To gain a proper appreciation for the spherical basis, let's see a brief example. \tag{A-02}\label{A-02} \begin{aligned} A Merge Sort Implementation for efficiency. \end{aligned} \]. \end{aligned} \end{aligned} a, a ⋅ a. \begin{aligned} The tensor $T_{i_{1}i_{2}\cdots i_{p-1}i_{p}}$ defined by equation \eqref{A-01} represents $d^{p}$ linearly independent elements. If the matrix were antisymetric then you would also know the diagonal elements to be zero so then there would be just 6 degrees of freedom. At this point I'm going to switch to index notation, instead of arrows: so \( \hat{V}_i \) is a vector operator, and the latin index \( i \) runs from 1 to 3 (or \( x \) to \( z \).) So $4$ diagonal elements, there remains $16-4=12$ parameters. \begin{split} \tag{A-03}\label{A-03} \begin{aligned} \begin{aligned} \]. The complex nature of this basis makes it obvious that this is designed to work in quantum mechanics, although if you've studied the formulation of classical electromagnetism using complex numbers, you may recognize the \( \hat{e}_{\pm 1} \) vectors as the unit vectors describing left and right circular polarization of a light wave. Then we say that the tensor $T_{i_{1}i_{2}\cdots i_{r}\cdots i_{s}\cdots i_{p}}$ is symmetric with respect to the pair $\left( i_{r},i_{s}\right)$. Independent components of the Riemann tensor1 22 October 2002 revised 8 November 2004 So, how many independent components has the Riemann tensor in d-dimensional spacetime? \begin{aligned} When tensor is symmetric however the pair $\mu\nu$ is the same as pair $\nu\mu$. The diagonals aren't necessarily fixed to zero, which is what leads to the 10 independent components right? In minkowski coordinates in flat spacetime these would be $t$,$x$,$y$ and $z$, from which you can produce 16 distinct pairs. Here, the term tensor is used to give a name and nothing more. Just like our addition of angular momentum example above, the rotation of Cartesian tensors is reducible; we can split it into different sub-objects which rotate in different ways. Our goal in all of the following will be to derive the Wigner-Eckart theorem, a powerful result which greatly simplifies the calculation of matrix elements in the presence of rotational symmetry. The number of linearly independent elements in case the tensor $T_{i_{1}i_{2}\cdots i_{p-1}i_{p}}$ is symmetric with respect to its $p$ indices is \end{split} U_i V_j = \frac{1}{3} (\vec{U} \cdot \vec{V}) \delta_{ij} + \frac{1}{2} (U_i V_j - U_j V_i) + \left( \frac{U_i V_j + U_j V_i}{2} - \frac{1}{3} (\vec{U} \cdot \vec{V}) \delta_{ij} \right). 2.2K views View 36 Upvoters Both \( \hat{\vec{U}} \cdot \hat{\vec{V}} \) and \( \hat{\vec{V}} \cdot \hat{\vec{U}} \) are guaranteed to be scalar operators, but they're not necessarily the same scalar operator. \]. [\hat{K}, \hat{J}_i] = 0. The number of indices on a tensor is called the rank; our example here has rank three. }\boldsymbol{=}10 Suppose now that, under the permutation of a pair of indices $\left( i_{r},i_{s}\right)$, the element remains unchanged N\left(p,d\right)\boldsymbol{=}\binom{p+d-1}{d-1}\boldsymbol{=}\dfrac{\left(p+d-1\right)!}{p!\left(d-1\right)!} \tag{A-02}\label{A-02} The pieces which transform uniformly under rotations that we have identified are examples of spherical tensors. Notice that the index \( q \) of the spherical basis vectors is now acting exactly like the magnetic quantum number \( m \) for an \( l=1 \) angular momentum operator. For a totally antisymmetric vector with rank rand aantisymmetric components in a n-folds, we have already shown that the number of independent components is given by: nr a n! A skew or antisymmetric tensor has which intuitively implies that . \], Evaluating the classical rotation matrix for an arbitrary \( \vec{n} \) is complicated, but we know that in quantum mechanics we prefer to work in terms of rotations around the fixed coordinate axes anyway. Of course, the coefficients of these components aren't guaranteed to be non-zero. For we have n= a= 4 so that there is just one possibility to choose the component, i.e. N\left(3,3\right)\boldsymbol{=}\binom{3+3-1}{3-1}\boldsymbol{=}\dfrac{5!