Self-verifying theories

The Wigner–Eckart theorem is a theorem of representation theory and quantum mechanics. It states that matrix elements of spherical tensor operators on the basis of angular momentum eigenstates can be expressed as the product of two factors, one of which is independent of angular momentum orientation, and the other a Clebsch-Gordan coefficient. The name derives from physicists Eugene Wigner and Carl Eckart who developed the formalism as a link between the symmetry transformation groups of space (applied to the Schrödinger equations) and the laws of conservation of energy, momentum, and angular momentum.

The Wigner–Eckart theorem reads:

$⟨jm\left|T_q^k\right|j^{\prime }{m}^{\prime }⟩=⟨j\left|\right|T^k\left|\right|j^{\prime }⟩C_{kq{j}^{\prime }{m}^{\prime }}^{jm}$

where T_q^k is a rank k spherical tensor, $|jm⟩$ and $|j^{\prime }{m}^{\prime }⟩$ are eigenkets of total angular momentum J² and its z-component J_z, $⟨j\left|\right|T^k\left|\right|j^{\prime }⟩$ has a value which is independent of m and q, and $C_{kq{j}^{\prime }{m}^{\prime }}^{jm}=⟨j^{\prime }{m}^{\prime };kq|jm⟩$ is the Clebsch-Gordan coefficient for adding j′ and k to get j.

In effect, the Wigner–Eckart theorem says that operating with a spherical tensor operator of rank k on an angular momentum eigenstate is like adding a state with angular momentum k to the state. The matrix element one finds for the spherical tensor operator is proportional to a Clebsch-Gordan coefficient, which arises when considering adding two angular momenta.

Proof

Starting with the definition of a spherical tensor, we have that

$[J_{\pm },T_q^{(k)}]=\hslash \sqrt{(k\mp q)(k\pm q+1)}{T}_{q\pm 1}^{(k)}$

which we use to then calculate

$⟨{\alpha }^{\prime },j^{\prime }{m}^{\prime }|[J_{\pm },T_q^{(k)}]|\alpha ,jm⟩=\hslash \sqrt{(k\mp q)(k\pm q+1)}⟨{\alpha }^{\prime },j^{\prime }{m}^{\prime }\left|T_{q\pm 1}^{(k)}\right|\alpha ,jm⟩$.

If we expand the commutator on the LHS by calculating the action of the J_± on the bra and ket, then we get

$\begin{array}{rl}⟨{\alpha }^{\prime },j^{\prime }{m}^{\prime }|[J_{\pm },T_q^{(k)}]|\alpha ,jm⟩& =\sqrt{(j^{\prime }\pm m^{\prime })(j^{\prime }\mp m^{\prime }+1)}⟨{\alpha }^{\prime },j^{\prime }{m}^{\prime }\mp 1\left|T_q^{(k)}\right|\alpha ,jm⟩\\ & -\sqrt{(j\mp m)(j\pm m+1)}⟨{\alpha }^{\prime },j^{\prime }{m}^{\prime }\left|T_q^{(k)}\right|\alpha ,jm\pm 1⟩\end{array}$.

We may combine these two results to get

$\begin{array}{rl}\sqrt{(j^{\prime }\pm m^{\prime })(j^{\prime }\mp m^{\prime }+1)}⟨{\alpha }^{\prime },j^{\prime }{m}^{\prime }\mp 1\left|T_q^{(k)}\right|\alpha ,jm⟩& =\sqrt{(j\mp m)(j\pm m+1)}⟨{\alpha }^{\prime },j^{\prime }{m}^{\prime }\left|T_q^{(k)}\right|\alpha ,jm\pm 1⟩\\ & +\sqrt{(k\mp q)(k\pm q+1)}⟨{\alpha }^{\prime },j^{\prime }{m}^{\prime }\left|T_{q\pm 1}^{(k)}\right|\alpha ,jm⟩\end{array}$.

This recursion relation for the matrix elements closely resembles that of the Clebsch-Gordan coefficient. In fact, both are of the form $\sum _j{a}_{ij}{x}_j=0$. We therefore have two sets of linear homogeneous equations

$\sum _j{a}_{ij}{x}_j=0,\sum _j{a}_{ij}{y}_j=0$

one for the Clebsch-Gordan coefficients (x_j) and one for the matrix elements (y_j). It is not possible to exactly solve for the x_j. We can only say that the ratios are equal, that is

$\frac{x_j}{x_k}=\frac{y_j}{y_k}$

or that x_j = cy_j, where c is a coefficient of proportionality independent of the indices. Hence, by comparing recursion relations, we can identify the Clebsch-Gordan coefficient $⟨j_1{j}_2;m_1,m_2\pm 1|j_1{j}_2;jm⟩$ with the matrix element $⟨{\alpha }^{\prime },j^{\prime }{m}^{\prime }\left|T_{q\pm 1}^{(k)}\right|\alpha ,jm⟩$, then we may write

$⟨{\alpha }^{\prime },j^{\prime }{m}^{\prime }\left|T_{q\pm 1}^{(k)}\right|\alpha ,jm⟩=\text{(proportionality constant)}⟨jk;mq\pm 1|jk;j^{\prime }{m}^{\prime }⟩$.

By convention the proportionality constant is written as $⟨{\alpha }^{\prime }{j}^{\prime }\left|\right|T^{(k)}\left|\right|\alpha j⟩\frac{1}{\sqrt{2j+1}}$, where the denominator is a normalizing factor.

Example

Consider the position expectation value $⟨njm\left|x\right|njm⟩$. This matrix element is the expectation value of a Cartesian operator in a spherically-symmetric hydrogen-atom-eigenstate basis, which is a nontrivial problem. However, using the Wigner–Eckart theorem simplifies the problem. (In fact, we could obtain the solution quickly using parity, although a slightly longer route will be taken.)

We know that x is one component of Template:Vec, which is a vector. Vectors are rank-1 tensors, so x is some linear combination of T¹_q for q = -1, 0, 1. In fact, it can be shown that

$x=\frac{T_{-1}^1-T_1^1}{\sqrt{2}},$

where we defined the spherical tensors T¹₀ = z and

$T_{\pm 1}^1=\mp (x\pm iy)/\sqrt{2}$

(the pre-factors have to be chosen according to the definition of a spherical tensor of rank k. Hence, the T¹_q are only proportional to the ladder operators). Therefore

$⟨njm\left|x\right|n^{\prime }{j}^{\prime }{m}^{\prime }⟩=⟨njm\left|\frac{T_{-1}^1-T_1^1}{\sqrt{2}}\right|n^{\prime }{j}^{\prime }{m}^{\prime }⟩=\frac{1}{\sqrt{2}}⟨nj\left|\right|T^1\left|\right|n^{\prime }{j}^{\prime }⟩(C_{1(-1)j^{\prime }{m}^{\prime }}^{jm}-C_{11j^{\prime }{m}^{\prime }}^{jm})$

The above expression gives us the matrix element for x in the $|njm⟩$ basis. To find the expectation value, we set n′ = n, j′ = j, and m′ = m. The selection rule for m′ and m is $m\pm 1=m^{\prime }$ for the $T_{\mp 1}^{(1)}$ spherical tensors. As we have m′ = m, this makes the Clebsch-Gordan Coefficients zero, leading to the expectation value to be equal to zero.

External links

• J. J. Sakurai, (1994). "Modern Quantum Mechanics", Addison Wesley, ISBN 0-201-53929-2.
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• Wigner–Eckart theorem
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