• View
• Edit
• History
• Print

FMuF

< A special case: a muon with few nuclei | Index | FMu >

---

The dipolar Hamiltonian in zero external field

Let us consider a linear F-Mu-F molecule of total length {$2r$}. The two fluorine nuclei and the muon all have spin 1/2. Since {$\gamma_F\approx\gamma_\mu/3$} and the muon-fluorine distance {$r$} is half the fluorine-fluorine distance, the dipolar interaction between fluorines is {$\omega_d^{FF}\approx\frac 1 {24}\omega_d$} that between the muon and each fluorine. Considering the classical dipole-dipole interaction among two spin {$ \mathbf I$} and {$\mathbf F$} -{$ \omega_d \left(3 \mathbf I \cdot \hat r \mathbf F \cdot \hat r - \mathbf I \cdot \mathbf F \right) , $}

exploiting the rotational invariance to choose {$\hat r =\hat z$} and ignoring the F-F interaction, the zero field Hamiltonian can be written

{$ \frac {\cal H} \hbar = -\omega_d (2 I_zF_z-I_xF_x-I_yF_y+ 2 I_zL_z-I_xL_x-I_yL_y) $}

where {$ \omega_d=\mu_0\gamma_F\gamma_\mu h/(2 r^3) $}, {$I_\alpha, \alpha=x,y,z$} are the muon spin operators and {$F_\alpha,L_\alpha$} those of the fluorine and the gyromagnetic ratios are {$\gamma=\nu/B$}.

In the tensor space of {$F\otimes I\otimes L$}, with the basis of {$F_z,I_z,L_z$} one has

{$\frac {\cal H} \hbar=\omega_d \left( \begin{array} \quad -1\quad & \quad 0\quad & \quad 0\quad & \quad 0\quad & \quad 0\quad & \quad 0\quad & \quad 0\quad & \quad 0\quad \\ 0& 0& \frac 1 2& 0& 0& 0& 0& 0\\ 0& \frac 1 2& 1& 0& \frac 1 2& 0& 0& 0\\ 0& 0& 0& 0& 0& \frac 1 2& 0& 0\\ 0& 0& \frac 1 2& 0& 0& 0& 0& 0\\ 0& 0& 0& \frac 1 2& 0& 1& \frac 1 2& 0\\ 0& 0& 0& 0& 0& \frac 1 2& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& -1\end{array} \right)$}

(the order in the tensor product is outer to inner, from left to right, i.e. {$F_z$} is an 8x8 matrix with all 1/2 along the top half of the diagonal and -/12 along the bottom half, whereas {$L_z$} alternates +1/2 and -1/2 along the diagonal).

Top

---

The eigenvalues

The Hamiltonian has doubly degenerate eigenvectors and one is clearly {$-\omega_d$} with eigenvectors

{$|1\rangle = \left( \begin{array} 1\\0\\0\\0\\0\\0\\0\\0 \end{array}\right); \qquad\qquad |8\rangle = \left( \begin{array} 0\\0\\0\\0\\0\\0\\0\\1\end{array}\right) $}.

The others may be found by considering the block of the 2nd, 3rd and 5th rows and columms of {$\cal H$}

{$ \left( \begin{array} 0& \frac 1 2& 0\\ \quad\frac 1 2\quad& \quad 1\quad & \quad \frac 1 2\quad \\ 0& \frac 1 2& 0 \end{array}\right) $}

which has eigenvalues {$\omega_2=0$} and {$\omega_{3,4}=\omega_d\frac {1\pm \sqrt 3} 2$}. The identical bloc of the 4th, 6th and 7th rows and columns yields the same solutions. Summarizing the eight eigenvalues of the Hamiltonian are: {$ \omega_d (-1, 0,\frac {1 + \sqrt 3} 2,\frac {1 - \sqrt 3} 2, 0, \frac {1 + \sqrt 3} 2,\frac {1 - \sqrt 3} 2,-1) $}

Top

---

The eigenvectors

To find the eigenvector of the block one must solve the three simultaneous equations

{$ \left\{ \begin{array} \frac b 2 = \omega_j a \\ \frac a 2 + b + \frac c 2 = \omega_j b \\ \frac b 2 = \omega_j c \end{array} \right. $}

For {$\omega_2=0$} one has {$ a=-c$} and {$b=0$}. Considering the full Hamiltonian there are two such states (as well as the linear combinations of these two), e.g.:

{$|2\rangle = \frac 1 {\sqrt 2} \,\left( \begin{array} \quad 0\\ \quad 1\\ \quad 0\\0\\ -1\\ \quad 0\\ \quad 0\\ \quad 0 \end{array}\right); \qquad\qquad |5\rangle = \frac 1 {\sqrt 2} \,\left( \begin{array} \quad 0\\ \quad 0\\ \quad 0\\-1\\ \quad 0\\ \quad 0\\ \quad 1\\ \quad 0\end{array}\right) $}.

