Talk:Tensor operator

Spill from spherical basis

It seems the two concepts spherical basis + spherical tensor, and tensor operators + spherical tensor operators, are really very closely intertwined through the spherical harmonics, and also appear to be two separate things.

This article on tensor operators should include spherical tensor operators as a subset. The other article on spherical basis should discuss them then go on to explain spherical tensors. This arrangement of articles seems to nicely run with the terminology, as far as I can tell.

If there are big objections, which so far there aren't yet at wikiproject mathematics, then the changes can all be reverted easily. M∧Ŝc²ħεИ_τlk 08:54, 13 June 2013 (UTC)

Plan

I will be looking to improve upon the Wiki-page 'Tensor Operator',

• Defining Tensor Operators.
• Examples of Scalar Operators, Vector Operators and Tensor Operators.
• Applications of the Spherical Basis in the problem of dipole radiative transitions in a single-electron atom eg hydrogen or an alkali.
• Reducible and Irreducible spaces of operators.
• Magnetic Resonance.

I have added the general definition of the Rotation of an operator in terms of the expectation value as,

We define the Rotation of an operator by requiring that the expectation value of the original operator A with respect to the initial state be equal to the expectation value of the rotated operator with respect to the rotated state,

$\langle \psi '|A'|\psi '\rangle =\langle \psi |A|\psi \rangle$

Now as,

$|\psi \rangle$ → $|\psi '\rangle =U(R)|\psi \rangle \,,\quad \langle \psi |$ → $\langle \psi '|=\langle \psi |U^{\dagger }(R)$

we have,

$\langle \psi |U^{\dagger }(R)A'U(R)|\psi \rangle =\langle \psi |A|\psi \rangle$

since, $|\psi \rangle$ is arbitrary,

$U^{\dagger }(R)A'U(R)=A$

For suggestions,

Tshibli (talk)

Example of a Tensor operator,

• The Quadrupole moment operator,

$Q_{ij}=\sum _{\alpha }q_{\alpha }(3r_{\alpha i}r_{\alpha j}-r_{\alpha }^{2}\delta _{ij})$

• Two Tensor operators can be multiplied to give another Tensor operator.

$T_{ij}=V_{i}W_{j}$

in general,

$T_{i_{1}i_{2}\cdots j_{1}j_{2}\cdots }=V_{i_{1}i_{2}\cdots }W_{j_{1}j_{2}\cdots }$

NOTE

This is just an example, in general, a tensor operator cannot be written as the product of two Tensor operators.

dipole radiative transitions in a single-electron atom (alkali)

The transition amplitude is proportional to matrix elements of the dipole operator between the initial and final states. We use an electrostatic, spinless model for the atom, as in Notes 16, and we consider the transition from the initial energy level E_{nℓ} to final level E_{n′ℓ′}. These levels are degenerate, since the energy does not depend on the magnetic quantum number m or m′. The wave functions have the form,

$\psi _{nlm}(r,\theta ,\phi )=R_{nl}(r)Y_{lm}(\theta ,\phi )$

Wrong formula?

In the section "Spherical tensor operators" the formula

${U(R)}^{\dagger }{\widehat {T}}_{q'}^{(k)}U(R)=\sum _{q}D_{qq'}^{(k)}{\widehat {T}}_{q}^{(k)}$

seems to be inconsistent with the previous one:

${U(R)}^{\dagger }{\widehat {T}}_{q}^{(2)}U(R)=\sum _{q'}{D(R)}_{qq'}^{(2)}{\widehat {T}}_{q'}^{(2)}$

Shouldn't the first one be:

${U(R)}^{\dagger }{\widehat {T}}_{q}^{(k)}U(R)=\sum _{q'}{D(R)}_{qq'}^{(k)}{\widehat {T}}_{q'}^{(k)}$

? Do you agree with the change? 93.41.234.220 (talk) 18:11, 9 November 2014 (UTC)

Yes, you're correct, I must have copy-pasted the indices wrong or something. Will update. Thanks, M∧Ŝc²ħεИ_τlk 21:39, 9 November 2014 (UTC)

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