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Chemistry LibreTexts

265: Using the Trace Function to Calculate Expectation Values

  • Page ID
    138798
  • Starting with the traditional expression for the calculation of the expectation value, the identity operator is inserted between the measurement operator and the ket containing the wave function. Rearranging terms gives the trace function operating on the product of the state's density operator and the measurement operator.

    \[ \begin{align*} \langle \psi | \hat{O} | \psi \rangle &= \sum_i \langle \psi | \hat{O} | i \rangle \langle i | \psi \rangle \\[4pt] &= \sum_i \langle i | \psi \rangle \langle \psi | \hat{O} | i \rangle \\[4pt] &= Trace \left( | \psi \rangle \langle \psi | \hat{O} \right) \end{align*}\]

    where

    \[ \sum_i | i \rangle \langle i | = identity\]

    Next this transition is carried out in detail using matrix algebra.

    \[ \begin{pmatrix}
    a & b
    \end{pmatrix}\begin{pmatrix}
    0 & 1\\
    1 & 0
    \end{pmatrix}\begin{pmatrix}
    a\\
    b
    \end{pmatrix} = \begin{pmatrix}
    a & b
    \end{pmatrix}\begin{pmatrix}
    0 & 1\\
    1 & 0
    \end{pmatrix}\begin{pmatrix}
    1 & 0\\
    0 & 1
    \end{pmatrix}\begin{pmatrix}
    a\\
    b
    \end{pmatrix} = \begin{pmatrix}
    a & b
    \end{pmatrix}\begin{pmatrix}
    0 & 1\\
    1 & 0
    \end{pmatrix} \left[ \begin{pmatrix}
    1\\
    0
    \end{pmatrix}\begin{pmatrix}
    1 & 0
    \end{pmatrix} + \begin{pmatrix}
    0\\
    1
    \end{pmatrix}\begin{pmatrix}
    0 & 1
    \end{pmatrix} \right] \begin{pmatrix}
    a\\
    b
    \end{pmatrix} = 2ab\]

    \[ \begin{pmatrix}
    a & b
    \end{pmatrix} \begin{pmatrix}
    0 & 1\\
    1 & 0
    \end{pmatrix} \begin{pmatrix}
    1\\
    0
    \end{pmatrix}\begin{pmatrix}
    1 & 0
    \end{pmatrix} \begin{pmatrix}
    a\\
    b
    \end{pmatrix} = \begin{pmatrix}
    a & b
    \end{pmatrix} \begin{pmatrix}
    0 & 1\\
    1 & 0
    \end{pmatrix} \begin{pmatrix}
    0\\
    1
    \end{pmatrix}\begin{pmatrix}
    0 & 1
    \end{pmatrix}\begin{pmatrix}
    a\\
    b
    \end{pmatrix} = 2ab\]

    \[ \begin{pmatrix}
    1 & 0
    \end{pmatrix} \begin{pmatrix}
    a \\
    b
    \end{pmatrix} \begin{pmatrix}
    a & b
    \end{pmatrix}\begin{pmatrix}
    0 & 1\\
    1 & 0
    \end{pmatrix} \begin{pmatrix}
    1\\
    0
    \end{pmatrix} + \begin{pmatrix}
    0 & 1
    \end{pmatrix} \begin{pmatrix}
    a\\
    b
    \end{pmatrix} \begin{pmatrix}
    a & b
    \end{pmatrix} \begin{pmatrix}
    0 & 1\\
    1 & 0
    \end{pmatrix} \begin{pmatrix}
    0\\
    1
    \end{pmatrix} = 2ab\]

    \[ \begin{pmatrix}
    1 & 0
    \end{pmatrix} \begin{pmatrix}
    a^2 & ab \\
    ab & b^2
    \end{pmatrix} \begin{pmatrix}
    0 & 1\\
    1 & 0
    \end{pmatrix} \begin{pmatrix}
    1\\
    0
    \end{pmatrix} + \begin{pmatrix}
    0 & 1
    \end{pmatrix} \begin{pmatrix}
    a^2 & ab\\
    ab & b^2
    \end{pmatrix} \begin{pmatrix}
    0 & 1\\
    1 & 0
    \end{pmatrix} \begin{pmatrix}
    0\\
    1
    \end{pmatrix} = 2ab\]

