Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

PHY456H1F: Quantum Mechanics II. Lecture 20 (Taught by Prof J.E. Sipe). Spherical tensors.

Posted by peeterjoot on November 23, 2011

[Click here for a PDF of this post with nicer formatting and figures if the post had any (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]


Peeter’s lecture notes from class. May not be entirely coherent.

Spherical tensors (cont).

READING: section 29 of [1].

definition. Any (2k + 1) operator T(k, q), q = -k, \cdots, k are the elements of a spherical tensor of rank k if

\begin{aligned}U[M] T(k, q) U^{-1}[M]= \sum_{q'} T(k, q') D^{(k)}_{q q'}\end{aligned} \hspace{\stretch{1}}(2.1)

where D^{(k)}_{q q'} was the matrix element of the rotation operator

\begin{aligned}D^{(k)}_{q q'} = {\left\langle {k q'} \right\rvert} U[M] {\left\lvert {k q''} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(2.2)

So, if we have a Cartesian vector operator with components V_x, V_y, V_z then we can construct a corresponding spherical vector operator

\begin{aligned}\begin{array}{l l l}T(1, 1) &= - \frac{V_x + i V_y}{\sqrt{2}} &\equiv V_{+1} \\ T(1, 0) &= V_z &\equiv V_0 \\ T(1, -1) &= - \frac{V_x - i V_y}{\sqrt{2}} &\equiv V_{-1}\end{array}.\end{aligned} \hspace{\stretch{1}}(2.3)

By considering infinitesimal rotations we can come up with the commutation relations between the angular momentum operators

\begin{aligned}\left[{J_{\pm}},{T(k, q)}\right] &= \hbar \sqrt{(k \mp q)(k \pm q + 1)} T(k, q \pm 1) \\ \left[{J_{z}},{T(k, q)}\right] &= \hbar q T(k, q)\end{aligned} \hspace{\stretch{1}}(2.4)

Note that the text in (29.15) defines these, whereas in class these were considered consequences of 2.1, once infinitesimal rotations were used.

Recall that these match our angular momentum raising and lowering identities

\begin{aligned}J_{\pm} {\left\lvert {k q} \right\rangle} &= \hbar \sqrt{(k \mp q)(k \pm q + 1)} {\left\lvert {k, q \pm 1} \right\rangle} \\ J_{z} {\left\lvert {k q} \right\rangle} &= \hbar q {\left\lvert {k, q} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(2.6)

Consider two problems

\begin{aligned}\begin{array}{l l l}T(k, q)						& \right\rangle} \\ \left[{J_{\pm}},{T(k, q)}\right] 		&\leftrightarrow &J_{\pm} {\left\lvert {k q} \right\rangle} \\ \left[{J_{z}},{T(k, q)}\right] 			&\leftrightarrow &J_{z} {\left\lvert {k q} \right\rangle}\end{array}\end{aligned} \hspace{\stretch{1}}(2.8)

We have a correspondence between the spherical tensors and angular momentum kets

\begin{aligned}\begin{array}{l l l l}T_1(k_1, q_1)&\qquad q_1 = -k_1, \cdots, k_1 		& \qquad {\left\lvert {k_1 q_1} \right\rangle} 		\right\rangle} \\ T_2(k_2, q_2)&\qquad q_2 = -k_2, \cdots, k_2		& \qquad q_1 = -k_1, \cdots k_1 	& q_2 = -k_2, \cdots k_2 \\ \end{array}\end{aligned} \hspace{\stretch{1}}(2.9)

So, as we can write for angular momentum

\begin{aligned}{\left\lvert {kq} \right\rangle} &= \sum_{q_1, q_2} {\left\lvert {k_1, q_1} \right\rangle}{\left\lvert {k_2, q_2} \right\rangle}\underbrace{\left\langle{{ k_1 q_1 k_2 q_2 }} \vert {{ k q}}\right\rangle}_{\text{These are the C.G coefficients}}  \\ {\left\lvert {k_1 q_1 ; k_2 q_2} \right\rangle}&=\sum_{k, q'}{\left\lvert {k q'} \right\rangle} \left\langle{{ k q'}} \vert {{ k_1 q_1 k_2 q_2 }}\right\rangle \end{aligned}

We also have for spherical tensors

\begin{aligned}T(k, q) &= \sum_{q_1, q_2} T_1(k_1, q_1)T_2(k_2, q_2)\left\langle{{ k_1 q_1 k_2 q_2 }} \vert {{ k q}}\right\rangle	\\ T_1(k_1, q_1)T_2(k_2, q_2)&=\sum_{k, q'}T(k, q') \left\langle{{ k q'}} \vert {{ k_1 q_1 k_2 q_2 }}\right\rangle &\end{aligned}

Can form eigenstates {\left\lvert {kq} \right\rangle} of (\text{total angular momentum})^2 and (z-comp of the total angular momentum).
FIXME: this won’t be proven, but we are strongly suggested to try this ourselves.

\begin{aligned}\begin{array}{l l l}\text{spherical tensor (3)} 				&\leftrightarrow &\text{Cartesian vector (3)} \\ (\text{spherical vector})(\text{spherical vector})	&		 &\text{Cartesian tensor}\end{array}\end{aligned} \hspace{\stretch{1}}(2.10)

We can check the dimensions for a spherical tensor decomposition into rank 0, rank 1 and rank 2 tensors.

