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I've never taken a quantum mechanics class, but I find myself now using principles developed in the quantum theory of angular momentum. One particularly confusing aspect that I'm struggling with is the notation. Could someone please explain the meaning of a ket (or bra) with two arguments, i.e.: $\left|j_1m_1\right>$? I'm mainly interested in the mathematical meaning (i.e. is it a vector with two components?, are $j_1$ and $m_1$ multiplied), but a physical explanation would be helpful too.

I have seen other similar notation as well which perplexes me. For instance, in the definition of the Clebsch-Gordan coefficients: $C_{j_1m_1j_2m_2}^{jm}=\left<j_1m_1j_2m_2\mid j_1j_2jm\right>=\left<j_1j_2jm\mid j_1m_1j_2m_2\right>$

what does a bra (or ket) with four arguments written adjacent to each other mean?

Similarly I've seen $\left|S_x;+\right>$ which, as far as I can tell indicates that the spin component in the x-direction is $+\frac{1}{2}$, but could this be equivalently written simply as $\left|+\right>$?

Please any help would be very illuminating, every text and resource seems to use similar but slightly different notation and they all seem to assume you already know the conventions.

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It is just a label. Each element of the set $\left|j_1m_1\right>$ is a unique state vector. For example, $\left|1,1\right>$ and $\left|1,-1\right>$ are two independent angular momentum state vectors, just like any other state vector. Whenever you see the indices $j$, $m$, etc., see them as referring to a set of state vectors each labelled with particular values and put some numbers into the $j$s and $m$s in an equation to get an idea of what one of the equations belonging to that set is like.

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    $\begingroup$ Dear @okj, unless stated otherwise, an extra comma or semicolon separating the labels should make no difference. ;-) Otherwise, $|S_x,+\rangle$ surely can't be simplified to $|+\rangle$ because the normal component that the spin measurements refer to is $S_z$, not $S_x$, so the $S_x$ refinement is quite important to make it clear what state one is talking about. The $|S_z,+\rangle$ state is an eigenstate of $S_x$ which is a linear combination of both eigenstates of $S_z$. $\endgroup$
    – Luboš Motl
    May 24, 2011 at 16:42
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    $\begingroup$ To this should be added that the list of indices or values in the brackets are the eigenvalues of some list of commuting operators. The list should preferably be maximal, so that there is no other operator available that commutes with all the operators already in the list. A different list of commuting operators will result in a different basis for the Hilbert space. The $\left|S_x;+\right>$ formalism tells you what the operator is that has the eigenvalue $+$, only equivalently to $\left|+\right>$ if the $S_x$ can be inferred from context. $\endgroup$
    – Peter Morgan
    May 24, 2011 at 16:44
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    $\begingroup$ Otherwise, $|jm\rangle$ is a state in the $(2j+1)$-dimensional representation of $SU(2)=Spin(3)$ whose eigenvalue of $J_z$ is equal to $m$. So the states $|jm\rangle$ for $m=-j,-j+1,\dots j$ form an irreducible representation of the $SU(2)$ group. For different values of $j$, we get different representations. $|j_1 m_1 j_2 m_2\rangle$ is just a basis vector in the tensor product linear space of two angular momenta, it's just $|j_1m_1\rangle\otimes |j_2m_2\rangle$. $\endgroup$
    – Luboš Motl
    May 24, 2011 at 16:47
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    $\begingroup$ $\langle x|y\rangle$ is what mathematicians would write as $b(y,x)$ - at any rate, the $x$ is being complex conjugated when you write it in terms of components. The vector $|j_1j_2 jm\rangle$ is a basis vector in the same tensor product space $H_1\otimes H_2$ which has eigenvalue of $(\vec J_1+\vec J_2)^2$ equal to $j(j+1)$ and the eigenvalue of $J_{1z}+J_{2z}$ equal to $m$. The Clebsch-Gordan coefficients are translating the two bases into one another. $\endgroup$
    – Luboš Motl
    May 24, 2011 at 16:49
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    $\begingroup$ @okj: actually it's two different bases of the same Hilbert space $H_1\otimes H_2$. One basis is formed by taking tensor products of the individual basis elements of $H_1$ and $H_2$, i.e. if we denote a basis ket of $H_1$ by $|j_1m_1\rangle$ and a basis ket of $H_2$ by $|j_2m_2\rangle$, then all kets of the form $|j_1j_2m_1m_2\rangle = |j_1m_1\rangle\otimes|j_2m_2\rangle$ form a basis of $H_1\otimes H_2$. (to be continued) $\endgroup$
    – David Z
    May 24, 2011 at 17:35

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