- #1

- 402

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## Main Question or Discussion Point

I have angular momenta [tex]S=\frac{1}{2}[/tex] for spin, and [tex]I=\frac{1}{2}[/tex]

for nuclear angular momentum, which I want to add using the Clebsch-Gordon basis, so the conversion looks like:

$$

\begin{align}

\lvert 1,1\rangle &= \lvert\bigl(\tfrac{1}{2}\tfrac{1}{2}\bigr)\tfrac{1}{2}\tfrac{1}{2} \rangle ,\tag{4.21a} \\

\lvert 1,0\rangle &= \frac{1}{\sqrt{2}}\biggl(\lvert\bigl( \tfrac{1}{2} \tfrac{1}{2} \bigr) \tfrac{1}{2},- \tfrac{1}{2}\rangle + \lvert\bigl( \tfrac{1}{2} \tfrac{1}{2}\bigr) ,-\tfrac{1}{2}\tfrac{1}{2}\rangle\biggr),\tag{4.21b} \\

\lvert 1,-1\rangle &= \lvert\bigl( \tfrac{1}{2} \tfrac{1}{2}\bigr) , -\tfrac{1}{2} , -\tfrac{1}{2}\rangle , \tag{4.21c} \\

\lvert 0,0\rangle &= \frac{1}{\sqrt{2}}\biggl(\lvert\bigl( \tfrac{1}{2}\tfrac{1}{2}\bigr) \tfrac{1}{2},- \tfrac{1}{2} \rangle - \lvert\bigl( \tfrac{1}{2} \tfrac{1}{2}\bigr),- \tfrac{1}{2} \tfrac{1}{2}\rangle\biggr) , \tag{4.21d}

\end{align}

$$

where [tex]F=I+S[/tex], so this is the basis [tex]\lvert F m_F \rangle = \sum_m \lvert\bigl(I S\bigr),m_I m_S\rangle [/tex].

Now since adding angular momenta is commutative, the exchange between [tex]I[/tex] and [tex]S[/tex] shouldn't mathematically introduce any kind of difference.

In other words, in the basis described in those equations, it shouldn't matter whether I write it as [tex]\lvert\bigl(I S\bigr),m_I m_S\rangle[/tex] or [tex]\lvert\bigl(S I\bigr),m_S m_I\rangle[/tex], right?

Now the problem is the following: I have created the Hamiltonian matrix [tex]H=-\vec{\mu}\cdot \vec{B} = -2 \mu B_z S_z/\hbar[/tex] in the [tex]\lvert F m_F \rangle[/tex] representation, and actually the result depends on how you call those angular momenta, so the result could be

$$H = \begin{pmatrix}

\mu_B B & 0 & 0 & 0 \\

0 & - \mu_B B & 0 & 0 \\

0 & 0 & 0 &\mu_B B \\

0 & 0 & \mu_B B & 0

\end{pmatrix}$$

Or could be

$$H = \begin{pmatrix}

\mu_B B & 0 & 0 & 0 \\

0 & - \mu_B B & 0 & 0 \\

0 & 0 & 0 &-\mu_B B \\

0 & 0 & -\mu_B B & 0

\end{pmatrix}$$

Depending on how you "label" them, [tex]I[/tex] or [tex]S[/tex]... which is very confusing!

This happens because the off-diagonal terms

$$\left\langle 1 0 \right| S_z \left| 0 0 \right\rangle = \frac{1}{2} \left( \left\langle (\frac{1}{2} \frac{1}{2}) \frac{1}{2} -\frac{1}{2} \right| + \left\langle (\frac{1}{2} \frac{1}{2}) -\frac{1}{2} \frac{1}{2} \right| \right) S_z \left( \left| (\frac{1}{2} \frac{1}{2}) \frac{1}{2} -\frac{1}{2} \right\rangle - \left| (\frac{1}{2} \frac{1}{2}) -\frac{1}{2} \frac{1}{2} \right\rangle \right)$$

will be either [tex]\hbar/2[/tex] or [tex]-\hbar/2[/tex] depending on your convention whether it's [tex]\lvert\bigl(I S\bigr),m_I m_S\rangle[/tex] or [tex]\lvert\bigl(S I\bigr),m_S m_I\rangle[/tex].

How can I understand this physically and mathematically? Shouldn't the addition be commutative and the process be blind to which labels I use?