The other two cases yield {$ b= 2\omega_j a$} and {$c=a$}. Considering again the full Hamiltonian this yields the remaining four states which can be written as:

{$ a \left( \begin{array} \quad 0\\ \quad 1\\ 2\omega_j/\omega_d \\ 0\\ \quad 1\\ \quad 0\\ \quad 0\\ \quad 0 \end{array}\right); \qquad\qquad a \left( \begin{array} \quad 0\\ \quad 0\\ \quad 0 \\ \quad 1\\ \quad 0\\ 2 \omega_j/\omega_d \\ \quad 1\\ \quad 0 \end{array}\right).$}.

Imposing the normalization one obtains {$ a=\frac 1 2 \sqrt{1\mp \frac 1 {\sqrt3}} $}, respectively, for the two eigenvalues {$\omega_j=\frac{\sqrt 3 \pm 1} 2$}, that is

{$|3\rangle = \frac 1 2 \sqrt{1-\frac 1{\sqrt 3}} \left( \begin{array} \quad 0\\ \quad 1 \\ \sqrt 3 +1 \\ \quad 1 \\ \quad 0 \\ \quad 0\\ \quad 0 \end{array}\right); \qquad\qquad |4\rangle = \frac 1 2 \sqrt{1+\frac 1{\sqrt 3}}\left( \begin{array} \quad 0\\ \quad 1\\ \sqrt 3 -1 \\ \quad 0\\ \quad 1\\ \quad 0 \\ \quad 0\\ \quad 0 \end{array}\right)$}

and {$ |6\rangle = \frac 1 2 \sqrt{1-\frac 1{\sqrt 3}} \left( \begin{array} \quad 0\\ \quad 0\\ \quad 0 \\ \quad 1\\ \quad 0\\ \sqrt 3 +1 \\ \quad 1\\ \quad 0 \end{array}\right); \qquad\qquad |7\rangle = \frac 1 2 \sqrt{1+\frac 1{\sqrt 3}} \left( \begin{array} \quad 0\\ \quad 0\\ \quad 0 \\ \quad 1\\ \quad 0\\ \sqrt 3 -1 \\ \quad 1\\ \quad 0 \end{array}\right) $}.

Top

---

The muon polarization

Now {$\left\lbrace |1\rangle,\cdots,|8\rangle \right\rbrace $} form an orthonormal basis of eigenvectors of {$\cal H$}. The eigenvalues do not depend on the orientation of the molecular axis, i.e. the Hamiltonian is invariant under rotations, but it does not commute with the muon polarization observable, the muon spin projection along the detector axis (which we assume to coincide with the initial muon spin direction). Let us call {$\zeta$} this direction. The observable is then {$\sigma_\zeta={\cal 1}_F \otimes\sigma_{\zeta,\mu}\otimes{\cal 1}_L$} and the muon polarization along {$\zeta$} is: {$ P_{\zeta\zeta}(t)= \frac 1 8 Tr \left( e^{i{\cal H}t/\hbar} \sigma_\zeta e^{-i{\cal H}t/\hbar} \sigma_\zeta\right) = \frac 1 8 \sum_{m,n} e^{i(\omega_m-\omega_n)t}\, |\langle m|\sigma_\zeta|n\rangle|^2$} The expression is easily evaluated in the same orthonormal basis, considering that {$\sigma_\zeta=\sigma_x\cos\theta + \sigma_x\sin\theta$}, according to the rotation of Pauli matrices

The direction {$\zeta$} is parallel to both the initial muon spin and the detector axis

This is the correct expression for a crystalline array of equally aligned FMuF molecules. For a polycrystalline sample we must take the average over {$\theta$}

{$\overline{|\langle m|\sigma_\zeta|n\rangle|^2} = \frac 1 3 \left(|\langle m|\sigma_z|n\rangle|^2 + 2 |\langle m|\sigma_x|n\rangle|^2\right) $}

Top

---

The matrix elements

We summarize below the three relevant matrices: {$v$}, with eigenstates along columns, and the two observables

