Suppose we have quantum register of $n$ qubits in a pure state

$\Psi=\sum_i\alpha_i\left|i\right>= \alpha_0\left|00...0\right> + \alpha_1\left|00...1\right> + ... + \alpha_{2^n-1}\left|11...1\right>$

Now regard single qubit $i$. In general case it is entangled with other qubits and hence it is not in a pure state. It is in mixed state. Then what is density matrix


of it's mixed state? I guess, this matrix should be $2\times2$ right? And it should be possible to compute entire $\Psi$ from it? How to do this?


1 Answer 1


The density matrix of a single qubit can be obtained from the density matriz of the whole system through the partial trace operation:

\begin{align} \rho_i = \text{tr}_{\neq i}(\rho). \end{align}

In order to do this operation, you must first find a basis $\{\vert \alpha\rangle\}$ for the Hilbert space without the qubit. Then the partial trace can be written as \begin{align} \text{tr}_{\neq i}(\rho) = \sum_{\{\vert \alpha\rangle\}} \langle \alpha\vert \rho \vert \alpha \rangle \end{align}

To calculate this, you should just notice that every ket or bra of the basis of the full Hilbert space can be written as a tensor product of a ket (bra) of the qubit and a ket (bra) of the rest of the system: $\vert n_1, n_2... n_i ... n_N \rangle = \vert n_1... n_{i-1}, n_{i+1},...n_N\rangle \otimes \vert n_i \rangle $.

After doing this you obtain $\rho_i$, which is a $2\times2$ matrix, as you said.

As for your final question, no, in general you cannot obtain $\vert\psi\rangle$ from the $\rho_i$s, because the set of these matrices contains less information than the full $\rho$. We can see this from the following example:

Consider the pure states

\begin{align} \vert + \rangle &= \frac{1}{\sqrt{2}} \left(\vert \uparrow \downarrow \rangle + \vert \downarrow \uparrow \rangle \right) \\ \vert - \rangle &= \frac{1}{\sqrt{2}} \left( \vert \uparrow \downarrow \rangle - \vert \downarrow \uparrow \rangle \right). \end{align}

The density matrices which describes these pure states are then \begin{align} \rho^{\pm} = \vert \pm \rangle \langle \pm \vert \end{align}

The reduced density matrices are given by

\begin{align} \rho^{\pm}_1 &= {}_2\langle \uparrow \vert \rho^{\pm} \vert \uparrow \rangle_{2} + {}_2\langle \downarrow \vert \rho^{\pm} \vert \downarrow \rangle_{2} \\ &= \frac{1}{2} \left( (\pm \vert \downarrow \rangle_{1}) (\pm {}_{1} \langle \downarrow \vert) + (\vert \uparrow \rangle_{1}) ({}_{1} \langle \uparrow \vert) \right) \\ &= \frac{1}{2} \left( \vert \downarrow \rangle \langle \downarrow \vert_{1} + \vert \uparrow \rangle \langle \uparrow \vert_{1} \right) \\ \end{align}


\begin{align} \rho^{\pm}_2 &= {}_1\langle \uparrow \vert \rho^{\pm} \vert \uparrow \rangle_{1} + {}_1\langle \downarrow \vert \rho^{\pm} \vert \downarrow \rangle_{1} \\ &= \frac{1}{2} \left( (\vert \downarrow \rangle_{2}) ({}_{2} \langle \downarrow \vert) + ( \pm\vert \uparrow \rangle_{2}) ( \pm{}_{2} \langle \uparrow \vert) \right) \\ &= \frac{1}{2} \left( \vert \downarrow \rangle \langle \downarrow \vert_{2} + \vert \uparrow \rangle \langle \uparrow \vert_{2} \right) \\ \end{align}.

So you can see that although $\rho^{+}$ and $\rho^{-}$ represent distinct bipartite states, their reduced density matrices are the same. This means that with only the reduced density matrices we cannot tell which bigger state they came from, $\rho^{+}$ or $\rho^{-}$.

  • $\begingroup$ I meant compute $\psi$ from ALL $\rho_i$ (known for each $i$). $\endgroup$
    – Dims
    Jun 15, 2021 at 11:43
  • 1
    $\begingroup$ I see. Even so, the set of ALL $\rho_i$s has less information than the full $\rho$ because by taking the partial trace you are throwing information away on the coherences between the partitions. I will edit the answer and elaborate on this. $\endgroup$ Jun 15, 2021 at 17:19

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