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Combining spins of two particles of spin |1,0> and |1/2, 1/2>

Physics Asked by helpwithphysics on July 15, 2021

I am trying to combine the spin of two particles.

Their individual spins are:

$|1,0rangle $ and $left|frac{1}{2},frac{1}{2} right>$

Now I am told that they combine to give a total spin state of:

$$left|frac{1}{2},frac{1}{2} right>$$

However, I am confused as to how this works. Thinking physically I am confused as:

an image showing how the first state has a magnitude of one, just not in the measurement direction (z), and the second has a magnitude of 1/2 but in the z-direction. Hence combining the two gives a measured value of 1/2 in the z-direction, however, I am confused as to how the total vector has a magnitude of 1/2? combining the perpendicular vectors would suggest a magnitude of sqrt(5)/2.

I understand that this doesn’t really make sense as spin is quantized, however, I am confused as to how the states combine if this is not the case.

2 Answers

$ newcommand{ket}[1]{{textstyleleft|{#1}right>}} newcommand{sqrtfrac}[2]{{color{lightblue}{textstylesqrt{frac{#1}{#2}}}}} %%% $There is no way to combine $ket{1,0}$ and $ket{frac12, frac12}$ to get a pure $j=frac12$ state. The only possible value for the $z$-axis projection $m$ is $0 + frac12 = frac12$, but the total angular momentum can take either of $frac12$ or $frac32$.

The orthonormal combinations are given by the Clebsch-Gordan coefficients, which are usually presented in horrible tables like

wikipedia screenshot

The way to read this horrible table is that, if you wanted to construct the composite state like $ket{frac32,frac12}$ or $ket{frac12,frac12}$, you would read down the columns of the second table:

begin{align} ket{frac32,frac12} &= sqrtfrac13 ket{1,1}ket{frac12,{-frac12}} + sqrtfrac23 ket{1,0}ket{frac12,{+frac12}} ket{frac12,frac12} &= sqrtfrac23 ket{1,1}ket{frac12,{-frac12}} - sqrtfrac13 ket{1,0}ket{frac12,{+frac12}} end{align}

If your constituent particles are in a pure state $ket{1,0}ket{frac12,frac12}$, your composite system is in a superposition of $ket{frac12,frac12}$ and $ket{frac32,frac12}$. You should convince yourself that you can find its coefficients either by solving the system of equations above for $ket{1,0}ket{frac12, frac12}$, or by reading across the Clebsch-Gordan table.

Correct answer by rob on July 15, 2021

This can be done as follows: $$1otimes frac{1}{2}=frac{3}{2}oplusfrac{1}{2}$$ Using the notation $|j_1m_1,j_2m_2rangle $ for product ket and $|jmrangle$ for sum ket.

$$|j_1m_1,j_2m_2rangle =sum_{j,m}|jmrangle langle jm|j_1m_1,j_2m_2rangle $$ $$left|10,frac{1}{2}frac{1}{2}rightrangle=leftlangle frac{1}{2}frac{1}{2}left|10,frac{1}{2}frac{1}{2}right.rightrangle left|frac{1}{2}frac{1}{2}rightrangle+ leftlangle frac{3}{2}frac{1}{2}left|10,frac{1}{2}frac{1}{2}right.rightrangle left|frac{3}{2}frac{1}{2}rightrangle$$ Putting the value of Clebsch-Gordan coefficient from here.

$$left|10,frac{1}{2}frac{1}{2}rightrangle=sqrt{frac{2}{3}}left|frac{3}{2}frac{1}{2}rightrangle-sqrt{frac{1}{3}}left|frac{1}{2}frac{1}{2}rightrangle$$

Answered by Young Kindaichi on July 15, 2021

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