Measurement

To determine the ``answer'' of a quantum computation it is necessary to make a measurement. As noted earlier, the technology for making a projective measurement of individual nuclear spins does not yet exist. In liquid-state NMR, instead of using just one molecule to define a single quantum register, we use a large ensemble of molecules in a test tube. Ideally, their nuclear spins are all placed in the same initial state, and the subsequent RF pulses affect each molecule in the same way. As a result, weak magnetic signals from (say) the proton spins in TCE add to form a detectable magnetic field called the ``bulk magnetization''. The signal that is measured in high-field NMR is the magnetization in the $xy$-plane, which can be picked up by coils whose axes are placed transversely to the external field. Because the interaction of any given nuclear spin with the coil is very weak, the effect of the coil on the quantum state of the spins is negligible in most NMR experiments. As a result, it is a good approximation to think of the generated magnetic fields and their detection classically. In this approximation, each nuclear spin behaves like a tiny bar magnet and contributes to the bulk magnetization. As the nuclear spins precess, so does the magnetization. As a result, an oscillating current is induced in the coil, provided it is electronically configured to be ``tuned'' to the precession frequency. By observing the amplitude and phase of this current over time, we can keep track of the absolute magnetization in the plane and its phase with respect to the rotating frame. This process yields information about the qubit states represented by the state of the nuclear spins.

To see how one can use the bulk magnetization to learn about the qubit states, consider the TCE molecule with three spin-${1\over 2}$ nuclei used for information processing. The bulk magnetizations generated by the protons and the carbons precess at $500{\mathchoice{\mbox{Mhz}}{\mbox{Mhz}}{\mbox{\small Mhz}}{\mbox{\tiny Mhz}}}\,$ and $125{\mathchoice{\mbox{Mhz}}{\mbox{Mhz}}{\mbox{\small Mhz}}{\mbox{\tiny Mhz}}}\,$, respectively. The proton and carbon contributions to the magnetization are detected separately with two coils tuned to $500{\mathchoice{\mbox{Mhz}}{\mbox{Mhz}}{\mbox{\small Mhz}}{\mbox{\tiny Mhz}}}\,$ (proton magnetization) and $125{\mathchoice{\mbox{Mhz}}{\mbox{Mhz}}{\mbox{\small Mhz}}{\mbox{\tiny Mhz}}}\,$ (carbon magnetization). For simplicity, we restrict our attention to the two carbons and assume that the protons are not interacting with the carbons. (It is possible to actively remove such interactions by using a technique called ``decoupling''.)

At the end of a computation, the qubit state of the two nuclear spins is given by a density matrix $\rho_q$. We can assume that this state is the same for each molecule of TCE in the sample. As we mentioned earlier, the density matrix is relative to logical frames for each nuclear spin. The current phases for the two logical frames with respect to a rotating reference frame at the precession frequency of the first carbon are known. If we learn something about the state in the reference frame, that information can be converted to the desired logical frame by a rotation around the $z$-axis. Let $\rho(0)$ be the state of the two nuclear spins in the reference frame. In this frame, the state evolves in time as $\rho(t)$ according to a Hamiltonian $H$ that consists of a chemical shift term for the difference in the precession frequency of the second carbon and of a coupling term. To a good approximation,

$$H = \pi900{\mathchoice{\mbox{Hz}}{\mbox{Hz}}{\mbox{\small Hz... ...}}}\,{\sigma_z}^{({\mathsf {1}})}{\sigma_z}^{({\mathsf {2}})}.$$ (4)

The magnetization detected in the reference $x$-direction at time $t$ is given by

$$M_x(t)= m\;\mbox{tr}\left(\rho(t)({\sigma_x}^{({\mathsf {1}})}+{\sigma_x}^{({\mathsf {2}})})\right),$$ (5)

where $\mbox{tr}(\sigma)$ denotes the trace, that is, the sum of the diagonal elements of the matrix $\sigma$. Eq. 5 links the magnetization to the Bloch sphere representation. The constant of proportionality $m$ depends on the size of the ensemble and the magnetic moments of the nuclei. From the point of view of NMR, $m$ determines a scale whose absolute size is not relevant. What matters is how strong this signal is compared to the noise in the system. For the purpose of the following discussion, we set $m=1$.

We can also detect the magnetization $M_y(t)$ in the $y$-direction and combine it with $M_x(t)$ to form a complex number representing the planar magnetization.

$$M(t) = M_x(t)+iM_y(t)$$ (6)

$$= \mbox{tr}\left(\rho(t)({\sigma_+}^{({\mathsf {1}})}+{\sigma_+}^{({\mathsf {2}})})\right),$$ (7)

where we defined $\sigma_+=\sigma_x+i\sigma_y = \left(\begin{array}{cc}0&2\\ 0&0\end{array}\right)$. What can we infer about $\rho(0)$ from observing $M(t)$ over time? For the moment, we neglect the coupling Hamiltonian. Under the chemical shift Hamiltonian $H_{CS}=\pi900{\mathchoice{\mbox{Hz}}{\mbox{Hz}}{\mbox{\small Hz}}{\mbox{\tiny Hz}}}\,{\sigma_z}^{({\mathsf {2}})}$, $M(t)$ evolves as

$$\begin{array}[b]{rcll} M(t) &=& \mbox{tr}\left( e^{-iH_{CS}t... ... {2}})}\right) &\mbox{because the trace is linear.} \end{array}$$ (8)

Thus the signal is a combination of a constant signal given by the first spin's contribution to the magnetization in the plane, and a signal oscillating with a frequency of $900{\mathchoice{\mbox{Hz}}{\mbox{Hz}}{\mbox{\small Hz}}{\mbox{\tiny Hz}}}\,$ with amplitude given by the second spin's contribution to the planar magnetization. The two contributions can be separated by Fourier transforming $M(t)$, which results in two distinct peaks, one at $0{\mathchoice{\mbox{Hz}}{\mbox{Hz}}{\mbox{\small Hz}}{\mbox{\tiny Hz}}}\,$ and a second at $900{\mathchoice{\mbox{Hz}}{\mbox{Hz}}{\mbox{\small Hz}}{\mbox{\tiny Hz}}}\,$. See Fig. 7.

FIG. 7: Simulated magnetization signals (left) and spectra (right). (a) The $x$-magnetization signal is shown as a function of time for a pair of uncoupled spins with a relative chemical shift of $900{\mathchoice{\mbox{Hz}}{\mbox{Hz}}{\mbox{\small Hz}}{\mbox{\tiny Hz}}}\,$. The initial spin directions are along the $x$-axis. The signal (called the ``free induction decay'') decays with a halftime of $0.0385s$ because of simulated relaxation processes. Typically, the halftimes are much longer. A short one was chosen to broaden the peaks for visual effect. (b) This shows the spectrum for the signal in (a), that is, the Fourier transform of the combined $x$- and $y$-magnetization. The spectrum has peaks at frequencies of $0{\mathchoice{\mbox{Hz}}{\mbox{Hz}}{\mbox{\small Hz}}{\mbox{\tiny Hz}}}\,$ (spin $\mathsf {1}$'s peak) and $900{\mathchoice{\mbox{Hz}}{\mbox{Hz}}{\mbox{\small Hz}}{\mbox{\tiny Hz}}}\,$ (spin $\mathsf {2}$'s peak) because of the independently precessing pair of spins. (c) This is the $x$-magnetization signal when the two spins are coupled as described in the text. (d) This shows the spectrum for the signal in (c) obtained from the combined $x$- and $y$-magnetization. Each spin's peak from the previous spectrum ``splits'' into two. The left and right peaks of each pair are associated with the other spin being in the state $\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\mathfrak{1}}\mbox{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}$ and $\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\mathfrak{0}}\mbox{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}$, respectively. The vertical axis units are relative intensity with the same constant of proportionality for the two spectra.

To see how the coupling affects the observed magnetization, we rewrite the expression for $M(t)$ to take advantage of the fact that the up/down states are invariant under the full Hamiltonian.

$$M(t) = \mbox{tr}\left(\rho(t){\sigma_+}^{({\mathsf {1}})}\right) + \mbox{tr}\left(\rho(t){\sigma_+}^{({\mathsf {2}})}\right)$$

$$= \mbox{tr}\left(\rho(t){\sigma_+}^{({\mathsf {1}})}{{\mathchoice {... ...u l} {\rm 1\mskip-5mu l}}}^{({\mathsf {1}})}{\sigma_+}^{({\mathsf {2}})}\right)$$

$$= \mbox{tr}\left(\rho(t){\sigma_+}^{({\mathsf {1}})}({e_\uparrow}^{... ...{1}})} + {e_\downarrow}^{({\mathsf {1}})}) {\sigma_+}^{({\mathsf {2}})} \right)$$ (9)

where $e_\uparrow=\left(\begin{array}{cc}1&0\\ 0&0\end{array}\right)$ and $e_\downarrow=\left(\begin{array}{cc}0&0\\ 0&1\end{array}\right)$. Using a similar calculation to the one leading to Eq. 8, the first term can be written as

$$M_1(t) = \mbox{tr}\left( e^{-iH\;t}\rho(0)e^{iHt} {\sigma_+}^{({\mathsf {1}})}({e_\uparrow}^{({\mathsf {2}})} + {e_\downarrow}^{({\mathsf {2}})})\right)$$ (10)

$$= e^{i2\pi50{\mathchoice{\mbox{Hz}}{\mbox{Hz}}{\mbox{\small Hz}}{\m... ...ft(\rho(0){\sigma_+}^{({\mathsf {1}})}{e_\downarrow}^{({\mathsf {2}})})\right),$$ (11)

and similarly for the second term, but with an offset frequency of $900{\mathchoice{\mbox{Hz}}{\mbox{Hz}}{\mbox{\small Hz}}{\mbox{\tiny Hz}}}\,$ because of the chemical shift. It can be seen that the zero-frequency signal splits into two signals with frequencies of $-50{\mathchoice{\mbox{Hz}}{\mbox{Hz}}{\mbox{\small Hz}}{\mbox{\tiny Hz}}}\,$ and $50{\mathchoice{\mbox{Hz}}{\mbox{Hz}}{\mbox{\small Hz}}{\mbox{\tiny Hz}}}\,$, respectively. The difference between the two frequencies is the coupling constant. The amplitudes of the different frequency signals can be used to infer the expectations of operators such as ${\sigma_+}^{({\mathsf {1}})}{e_\uparrow}^{({\mathsf {2}})}$, given by $\mbox{tr}\left(\rho(0){\sigma_+}^{({\mathsf {1}})}{e_\uparrow}^{({\mathsf {2}})}\right)$. For $n$ spin-${1\over 2}$ nuclei, the spectral peak of a nucleus splits into a group of $2^{n-1}$ peaks, each associated with operators like ${\sigma_+}^{({\mathsf {a}})}{e_\uparrow}^{({\mathsf {b}})} {e_\downarrow}^{({\mathsf {c}})}{e_\downarrow}^{({\mathsf {d}})}\ldots$. Fig. 12 shows a simulated peak group for a nuclear spin coupled to three other spins. Expectations of the single spin operators ${\sigma_x}^{({\mathsf {a}})}$ and ${\sigma_y}^{({\mathsf {a}})}$ can be obtained from the real and imaginary parts of the total signal in a peak group for a nucleus. The positions of the $2^{n-1}$ peaks depend on the couplings. If the peaks are all well separated, we can infer expectations of product operators with only one $\sigma_x$ or $\sigma_y$, such as ${\sigma_x}^{({\mathsf {a}})}{\sigma_z}^{({\mathsf {b}})}{{\mathchoice {\rm 1\ms... ...ip-4.5mu l} {\rm 1\mskip-5mu l}}}^{({\mathsf {c}})}{\sigma_z}^{({\mathsf {d}})}$ by taking linear combinations with appropriate coefficients of the peak amplitudes in a peak group.

In addition to the unitary evolution due to the internal Hamiltonian, relaxation processes tend to decay $\rho(t)$ toward the equilibrium state. In liquid state, the equilibrium state $\rho_{\mbox{\tiny thermal}}$ is close to ${\mathchoice {\rm 1\mskip-4mu l} {\rm 1\mskip-4mu l} {\rm 1\mskip-4.5mu l} {\rm 1\mskip-5mu l}}/N$ where $N$ is the total dimension of the state space. The difference between $\rho_{\mbox{\tiny thermal}}$ and ${\mathchoice {\rm 1\mskip-4mu l} {\rm 1\mskip-4mu l} {\rm 1\mskip-4.5mu l} {\rm 1\mskip-5mu l}}/N$ is the equilibrium ``deviation'' density matrix and has magnetization only along the $z$-axis (see Sect. 2.6). Because the only observed magnetization is planar, the observed signal decays to zero as the state relaxes to equilibrium. To a good approximation we can write

$$\rho(t) = {1\over N}{\mathchoice {\rm 1\mskip-4mu l} {\rm 1\... ...skip-5mu l}}+ e^{-\lambda t} \rho'(t) + (\mbox{not observed}),$$ (12)

where $\rho'(t)$ has trace zero and evolves unitarily under the Hamiltonian. The effect of the relaxation process is that $M(t)$ has an exponentially decaying envelope, explaining the conventional name for $M(t)$, namely, the ``free induction decay'' (FID). Typical half-times for the decay are $.1{\mathchoice{\mbox{s}}{\mbox{s}}{\mbox{\small s}}{\mbox{\tiny s}}}\,$ to $2{\mathchoice{\mbox{s}}{\mbox{s}}{\mbox{\small s}}{\mbox{\tiny s}}}\,$ for nuclear spins used for QIP. A normal NMR observation consists of measuring $M(t)$ at discrete time intervals until the signal is too small. The acquired FID is then Fourier transformed to visualize the amplitudes of the different frequency contributions. The shape of the peaks in Fig. 7 reflects the decay envelope. The width of the peaks is proportional to the decay rate $\lambda$.

For QIP, we wish to measure the probability $p$ that a given qubit, say the first, labeled $\mathsf {1}$, is in the state $\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\mathfrak{1}}\mbox{$\rangl... ...e*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}_{{}_{\!\!\scriptstyle{\mathsf {1}}}}$. We have $1-2p=\mbox{tr}(\rho {\sigma_z}^{({\mathsf {1}})})$, which is the expectation of ${\sigma_z}^{({\mathsf {1}})}$. One can measure this expectation by first applying a $90^\circ$ $y$-pulse to qubit $\mathsf {1}$, thus changing the state to $\rho'$. This pulse has the effect of rotating initial, unobservable $z$-magnetization to observable $x$-magnetization. From $M(t)$ one can then infer $\mbox{tr}(\rho'{\sigma_x}^{({\mathsf {1}})})$, which is the desired number. For the coupled pair of carbons, $\mbox{tr}(\rho'{\sigma_x}^{({\mathsf {1}})})$ is given by the sum of the real components of the amplitudes of the $50{\mathchoice{\mbox{Hz}}{\mbox{Hz}}{\mbox{\small Hz}}{\mbox{\tiny Hz}}}\,$ and the $-50{\mathchoice{\mbox{Hz}}{\mbox{Hz}}{\mbox{\small Hz}}{\mbox{\tiny Hz}}}\,$ contributions to $M(t)$. However, the problem is that these amplitudes are determined only up to a scale. A second problem is that the available states $\rho$ are highly mixed (close to ${\mathchoice {\rm 1\mskip-4mu l} {\rm 1\mskip-4mu l} {\rm 1\mskip-4.5mu l} {\rm 1\mskip-5mu l}}/N$). The next section discusses how to compensate for both problems.

As a final comment on NMR measurement, note that the ``back reaction'' on the nuclear spins due to the emission of electromagnetic energy is weak. This is what enables us to measure the bulk magnetization over some time. The ensemble nature of the system gives us direct, if noisy, access to expectations of observables such as $\sigma_z$, rather than a single answer--$\mathfrak{0}$ or $\mathfrak{1}$. For algorithms that provide a definite answer, having access only to expectations is not a problem, because it is easy to distinguish the answer from the noise. However, using expectations can increase the need for quantum resources. For example, Shor's factoring algorithm includes a significant amount of classical post-processing based on highly random answers from projective measurements. In order to implement the algorithm in an ensemble setting, the post-processing must be performed reversibly and integrated into the quantum computation to guarantee a definite answer. This post-processing can be done with polynomial additional quantum resources.