# What is the Reduced Density Matrix?

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?

• What exactly is your question, and how does it relate to reduced density matrices? – joshphysics Mar 21 '14 at 7:02
• sorry, added question – user_user Mar 21 '14 at 7:19
• I cannot understand your question. Please try to be clearer. – Valter Moretti Mar 21 '14 at 7:22
• How can i find density matrix of Bell state $\frac{|00\rangle+|11\rangle}{\sqrt{2}}$ ? – user_user Mar 21 '14 at 7:44
• Your $\rho$ is a pure state, not mixed, hence the trace of $\rho^2$ is 1. – Arnold Neumaier Mar 21 '14 at 16:33

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

Be $$A$$, $$B$$ finite dimensional Hilbert spaces and be $$T$$ $$\in$$ $$L(A \otimes B)$$ (Linear operators on $$A \otimes B$$), then the partial trace of T is defined as $$\rm{Tr}_B [T]$$ in $$L(A)$$ is defined by

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

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

Finally, note that the reduced matrix isn't the correct way of the describing a quantum state, is just a way to describe it as seen by looking just at a subsystem. This usually involves ignoring part of the information of the state and 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.

• 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. – alexvas Apr 8 '14 at 8:10
• 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. – giulio bullsaver Apr 8 '14 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.