Undergrad C-G coefficients and Angular momentum

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The discussion focuses on deriving the C-G coefficients in angular momentum coupling, specifically how to express the combined states |j, m⟩ from individual angular momentum states |l, m_l⟩ and |s, m_s⟩. It emphasizes that the total z-component of angular momentum must be conserved, leading to the equation m_l + m_s = m_j. The participants clarify that the values of j in the linear combination must satisfy the triangle inequality, allowing for multiple j values between j_min and j_max. The process involves using lowering operators to derive coefficients for various combinations of |j, m⟩ states, ensuring orthogonality and normalization to find all possible states. This methodical approach allows for a comprehensive understanding of angular momentum coupling in quantum mechanics.
QuarkDecay
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Can someone explain to me how we find it?
Examples

Y10X-=
= |1,0>|1/2,-1/2> = √2/3 |3/2,-1/2> + √1/3|1/2,-1/2>

Y11X-=
= |1,1>|1/2,-1/2> = √1/3|3/2,1/2> + √2/3|1/2,1/2>

Y2-1X- =
= √4/5 | 5/2, -3/2> + √1/5 | 3/2, -3/2>

I understand it goes like YlmX± = |l,m>|s,ms> = a |jmax, ?> + b |jmin, ?> and a,b is from the (C-G) coefficients, but I can't find how the ? comes up. I thought it was m-ms or ms-m but it doesn't match all of the examples.
 
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When you add angular momenta, say ##l## and ##s## so that ##\vec l+\vec s = \vec j##, you have terms like ##| l, m_l\rangle |s, m_s\rangle## that you want to write as a linear combination of the ## |j, m_j\rangle## states. First note that the terms that appear in the linear combination may have all the values of ##j## that satisfy the triangle inequality, i.e. ##j_{min} \leq j \leq j_{max}##, not just the two extremal values. Also, the component of the total angular momentum along the quantization axis must be conserved, i.e. ##m_l+m_s=m_j##. In all the examples you have posted the ? is that sum.
 
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QuarkDecay said:
Can someone explain to me how we find it?

Let ##|j_1, m_1\rangle |j_2, m_2\rangle## be the state in which the first system has angular momentum ##j_1## and z-component ##m_1##, and the second system has angular momentum ##j_2## and z-component ##m_2##. Let ##|j, m\rangle## in which the total angular momentum is ##j## and the total z-component is ##m##.

Since the z-component of angular momentum, ##m##, is additive, we know that

##|j_1, m_1\rangle |j_2, m_2\rangle = ## some combination of terms of the form ##|j, m\rangle## where
##|j_1 - j_2| \leq j \leq j_1 + j_2## and
##-j \leq m \leq +j##

To start with, it's clear that if ##m_1 = j_1## and ##m_2 = j_2##, then there is only one possibility:

1. ##|j_1, j_1\rangle |j_2, j_2\rangle = |(j_1 + j_2), (j_1 + j_2)\rangle## (The case with ##j = j_1 + j_2## and ##m = m_1 + m_2##)

Now, we use the lowering operators:

2. ##J^- = J_1^{-} + J_2^{-}##

You use the fact that
3. ##J^{-} |j, m\rangle = \sqrt{j (j+1) - m (m-1)}|j, m-1\rangle##
4. ##J_1^{-} |j_1, m_1\rangle |j_2, m_2\rangle = \sqrt{j_1 (j_1+1) - m_1 (m_1-1)}|j_1, m_1 - 1\rangle |j_2, m_2\rangle##
5. ##J_2^{-} |j_1, m_1\rangle |j_2, m_2\rangle = \sqrt{j_2 (j_2+1) - m_2 (m_2-1)}|j_1, m_1\rangle |j_2, m_2 - 1\rangle##

(##J_1^{-}## doesn't affect ##|j_2, m_2\rangle## and ##J_2^{-}## doesn't affect ##|j_1, m_1\rangle##)

So applying ##J_1^{-} + J_2^{-}## to the left side of equation 1, and applying ##J^{-}## to the right side gives us:

6. ##\sqrt{j_1 (j_1+1) - m_1 (m_1-1)} |j_1, j_1 - 1\rangle |j_2, j_2\rangle ##
##+ \sqrt{j_2 (j_2+1) - m_2 (m_2-1)} |j_1, j_1\rangle |j_2, j_2 - 1\rangle##
##= \sqrt{(j_1 + j _1)(j_1+j_2 +1) - (j_1 + j_2) (j_1 + j_2 -1)} |(j_1 + j_2), (j_1 + j_2 - 1)\rangle##

You can continue using ##J^{-}## to get all the coefficients for ##|j, m\rangle## with ##j = j_1 + j_2##.

Now, let's look at the case where ##j = j_1 + j_2 - 1##. How do you figure out that case? Well, let's look at the case ##j = j_1 + j_2 - 1## and ##m = j_1 + j_2 - 1##. We know that that has to be some combination of ##m_1 = j_1, m_2 = j_2 -1## and ##m_1 = j_1, m_2 = j_2##. Those are the only two ways to get ##m = m_1 + m_2##. So we know that there must be coefficients ##\alpha## and ##\beta## such that:

##\alpha |j_1, j_1 - 1\rangle |j_2, j_1\rangle + \beta |j_1, j_1\rangle |j_2, m_2 - 1\rangle = |(j_1 + j_2 - 1), (j_1 + j_2 - 1)\rangle##

So we have to figure out what ##\alpha## and ##\beta## must be. That's two unknowns. But we also have two constraints: (1) This state must be orthogonal to the state ##|(j_1 + j_2), (j_1 + j_2 - 1)\rangle##, and (2) ##|\alpha|^2 + |\beta|^2 = 1## (it has to be normalized). Those two constraints uniquely determine ##\alpha## and ##\beta## up to an unknown phase factor. (I'm not sure if there is some standard convention for picking the phase).

Once you know ##|j, m\rangle## for the case ##j = j_1 + j_2 - 1##, ##m = j_1 + j_2 - 1##, you can again use the lowering operators to find the states for all other values of ##m##.

Then you can again use orthogonality to find the state ##|j, m\rangle## with ##j = j_1 + j_2 - 2## and ##m = j_1 + j_2 - 2##.

Continuing in this way, you can find all the possibilities for ##|j, m\rangle##.
 
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Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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