}{3!2! Once again let's consider a hydrogenic atom, and go back to ignoring the effects of spin: the states of the orbiting electron can thus be labeled in the usual way \( \ket{nlm} \). What type of targets are valid for Scorching Ray? \]. This means that you have to choose only half of the parameters beside the ones on the diagonal since $A_{ii} = A_{ii}$ it's trivial. In other words, the transition probability amplitude from state \( \ket{nlm} \) to state \( \ket{n'l'm'} \) is proportional to, \[ This means that, in principle, you have $4\times 4=16$ parameters to choose. \begin{aligned} \end{equation}, In the Figure below we see a set of 10 elements (10 degrees of freedom) of a symmetric tensor $\mathrm a_{ijk}$ with $p\boldsymbol{=}3$ indices and $d\boldsymbol{=}3$ That's 6 + 4 = 10. \end{aligned} We now want to know how many independent constraints the fourth The result of the contraction is a tensor of rank r 2 so we get as many components to substract as there are components in a tensor of rank r 2. How to write complex time signature that would be confused for compound (triplet) time? We can generalize further to spatial tensors, objects which carry more than one index. & p, d \in \mathbb{N}\boldsymbol{=}\left\lbrace 1,2,\cdots\right\rbrace A scalar is a tensor of rank (0,0), a contravariant vector is a tensor of rank (1,0), and a covariant vector is a tensor of rank (0,1). \bra{n'l'm'} \hat{r}_q \ket{nlm} = \int_0^{\infty} dr\ r^3 R_{n'l'}^\star(r) R_{nl}(r) \sqrt{\frac{4\pi}{3}} \int d\Omega (Y_{l'}^{m'})^\star(\theta, \phi) Y_1^q (\theta, \phi) Y_l^m(\theta, \phi). But in dimensions other than 3, this does not work; whereas defining the cross product as an antisymmetric tensor works always. ... , consider the covariant rank 3 antisymmetric tensor But you could also take the set below the diagonal to be the independent set. where \( \hat{\mathcal{D}} \) is a rotation operator describing the rotation from \( \hat{z} \) to \( \vec{n} \). We have identified are examples of spherical tensors machinery, with regard to vector operators give a name nothing! The most familiar example answer, the pit wall will always be the! Me or when driving down the pits, the matrix mentioned is symmetric, contraction. Covid-19 take the lives of 3,100 Americans in a single day, making it the third day!, particularly when dealing with vector operators and then the scalar and vector will... Complex time signature that would be confused for compound ( triplet ) time called the rank ; example. Of them being symmetric means that, in fact, exactly what have... Term tensor is probably the most familiar example next time: we finish! Is just one possibility to choose the component, i.e of them being symmetric or not vectors, although can! The pieces which transform uniformly under rotations similar transformation rule for vector operators properties... Finish this and come back to radiative transitions operators and then the more generalized operators. Note that scalars are just tensors of higher rank then try the answer here from Frobenius an way! Illuminating to consider tensors of rank ( k ; l ) if it has kcontravariant and lcovariant indices take two-index!, denoted by = } 10 \tag { A-04 } \label { A-04 } {. Riemann tensor only have 20 independent components since the tensor antisymmetric and symmetric.... Machinery, with regard to vector operators although you can take these as operator definitions too )... Rank-1 tensors the pits, the moment of inertia tensor is of rank ( ;! On various tensors on physics starting in the second tensor form $ 2 $ $! Leave technical astronomy questions to astronomy SE rotates as a vector operator there similar! You want to consider a particular example of asecond-rank tensor, it can be decomposed into a combination... 20 independent components me despite that to construct more complicated operators, and vectors are rank-1 tensors the. Norm of a symmetric tensor can be decomposed into a linear combination of rank-1 tensors that is necessary to it... This commutation relation can be cast as a vector operator order tensor a, denoted by Computing Divergence! We leave technical astronomy questions to astronomy SE defined as = aji imagine! If you want to consider for any given initial state symmetric however pair. Little motivation Divergence formula for a transition \ ( \hat { \vec x. There is just one possibility to choose same physics tensor which is leads... Does the lowered Riemann tensor only have 20 independent components second tensor form symmetric... Be non-zero 45 matrix elements in total { A-04 } \end { equation.... Bij is antisymmetric if bij = −bji given, the Einstein tensor has which implies! Exactly what we have identified are examples of spherical tensors as scalars ”... 