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2.4.3 The reduced density operator

When it comes to the analysis of composite quantum systems, the reduced density operator is indispensible. (適合用來描述 subsystems)

• Suppose we have physical systems \(A\) and \(B\), whose state is described by a density operator \(\rho^{AB}\). The reduced density operator for \(A\) is defined by

  \[ \rho^A \equiv \tr_B(\rho^{AB}) \]

  , where \(\tr_B\) is a map of operators known as the partial trace over system \(B\), which is defined by:

  \[ tr_B(\ket{a_1}\bra{a_2} \otimes \ket{b_1}\bra{b_2}) \equiv \ket{a_1}\bra{a_2} \tr(\ket{b_1}\bra{b_2}) \]

  , where \(\ket{a_1}\) and \(\ket{a_2}\) are any two vectors in the state space of \(A\), and similar for the \(\ket{b}\)s.

• 注意到 trace 的特徵, 就是 \(\tr(\ket{b_1}\bra{b_2}) = \braket{b_2}{b_1}\).

• Why do we define partial trace this way? One can first think simply that the partial trace over \(A\) for system \(\rho^{AB} = \rho\otimes\sigma\) is \(\tr_B(\rho\otimes\sigma) = \rho\tr(\sigma) = \rho\), and vice versa for partial trace over \(B\). (雖然這只在 \(AB\) 是 product state 的時候才成立).

A less trivial example is the Bell state \(\frac{\ket{00}+\ket{11}}{\sqrt2}\), its density operator is:

\[ \rho = \bigg( \frac{\ket{00}+\ket{11}}{\sqrt2} \bigg) \bigg( \frac{\bra{00}+\bra{11}}{\sqrt2} \bigg) \]

, which is a pure state. Now we find a surprising 違反直覺的特性, that is, for example,

\[ \begin{align*} \rho^1 &= \tr_2(\rho) \\ &= \frac{\tr_2(\ket{00}\bra{00}) + \tr_2(\ket{00}\bra{11}) + \tr_2(\ket{11}\bra{00}) + \tr_2(\ket{11}\bra{11})}{2} \\ &= \frac{\ket0\bra0\braket{0}{0} + \ket0\bra1\braket{0}{1} + \ket1\bra0\braket{1}{0} + \ket1\bra1\braket{1}{1}}{2} \\ &= \frac{\ket0\bra0 + \ket1\bra1}{2} \\ &= \frac{I}{2} \end{align*} \]

, is a mixed state instead of a pure state! (可以由 \(\tr((\frac{I}{2})^2)<1\) 確認). 換句話說, 我們完全知道 joint system 的資訊, 卻不確定 individual qubit 的資訊, this is another hallmark of quantum entanglement.

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Exercise 2.74

Suppose a composite of systems \(A\) and \(B\) is in the state \(\ket a\ket b\) (也就是 \(\ket a \otimes \ket b\)), where \(\ket a\) is a pure state of sysem \(A\), and \(B\) is a pure state of system \(B\). Show that the reduced density operator of system \(A\) alone is a pure state.

solution

\[ \begin{align*} \rho^A &= \tr_B(\rho^{AB}) \\ &= \tr_B(\ket a \ket b \bra a \bra b) \\ &= \tr_B(\ket a \bra a\otimes \ket b \bra b) \\ &= \ket a\bra a \tr(\ket b\bra b) \\ &= \ket a\bra a. \end{align*} \]

Check if it's a pure state: \(\tr((\rho^A)^2) = \tr(\ket a\bra a\ket {a}\bra a) = 1\), hence also a pure state after partial tracing.

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Exercise 2.75

For each of the four Bell states, find the reduced density operator for each qubit.

solution

to be continued

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