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In an early linear algebra class of mine, I learnt that a linear map $\mathcal{A}$ acting on a vector space could be represented by a matrix $A_{ij}$ according to the rule:

$$\mathcal{A}({e_j}) = A_{ij} e_i \,,$$

where $e_i$ was some chosen basis. In quantum mechanics, the rule for getting the matrix elements of a particular linear operator is seemingly very different:

$$ A_{ij} = \langle i | \mathcal{A} | j \rangle $$

I was quite satisfied when I convinced myself that for orthonormal bases, these two definitions coincide. My question is then essentially this: is the above rule for quantum operators valid for orthonormal bases only? The problem I see, if it's valid for all bases, is that the two definitions do not coincide. So when I proved results in my linear algebra class like "the trace of a linear map is a meaningful object, since the trace of its matrix representation is basis-independent", this was using the first definition, and so then surely it won't be true if we use the second definition?

According to this Wikipedia page on density matrices, in the third equation on the page, it suggests that we can write the trace of an operator $\rho A$ like so

$$ \mathrm{tr}(\rho A) = \sum_n \langle u_n | \rho A |u_n \rangle = \sum_n (\rho A)_{nn} \qquad (\mathrm{second\ rule})$$

even if the basis $\{|u_n\rangle\}$ is not orthonormal. Will this not give a different trace to that evaluated using the first rule for the matrix elements of a linear map? Thank you.

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2 Answers 2

up vote 4 down vote accepted

In general, it is valid for orthonormal bases only. In the rest of this post I consider the finite dimensional case only, though everything can be extended to the infinite dimensional case under suitable hypotheses. In general you have the completely general relation, valid for every sort of basis: $$A^i\:_{j} = e^{*i}(Ae_j)$$ where $\{e^{*i}\}$ is the basis in the dual space $\cal H'$ associated with the basis $\{e_k\}$ initial one $\cal H$, completely defined by the requirements: $$e^{*i}(e_k) = \delta^i_{k}\:.$$ In this case: $$A = A^i\:_{j} $$ The notion of trace of a linear operator $A:\cal H \to \cal H$ (not a quadratic form!) does not need the existence of a scalar product to be defined. It is consequence of the notion of contraction of tensors. In components, you easily see that $tr(A) := \sum_i A^i\:_i$ does not depend on the choice of the basis.

In the presence of a scalar product, using orthonormal bases, all general formalism I have briefly introduced specializes to the standard one as you correctly wrote.

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Thanks --- so does this mean that the Wikipedia article I linked is incorrect? –  gj255 Mar 20 '14 at 13:40
I have no time to check, however if it states that the trace can be computed using the scalar product AND a basis which is not orthonormal, it is evidently wrong! –  Valter Moretti Mar 20 '14 at 17:41

I) In a finite-dimensional vector space $V$, the trace ${\rm Tr}A=\sum_{i}A^i{}_i$ of a linear map $A:V\to V$ can be evaluated wrt an arbitrary basis $(e_i)_{i=1,\ldots, n}$. This is because the trace is invariant under general similarity transformations.

II) Note In particular, that the concept of inner product, norm, ortogonality, etc, is not necessary for the definition of a trace.

III) There exist pertinent generalizations for infinite-dimensional vector spaces.

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