Given that the Hamiltonian for Muonium spin in zero magnetic field is

$$\hat{H} = a \vec I \cdot \vec J$$

where $\vec I$ is the spin of a muon, and $\vec J$ is the spin of the electron, what is the commutation relation between $[H,I^2]$ ?

What I initially thought was that the electron spin could be written as
$$\vec J = (\sigma_x , \sigma_y , \sigma_z ) $$
where $\sigma$ are the Pauli matrices. However I'm not sure if that is right because I have no idea how to do the same for muon spin $\vec I$.


Since it appears you are actively working on the problem I will offer a series of hints. Please comment if you need further guidance.

Hint 1: Muonium is a bound state of $\mu^+ e^-$. Note that they are both spin 1/2 particles, and have their own Hilbert spaces $\mathcal{H}_\mu$ and $\mathcal{H}_e$. Your operator $\hat{H}$ lives in the Hilbert space $\mathcal{H}=\mathcal{H}_\mu \otimes \mathcal{H}_e$. (If this notation frightens you, don't worry I won't use it but I mention it if you're curious about the machinery).

First let us write

$$ H = a\sum_{i=k}^3 \hat{I}_k \hat{J}_k \equiv a\sum_{i=k}^3 \hat{S}_k^{(\mu)} \hat{S}^{(e)}_k $$

where I just just relabeled the spin operators in a notation that I think may be familiar to you. I haven't actually done anything yet.

Hint 2: An implication of the muon and the electron living in two different hilbert spaces is that they each separately obey their own commutation relations and are completely independent of one another. In the language of commutators this means that

$$[\hat{S}^{(e)}_i, \hat{S}^{(e)}_j] = i\hbar\epsilon_{ijk} \hat{S}^{(e)}_k $$

$$[\hat{S}^{(\mu)}_i, \hat{S}^{(\mu)}_j] = i\hbar\epsilon_{ijk} \hat{S}^{(\mu)}_k $$

$$[\hat{S}^{(e)}_i,\hat{S}^{(\mu)}_j] = 0$$

If you need more hints just shoot me a comment.

  • $\begingroup$ If I substituted the $\hat S$ with the corresponding Matrix that shows that $[H,\vec I] = 0$, that is what I am getting $\endgroup$ – Tman Jan 5 '19 at 8:47
  • $\begingroup$ can you type out your work $\endgroup$ – InertialObserver Jan 5 '19 at 8:49
  • $\begingroup$ $[H,\vec I] = \sum_{i=1}^3 \hat S_i^{(\mu)} \hat S_i^{(e)} \hat S_i^{(\mu)} - \sum_{i=1}^3 \hat S_i^{(\mu)} \hat S_i^{(\mu)} \hat S_i^{(e)}$ so expand the sum and sub in $$S_1 = \begin{pmatrix} 0&1 \\ 1&0 \end{pmatrix}}$$, $$S_2 = \begin{pmatrix} 0&i \\ -i&0 \end{pmatrix}}$$ and $$S_3 = \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix}}$$ and that should give zero $\endgroup$ – Tman Jan 5 '19 at 9:02
  • $\begingroup$ I meant in your question so everybody can track your progress $\endgroup$ – InertialObserver Jan 5 '19 at 9:32
  • $\begingroup$ Is the below answer correct? $\endgroup$ – Tman Jan 6 '19 at 8:10

$$[H,I^2] = \sum_{i=1}^3 \hat S_i^{(\mu)} \hat S_i^{(e)} \hat S_i^{2(\mu)} - \sum_{i=1}^3 \hat S_i^{2(\mu)} \hat S_i^{(\mu)} \hat S_i^{(e)}$$
Expand the sum, and sum in the matrices $$S_1 = \frac{\hbar} {2} \begin{pmatrix} 0&1 \\ 1&0 \end{pmatrix}$$ , $$S_2 = \frac{\hbar} {2} \begin{pmatrix} 0&i \\ -i&0 \end{pmatrix}$$ and $$S_3 = \frac{\hbar} {2} \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix}$$

Let's look when $i=1$, the equation will be $$\hat S_1^{(\mu)} \hat S_1^{(e)} \hat S_1^{2(\mu)} - \hat S_1^{2(\mu)} \hat S_1^{(\mu)} \hat S_1^{(e)}$$ In the first term sub in the matrix and for simplicity lets make $\hbar = 1$ $$\frac{1} {2} \begin{pmatrix} 0&1 \\ 1&0 \end{pmatrix} . \frac{1} {2} \begin{pmatrix} 0&1 \\ 1&0 \end{pmatrix} . \Biggl(\frac{1} {2} \begin{pmatrix} 0&1 \\ 1&0 \end{pmatrix}\Biggr)^2$$ Because these matrices are Hermitian $\hat S_1^{2(\mu)}$ will just become the identity matrix and so is $\hat S_1^{(\mu)} \hat S_1^{(e)}$ so the end result for the first term will be $\frac{1}{16}$, doing the same for the second term we will get $\frac{1}{16}$, then subtracting both sides we get zero. Doing the same for $i=2,3$ we will get zero as well, so this Hamiltonian commute with the muon spin.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.