Why is this important? Because in a computer program, when you add angular momenta, you don't look for their order, but only to their value! I wrote a Mathematica script to do this addition for me, and I got the second Hamiltonian which is different than the Hamiltonian in the book, which is the first one.

for nuclear angular momentum, which I want to add using the Clebsch-Gordon basis, so the conversion looks like:

$$

\begin{align}

\lvert 1,1\rangle &= \lvert\bigl(\tfrac{1}{2}\tfrac{1}{2}\bigr)\tfrac{1}{2}\tfrac{1}{2} \rangle ,\tag{4.21a} \\

\lvert 1,0\rangle &= \frac{1}{\sqrt{2}}\biggl(\lvert\bigl( \tfrac{1}{2} \tfrac{1}{2} \bigr) \tfrac{1}{2},- \tfrac{1}{2}\rangle + \lvert\bigl( \tfrac{1}{2} \tfrac{1}{2}\bigr) ,-\tfrac{1}{2}\tfrac{1}{2}\rangle\biggr),\tag{4.21b} \\

\lvert 1,-1\rangle &= \lvert\bigl( \tfrac{1}{2} \tfrac{1}{2}\bigr) , -\tfrac{1}{2} , -\tfrac{1}{2}\rangle , \tag{4.21c} \\

\lvert 0,0\rangle &= \frac{1}{\sqrt{2}}\biggl(\lvert\bigl( \tfrac{1}{2}\tfrac{1}{2}\bigr) \tfrac{1}{2},- \tfrac{1}{2} \rangle - \lvert\bigl( \tfrac{1}{2} \tfrac{1}{2}\bigr),- \tfrac{1}{2} \tfrac{1}{2}\rangle\biggr) , \tag{4.21d}

\end{align}

$$

where [tex]F=I+S[/tex], so this is the basis [tex]\lvert F m_F \rangle = \sum_m \lvert\bigl(I S\bigr),m_I m_S\rangle [/tex].

Now since adding angular momenta is commutative, the exchange between [tex]I[/tex] and [tex]S[/tex] shouldn't mathematically introduce any kind of difference.

In other words, in the basis described in those equations, it shouldn't matter whether I write it as [tex]\lvert\bigl(I S\bigr),m_I m_S\rangle[/tex] or [tex]\lvert\bigl(S I\bigr),m_S m_I\rangle[/tex], right?

Now the problem is the following: I have created the Hamiltonian matrix [tex]H=-\vec{\mu}\cdot \vec{B} = -2 \mu B_z S_z/\hbar[/tex] in the [tex]\lvert F m_F \rangle[/tex] representation, and actually the result depends on how you call those angular momenta, so the result could be

$$H = \begin{pmatrix}

\mu_B B & 0 & 0 & 0 \\

0 & - \mu_B B & 0 & 0 \\

0 & 0 & 0 &\mu_B B \\

0 & 0 & \mu_B B & 0

\end{pmatrix}$$

Or could be

$$H = \begin{pmatrix}

\mu_B B & 0 & 0 & 0 \\

0 & - \mu_B B & 0 & 0 \\

0 & 0 & 0 &-\mu_B B \\

0 & 0 & -\mu_B B & 0

\end{pmatrix}$$

Depending on how you "label" them, [tex]I[/tex] or [tex]S[/tex]... which is very confusing!

This happens because the off-diagonal terms

$$\left\langle 1 0 \right| S_z \left| 0 0 \right\rangle = \frac{1}{2} \left( \left\langle (\frac{1}{2} \frac{1}{2}) \frac{1}{2} -\frac{1}{2} \right| + \left\langle (\frac{1}{2} \frac{1}{2}) -\frac{1}{2} \frac{1}{2} \right| \right) S_z \left( \left| (\frac{1}{2} \frac{1}{2}) \frac{1}{2} -\frac{1}{2} \right\rangle - \left| (\frac{1}{2} \frac{1}{2}) -\frac{1}{2} \frac{1}{2} \right\rangle \right)$$

will be either [tex]\hbar/2[/tex] or [tex]-\hbar/2[/tex] depending on your convention whether it's [tex]\lvert\bigl(I S\bigr),m_I m_S\rangle[/tex] or [tex]\lvert\bigl(S I\bigr),m_S m_I\rangle[/tex].

How can I understand this physically and mathematically? Shouldn't the addition be commutative and the process be blind to which labels I use?

Why is this important? Because in a computer program, when you add angular momenta, you don't look for their order, but only to their value! I wrote a Mathematica script to do this addition for me, and I got the second Hamiltonian which is different than the Hamiltonian in the book, which is the first one.