\begin{aligned}\begin{array}{l l l}\text{spherical tensor rank latex 0$} & (1) & (\text{Cartesian vector})(\text{Cartesian vector}) \\ \text{spherical tensor rank 1} & (3) & (3)(3) \\ \text{spherical tensor rank 2} & (5) & 9 \\ \hline\text{dimension check sum} & 9 & \\ \end{array}\end{aligned} \hspace{\stretch{1}}(2.11)$

Or in the direct product and sum shorthand

\begin{aligned}1 \otimes 1 = 0 \oplus 1 \oplus 2\end{aligned} \hspace{\stretch{1}}(2.12)

Note that this is just like problem 4 in problem set 10 where we calculated the CG kets for the 1 \otimes 1 = 0 \oplus 1 \oplus 2 decomposition starting from kets {\left\lvert {1 m} \right\rangle}{\left\lvert {1 m'} \right\rangle}.

\begin{aligned}\begin{array}{l l l}{\left\lvert {22} \right\rangle}		&				& 		\\ {\left\lvert {21} \right\rangle}		\right\rangle} 			& 		\\ {\left\lvert {20} \right\rangle}		\right\rangle} 			\right\rangle} 	\\ {\left\lvert {2\overline{1}} \right\rangle}	} \right\rangle} 		& 		\\ {\left\lvert {2\overline{2}} \right\rangle}	&				&\end{array}\end{aligned} \hspace{\stretch{1}}(2.13)


How about a Cartesian tensor of rank 3?

\begin{aligned}A_{ijk}\end{aligned} \hspace{\stretch{1}}(2.14)

\begin{aligned}1 \otimes 1 \otimes 1  &=1 \otimes ( 0 \oplus 1 \oplus 2) \\ &=(1 \otimes 0) \oplus (1 \otimes 1) \oplus (1 \otimes 2) \\ &=\begin{array}{l l l l l l l l l l l l l l}1 &\oplus   &(0 &\oplus &1 &\oplus &2) &\oplus &(3  &\oplus & 2 &\oplus &1) \\ 3 &+        &1 &+      &3 &+      &5  &+       &7  &+      & 5 &+      &3 = 27\end{array}\end{aligned}

Why bother?

Consider a tensor operator T(k, q) and an eigenket of angular momentum {\left\lvert {\alpha j m} \right\rangle}, where \alpha is a degeneracy index.

Look at

\begin{aligned}T(k, q) {\left\lvert {\alpha j m} \right\rangle}U[M] T(k, q) {\left\lvert {\alpha j m} \right\rangle}&=U[M] T(k, q) U^\dagger[M] U[M] {\left\lvert {\alpha j m} \right\rangle} \\ &=\sum_{q' m'} D^{(k)}_{q q'} D^{(j)}_{m m'} T(k, q') {\left\lvert {\alpha j m'} \right\rangle} \end{aligned}

This transforms like {\left\lvert {k q} \right\rangle} \otimes {\left\lvert {j m} \right\rangle}. We can say immediately

\begin{aligned}{\left\langle {\alpha' j' m'} \right\rvert} T(k, q) {\left\lvert {\alpha j m} \right\rangle} = 0 \end{aligned} \hspace{\stretch{1}}(2.15)


\begin{aligned}{\left\lvert{k - j}\right\rvert} &\le j' \le k + j \\ m' &= m + q\end{aligned} \hspace{\stretch{1}}(2.16)

This is the “selection rule”.


\item Scalar T(0, 0)

\begin{aligned}{\left\langle {\alpha' j' m'} \right\rvert} T(0, 0) {\left\lvert {\alpha j m} \right\rangle} = 0 ,\end{aligned} \hspace{\stretch{1}}(2.18)

unless j = j' and m = m'.

\item V_x, V_y, V_z. What are the non-vanishing matrix elements?

\begin{aligned}V_x = \frac{ V_{-1} - V_{+1}}{\sqrt{2}}, \cdots\end{aligned} \hspace{\stretch{1}}(2.19)

\begin{aligned}{\left\langle {\alpha' j' m'} \right\rvert} V_{x, y} {\left\lvert {\alpha j m} \right\rangle} = 0 ,\end{aligned} \hspace{\stretch{1}}(2.20)


\begin{aligned}{\left\lvert{j - 1}\right\rvert} &\le j' \le j + 1  \\ m' &= m \pm 1\end{aligned} \hspace{\stretch{1}}(2.21)

\begin{aligned}{\left\langle {\alpha' j' m'} \right\rvert} V_{z} {\left\lvert {\alpha j m} \right\rangle} = 0 ,\end{aligned} \hspace{\stretch{1}}(2.23)


\begin{aligned}{\left\lvert{j - 1}\right\rvert} &\le j' \le j + 1  \\ m' &= m  \end{aligned} \hspace{\stretch{1}}(2.24)


Very generally one can prove (the Wigner-Eckart theory in the text section 29.3)

\begin{aligned}{\left\langle {\alpha_2 j_2 m_2} \right\rvert} T(k, q) {\left\lvert {\alpha_1 j_1 m_1} \right\rangle}={\left\langle {\alpha_2 j_2 } \right\rvert} T(k) {\left\lvert {\alpha_1 j_1} \right\rangle} \cdot\left\langle{{j_2 m_2}} \vert {{k q_1 ; j_1 m_1}}\right\rangle \end{aligned} \hspace{\stretch{1}}(2.26)

where we split into a “reduced matrix element” describing the “physics”, and the CG coefficient for “geometry” respectively.


[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

%d bloggers like this: