Mixtures and Density Operators

The measurement operation ``reads out'' information from qubits to pbits. What if we discard the pbit that contains the measurement outcome? The result is that the qubits are in a probabilistic ``mixture'' of two pure states. Such mixtures are a generalization of pure states. The obvious way to think about a mixture is that we have a probability distribution over pure quantum states. For example, after discarding the pbit and qubit $\mathsf {A}$ in Eq. 36, we can write the state of $\mathsf {B}$ as $\rho=\{\vert\alpha\vert^2{:}\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$... ...ngle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}_{{}_{\!\!{\mathsf {B}}}}\}$, using the notation for probability distributions introduced earlier.

Mixtures frequenty form when using irreversible operations such as measurement. Except for measurement, the quantum gates that we have introduced so far are reversible and therefore transform pure states to pure states, so that no mixtures can be formed. One of the fundamental results of reversible classical and quantum computation is that there is no loss in power in using only reversible gates. Specifically, it is possible to change a computation that includes irreversible operations to one that accomplishes the same goal, has only reversible operations and is efficient in the sense that it uses at most polynomial additional resources. However, the cost of using only reversible operations is not negligible. In particular, for ease of programming, and more importantly, when performing repetitive error-correction tasks (see [12]), the inability to discard or reset qubits can be very inconvenient. We therefore introduce additional operations that enable resetting and discarding.

Although resetting has a so-called ``thermodynamic'' cost (think of the heat generated by a computer), it is actually a simple operation. The $\mathbf{reset}$ operation applied to qubit $\mathsf {A}$ can be thought of as the result of first measuring $\mathsf {A}$, then flipping $\mathsf {A}$ if the measurement outcome is $\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\mathfrak{1}}\mbox{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}$, and finally discarding the measurement result. Using the notation of Eq. 36, the effect on a pure state $\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\psi}\mbox{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}_{{}_{\!\!{\mathsf {AB}}}}$ is given by:

$${\mathbf{reset}}^{({\mathsf {A}})}\mbox{$\vert\hspace*{-3pt}... ...t}\rangle\hspace*{-4.3pt}\rangle$}_{{}_{\!\!{\mathsf {B}}}}\}.$$ (39)

To apply $\mathbf{reset}$ to an arbitrary probability distribution, you apply it to each of the distribution's pure states and combine the results to form an expanded probability distribution. The ${\mathbf{discard}}^{({\mathsf {A}})}$ operation is ${\mathbf{reset}}^{({\mathsf {A}})}$ followed by discarding qubit $\mathsf {A}$. Therefore, in the expression for the state after ${\mathbf{reset}}^{({\mathsf {A}})}$, all the $\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\mathfrak{0}}\mbox{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}_{{}_{\!\!{\mathsf {A}}}}$ are removed. It is an important fact that every physically realizable quantum operation, whether reversible or not, can be expressed as a combination of $\mathbf{add}$ operations, gates from the universal set and $\mathbf{discard}$ operations.

The representation of mixtures using probability distributions over pure states is redundant. That is, there are many probability distributions that are physically indistinguishable. A non-redundant description of a quantum state can be obtained by using ``density operators''. The density operator for the mixture $\rho$ given in Eq. 39 is given by

$$\hat\rho= \vert\alpha\vert^2\mbox{$\vert\hspace*{-3pt}\vert\... ...}{\phi_1}\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}.$$ (40)

The general rule for calculating the density operator from a probability distribution is as follows: For each pure state $\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\phi}\mbox{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}$ in the distribution, calculate the operators $\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\phi}\mbox{$\rangle\hspace... ...ace*{-4.3pt}\langle$}{\phi}\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}$ and sum them weighted by their probabilities.

There is a way to apply gates to the density operators defined by states. If the gate acts by the unitary operator $U$, then the effect of applying it to $\hat\rho$ is given by $U\hat\rho U^\dagger$, where $U^\dagger$ is the conjugate transpose of $U$. (In the bra-ket expression for $U$, $U^\dagger$ is obtained by replacing all complex numbers by their conjugates, and terms such as $\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\phi}\mbox{$\rangle\hspace... ...*{-4.3pt}\langle$}{\varphi}\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}$ by $\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\varphi}\mbox{$\rangle\hsp... ...ace*{-4.3pt}\langle$}{\phi}\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}$.)

The relationship between a qubit's state space and a sphere can be explained in terms of the qubit's density operators. In matrix form, this operator is a $2\times 2$ matrix, which can be written uniquely as a sum $({\mathchoice {\rm 1\mskip-4mu l} {\rm 1\mskip-4mu l} {\rm 1\mskip-4.5mu l} {\rm 1\mskip-5mu l}}+x\sigma_x+y\sigma_y+z\sigma_z)/2$. One can check that if the density operator $\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\psi}\mbox{$\rangle\hspace... ...ace*{-4.3pt}\langle$}{\psi}\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}$ for a qubit's pure state is written as such a sum,

$$\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\psi}\m... ....5mu l} {\rm 1\mskip-5mu l}}+x\sigma_x+y\sigma_y+z\sigma_z)/2,$$ (41)

then the vector $(x,y,z)$ thus obtained is on the surface of the unit sphere in three dimensions. In fact, for every vector $(x,y,z)$ on the unit sphere, there is a unique pure state satisfying Eq. 41. Since the density operators for mixtures are arbitrary, convex (that is probabilistic) sums of pure states, the set of $(x,y,z)$ thus obtained for mixtures fills out the unit ball. The rotations introduced earlier modify the vector $(x,y,z)$ in the expected way, by rotation of the vector around the appropriate axis. See [11] for more details.