THE UNIVERSITY OF BRITISH COLUMBIA
 
Physics 210 Assignment # 9:
 
MATRIX MADNESS!
 
Tue. 16 Nov. 2010 - finish by Tue. 23 Nov.

Since the course descriptions headlines MatLab, let's do something truly "computational" with it. (I will describe the exercises in terms of MatLab, but you are welcome to use octave or python instead, as they will both do everything required for this Assignment at least as well as MatLab does. Just pick your favorite!)

As usual, create your /home2/phys210/$USER/a09/ directory to store your work in.

  1. MATLAB WARMUP: Remember the Fibonacci numbers from earlier Assignments? In a file fibmat.m, write a MatLab function to generate the Fibonacci numbers $ F_n$ and plot the resulting $ F_n$ as a function of $ n$ (up to at least $ n=10$ )1 so it will be easy to check your work. Store your plot in /home2/phys210/$USER/a09/fib.pdf (using ImageMagick's convert if necessary).

     

  2. PAULI MATRICES: The most important matrices in Physics (so say I) are the Pauli spin matrices, described accurately in the WikipediA2 as "a set of $ 2\times2$ complex Hermitian and unitary matrices . . . "

    $\displaystyle \sigma_1 = \left[\begin{matrix}0 & 1 \cr 1 & 0 \end{matrix}\right . . . 
 . . . \right]; \; \sigma_3 = \left[\begin{matrix}1 & 0 \cr 0 & -1 \end{matrix}\right]$ (1)

    which can represent (among other things) the three components ( $ \sigma_x \equiv \sigma_1$ , $ \sigma_y \equiv \sigma_2$ and $ \sigma_z \equiv \sigma_3$ ) of the vector spin operator $ \Vec{\sigma}$ for a spin-$ {1\over2}$ particle.3 Well, MatLab claims to be a "Matrix Laboratory", so it should be an ideal platform for verifying the essential properties of the Pauli matrices.4 Do so, for the list of properties listed on http://en.wikipedia.org/wiki/Pauli_matrices down to the beginning of the subject heading labelled "$ SU(2)$ ". Make sure you understand the meaning of all these properties thoroughly.5

    In this notation, the spin state of a spin-$ {1\over2}$ particle is represented by a 2-component column vector, like

    $\displaystyle \vert \! \uparrow \rangle = \left[ 1 \atop 0 \right] \qquad \hbox{\rm and} \qquad \vert \! \downarrow \rangle = \left[ 0 \atop 1 \right]$ (2)

    for "spin up" and "spin down" (along the $ \Hat{z}$ axis) respectively. Verify that operating on these column vectors from the left with the Pauli matrix $ \sigma_z$ yields $ +\vert \! \uparrow \rangle$ and $ -\vert \! \downarrow \rangle$ , respectively.6

    Construct a column vector $ \vert \!\! \rightarrow \rangle$ with the property that $ \sigma_x \vert \!\! \rightarrow \rangle =
+\vert \!\! \rightarrow \rangle$ (so that $ \vert \!\! \rightarrow \rangle$ represents a spin-$ {1\over2}$ particle with its spin in the $ +\Hat{x}$ direction).

    Similarly, construct a column vector $ \vert \otimes \rangle$ with the property that $ \sigma_y \vert \otimes \rangle =
+\vert \otimes \rangle$ (so that $ \vert \otimes \rangle$ represents a spin-$ {1\over2}$ particle with its spin in the $ +\Hat{y}$ direction).

  3. TWO SPIN-$ {1\over2}$ PARTICLES: Suppose you have two spin-$ {1\over2}$ particles, such as a proton ($ p$ ) and an electron ($ e$ ), whose magnetic moments $ \Vec{\mu}_p = \mu_p \, \Vec{\sigma}_p$ and $ \Vec{\mu}_e = -\mu_e \, \Vec{\sigma}_e$ interact with an external magnetic field $ \Vec{B}$ , each contributing its Zeeman energy $ E_Z = -\Vec{\mu}\cdot\Vec{B}$ . Then the Zeeman hamiltonian operator is

    $\displaystyle {\cal H}_Z = - \mu_p \, \Vec{\sigma}_p \cdot \Vec{B} + \mu_e \, \Vec{\sigma}_e \cdot \Vec{B} .$ (3)

    Again picking the $ \Hat{z}$ direction as the quantization axis, we have four possible fully-specified quantum states:
    $\displaystyle \vert \! \Uparrow \uparrow \rangle
= \left[ {1 \atop 0} \atop { . . . 
 . . .  \Downarrow \downarrow \rangle
= \left[ {0 \atop 0} \atop {0 \atop 1} \right]$     (4)

    where the $ \Updownarrow$ and $ \updownarrow$ symbols designate "spin up/down" (along the $ \Hat{z}$ axis) for the electron and the proton, respectively. In this basis, verify that the $ 4\times4$ matrix representations of the electron and proton spin operators are
      $\displaystyle \sigma_{e_1} = \left[\begin{matrix}
0 & 0 & 1 & 0 \cr
0 & 0 & 0 & 1 \cr
1 & 0 & 0 & 0 \cr
0 & 1 & 0 & 0
\end{matrix}\right];$ $\displaystyle \sigma_{p_1} = \left[\begin{matrix}
0 & 1 & 0 & 0 \cr
1 & 0 & 0 & 0 \cr
0 & 0 & 0 & 1 \cr
0 & 0 & 1 & 0 \cr
\end{matrix}\right]$ (5)


