[ NOTES/QCQI-01003] Trace and Partial Trace

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1.Some notation

We will work with Dirac bra ket notation.The elements of a vector space are denoted by kets, for example \(\ket{\psi}\).
The elements of the dual vector space will be denoted by bras, for example \(\bra{\psi}\).

An object written as 'outer product', \(\ket{\phi_1}\bra{\phi_2} \equiv \hat{X}\), is a linear operator in the vector space. The action of operator \(\hat{X}\) on an arbitrary vector \(\ket{\psi}\) is given by
\begin{equation}
\hat{X} \ket{\psi} = \innerproduct{\phi_2}{\psi}\, \ket{\phi_1}.
\end{equation}

If \{\(\ket{n}\)\} is an orthonormal basis in the vector space,
then it satisfies the following mathematical properties.
\begin{align}
Orthonormal property  \innerproduct{m}{n} = \delta_{mn}
Completeness property  \sum_n \ket{n}\bra{n} = \hat{I},
\end{align}
where \(\hat{I}\) is identity operator.

Using an orthonormal basis \(\{\ket{j}\}\), every linear operator, \(\hat{X}\), in a vector space can be represented by a matrix,\(\underline{X}\), with its elements
$(\underline{X})_{jk}$ given by
\begin{equation}\label{eq01}
(\underline{X})_{jk}=\bra{j}\hat{\underline{X}}\ket{k}~,\qquad
j,k=1,\cdots n
\end{equation}
where $\{\ket{j}, j=1,\cdots n\}$ is an orthonormal basis. The
matrix representing the operator depends on choice of the basis
Eq.(\ref{eq01}) is equivalent to operator relation
\begin{equation}\label{eq02}
\hat{X} = \sum_{j,k=1}^N X_{jk}\ket{j}\bra{k}.
\end{equation}
A useful quantity, trace of an operator $\hat{X}$, is defined to be
just the sum of diagonal elements of the matrix $\underline{X}$:
\begin{eqnarray}
tr(X) &=& \sum_{j=1}^N (\underline{X})_{jj}\nonumber\\
&=& \sum_{j=1} \bra{j}\hat{X}\ket{j}\label{eq03}
\end{eqnarray}
From (\ref{eq01}) and (\ref{eq02}) it is easy to verify that the
trace can be cast in the form
\begin{equation}\label{eq04}
tr(X) = \sum_{n=1}^N tr(X\ket{n}\bra{n})
\end{equation}
This relation becomes transparent on recalling that for an
orthonormal basis we have the completeness relation
$\displaystyle{\sum_{n=1}^N \ket{n}\bra{n}=I_n}$\\

2.Partial Trace:

Let $\mathcal{H}_A,
\mathcal{H}_B$ be two Hilbert spaces with orthonormal bases
$\{\ket{\nu A}|\nu=1\cdots N\}$ and $\{\ket{aB}|a=1\cdots M\}$
respectively. Then the tensor product Hilbert space has a basis
\begin{equation}\label{eq05}
\{\ket{\nu A}\otimes\ket{aB}\equiv \ket{\nu A;aB}|, \qquad\nu=1\cdots N,
a=1\cdots M\}\,.
\end{equation}
This basis will be orthonormal basis and can be used to construct a
representation of vectors and operators in
$\mathcal{H}_A\otimes\mathcal{H}_B$.\\
An operator $\widehat{\mathscr X}$ on the tensor product space will
then be represented by a matrix $\underline{\mathscr X}$ with
elements
\begin{equation}\label{eq06}
(\underline{\widehat{\mathscr X}})_{\nu a,\lambda b} = \bra{\nu
A,aB}\widehat{\mathscr X}\ket{\lambda A,bB}
\end{equation}
The trace of operator $\widehat{\mathscr X}$ is as usual defined to be
\begin{equation}\label{eq07}
Tr(\widehat{\mathscr X})=\sum_{a=1}^M\sum_{\nu=1}^N \bra{\nu
A,aB}\widehat{\mathscr X}\ket{\nu A,Ba}
\end{equation}
If we sum over only one of the two indices $a$ (or $\nu$) we get
partial trace of operator $\widehat{\mathscr X}$.
\begin{equation}\label{eq08}
\left(\widehat{\mathscr X}\bigl|_{tr B}\right)_{\nu\lambda} =
\sum_{a=}^M\bra{\nu A,aB}\widehat{\mathscr X}\ket{\lambda A,aB}
\end{equation}
the above expression defines an operator $\widehat{\mathscr X}_A$ in
$\mathcal{H}_A$ with matrix elements given by r.h.s. of \eqref{eq08}.
$\widehat{\mathscr X}_A$ will be called particle trace of $\widehat{\mathscr X}$ over
$B$.

Similarly $\widehat{\mathscr X}_B$, the result of partial of $\widehat{\mathscr X}$ over $A$
will be an operator in $\mathcal{H}_B$ and is defined in a smilar
manner.
\begin{eqnarray}
\widehat{\mathscr X}_A &=& \sum_{\nu\lambda} \left(\widehat{\mathscr X}\bigl|_{tr
B}\right)_{\nu\lambda} \ket{\nu}\bra{\lambda}\label{eq09}\\
\widehat{\mathscr X}_B &=& \sum_{a,b} \left(\widehat{\mathscr X}\bigl|_{tr
A}\right)_{ab} \ket{a}\bra{b}\label{eq10}
\end{eqnarray}

3.Remarks:

The trace of an operator, written
as $P=\ket{\phi}\bra{\psi}$, is easily computed. Let $\{\ket{n}\}$ be
an orthonormal basis. Then
\begin{eqnarray*}
Tr(P) &=& \sum_n\average{n}{P}
= \sum_n \innerproduct{n}{\phi}\innerproduct{\psi}{n}\\
&=&\sum_n \innerproduct{\psi}{n}\innerproduct{n}{\phi}\\
\therefore~Tr(\ket{\phi}\bra{\psi})&=& \innerproduct{\psi}{\phi}
\end{eqnarray*}
In general, for an operator, $P$, having a form
\begin{eqnarray*}
P &=& \sum_{i,j} p_{ij}\ket{\psi_i}\bra{\phi_j}\\
Tr(P)&=& \sum_{i,j} p_{ij} \innerproduct{\phi_j}{\psi_i}\,.
\end{eqnarray*}