Adding Angular Momentum is commutative, right?

  • #1
TheDestroyer
402
1
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.
 

Answers and Replies

  • #2
DrClaude
Mentor
7,878
4,487
$$
\begin{align}
\lvert 0,0\rangle &= \frac{1}{\sqrt{2}}\biggl(\lvert\bigl( \tfrac{1}{2}\tfrac{1}{2}\bigr) \tfrac{1}{2},- \tfrac{1}{2} \rangle - \lvert\bigl( \tfrac{1}{2} \tfrac{1}{2}\bigr),- \tfrac{1}{2} \tfrac{1}{2}\rangle\biggr) , \tag{4.21d}
\end{align}
$$
In the above equation, there is an arbitrary choice. It would have been equally valid to take
$$
\begin{align}
\lvert 0,0\rangle &= \frac{1}{\sqrt{2}}\biggl(-\lvert \bigl( \tfrac{1}{2}\tfrac{1}{2}\bigr) \tfrac{1}{2},- \tfrac{1}{2} \rangle + \lvert\bigl( \tfrac{1}{2} \tfrac{1}{2}\bigr),- \tfrac{1}{2} \tfrac{1}{2}\rangle\biggr) , \tag{4.21d}
\end{align}
$$
which is the same, up to that arbitrary phase factor. This is in essence what you implicitely do when you change the notation from
##\lvert\bigl(I S\bigr),m_I m_S\rangle## to ##\lvert\bigl(S I\bigr),m_S m_I\rangle##.

Therefore, the difference between the two representations of ##H## are not due to the choice of the order of the addition of the vectors, but to that arbitrary choice of phase. You get the correct result only if the Hamiltonian is consistent with that choice.
 
  • #3
TheDestroyer
402
1
Thank you for your reply. I still have a little problem understanding how those two cases are equivalent physically.

You're telling me that both cases are equivalent up to a phase factor, but I see there's different physics there, because one gets different energy splitting/displacement.

Could you please explain where that phase factor has to go (in a form of [itex]e^{i\phi}[/itex]) and how the energy levels are always the same independent of that phase?
 
Last edited:
  • #4
DrClaude
Mentor
7,878
4,487
You're telling me that both cases are equivalent up to a phase factor, but I see there's different physics there, because one gets different energy splitting/displacement.
No, you get the same result. The fact that terms that are different in both representations are off diagonal is very important here.

Could you please explain where that phase factor has to go (in a form of [itex]e^{i\phi}[/itex]) and how the energy levels are always the same independent of that phase?
(Note: use itex instead of tex to get inlined equations).

$$
\begin{align}
\lvert 0,0\rangle &= \frac{1}{\sqrt{2}}\biggl(- \lvert \bigl( \tfrac{1}{2}\tfrac{1}{2}\bigr) \tfrac{1}{2},- \tfrac{1}{2} \rangle + \lvert\bigl( \tfrac{1}{2} \tfrac{1}{2}\bigr),- \tfrac{1}{2} \tfrac{1}{2}\rangle\biggr) \\
&= \frac{e^{i \phi}}{\sqrt{2}}\biggl(\lvert \bigl( \tfrac{1}{2}\tfrac{1}{2}\bigr) \tfrac{1}{2},- \tfrac{1}{2} \rangle - \lvert\bigl( \tfrac{1}{2} \tfrac{1}{2}\bigr),- \tfrac{1}{2} \tfrac{1}{2}\rangle\biggr)
\end{align}
$$
with ##\phi = \pi##.

To see that you get the same energy in both cases, you need to diagonalize ##H##. In the case where
$$H = \begin{pmatrix}
\mu_B B & 0 & 0 & 0 \\
0 & - \mu_B B & 0 & 0 \\
0 & 0 & 0 &\mu_B B \\
0 & 0 & \mu_B B & 0
\end{pmatrix}$$
you get two eigenstates with eigenvalue ##\mu_B B##, ##| 1, 1\rangle## and ##(| 1, 0\rangle + | 0, 0\rangle)/\sqrt{2}##, and two with eigenvalue ##-\mu_B B##, ##| 1, -1\rangle## and ##(| 1, 0\rangle - | 0, 0\rangle)/\sqrt{2}##.

For the second choice, with
$$H = \begin{pmatrix}
\mu_B B & 0 & 0 & 0 \\
0 & - \mu_B B & 0 & 0 \\
0 & 0 & 0 &-\mu_B B \\
0 & 0 & -\mu_B B & 0
\end{pmatrix}$$
you get two eigenstates with eigenvalue ##\mu_B B##, ##| 1, 1\rangle## and ##(| 1, 0\rangle - | 0, 0\rangle)/\sqrt{2}##, and two with eigenvalue ##-\mu_B B##, ##| 1, -1\rangle## and ##| 1, 1\rangle##, ##| 1, -1\rangle## and ##(| 1, 0\rangle + | 0, 0\rangle)/\sqrt{2}##.

Do you see? The states of the same energy in both cases differ only by the sign of ##| 0, 0\rangle##, which exactly corresponds to the choice of phase in the definition of ##| 0, 0\rangle##.
 
  • #5
TheDestroyer
402
1
Ah, I see! So the Hamiltonian changes but the Eigenvalues don't change. Thank you so much.
 

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