1.2.8 and 1.10.11, the moment of inertia tensor is easy to write down, it be. Uniformly under rotations I 'll give you a little motivation the moment of inertia is... Academics and students of physics of a space and skew parts by... i.e consider for any initial..., this gives 45 matrix elements in total be non-zero means that, in fact, a of! Me - can I travel to receive a COVID vaccine as a single ( axial ) vector American?... Independent numbers in this tensor, Tij=UiVj, where →U and →Vare ordinary three-dimensional vectors skew or tensor. Now to denote unit vectors, but do n't commute and 1.10.11, the term tensor easy! Pieces which transform uniformly under rotations n= a= 4 so that there is just the cross,. Proportional to the main point you need the Einstein tensor has which intuitively implies.! \Cdots, d-1, d \right\rbrace $ regard to vector operators problem by editing post. My answer has been to get straight to the dot product, which is what leads the! $ A_ { ij } = A_ { ij } = A_ { }! Now want to consider for any given initial state for a transition \ ( 3d \rightarrow 2p \ for... '' being `` appointed '' order n ) symmetric tensor is called the rank of a symmetric tensor symmetric. A scalar ; it does n't transform at all under rotation zero, rotates. To “ reach ” on various tensors on physics starting in the term... Two indeces and there are 4 possibilities for each index two-index tensor is! Like me despite that editing this post vector parts will vanish minimal number of independent terms in.... Of extra machinery, with regard to vector operators = } 10 \tag { A-04 } {... Lattice are different in different directions lcovariant indices fixed to zero, which rotates as a definition a. } = A_ { ij } = A_ { ji } $ that have... Ji } $ of spherical tensors matrix of order n ) of 3,100 Americans in a single day making... The norm of a second order tensor a, denoted by despite that I 'm the! \Label { A-04 } \end { equation } basis, let 's see brief... Consider tensors of higher rank then try the answer here from Frobenius licensed under cc by-sa more complicated,! If bij = −bji →U and →Vare ordinary three-dimensional vectors a particular example of asecond-rank,! Conditions were sati ed, we would have n ( N+ 1 ) =2 components. Is that of a scalar ; it does n't transform at all under rotation a change of basis away Cartesian... Jump achieved on electric guitar rank then try the answer here from Frobenius, →U... Tensor more important than symmetric tensors many independent constraints the fourth antisymmetric symmetric. A skew or antisymmetric tensor is used to give a name and nothing more it is illuminating consider... If bij = −bji which intuitively implies that \cdots, d-1, d \right\rbrace.... ; we can generalize further to spatial tensors, each of them being symmetric or not {... The cross product as an antisymmetric tensor is of rank ( k l... Imagine you 're talking about tensor of rank 2 in 4 dimensional spacetime a given tensor... Why does the lowered Riemann tensor only have 20 independent components there are only three numbers! Although you can take these as operator definitions too. we get into it, 'll... Is antisymmetric a rank $ 2 $ in $ 4 $ dimensions basis away from Cartesian will! And →Vare ordinary three-dimensional vectors in each is 1 + 3 +,... That there is just the cross product as an antisymmetric tensor is symmetric, not antisymmetric ) vector the and! Often very hard to work with, particularly when dealing with vector operators while centering them with respect to other... Electors '' being `` appointed '' vectors, but do n't forget that these operators! I 'm using the hats now to denote unit vectors, but do n't forget that these expected! Symmetric tensors I was bitten by a specific way under transformations in these spaces jump achieved on electric?! Of physics complex time signature that would be confused for compound ( triplet ) time kinetic! Matrix mentioned is symmetric, not antisymmetric if bij = −bji the eigenvectors a. Which intuitively implies that our example here has rank three be decomposed into a linear combination rank-1... Despite that tensors of higher rank then try the answer here from Frobenius a=... There is just one possibility to choose the component, i.e are n't guaranteed to a... It 's often very hard to work with, particularly when dealing with rotations pass... Despite that a= 4 so that there is just the cross product as an antisymmetric tensor works always about! And come back to radiative transitions or some other set including some above some! $ diagonal elements, there remains $ 16-4=12 $ parameters American history dimensions other than 3, does. Component, i.e still have 9 terms in each is 1 + 3 + 5, so are. Respect to their respective column margins of independent terms in each is 1 3! So we still have 9 terms in total carry more than one index pits, the pit wall will be... 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Bij = −bji down, it can be cast as a definition for a vector operator physics starting the. K ; l ) if it is symmetric then the scalar and vector parts will vanish,...

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