      $\displaystyle \sigma_{e_2} = \left[\begin{matrix}
0 & 0 & -i & 0 \cr
0 & 0 & 0 & -i \cr
i & 0 & 0 & 0 \cr
0 & i & 0 & 0
\end{matrix}\right];$ $\displaystyle \sigma_{p_2} = \left[\begin{matrix}
0 & -i & 0 & 0 \cr
i & 0 & 0 & 0 \cr
0 & 0 & 0 & -i \cr
0 & 0 & i & 0 \cr
\end{matrix}\right]$  


      $\displaystyle \sigma_{e_3} = \left[\begin{matrix}
1 & 0 & 0 & 0 \cr
0 & 1 & 0 & 0 \cr
0 & 0 & -1 & 0 \cr
0 & 0 & 0 & -1 \cr
\end{matrix}\right];$ $\displaystyle \sigma_{p_3} = \left[\begin{matrix}
1 & 0 & 0 & 0 \cr
0 & -1 & 0 & 0 \cr
0 & 0 & 1 & 0 \cr
0 & 0 & 0 & -1 \cr
\end{matrix}\right]$  

    Given this information, write down the matrix representation of the full Zeeman hamiltonian for these two spins in an arbitrary magnetic field $ \Vec{B} = B_x \Hat{x} + B_y \Hat{y} + B_z \Hat{z}$ . Express your result in terms of $ \mu_p$ , $ \mu_e$ and the three components of $ \Vec{B}$ .

  4. THE CONTACT INTERACTION: Suppose your two spin-$ {1\over2}$ particles (e.g. the proton and the electron in a hydrogen atom) interact in a way that depends only on the scalar product of their spin vectors,7

    $\displaystyle {\cal H}_{\rm hf} = A \; \Vec{\sigma}_p \cdot \Vec{\sigma}_e \; { . . . 
 . . . ma_{e_1} + \sigma_{p_2} \sigma_{e_2} + \sigma_{p_3} \sigma_{e_3} \right) } \; ,$ (6)

    where $ {\cal H}_{\rm hf}$ is the Heisenberg hamiltonian operator and $ A$ is the strength of the interaction, in energy units. For simplicity, set $ A=1$ (i.e. measure all energies as multiples of $ A$ ) in this part.

    Express the Heisenberg spin hamiltonian (6) as a matrix in the 4-state basis (4) defined above, and show that it is not diagonal. Using MatLab, diagonalize it and describe the new basis in which it is diagonal.8

  5. BREIT-RABI DIAGRAM: [EXTRA CREDIT] We are now ready to solve the general problem of the spin hamiltonian (which governs everything the spins do!) of a hydrogen atom in an $ s$ state with orbital angular momentum $ \ell = 0$ .9 The Breit-Rabi hamiltonian is
    $\displaystyle {\cal H}_{\rm BR}$ $\displaystyle =$ $\displaystyle {\cal H}_{\rm hf} + {\cal H}_Z
\cr\cr$ (7)

    Express this hamiltonian in matrix form for the 4-state basis (4) and (using MatLab) diagonalize it for some particular choice of applied magnetic field, let's say $ \Vec{B} = (0.1\hbox{~T}) \Hat{z}$ . Once you have accomplished this, you can repeat the diagonalization for a succession of different values of $ \vert\Vec{B}\vert = B_z$ and plot the four energy eigenvalues as a function of field to get the famous Breit-Rabi diagram for hydrogen:

    Figure: : Breit-Rabi diagram showing the energy levels of a system of two spin-1/2 particles of opposite sign and different magnetic moments (e.g. the hydrogen atom) as functions of the reduced field $ x \equiv B/B_0$ where $ B_0$ (504.4 Oe for H in vacuum) is a characteristic hyperfine field. For the purpose of illustration, unphysical values of moments and coupling constants have been used.
    \begin{figure}\begin{center}
\epsfig{file=breit-rabi.ps,width=0.4\textwidth}
\end{center}\vskip -5mm
%% } %% {end Fig capt)
\end{figure}
-The actual hyperfine frequency $ \nu_0 \equiv \omega_0/2\pi \equiv A/h$ (where $ h$ is Planck's constant) has the value 1.42040575 GHz for hydrogen in vacuum. In consistent units, $ \mu_e/h = - 28.024953$  GHz/T and $ \mu_p/h = 0.042577482$  GHz/T.

In zero field the three triplet ($ J=1$ ) eigenstates $ \vert 1\rangle$ , $ \vert 2\rangle$ and $ \vert 3\rangle$ are degenerate and the singlet ($ J=0$ ) ground state $ \vert 4\rangle$ is $ \hbar\omega_0$ lower in energy.

At high reduced field ( $ x \to \infty$ ) the eigenstates are
$ \vert 1\rangle \to \vert\!\Uparrow\uparrow\rangle$ , $ \vert 2\rangle \to \vert\!\Uparrow\downarrow\rangle$ , $ \vert 3\rangle \to \vert\!\Downarrow\downarrow\rangle$ and $ \vert 4\rangle \to \vert\!\Downarrow\uparrow\rangle$ .
That is, the original basis!



Jess H. Brewer 2010-10-30