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I am trying to combine the spin of two particles.

Their individual spins are:

$|1,0\rangle $ 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.

Young Kindaichi
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helpwithphysics
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2 Answers2

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$ \newcommand{\ket}[1]{{\textstyle\left|{#1}\right>}} \newcommand{\sqrtfrac}[2]{{\color{lightblue}{\textstyle\sqrt{\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.

Photon
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rob
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  • thanks a lot i realised where i misunderstood my notes, this really helped! – helpwithphysics Mar 13 '21 at 10:14
  • You’re welcome! If an answer is useful to you, you can indicate this to the system by clicking the “upvote” arrow; if it solves your problem, you can use the green checkmark to indicate you have “accepted” an answer. Posts with upvoted or accepted answers are more likely to show up in searches for future readers. – rob Mar 13 '21 at 11:20
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This can be done as follows: $$1\otimes \frac{1}{2}=\frac{3}{2}\oplus\frac{1}{2}$$ Using the notation $|j_1m_1,j_2m_2\rangle $ for product ket and $|jm\rangle$ for sum ket.

$$|j_1m_1,j_2m_2\rangle =\sum_{j,m}|jm\rangle \langle jm|j_1m_1,j_2m_2\rangle $$ $$\left|10,\frac{1}{2}\frac{1}{2}\right\rangle=\left\langle \frac{1}{2}\frac{1}{2}\left|10,\frac{1}{2}\frac{1}{2}\right.\right\rangle \left|\frac{1}{2}\frac{1}{2}\right\rangle+ \left\langle \frac{3}{2}\frac{1}{2}\left|10,\frac{1}{2}\frac{1}{2}\right.\right\rangle \left|\frac{3}{2}\frac{1}{2}\right\rangle$$ Putting the value of Clebsch-Gordan coefficient from here.

$$\left|10,\frac{1}{2}\frac{1}{2}\right\rangle=\sqrt{\frac{2}{3}}\left|\frac{3}{2}\frac{1}{2}\right\rangle-\sqrt{\frac{1}{3}}\left|\frac{1}{2}\frac{1}{2}\right\rangle$$

Young Kindaichi
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