{$ v=\left( \begin{array}1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & c & a & a' & 0 & 0 & 0 & 0 \\ 0 & 0 & b & b' & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & a & a' & c & 0\\ 0 & -c & a & a' & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & b & b' & 0 & 0 \\ 0 & 0 & 0 & 0 & a & a' & -c & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\end{array}\right),\qquad \sigma_z=\left( \begin{array}1&0&0&0&0&0&0&0\\0&1&0&0&0&0&0&0\\0&0&-1&0&0&0&0&0\\0&0&0&-1&0&0&0&0\\0&0&0&0&1&0&0&0\\0&0&0&0&0&1&0&0\\ 0&0&0&0&0&0&-1&0\\0&0&0&0&0&0&0&-1 \end{array}\right),\qquad \sigma_x=\left( \begin{array} 0&0&1&0&0&0&0&0\\0&0&0&1&0&0&0&0\\1&0&0&0&0&0&0&0\\0&1&0&0&0&0&0&0\\0&0&0&0&0&0&1&0\\0&0&0&0&0&0&0&1\\ 0&0&0&0&1&0&0&0\\0&0&0&0&0&1&0&0 \end{array}\right) $}

where

{$ c=\frac 1 {\sqrt 2}, \qquad a,a'=\frac 1 2 \sqrt{1\mp \frac 1 {\sqrt 3} }, \qquad b,b'=\pm \sqrt{ \frac 1 2 \left(1\pm \frac 1 {\sqrt 3}\right) } $}

Hence we have the following non vanishing matrix elements for {$\sigma_z$}:

{$ \langle m |\sigma_z|k\rangle = \delta_{mk}, \qquad\qquad \mbox{for} \qquad m=1,2,5,8 $}

{$ \langle 3 |\sigma_z|3\rangle = \langle 6 |\sigma_z|6\rangle = 2 a^2-b^2= -\frac 1 {\sqrt3} $}

{$ \langle 4 |\sigma_z|4\rangle = \langle 7 |\sigma_z|7\rangle = 2 a'^2-b'^2= \frac 1 {\sqrt3} $}

{$ \langle 3 |\sigma_z|4\rangle = \langle 6 |\sigma_z|7\rangle = 2 aa'-bb'= \sqrt{\frac 2 3} $}

All but the last give rise to constant polarization in time. Since

{$\omega_{34}=\omega_3-\omega_4=\omega_6-\omega_7=\omega_d\left(\frac {1+\sqrt 3} 2 - \frac {1-\sqrt 3} 2\right) = \sqrt3 \omega_d$}

the last is the coefficent of two identical time dependent terms {$2\cos\omega_{34}t$}. The non vanishing matrix elements for {$\sigma_x$} are:

{$ \langle 1 |\sigma_x|3\rangle = \langle 6 |\sigma_x|8\rangle = b = \sqrt{ \frac 1 2 \left(1 + \frac 1 {\sqrt3}\right)} $}

{$ \langle 1 |\sigma_x|4\rangle = \langle 7 |\sigma_x|8\rangle = b' = \sqrt{ \frac 1 2 \left(1 - \frac 1 {\sqrt3}\right)} $}

{$ \langle 2 |\sigma_x|5\rangle = 1 $}

{$ \langle 3 |\sigma_x|6\rangle = 2a^2= \frac 1 2 \left( 1-\frac 1 {\sqrt3}\right) $}

{$ \langle 3 |\sigma_x|7\rangle = \langle 4 |\sigma_x|6\rangle = 2aa'= \frac 1 {\sqrt6} $}

{$ \langle 4 |\sigma_x|7\rangle = 2a'^2= \frac 1 2 \left( 1+\frac 1 {\sqrt3}\right) $}

which are related to the following transition frequencies

{$ \omega_{13}=\omega_{68}=\omega_d\left(\frac {1+\sqrt 3} 2 - (-1)\right)=\omega_d\frac {3-\sqrt 3} 2 $}

{$ \omega_{14}=\omega_{78}=\omega_d\left(\frac {1-\sqrt 3} 2 - (-1)\right)=\omega_d\frac {3+\sqrt 3} 2 $}

{$ \omega_{25}=\omega_{36}=\omega_{47}=0$}

{$\omega_{37}=-\omega_{46}=\omega_d\left(\frac {1+\sqrt 3} 2 - \frac {1-\sqrt 3} 2\right)=\sqrt3 \omega_d $}

Top

---

The result

Putting all this together one gets

{$ P_{\zeta\zeta}(t)=\frac 1 6 \left( 3 + \cos \sqrt 3 \omega_d t + (1+\frac 1 {\sqrt 3})\cos \frac{3+\sqrt 3} 2 \omega_d t + (1-\frac 1 {\sqrt 3})\cos \frac{3-\sqrt 3} 2 \omega_d t \right) $}

This is the result experimentally observed in LiF by J.H. Brewer et al.

---

< A special case: a muon with few nuclei | Index | FMu >