)$$ And using standard rule ("every element in row per every element in column") for matrix multiplication. ...

6

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 ...

5

Let $\{|i\rangle\}$ be a 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\rangle. ...

5

Isn't your Ex 3 such an example? In any event, you might have trouble producing a pure photon state $\left|\,001\right>$, but you should have no trouble creating a coherent state. Just turn on your laser pointer. The coherent state is a superposition of single photon states, summing over all single photon number states. You can't impose a cutoff. ...

5

Consider first the matrix representation of a single Grassmann number as $\theta = \left[\begin{matrix} a & b \\ c & d \end{matrix}\right]$ where $\theta^2 = \left[\begin{matrix} a^2+bc & b(a+d) \\ c(a+d) & d^2+bc \end{matrix}\right] = 0 \Leftrightarrow \theta = \left[\begin{matrix} i\sqrt{bc} & b \\ c & -i\sqrt{bc} ... 5 Well, learn linear algebra. An advanced text (on linear algebra over "field" number systems) is these lecture notes [pdf] from UC Davis. Once you get that done, you should study differential equations. Or if you want to skip ahead, perhaps Fourier analysis. A free reference would be my notes [pdf]. It's mildly physics-oriented, but connects the ideas back ... 5 There's no wonder you're confused - the author obviously was as well. First, the operations he's talking about are direct sum$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 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 = ...

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