Tensor operators

Tensor operators#

So far we have mostly discussed the action of transformations on quantum states. But transformations are also defined by their action on operators:

\[A \to U A U^{\dagger} = A'\]

In the case that the transformation is infintesimal, $U = {\bf 1} - i \eps K + \cO(\eps^2)$ with $K$ Hermitian, then to $\cO(\eps)$, we have

\[A' = A - i \eps [K,A]\]

We cna compare this to the change in classical observables under a transformation generated by an obserbale $K(p,q)$: for this %\delta A = - {K, A}$ where ${ , }$ is the Poisson bracket.

Now operators also form a vector space under addition. So we can expect them to form irreps of the transformation group. Here we will discuss such operators in the case of the rotation group.

Example: position operators#

To get our heads around the concept of the transformation of operators. To see how these transform, consider the matrix element

\[\begin{split}\begin{align} \bra{\chi} ({\hat x}^i)' \ket{\psi} & = \bra{\chi} U(R) {\hat x}^i U^{\dagger}(R) \ket{\psi}\\ & = \int d^3 x \chi^*(R^{-1} x) x^i \psi(R^{-1} x) \end{align}\end{split}\]

Now, we can change variables as ${\vec x} \to {\vec (Rx)}$. Because $R$ has determinant $1$, the measure factor does not change and we have```{math}

\[\bra{\chi} ({\hat x}^i)' \ket{\psi} = \int d^3 x \chi^*(x) R^i_j x^j \psi(x)\]

This is true for any pair of states, so we have $({\hat x}^i)' = R^i_j {\hat x}^j$; the operator transforms the same way as a vector in physical space.

In the case of infinitesimal transformations, we can expand out $R$. Instead, let us consider the simple case of an infinitesimal rotation about the $z$ axis. In this case, $U \sim {\bf 1} - \frac{i\phi}{\hbar} J_z$, and so $({\hat x}^i)' = {\hat x}^i - \frac{i\phi}{\hbar} [J_z, {\hat x}^i]$. Thus

\[\begin{split}\begin{align} {\hat x}' & = {\hat x} + \phi {\hat y}\\ {\hat y}' & = {\hat y} - \phi {\hat y}\\ {\hat z}' & = {\hat z} \end{align}\end{split}\]

This is precisely how a vector transforms under infinitesimal rotations about the $z$ axis. The first two transformations can be reassembled like so:

\[({\hat x} \pm i {\hat y})' = ({\hat x} \pm i {\hat y})(1 \pm i \phi)\]

Under finite rotations, $A' = e^{-i \phi J_z/\hbar} A e^{i\phi J_z/\hbar}$. To see how this changes with $\phi$, we compute

\[\frac{\del}{\del \phi} A' = \frac{\del}{\del \phi} \left(e^{-i \phi J_z/\hbar} A e^{i\phi J_z/\hbar}\right) = \frac{- i}{\hbar} \left(e^{-i \phi J_z/\hbar} [J_z,A] e^{i\phi J_z/\hbar}\right)\]

Now in the present case we have

\[\begin{split}\begin{align} \frac{\del}{\del\phi} (x + i y)' & = y' - i x' = i (x + i y)'\\ \frac{\del}{\del\phi} (x - i y)' & = - i (x - i y)'\\ \frac{\del}{\del\phi} z' = 0 \end{align}\end{split}\]

Integraing these equations from $\phi = 0$, we find

\[\begin{split}\begin{align} (x + i y)' & = (x + i y) e^{i\phi}\\ (x - i y)' & = (x + i y) e^{-i\phi}\\ z' & = z \end{align}\end{split}\]

In other words, under rotations, $x \pm i y$ transform as the $m = \pm 1$ components of a spin-1 irrep, and $z$ as the $m = 0$ component. We can show that the commutators of $J_{\pm}$ with $x \pm i y$ and $z$ raise and lower $m$ as expected.

Spherical tensor operators#

More generally, we denote the irreducible spherical tensor operators $\cO^j_m$ for fixed $j$ as operators which transform as a spin-$j$ irrep under commutators with ${\vec J}$. In particular, this means

\[\begin{split}\begin{align} [J_z, \cO^j_m] & = \hbar m \cO^j_m\\ [J_{\pm}, \cO^j_m] & = \hbar \sqrt{j(j+1) - m(m\pm 1)} \cO^j_{m\pm 1} \end{align}\end{split}\]

Under finite rotations, we have

\[U(R) \cO^j_m U(R)^{\dagger} = d^j_{m,m'} \cO^j_{m'}\]

so the operators transform just as the $\ket{j,m}$ states do.

The simples tensor operators have specific names:

An operator $\cO^0_0$ is called a scalar operator; it does not change under rotations. Thee include operators such as ${\hat {\vec p}}^2$, ${\vec J}^2$, and so on.
The operators $\cO^{1/2}_{\pm \half}$ are called spinor operators. Off the tiop of my head the examples I can think of appear in quantum field theory.
The operators $\cO^1_{\pm 1}$, $\cO^1_0$ are called vector operators. We can find linear combinations which transform the same way as a vector in $\CR^3$. Examples as ${\hat {\vec x}}$, ${\hat {\vec p}}$,
${\hat {\vec J}}$.

Relationship to cartesian tensors#

In general when people say “tensors” they have in mind objects with lots of indices that run form $1,\ldots d$ (where $d$ is the dimension fo spacetime). One example is the totally antisymmetric tensor $\eps_{ijk}$.

