next up previous
Next: Gradient Fields Up: Principles of Liquid-State NMR Previous: Measurement


The Initial State

Because the energy difference between the nuclear spins' up and down states is so small compared to room temperature, the equilibrium distribution of states is nearly random. In the liquid samples used, equilibrium is established after $10{\mathchoice{\mbox{s}}{\mbox{s}}{\mbox{\small s}}{\mbox{\tiny s}}}\,$- $40{\mathchoice{\mbox{s}}{\mbox{s}}{\mbox{\small s}}{\mbox{\tiny s}}}\,$ if no RF fields are being applied. As a result, all computations start with the sample in equilibrium. One way to think of this initial state is that every nuclear spin in each molecule begins in the highly mixed state $(1-\epsilon){\mathchoice {\rm 1\mskip-4mu l} {\rm 1\mskip-4mu l} {\rm 1\mskip-4...
...3pt}\langle$}{\mathfrak{0}}\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}$, where $\epsilon$ is a small number (of the order of $10^{-5}$). This is a nearly random state with a small excess of the state $\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\mathfrak{0}}\mbox{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}$. The expression for the initial state derives from the fact that the equilibrium state $\rho_{\mbox{\tiny thermal}}$ is proportional to $e^{-H/kT}$, where $H$ is the internal Hamiltonian of the nuclear spins in a molecule (in energy units), $T$ is the temperature and $k$ is the Boltzman constant. In our case, $H/kT$ is very small and the coupling terms are negligible. Therefore

$\displaystyle e^{-H/kT}$ $\textstyle \approx$ $\displaystyle e^{-\epsilon_1{\sigma_z}^{({\mathsf {1}})}/kT}e^{-\epsilon_2{\sigma_z}^{({\mathsf {2}})}/kT}\ldots$ (13)
$\displaystyle e^{-\epsilon_1{\sigma_z}^{({\mathsf {1}})}/kT}$ $\textstyle \approx$ $\displaystyle {\mathchoice {\rm 1\mskip-4mu l} {\rm 1\mskip-4mu l} {\rm 1\mskip-4.5mu l} {\rm 1\mskip-5mu l}}- \epsilon_1{\sigma_z}^{({\mathsf {1}})}/kT$ (14)
$\displaystyle e^{-H/kT}$ $\textstyle \approx$ $\displaystyle {\mathchoice {\rm 1\mskip-4mu l} {\rm 1\mskip-4mu l} {\rm 1\mskip...
...ma_z}^{({\mathsf {1}})}/kT - \epsilon_2{\sigma_z}^{({\mathsf {2}})}/kT - \ldots$ (15)

where $\epsilon_l$ is half of the energy difference between the up and down states of the $l$'th nuclear spin.

Clearly the available initial state is very far from what is needed for standard QIP. However, it can still be used to perform interesting computations. The main technique is to use available NMR tools to change the initial state to a ``pseudopure'' state, which for all practical purposes behaves like the initial state required by QIP. The technique is based on three key observations. First, only the trace-less part of the density matrix contributes to the magnetization. Suppose that we are using $n$ spin-$1\over 2$ nuclei in a molecule and the density matrix is $\rho$. Then the current magnetization is proportional to $\mbox{tr}(\rho \hat m)$, where $\hat m$ is a traceless operator (see Eq. 9). Therefore the magnetization does not depend on the part of $\rho$ proportional to the identity matrix. A ``deviation density matrix'' for $\rho$ is any matrix $\delta$ such that $\delta-\rho=\lambda{\mathchoice {\rm 1\mskip-4mu l} {\rm 1\mskip-4mu l} {\rm 1\mskip-4.5mu l} {\rm 1\mskip-5mu l}}$ for some $\lambda$. For example, $\epsilon\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\mathfrak{0}}\mbox...
...3pt}\langle$}{\mathfrak{0}}\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}$ is a deviation for the equilibrium state of one nuclear spin. We have

$\displaystyle \mbox{tr}(\delta\hat m)$ $\textstyle =$ $\displaystyle \mbox{tr}((\rho+\lambda{\mathchoice {\rm 1\mskip-4mu l} {\rm 1\mskip-4mu l} {\rm 1\mskip-4.5mu l} {\rm 1\mskip-5mu l}})\hat m)$  
  $\textstyle =$ $\displaystyle \mbox{tr}(\rho\hat m) +\mbox{tr}(\hat m)$  
  $\textstyle =$ $\displaystyle \mbox{tr}(\rho\hat m).$ (16)

The second observation is that all the unitary operations used, as well as the non-unitary ones to be discussed below, preserve the completely mixed state ${\mathchoice {\rm 1\mskip-4mu l} {\rm 1\mskip-4mu l} {\rm 1\mskip-4.5mu l} {\rm 1\mskip-5mu l}}/2^n$.1 Therefore, all future observations of magnetization depend only on the initial deviation.

The third observation is that all the scales are relative. In particular, as will be explained, the probability that the final answer of a quantum computation is $\mathfrak{1}$ can be expressed as the ratio of two magnetizations. It follows that one can arbitrarily rescale a deviation density matrix. For measurement, the absolute size of the magnetizations is not important; the most important issue is that the magnetizations are strong enough to be observable over the noise.

