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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 ?

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

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  • $\begingroup$ Thank you for your answer ! I still feel a bit confused over some points though. Would you care to elaborate a bit ? I'm thinking of the computation, I'm not sure why we can "switch" terms in the product, between the first and the second equality of your example. Also if you have any element of answer on the justification of this shortcut notation it would be great. At least it makes sense "dimensionally" to me now but I can't see why anyone would think using this shortcut is a good idea. $\endgroup$ – nathan raynal Mar 12 at 16:45
  • $\begingroup$ 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 $H$. $\endgroup$ – Danylo Y Mar 12 at 17:11
  • $\begingroup$ Thanks a lot it's much clearer now. Can you add it to your answer so I can mark it as accepted ? About the Wikipedia article, is this shortcut used enough to not mention it or should it be mentioned explicitly somewhere, in your opinion ? I'm thinking about suggesting a change on the talk page about it, would be great for non-expert like me. $\endgroup$ – nathan raynal Mar 12 at 17:44
  • $\begingroup$ Wikipedia is not supposed to be a learning material, it's a catalogue of facts that give overview of the topic. Studying through Wikipedia is not a great way of learning. Courses and books on QM explain formulas in details and give a structural and complete presentation of the material. $\endgroup$ – Danylo Y Mar 13 at 8:55
  • $\begingroup$ ok thank you for your insights on the matter $\endgroup$ – nathan raynal Mar 13 at 12:40

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