In general, a tensor in $\CR^3$ is an object of the form

\[T^{i_1,\ldots,i_m}{}_{j_1,\ldots,j_n}\ ,\]

where $i_k, j_k \in \{1,\ldots,d\}$, that transforms as follows under rotations $R$:

\[T^{i_1,\ldots,i_m}{}_{j_1,\ldots,j_n} \to R^{i_1}_{i'_1}\ldots E^{i_m}_{i'_m} (R^{-1})^{j'_1}_{j_1}\ldots (R^{-1})^{j'_m}_{j_m}T^{i'_1,\ldots,i'_m}{}_{j'_1,\ldots,j'_n}\]

The upper indices are called contravariant indices and the lower indices covariant indices. The reason for this nomenclature is based on how the transform relative to the gradient operator $\frac{\del}{\del x^i}$; the chain rule shows that

\[\frac{\del}{\del x^i} \to (R^{-1})^k_i \frac{\del}{\del x^k}\]

Thus the upper indices transform oppositely to this (thus contravariant), and the lower indices transform in the same way (thus covariant).

Note that the coordinates $x^i$ are thus contravariant vectors. As it happens, the momenta $p_i$ are covariant vectors. This is compatible with their representation in quantum mechanics ${\hat p}_i = \frac{\hbar}{i} \frac{\del}{\del x^i}$.

Such tensors form reducible representations of the rotation group. We can see they behave like tensor products of the spin-$1$ irrep. From these, we can try to work out combinations that transform as irreps. The general theory is complicated; we will work out a simple example.

Consider a 2-index contravariant tensor $T^{ij}$. This has 9 total components. Since under the rules for tensor products of irreps of the rotation group, $D_1\otimes D_1 = D_2 \oplus D_1 \oplus D_0$, we expect $T^{ij}$ to be written in terms of a scalatr operator, a vector operator, and a $j = 2$ tensor operator. To see this note that the total angular omentum in $D_1\otimes D_2$ is invariant under exchange of these two factors. For the tensor, this corresponds to the exchange of the two indices. Therefore, we expect a given irrep to correspond to tensors which come back to themselves up to $\pm 1$ under the exchange of indices.

First, we can form $T = \frac{1]{3} \delta_{ij} T^{ij}$. Orthogonality of the rotation matrices guarantees that this is invariant under rotations; so this is a scalar operator.

Next, we can consider $T_{as}^{ij} = \half \left(T^{ij} - T^{ji}\right)$. This has three components. The following object

\[T_i = \frac{1}{2} \epsilon_{ijk} T^{jk}\]

depends on the same three components; it transforms as a (covariant) vector. So the antisymmetric part of $T$ is a vector opwerator.

Finally, this leaves $T_s^{ij} = \half(T^{ij} + T^{ji}) - \delta_{ij} T$. This is traceless and symmetric. By removing the trace, the resulting symmetric tensor has 5 independent parameters. Thus the components transform as operators of the form $\cO^2_m$, $m \in \{-2,1,0,1,2\}$.

Action on states and the Wigner-Eckart theorem#

Let us consider a spherical temsor operator $\cO^{j_1}_{m_1}$ acting on a state $\ket{\alpha, j_2 m_2}$. where $\alpha$ denote sadditional quantum numbers. Under finite rotations, this transforms as

\[\begin{split}\begin{align} \cO^{j_1}_{m_1}\ket{\alpha, j_2 m_2} & \to U(R) \cO^{j_1}_{m_1}\ket{\alpha, j_2 m_2}\\ & = U(R) \cO^{j_1}_{m_1}U(R)^{\dagger} U(R) \ket{\alpha, j_2 m_2}\\ & = d^{j_1}_{m_1,m_1'} \cO^{j_1}_{m_1'} d^{j_2}_{m_2,m_2'}\ket{\alpha, j_2 m_2'} \end{align}\end{split}\]

But this behaves exactly like s state in the tensor product $D_{j_1}\otimes D_{j_2}$. Furthermore, if we act on this state with the operator $J_z$, using $[J_z,\cO^j_m] = \hbar m \cO^j_m$, we have

\[J_z \cO^{j_1}_{m_1}\ket{\alpha, j_2 m_2} = \hbar(m_1 + m_2) \cO^{j_1}_{m_1}\ket{\alpha, j_2 m_2}\]

Thus,

\[\cO^{j_1}_{m_1}\ket{j_2 m_2,\alpha} = \oplus_{j = |j_1 - j_2|}^{j = j_1 + j_2} c_j^{\beta} \ket{\beta, j, m_1 + m_2}\]

This leads to the following important theorem, the Wigner-Eckart theorem. It comes from taking matrix elements of the commutators of $J_z, J_{\pm}$ with $\cO$. We will not prove it here, but simply state it:

\[\bra{\alpha_1, j_1, m_1} \cO^{j_2}_{m_2} \ket{\alpha_3, j_3, m_3} = \langle \alpha_1, j_1 || \cO^{j_2} || \alpha)3, j_3 \rangle \brket{j_1,m_2; j_2,m_2}{j_3,m_3}\]

Here $\brket{j_1,m_2; j_2,m_2}{j_3,m_3}$ is a Clebsch-Gordon coefficient for the expansion of $D_{j_1}\otimes D_{j_2}$ into irreps with angular momentum $j_3$; while $ \langle \alpha_1, j_1 || \cO^{j_2} || \alpha_3, j_3 \rangle$ is a reduced matrix element. The essential point is that it is independent of $m_k$; all of the $m_k$ dependence, and the information about which $j_3$ appear at all, are contained in the Clebsch-Gordon coefficients. This is kinematic information, goverened entirely by the properties under rotation. The dynamical informatio, dependent on the detailed nature of the system, is contained in the reduced matrix element. To calculate this, we need merely compute the matrix element for specific set of $m_k$s. We can then divide by the Clebsch-Gordon coefficient for that set, to extract the reduced matrix element; all other matrix elements as above are then determined by the reduced matrix element and the Clebsch-Gordon coefficients which are known (and can be looked up).