Computation of a reduced density matrix

Question

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 ?

Answer 1

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 $$ \langle 0|_B \big(- |0\rangle_A |1\rangle_B \langle 1|_A \langle 0|_B \big) |0\rangle_B = \big(I_A \otimes\langle 0|_B \big) \cdot \big(- |0\rangle_A \langle 1|_A \otimes |1\rangle_B \langle 0|_B \big) \cdot \big(I_A \otimes |0\rangle_B\big) = - |0\rangle_A \langle 1|_A \otimes \langle 0|_B |1\rangle_B \langle 0|_B|0\rangle_B = - |0\rangle_A \langle 1|_A \otimes 0 = 0 $$