# Talk:Tensor operator

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## 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∧Ŝc2ħεИτlk 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,

Example of a Tensor operator,

$Q_{ij} = \sum_{\alpha}q_{\alpha} (3 r_{\alpha i}r_{\alpha j} - r_{\alpha}^{2} \delta_{ij})$
• 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 and we consider the transition from the initial energy level Enℓ to final level En′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,

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

The dipole operator is proportional to the position operator of the electron, so we must evaluate matrix elements of the form,

$\langle n'l'm'|\mathbf{r}|nlm \rangle$

where, the initial state is on the right and the final one on the left. The position operator r has three components, and the initial and final levels consist of 2ℓ + 1 and 2ℓ′ + 1 degenerate states, respectively. Therefore if we wish to evaluate the intensity of a spectral line as it would be observed, we really have to evaluate 3(2ℓ′+ 1)(2ℓ+ 1) matrix elements, for example, 3×3×5 = 45 in a 3d → 2p transition. This is actually an exaggeration, as we shall see, because many of the matrix elements vanish, but there are still many non-vanishing matrix elements to be calculated.

A great simplification can be achieved by expressing the components of r, not with respect to the Cartesian basis, but with respect to the spherical basis. First we define,

$r_{q} = \hat{\mathbf{e}}_{q}\cdot r$

Next, by inspecting a table of the Yℓm’s, we find that for ℓ = 1 we have,

$r Y_{11}(\theta,\phi)= -r \sqrt{\frac{3}{8\pi}}sin\theta e^{i\phi}=\sqrt{\frac{3}{4\pi}}\left(-\frac{x+iy}{\sqrt{2}}\right)$
$r Y_{10}(\theta,\phi)= r \sqrt{\frac{3}{\pi}}cos\theta =\sqrt{\frac{3}{4\pi}}z$
$r Y_{1-1}(\theta,\phi)= r \sqrt{\frac{3}{8\pi}}sin\theta e^{-i\phi}=\sqrt{\frac{3}{4\pi}}\left(\frac{x-iy}{\sqrt{2}}\right)$

where, we have multiplied each Y1m by the radius r. On the right hand side we see the spherical components rq of the position vector r. The results can be summarized by,

$r Y_{1q}(\theta,\phi) = \sqrt{\frac{3}{4 \pi}} r_{q}$

for q = 1, 0, −1, where q appears explicitly as a magnetic quantum number. This equation reveals a relationship between vector operators and the angular momentum value ℓ = 1, something we will have more to say about presently. Now the matrix elements become a product of a radial integral times an angular integral,

$\langle n'l'm'|\mathbf{r}|nlm \rangle = \left(\int_0^{\infty}r^2 dr R_{n'l'}^* (r)r R_{nl}(r)\right) \left(\sqrt{\frac{4 \pi}{3}}\int d\Omega Y_{n'l'}^* (\theta,\phi)Y_{1q}(\theta,\phi)Y_{lm}(\theta,\phi)\right)$

We see that all the dependence on the three magnetic quantum numbers (m′,q,m) is contained in the angular part of the integral. Moreover, the angular integral can be evaluated by the three-Yℓm formula, whereupon it becomes proportional to the Clebsch-Gordan coefficient,

$\langle l'm'|l1mq\rangle$

The radial integral is independent of the three magnetic quantum numbers (m′, q, m), and the trick we have just used does not help us to evaluate it. But it is only one integral, and after it has been done, all the other integrals can be evaluated just by computing or looking up Clebsch-Gordan coefficients.

The selection rule m′ = q + m in the Clebsch-Gordan coefficient means that many of the integrals vanish, so we have exaggerated the total number of integrals that need to be done. But had we worked with the Cartesian components ri of r, this selection rule might not have been obvious. In any case, even with the selection rule, there may still be many nonzero integrals to be done (nine, in the case 3d → 2p). The example we have just given of simplifying the calculation of matrix elements for a dipole transition is really an application of the Wigner-Eckart theorem, which we take up later in these notes.

## Wrong formula?

In the section "Spherical tensor operators" the formula

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

seems to be inconsistent with the previous one:

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

Shouldn't the first one be:

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

? 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∧Ŝc2ħεИτlk 21:39, 9 November 2014 (UTC)

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