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  Why is there a phase factor when the two composite angular momentum is exchanged in Clebsch–Gordan coefficients

+ 3 like - 0 dislike

An identity exists for CG coefficients: $$\langle j_1 m_1 j_2 m_2 |J M \rangle = (-1)^{j_1+j_2-J} \langle j_2 m_2 j_1 m_1|J M\rangle,$$

But why is there a phase factor $(-1)^{j_1+j_2-J}$?

It seems to me that $$|JM\rangle =\sum_{m_{1},m_{2}}|j_{1}m_{1}\rangle\otimes|j_{2}m_{2}\rangle\langle j_{1}m_{1}j_{2}m_{2}|JM\rangle =\sum_{m_{1},m_{2}}|j_{2}m_{1}\rangle\otimes|j_{1}m_{2}\rangle\langle j_{2}m_{2}j_{1}m_{1}|JM\rangle $$

And since $|j_{1}m_{1}\rangle\otimes|j_{2}m_{2}\rangle$ and $|j_{2}m_{1}\rangle\otimes|j_{1}m_{2}\rangle$ are the same physical state, there should be no difference between $\langle j_1 m_1 j_2 m_2 |J M \rangle$ and $\langle j_2 m_2 j_1 m_1|J M\rangle$. What do I get wrong?

This post imported from StackExchange Physics at 2014-04-01 05:48 (UCT), posted by SE-user C.R.
asked Jun 2, 2012 in Theoretical Physics by C.R. (15 points) [ no revision ]

2 Answers

+ 2 like - 0 dislike

Your mistake is simply the statement that the state $|j_1 m_1\rangle$ and $|j_2 m_2\rangle$ are the same physical state: these are abstract angular momentum labels, they aren't full descriptions of the state. The two states would only be the same when the two objects carrying these quantum numbers are indistinguishable in every way, like two spin-1/2 electrons in an S-wave in the He atom ground-state. In this case, only the antisymmetric spin-0 combination survives, where the phase factor $(-1)^{j_1+j_2 + J}$ is -1. The two electrons are fermions, so you see the two states need to get a minus sign, and this is only possible when they are a spin 0 combination. There is no spin-1 version of the He4 ground state, because the two electrons can't have the spin aligned since they are fermions.

The way to understand the phase factor is through a few examples, and a better SU(2) representation theory. The examples are the vector combination law:

$$ A \cdot B $$ $$ A \times B $$ $$ (AB)_{ij} = {1\over 2}(A_i B_j + B_j A_i) - {1\over 3} A\cdot B \delta_{ij}$$

These are the spin-0, spin-1, and spin-2 parts of the product of two vectors, in SO(3) index notation. You can see that the spin-1 part is antisymmetric under interchange, and the spin-2 and spin-0 part are symmetric.

Likewise, the antisymmeric spin-1/2 spin-1/2 combination is the singlet, and the symmetric one is the triplet. This is easiest to see in SU(2) index notation, where

$$ \epsilon_{ij} a^i b^j $$

is the singlet formed from SU(2) vectors $a^i$ and $b^i$, while the triplet is

$$ (AB)^{ij} = (a^i b^j + a^j b^i )$$

Which is symmetric. Beware that $AB^{12}$ is not normalized properly as compared to the usual textbook $|j,m\rangle$ presentation. One of the states of the 2-tensor (2-tensor of SU(2), the spin-1 object) is

$$ |1,0\rangle $$

when represented as an SU(2) tensor, this has components

$$ (AB)^{12} = (AB)^{21} = {1\over \sqrt 2 }$$

These numbers are determined by making sure that the tensor is normalized, and give you square-root of integer factors. The other states don't have these annoying factors, but these build up the Clebsch-Gordon coefficients.

The representation theory of SU(2), when expressed in tensors, makes the $J=j_1+j_2$ representation by multiplying the two tensors for $j_1$ and $j_2$ without any epsilons. This makes a completely symmetric thing, where you get rid of the antisymmetric parts.