To explain the relativity of the scales and introduce ``pseudopure'' states for QIP, we begin with one spin-${1\over 2}$ qubit. Its equilibrium state has as a deviation $\delta=\epsilon\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\mathfrak{0...
...3pt}\langle$}{\mathfrak{0}}\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}$. If $U$ is the total unitary operator associated with a computation, then $\delta$ is transformed to $\delta'=\epsilon
U\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\mathfra...
...le$}{\mathfrak{0}}\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}U^\dagger$. For QIP purposes, the goal is to determine what the final probability $p_{\mathfrak{1}}$ of measuring $\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\mathfrak{1}}\mbox{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}$ is, given that $\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\mathfrak{0}}\mbox{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}$ is the initial state. This probability can be computed as follows:

$\displaystyle p_{\mathfrak{1}}$ $\textstyle =$ $\displaystyle \mbox{$\langle\hspace*{-4.3pt}\langle\hspace*{-4.3pt}\langle$}{\m...
...t$}{\mathfrak{1}}\mbox{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}$  
  $\textstyle =$ $\displaystyle \mbox{tr}\left(U\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\ver...
...ngle$}{\mathfrak{1}}\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}\right)$  
  $\textstyle =$ $\displaystyle \mbox{tr}\left(U\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\ver...
...rm 1\mskip-4mu l} {\rm 1\mskip-4.5mu l} {\rm 1\mskip-5mu l}}-\sigma_z)\right)/2$  
  $\textstyle =$ $\displaystyle \left(\mbox{tr}(U\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\ve...
...\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}U^\dagger\sigma_z)\right)/2$  
  $\textstyle =$ $\displaystyle \left(1-\mbox{tr}(U\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\...
...mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}U^\dagger\sigma_z)\right)/2.$ (17)

Thus, the probability can be determined by measuring the expectations of $\sigma_z$ for the initial and final states (in different experiments), which yields the quantities $a =
\mbox{tr}(\delta\sigma_z)=\epsilon$ and $a' = \mbox{tr}(\delta'\sigma_z) = \epsilon\,
\mbox{tr}\left( U\mbox{$\vert\hspa...
...0}}\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}U^\dagger\sigma_z\right)$, respectively. The desired answer is $p_{\mathfrak{1}}=(1-(a/a'))/2$ and does not depend on the scale $\epsilon$.

The method presented in the previous paragraph for determining the probability that the answer of a quantum computation is $\mathfrak{1}$ generalizes to many qubits. The goal is to determine the probability $p_{\mathfrak{1}}$ of measuring $\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\mathfrak{1}}\mbox{$\rangl...
...e*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}_{{}_{\!\!\scriptstyle{\mathsf {1}}}}$ in a measurement of the first qubit after a computation with initial state $\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\mathfrak{0}\ldots\mathfrak{0}}\mbox{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}$. Suppose we can prepare the spins in an initial state with deviation $\delta=\epsilon\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\mathfrak{0...
...hfrak{0}\ldots\mathfrak{0}}\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}$. A measurement of the expectations $a$ and $a'$ of ${\sigma_z}^{({\mathsf {1}})}$ for the initial and final states then yields $p_{\mathfrak{1}}$ as before, by the formula $p_{\mathfrak{1}}=(1-(a/a'))/2$.

A state with deviation $\epsilon\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\psi}\mbox{$\rangl...
...ace*{-4.3pt}\langle$}{\psi}\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}$ is called a ``pseudopure'' state, because this deviation is proportional to the deviation of the pure state $\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\psi}\mbox{$\rangle\hspace...
...ace*{-4.3pt}\langle$}{\psi}\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}$. With respect to scale-independent NMR observations and unitary evolution, a pseudopure state is equivalent to the corresponding pure state. Because NMR QIP methods are scale independent, we now generalize the definition of deviation density matrix: $\delta$ is a deviation of the density matrix $\rho$ if $\epsilon\delta =
\rho+\lambda{\mathchoice {\rm 1\mskip-4mu l} {\rm 1\mskip-4mu l} {\rm 1\mskip-4.5mu l} {\rm 1\mskip-5mu l}}$ for some $\lambda$ and $\epsilon$.

Among the most important enabling techniques in NMR QIP are the methods that can be used to transform the initial thermal equilibrium state to a standard pseudopure state with deviation $\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\mathfrak{0}\ldots\mathfra...
...hfrak{0}\ldots\mathfrak{0}}\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}$. An example of how that can be done will be given as the second algorithm in Sect. 3. The basic principle for each method is to create, directly or indirectly by summing over multiple experiments, a new initial state as a sum $\rho_0=\sum_iU_i\rho_{\mbox{\tiny thermal}}U_i^\dagger$, where the $U_i$ are carefully and sometimes randomly chosen [10,11,20,21] to ensure that $\rho_0$ has a standard pseudopure deviation. Among the most useful tools for realizing such sums are pulsed gradient fields.


next up previous
Next: Gradient Fields Up: Principles of Liquid-State NMR Previous: Measurement