Skip to main content
Physics LibreTexts

5.1: Composite Systems

  • Page ID
    56549
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    Suppose we have two systems, described by Hilbert spaces \(\mathscr{H}_{1}\) and \(\mathscr{H}_{2}\), respectively. We can choose orthonormal bases for each system:

    \[\mathscr{H}_{1}:\left\{\left|\phi_{1}\right\rangle,\left|\phi_{2}\right\rangle, \ldots,\left|\phi_{N}\right\rangle\right\} \quad \text { and } \mathscr{H}_{2}:\left\{\left|\psi_{1}\right\rangle,\left|\psi_{2}\right\rangle, \ldots,\left|\psi_{M}\right\rangle\right\}\tag{5.1}\]

    The respective dimensions of \(\mathscr{H}_{1}\) and \(\mathscr{H}_{2}\) are \(N\) and \(M\). We can construct \(N \times M\) basis states for the composite system via \(\left|\phi_{j}\right\rangle\) and \(\left|\psi_{k}\right\rangle\). This implies that the total Hilbert space of the composite system can be spanned by the tensor product

    \[\left\{\left|\phi_{j}\right\rangle \otimes\left|\psi_{k}\right\rangle\right\}_{j k} \quad \text { on } \quad \mathscr{H}_{1+2}=\mathscr{H}_{1} \otimes \mathscr{H}_{2}\tag{5.2}\]

    An arbitrary pure state on \(\mathscr{H}_{1+2}\) can be written as

    \[|\Psi\rangle=\sum_{j k} c_{j k}\left|\phi_{j}\right\rangle \otimes\left|\psi_{k}\right\rangle \equiv \sum_{j k} c_{j k}\left|\phi_{j}, \psi_{k}\right\rangle\tag{5.3}\]

    For example, the system of two qubits can be written on the basis \(\{|0,0\rangle,|0,1\rangle,|1,0\rangle,|1,1\rangle\}\). If system 1 is in state \(|\phi\rangle\) and system 2 is in state \(|\psi\rangle\), the partial trace over system 2 yields

    \[\operatorname{Tr}_{2}(|\phi, \psi\rangle\langle\phi, \psi|)=\operatorname{Tr}_{2}(|\phi\rangle\langle\phi|\otimes| \psi\rangle\langle\psi|)=|\phi\rangle\langle\phi|\operatorname{Tr}(|\psi\rangle\langle\psi|)=| \phi\rangle\langle\phi|,\tag{5.4}\]

    since the trace over any density operator is 1. We have now lost system 2 from our description! Therefore, taking the partial trace without inserting any other operators is the mathematical version of forgetting about it. This is a very useful feature: you often do not want to deal with every possible system you are interested in. For example, if system 1 is a qubit, and system two is a very large environment the partial trace allows you to “trace out the environment”.

    However, tracing out the environment will not always leave you with a pure state as in Eq. (5.4). If the system has interacted with the environment, taking the partial trace generally leaves you with a mixed state. This is due to entanglement between the system and its environment.


    This page titled 5.1: Composite Systems is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Pieter Kok via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.