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My intuition tells me that the partial trace should be congruent under a change of basis. That is, if I have some matrix $A$ in the space of linear operators acting on a joint hilbert space: $A \in \mathbb{L}(\mathbb{H}_1 \otimes \mathbb{H}_2)$, then for every invertible matrix $U \in \mathbb{L}(\mathbb{H}_1 \otimes \mathbb{H}_2)$ there should exist some invertible matrix $G \in \mathbb{L}(\mathbb{H}_1)$ such that $G Tr_2(A) G^{-1}=Tr_2(U A U^{-1})$. The reason I believe it to be true is because the action of the operator should not change under a change of basis. Is this true? Is there a simple expression which relates $G$ to $U$? I have a physics education, so please forgive me if my notation is maybe too physicsy or assumes too much.

Thanks!

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I think you've got the quantifiers wrong -- certainly there exist such matrices, namely the respective identities -- I think you mean that for each $G$ there should exist $U$ and/or vice versa? –  joriki Mar 10 '13 at 19:07
    
You're right, that is what I mean to say. I'll edit it. –  Joel Klassen Mar 10 '13 at 19:08
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You demand that $U$ be invertible, but you use $U^\dagger$, not $U^{-1}$ -- is $U$ meant to be unitary? –  joriki Mar 10 '13 at 19:12
    
For my purposes I would like it to be unitary. But I suppose the spirit of the question doesn't require that it be unitary. I suppose the constraint of unitarity would follow from the relationship between $G$ and $U$. That is, if it were true that the partial trace is congruent under a change of basis, then is it the case that if $U$ were unitary, could we demand that $G$ also be unitary. –  Joel Klassen Mar 10 '13 at 19:24
    
But Tr$_2$ definitely depends on the basis, or rather it depends on the decomposition โ„$_1\otimes\,$โ„$_2$. You conjecture is certainly true for $U = U_1 \otimes U_2$, but I don't see why it should be true for general $U$ โˆˆ ๐•ƒ(โ„$_1\otimes\,$โ„$_2$) –  Peter Shor Mar 12 '13 at 1:08

1 Answer 1

up vote 2 down vote accepted

This is incorrect.

There is a $U$ in $\mathbb{L}(\mathbb{H}_1\otimes\mathbb{H}_2)$ which takes a tensor product state $e_1\otimes e_2$ to an entangled state $U(e_1\otimes e_2) = \frac{1}{\sqrt{2}} (e_1 \otimes e_2 + f_1 \otimes f_2)$. Now, let $A = e_1^\phantom{\dagger} e_1^\dagger \otimes e_2^\phantom{\dagger} e_2^\dagger.$ Then $G\,\mathrm{Tr}_2(A)G^{โˆ’1}$ has rank 1, but $\mathrm{Tr}_2(UAU^{โˆ’1})$ has rank 2.

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Thanks Peter. Some friends and I worked this out yesterday evening and I was planning on posting it today, but it seems you've beat me to it. –  Joel Klassen Mar 13 '13 at 11:01
    
So for me this reveals an uncomfortable feature of the partial trace. The domain of the partial trace is not only restricted to operators in $L(\mathbb{H}_1 \otimes \mathbb{H}_2)$, but those operators must also be EXPRESSED in such a way that their basis is separable in this decomposition. This seems unusual to me. Should this be making me as uncomfortable as it is? Are there any simpler examples of operator valued functions which have this restriction? –  Joel Klassen Mar 13 '13 at 12:29
    
@Joel: The partial trace is independent of which bases you pick for $\mathbb{H}_1$ and for $\mathbb{H}_2$, but not of the decomposition $\mathbb{H} = \mathbb{H}_1 \otimes \mathbb{H}_2$. Since it's a map from $\mathbb{L}(\mathbb{H})$ to $\mathbb{L}(\mathbb{H}_1)$, I don't see why this should feel uncomfortable. The partial trace depends on the tensor product structure of the system. If there's no tensor product structure, you shouldn't be taking partial traces. However, quantum computing, quantum information, and so on, are all very much related to the tensor product structure of the system. –  Peter Shor Mar 13 '13 at 22:51

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