As you step down J, you get an $\epsilon$ tensor each time you go down, changing the symmetric/antisymmetric character. This way of doing the representation theory is the simplest way, it allows you to carry the Clebsch-Gordon coefficients in your head. It is described in detail in this answer: Mathematically, what is color charge? .

This post imported from StackExchange Physics at 2014-04-01 05:48 (UCT), posted by SE-user Ron Maimon
answered Jun 2, 2012 by Ron Maimon (7,720 points) [ no revision ]
I got it when you said "two spin-1/2 electrons in an S-wave in the He atom ground-state..."

This post imported from StackExchange Physics at 2014-04-01 05:48 (UCT), posted by SE-user C.R.
Well, $|j_1m_1\rangle$ surely is a full description of the state assuming that by "state", we mean a vector in the Hilbert space up to a phase or normalization (the physical definition of a state). So they're full descriptions of the state up to a phase - a non-degenerate eigenstate of a complete set of commuting observables is unique up to a phase. But the phase is exactly what is allowed to change under permutations etc.

This post imported from StackExchange Physics at 2014-04-01 05:48 (UCT), posted by SE-user Luboš Motl
@LubošMotl: That's true, but the thing to explain is why the phase choice under interchange of the spin states must be taken +/- 1 according to the three representations involved. You could of course stuff in arbitrary extra phases in the states, but then the raising and lowering operators won't work properly (without phases) on either the small representations or the large ones. The consistent phase choice is best explained by counting $\epsilon$ tensors, and this is not how it is done in elementary QM books, but it's much easier.

This post imported from StackExchange Physics at 2014-04-01 05:48 (UCT), posted by SE-user Ron Maimon
+ 1 like - 0 dislike

The actual overall phase of a Clebsch-Gordan is a matter of convention. This is easy to see in the simplest example: the coupling of two spin-1/2 particles. The state with angular momentum L=0 and M=0 is antisymmetric and can be written (up to a normalization) as $$|\textstyle\frac{1}{2},\frac{1}{2}\rangle_1|\frac{1}{2},-\frac{1}{2}\rangle_2 - |\frac{1}{2},-\frac{1}{2}\rangle_1|\frac{1}{2},\frac{1}{2}\rangle_2$$ The relative phase is essential if this state is to be orthogonal to the L=1,M=0 state, but we could also have taken $$-|\textstyle\frac{1}{2},\frac{1}{2}\rangle_1|\frac{1}{2},-\frac{1}{2}\rangle_2 + |\frac{1}{2},-\frac{1}{2}\rangle_1|\frac{1}{2},\frac{1}{2}\rangle_2$$ as the $L=0, M=0$ state: this second choice is still annihilated by $L_+$ and still an eigenstate of $L_z$ with eigenvalue $0$, thus is an equally good choice of $L=0,M=0$ state.

In more general cases, whenever $L$ is not $L_1+L_2$, the Clebsch's must contain some negative signs to enforce orthogonality between states of different $L$'s but same values of $M$. The relative position of these signs is fully determined by orthogonality with other states, but the overall sign is a matter of convenience: what we call the first or the second "system" is purely a matter of convention.

There are several conventions used to choose the overall sign. In angular momentum theory, the most common is the Condon-Shortley (but it is not the only one). The encyclopedic text by Varshalovich et al gives a summary of the conventions used by several authors.

One of the features of the CS phase convention is precisely that it produces phase relations of the type given above when we interchange two labels. This choice is therefore convenient for tabulation purposes: basically one only needs to tabulate the coefficients when $L_1\ge L_2$ since the cases where $L_2>L_1$ are obtained by permutation and related to the original case by a phase change.

This post imported from StackExchange Physics at 2014-04-01 05:48 (UCT), posted by SE-user ZeroTheHero
answered Dec 22, 2013 by ZeroTheHero (70 points) [ no revision ]

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