# Computation of a reduced density matrix

On wikipedia, the article on quantum entanglement gives an example of the computation of a reduced density matrix. I would like to understand precisely what's going on with the computation.

First the context. We consider two systems A and B respectively belonging to Hilbert spaces $$\mathcal{H}_A$$ and $$\mathcal{H}_B$$. We consider the state of the composite system $$| \Psi\rangle \in \mathcal{H}_A \otimes\mathcal{H}_B$$.

We are interested by the reduced density matrix on the subsystem A, given by the following formula :

$$\rho_A = \sum_j \langle j |_B ( | \Psi\rangle \langle\Psi| ) |j\rangle_B$$

Then the article gives the following example . Consider the entangled state :

$$| \Psi \rangle =\frac{1}{\sqrt{2}} ( |0\rangle_A |1\rangle_B - |1\rangle_A |0\rangle_B )$$

the article says that the reduced density of A is then :

$$\rho_A = \frac{1}{2} (|0\rangle_A \langle 0|_A + |1\rangle_A \langle 1|_A)$$

I wanted to compute it myself using the definition above but I'm having trouble with it.

The term $$| \Psi\rangle \langle\Psi|$$ makes sense to me :

$$| \Psi\rangle \langle\Psi| =\frac{1}{2} ( |0\rangle_A |1\rangle_B \langle 0|_A \langle 1|_B - |0\rangle_A |1\rangle_B \langle 1|_A \langle 0|_B - |1\rangle_A |0\rangle_B \langle 0|_A \langle 1|_B + |1\rangle_A |0\rangle_B \langle 1|_A \langle 0|_B )$$

but then let's consider the first term of the sum in the definition of the reduced density matrix :

$$\langle 0|_B | \Psi\rangle \langle\Psi| |0\rangle_B$$

we then have term such as

$$\langle 0|_B |0\rangle_A$$

and I can't make sense of it. If $$\mathcal{H}_A = \mathcal{H}_B$$ it's the inner product on $$\mathcal{H}_A$$ but here I'm confused. Dropping the subscripts I have :

$$\langle 0| | \Psi\rangle \langle\Psi| |0\rangle + \langle 1| | \Psi\rangle \langle\Psi| |1\rangle = - \frac{1}{2} (|0\rangle \langle 0| + |1\rangle \langle 1|)$$

So here there is this minus sign that is not in the given answer of Wikipedia plus I "lost" the information on the states. Could you help me doing the computation correctly ?

1. $$\mathcal{H}_A \neq \mathcal{H}_B$$, they are two distinct physical systems (even if they have the same dimension).
2. That formula for a reduced density matrix is a shortcut for $$\rho_A = \sum_j \big(I_A \otimes \langle j |_B\big) \cdot | \Psi\rangle \langle\Psi| \cdot \big(I_A \otimes |j\rangle_B \big)$$ You can check the dimensions. If $$d_A = \text{dim}H_A, d_B = \text{dim}H_B$$, then $$\langle j |_B$$ has size $$1 \times d_B$$ and $$| \Psi\rangle$$ has size $$d_Ad_B \times 1$$. You can't multiply $$1 \times d_B$$ sized matrix on a $$d_Ad_B \times 1$$ sized matrix. But $$I_A \otimes \langle j |_B$$ has size $$d_A \times d_Ad_B$$, so it fits.
So, for example, you can compute $$\langle 0|_B \big(- |0\rangle_A |1\rangle_B \langle 1|_A \langle 0|_B \big) |0\rangle_B = \big(I_A \otimes\langle 0|_B \big) \cdot \big(- |0\rangle_A \langle 1|_A \otimes |1\rangle_B \langle 0|_B \big) \cdot \big(I_A \otimes |0\rangle_B\big) = - |0\rangle_A \langle 1|_A \otimes \langle 0|_B |1\rangle_B \langle 0|_B|0\rangle_B = - |0\rangle_A \langle 1|_A \otimes 0 = 0$$
• 1. It's a mixed product property: $(A \otimes B)(C \otimes D) = AC \otimes DB$, see en.wikipedia.org/wiki/… 2. This shortcut is somewhat natural. If, for example, we have a tripartite system $H = H_A \otimes H_B \otimes H_C$ and we state that some operator $O_B$ acts on $H_B$, then it is natural to consider its extension $O = I_A \otimes O_B \otimes I_C$ that acts on the whole system $H$. – Danylo Y Mar 12 at 17:11