    \[ \begin{pmatrix}
    1 & 0
    \end{pmatrix} \begin{pmatrix}
    ab & a^2 \\
    b^2 & ab
    \end{pmatrix} \begin{pmatrix}
    1\\
    0
    \end{pmatrix} + \begin{pmatrix}
    0 & 1
    \end{pmatrix} \begin{pmatrix}
    ab & a^2\\
    b^2 & ab
    \end{pmatrix} \begin{pmatrix}
    0\\
    1
    \end{pmatrix} = ab + ab = Trace \begin{pmatrix}
    ab & a^2 \\
    b^2 & ab\\
    \end{pmatrix}\]

    \[ \begin{pmatrix}
    a & b
    \end{pmatrix} \begin{pmatrix}
    0 & 1 \\
    1 & 0
    \end{pmatrix} \begin{pmatrix}
    a\\
    b
    \end{pmatrix} \rightarrow 2ab ~~~ \begin{pmatrix}
    a & b
    \end{pmatrix} \begin{pmatrix}
    0 & 1\\
    1 & 0
    \end{pmatrix} \begin{pmatrix}
    1 & 0\\
    0 & 1
    \end{pmatrix} \begin{pmatrix}
    a\\
    b
    \end{pmatrix} \rightarrow 2ab\]

    \[ \begin{pmatrix}
    1\\
    0
    \end{pmatrix} \begin{pmatrix}
    1 & 0
    \end{pmatrix} + \begin{pmatrix}
    0\\
    1
    \end{pmatrix} \begin{pmatrix}
    0 & 1
    \end{pmatrix} \rightarrow \begin{pmatrix}
    1 & 0\\
    0 & 1
    \end{pmatrix}\]

    \[ \begin{pmatrix}
    a & b
    \end{pmatrix} \begin{pmatrix}
    0 & 1\\
    1 & 0
    \end{pmatrix} \begin{pmatrix}
    1\\
    0
    \end{pmatrix} \begin{pmatrix}
    1 & 0
    \end{pmatrix} \begin{pmatrix}
    a\\
    b
    \end{pmatrix} + \begin{pmatrix}
    a & b
    \end{pmatrix} \begin{pmatrix}
    0 & 1\\
    1 & 0
    \end{pmatrix} \begin{pmatrix}
    0 \\
    1
    \end{pmatrix} \begin{pmatrix}
    0 & 1 \end{pmatrix} \begin{pmatrix}
    a \\
    b\end{pmatrix} \rightarrow 2ab\]

    \[ \begin{pmatrix}
    1 & 0
    \end{pmatrix} \begin{pmatrix}
    a \\
    b
    \end{pmatrix} \begin{pmatrix}
    a & b
    \end{pmatrix}
    \begin{pmatrix}
    0 & 1\\
    1 & 0
    \end{pmatrix} \begin{pmatrix}
    1\\
    0
    \end{pmatrix} + \begin{pmatrix}
    0 & 1
    \end{pmatrix} \begin{pmatrix}
    a\\
    b
    \end{pmatrix} \begin{pmatrix}
    a & b
    \end{pmatrix} \begin{pmatrix}
    0 & 1\\
    1 & 0
    \end{pmatrix} \begin{pmatrix}
    0 \\
    1
    \end{pmatrix} \rightarrow 2ab\]

    \[ tr \left[ \begin{pmatrix}
    a \\
    b
    \end {pmatrix} \begin{pmatrix} a & b \end{pmatrix} \begin{pmatrix}
    0 & 1 \\
    1 & 0 \\
    \end{pmatrix} \right] \rightarrow 2ab ~~~ tr \left[ \begin{pmatrix}
    0 & 1 \\
    1 & 0 \\
    \end{pmatrix} \begin{pmatrix}
    a \\
    b
    \end {pmatrix} \begin{pmatrix} a & b \end{pmatrix} \right] \rightarrow 2ab\]

    The last calculation on the right is justified by the following:

    \[ \langle \psi | \hat{O} | \psi \rangle = \sum_i \langle \psi | i \rangle \langle i | \hat{O} | \psi \rangle = \sum_i \langle i | \hat{O} | \psi \rangle \langle \psi | i \rangle = Trace \left( \hat{O} | \psi \rangle \langle \psi | \right)\]