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QuantumSpinPrecessionWithDensityMatrix

< Spin precession from a quantum viewpoint | Index | A note on rotations >

The same result may be obtained with the density matrix. This may be a tedious exercise for such a plain thing as spin precession, which we have already calculated classically and quantum-mechanically. However it is perhaps better to recall and clarify the use of this method on a very easy task, before we dive into some slightly more complex calculations.

The density matrix {$\rho$} for our spin one-half system will be a simple 2x2 matrix, with elements given in terms of the coefficents of the starting state. Suppose that we start from the state

{$$ |\psi\rangle=\sum_{k=\pm} c_k |k\rangle $$}

then the density matrix is given by:

{$$ \rho=\begin{bmatrix} |c_-|^2 & c_-^* c_+ \\ c_+^* c_- & |c_+|^2 \end{bmatrix} $$}

Note that the coefficients could either represent a coherent quantum superposition of states or the statistical probability for the occupation of the two eigenstates in an ensamble, so the formalism we are demonstrating can smoothly bridge these two cases. In an ergodic statistical mixture, where the coefficents are uniform averages of all available states, the two offdiagonal values would vanish, {$\overline{c_\pm^* c_\mp}=0$}. However statistical averages over states evolving in time from a well defined initial quantum state could yield off-diagonal non-vanishing terms.

Hence for our initial state {$|+\rangle_x$} where {$c_\pm=1/\sqrt 2 $} (see Eq. (2), previous page) we get:

{$$\rho=\frac 1 2 \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} $$}

The virtue of this operator is that the expectation value of any other observable {$ O$} on the same space is obtained as

{$$ \langle O \rangle = Tr \rho(O) $$}

In order to calculate time dependent expectation value the time evolution of the density matrix (in the sense of the Heisenberg representation) must be used, which is given by:

{$$\begin{equation} \rho(t) = e^{iHt/\hbar}\rho(0) e^{-iHt/\hbar} \end{equation}$$},

So all we have to do now is to calculate:

{$$ \langle I_x(t) \rangle = Tr(e^{i\omega t I_z}\rho(0) e^{-i\omega t I_z} I_x)$$}

Note that {$\rho(0)=\mathbf{1}/2+ I_x$}, that any operator commutes with the identity {$\mathbf{1}$} and that {$I_x$} is traceless. We may then simply drop the identity term in the density matrix.

The easy way then, with 2x2 matrices, is to use their explicit expression, recalling that

{$$ e^{i\omega t I_z}= \begin{bmatrix} e^{i\omega t /2} & 0 \\ 0 & e^{-i\omega t /2} \end{bmatrix} $$}

(this route would not be recommendable with larger spaces). Substituting the expressions for the four matrices we get

{$$ \langle I_x(t) \rangle = \frac 1 4 Tr(\begin{bmatrix} e^{i\omega t /2} & 0 \\ 0 & e^{-i\omega t /2} \end{bmatrix} \,\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\, \begin{bmatrix} e^{-i\omega t /2} & 0 \\ 0 & e^{i\omega t /2} \end{bmatrix}\,\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}) $$}

which reduces again to {$ \langle I_x(t) \rangle = \cos\omega t /2$} after direct multiplication. A similar calculation may be performed as exercise for {$ \langle I_y(t) \rangle$}

< Spin precession from a quantum viewpoint | Index | A note on rotations >