The difference between pure and mixed states is the difference in their density matrix structure.

For density matrix $\rho$ of mixed state the trace of $\rho^{2}$ should be less than 1. For pure state corresponding trace $Tr(\rho^{2}) = 1$.

But when I tried to check the Bell two-qubit state, i got: $$ \rho = \frac{1}{2}\begin{pmatrix} 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1\end{pmatrix}$$ $$ \rho^{2} = \frac{1}{4}\begin{pmatrix} 2 & 0 & 0 & 2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 2 & 0 & 0 & 2 \end{pmatrix}$$ Trace of which is equal to 1. As I understand, reduced density matrix is the right describing of bell states. But my matrix is not reduced. Can you explain me how to find reduced matrix of bell state?

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    $\begingroup$ What exactly is your question, and how does it relate to reduced density matrices? $\endgroup$ Commented Mar 21, 2014 at 7:02
  • $\begingroup$ sorry, added question $\endgroup$
    – user_user
    Commented Mar 21, 2014 at 7:19
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    $\begingroup$ I cannot understand your question. Please try to be clearer. $\endgroup$ Commented Mar 21, 2014 at 7:22
  • $\begingroup$ How can i find density matrix of Bell state $\frac{|00\rangle+|11\rangle}{\sqrt{2}}$ ? $\endgroup$
    – user_user
    Commented Mar 21, 2014 at 7:44
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    $\begingroup$ Your $\rho$ is a pure state, not mixed, hence the trace of $\rho^2$ is 1. $\endgroup$ Commented Mar 21, 2014 at 16:33

3 Answers 3


The reduced matrix is defined as the partial trace of the density matrix.

Let $A$, $B$ be finite dimensional Hilbert spaces, and let there be a $T$ such that $T \in$ $L(A \otimes B)$ (i.e., $T$ is a linear operator on $A \otimes B$), then the partial trace of T, represented as $\rm{Tr}_B [T]$ in $L(A)$, is defined by:

\begin{equation} \langle a | \rm{Tr}_B [T]| b \rangle = \sum_n \langle a | \langle n | T| n\rangle | b \rangle \end{equation}

where $| n \rangle$ is an orthonormal basis in $B$, and $|a\rangle, |b\rangle$ are vectors in $A$.

Finally, note that the reduced matrix isn't the correct way of describing a quantum state, it is just a way to describe it as seen by looking only at a subsystem. This usually involves ignoring part of the information of the state; therefore, the reduced density matrix of a pure state may be a mixed state. This is spectacular for the Bell states, as their reduced matrix is $\rm{Id}/2$, the most disordered state.

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    $\begingroup$ Is there a reason you specified that $A$ and $B$ have to be finite dimensional? Where does the formalism break if you have infinite-dimensional spaces (for example, all possible energy states of the harmonic oscillator). Note: the trace might still converge. $\endgroup$
    – alexvas
    Commented Apr 8, 2014 at 8:10
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    $\begingroup$ Surely exists an extension of what I've wrote to non finite dimensional systems, but I've never studied deeply it so I preferred not to consider this situation. For instance you would not consider $L(A)$ (all the linear operators) but just the \emph{Trace class operators}, those for which the trace is well defined. If you want to study this situation I may suggest you the lectures of Walter Moretti. $\endgroup$ Commented Apr 8, 2014 at 12:35

The reduced density matrix can be found by taking the trace over the subspaces of the Hilbert space that represent systems you're not interested in. For the Bell state the density matrix of the whole system is $$\tfrac{1}{2}(|00\rangle+|11\rangle)(\langle 00|+\langle 11|)\\ = \tfrac{1}{2} (|00\rangle\langle 00|+|00\rangle\langle 11|+|11\rangle\langle 00|+|11\rangle\langle 11|)\\ \tfrac{1}{2}(|0\rangle\langle 0|\otimes|0\rangle\langle 0|+|0\rangle\langle 1|\otimes|0\rangle\langle 1|+|1\rangle\langle 0|\otimes|1\rangle\langle 0|+|1\rangle\langle 1|\otimes|1\rangle\langle 1|)$$.

So to get the reduced density matrix for the first qubit you take the trace over the Hilbert space for the second qubit. You take all the terms that have $|0\rangle\langle 0|$ or $|1\rangle\langle 1|$ for the second qubit, throw out the rest and then just take the parts of those terms that refer to the first qubit. This gives $$\rho_1 = \tfrac{1}{2}(|0\rangle\langle 0|+|1\rangle\langle 1|),$$ and $tr(\rho_1^2)<1$.

For more detail and rigour, you should read "Quantum Computation and Quantum Information" by Nielsen and Chuang.


The property you are referencing, $Tr(\rho^2)$, is called the purity of the density matrix [1]. Purity is a measure of how "mixed" a density matrix is. If the density matrix is not mixed and represents a pure state, then the purity will be unity. Since a Bell state is a pure state you are receiving $Tr(\rho^2) = 1$.

The property I believe you wish to be using is called concurrence [2]. The concurrence gives a measure of entanglement for a quantum state; A bell state has a concurrence of unity while a separable state has concurrence zero. Concurrence utilizes the reduced density matrix as you allude to. It is defined as,

$$ \text{Concurrence} = \sqrt{2(1-\text{Tr}(\tilde{\rho}^2)} $$

where $\tilde{\rho}$ is defined as the reduced density matrix. In a two qubit system, the reduced density matrix can be obtained by performing a partial trace over one of the qubits.

For example, suppose we construct a Bell state with two qubits, A and B.

$$ |\psi\rangle = \frac{|\uparrow\rangle_A|\downarrow\rangle_B+|\downarrow\rangle_A|\uparrow\rangle_B}{\sqrt{2}} $$

The density matrix for the both systems, $\rho = |\psi\rangle\langle \psi|$, is related to the reduced density matrix, $\tilde{\rho}$ for qubit B, by the relation,

$$ \tilde{\rho} = \text{Tr}_A(\rho) = (\langle \uparrow |_A \otimes \text{Id}_B) \ \rho\ (|\uparrow\rangle_A \otimes \text{Id}_B)+(\langle \downarrow |_A \otimes \text{Id}_B) \ \rho\ (|\downarrow\rangle_A \otimes \text{Id}_B) $$

where we have "traced out" system A. I apologize for the atrocious formatting. For our initial Bell state, this yields a reduced density matrix of,

$$ \tilde{\rho} = \frac{|\downarrow\rangle_B\langle\downarrow|_B+|\uparrow\rangle_B\langle\uparrow|_B}{2} $$

It is easy to see this yields a concurrence of unity.


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