Positive SemiDefiniteness of partial trace N.3.2

Codebook
1

In the Trace it and Forget it notebook, there is an attempt to prove that the partial trace preserves positive semidefiniteness in the theoretic exercise n.3.2.

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In this step, it is not clear to me how the outer-most terms are reduced from a tensor product to a single vector: in the case of a system of two qubits, phi and I are respectively a vector of length two and a 2x2 matrix, that when tensored result in a 2x4 matrix (or 4x2), which cannot be multiplied by rho_A as it is a 2x2 matrix (since the 2x4 matrix is on the left side and the 4x2 on the right).

What is the mathematical property that allows us to reduce these terms to a single vector? I have looked everywhere but there is nothing that suggests this is a rigorous approach instead of intuitively saying that B is traced out and thus we ignore the term, which honestly does not satisfy me.

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Hi @Tester_Loller, welcome to the forum.
Yeah, I see your point. I am thinking on an alternative way to approach this proof.
Let me know what you think about this.
Start from \langle \varphi | \rho_A | \varphi \rangle, with | \varphi \rangle \in \mathcal{H}_A.
And then we substitute the definition of \rho_A in that expectation value.

\rho_A = \sum_i (I_A \otimes \langle i |) \rho_{AB} (I_A \otimes |i\rangle)
\langle \varphi | \sum_i (I_A \otimes \langle i |) \rho_{AB} (I_A \otimes |i\rangle) | \varphi \rangle

Then we can move the bra and ket inside the sum since the sum is only over the subsystem B.

\sum_i (\langle \varphi |I_A \otimes \langle i |) \rho_{AB} (I_A | \varphi \rangle \otimes |i\rangle)

and arrive to:

\sum_i (\langle \varphi |\otimes \langle i |) \rho_{AB} ( | \varphi \rangle \otimes |i\rangle)

Then we can define a bipartite state vector like the one they used in the Codebook | \psi_i \rangle=| \varphi \rangle \otimes |i\rangle and finalize by using the property that \rho_{AB} is positive semidefinite to conclude:

\langle \varphi | \rho_A | \varphi \rangle= \sum_i \langle \psi_i|\rho_{AB} | \psi_i \rangle \geq 0
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Hi Daniela, thank you for your reply.

What eludes me now is how the following identity is obtained:

\langle\phi |(I_A \otimes \langle i|)= \langle \phi |I_A \otimes \langle {i}|

and equivalently:

(I_A \otimes |i\rangle) |\phi \rangle= I_A |\phi\rangle \otimes |i\rangle

I know that the tensor product is distributive according to matrix multiplication, thus

(A \otimes B)(C \otimes D) = (AC \otimes BD)

When we are multiplying phi, I am assuming that an identity operation is performed on system B (as it is done in the original proof), for example in the ket case:

(I_A \otimes |i \rangle)(|\phi\rangle \otimes I_B)

But due to the distributive property this results in

| \phi \rangle \otimes |i\rangle I_B

where the second term is not multipliable due to the different dimensions. (this also happens in the bra case, and is also the reason why the explanation in the notebook does not seem to be valid).

While playing around I thought that I_B instead of being the identity matrix should be the scalar [1] which makes the dimensions coherent and the reduction from the tensor product of phi and I into a single vector evident, as |\phi\rangle \otimes 1 is still phi and by the distributive property |i\rangle \otimes 1 is still |i \rangle

However, this allows us to decompose |\phi\rangle into both |\phi\rangle \otimes 1 and 1 \otimes|\phi\rangle , where only the former is correct due to the same dimensionality issue, but even the latter is supposed to be correct as it is still the same vector phi, which is another issue.

Am I missing some fundamental property that allows use to commute I_B and |i\rangle?

Sorry for the long message but I can’t quite seem to clear this issue.

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Hi!

I think that in my proof I used what you mentioned. I am including a tensor product with the scalar 1: |\varphi\rangle \otimes 1. But that doesn’t mean that there is ambiguity, because there are two different vector spaces. I didn’t write this explicitly as a subscript, but I did mentioned that |\varphi\rangle was part of the vector space \mathcal{H}_A. So it wouldn’t be the same to say 1 \otimes|\phi\rangle_B and |\phi\rangle_A\otimes 1. The proof from the Codebook would be correct if we use this too. Thanks for bringing this up.
I wanted to highlight something else and it is that from the matrix perspective, the dimensions are correct in the alternative proof (the one in the previous message) at all stages (of course, the scalar 1 is implicit).

Let me know if this resolves your doubts and thanks again for the discussion.