Physicists define the trace of an operator $\rho$ as the follows,

$Tr(\rho)=\sum\limits_{|s\rangle \in B} \langle s| \rho |s\rangle$

where B is some orthonormal basis, and this quantity is basis independent.

If we swapped B with a non-orthogonal basis,C , which, if any, of the properties of the trace will be preserved? In particular,

  • 1) Is it now basis dependent? (The intuitive answer seems to be YES, but can we do better?)
  • 2) Under what conditions (on C and $\rho$) will this value exceed the value of the actual trace? I will settle for an answer assuming $\rho$ is a density operator .



The trace defined as you did in the initial equation in your question is well defined, i.e. independent from the basis when the basis is orthonormal. Otherwise that formula gives rise to a number which depends on the basis (if non-orthonormal) and does not has much interest in physics. If you want to use non-orthonormal bases, you should adopt a different definition involving the dual basis: if $\{\psi_n\}$ is a generic basis, its dual basis is defined as another basis $\{\phi_n\}$ with $\langle \phi_n|\psi_m\rangle = \delta_{nm}$. In this case, the trace of $\rho$, the same obtained with your formula for orthonormal basis (in that case $\phi_n=\psi_n$) is: $$tr \rho = \sum_{n} \langle \phi_n | \rho \psi_n \rangle\:.$$ Everything I wrote holds for finite dimension, otherwise some further requirements on operators have to hold true.

ADDENDUM. (I am still considering the finite dimensional case with dimension $N>1$.) Using your definition of trace, let us indicate it by $Tr_C(\rho)$, you can always find $\rho$ and a non-orthogonal basis $C$ with $Tr_C(\rho)> tr(\rho)$. As an example pick out $\rho = |\psi \rangle \langle \psi |$ with $||\psi||=1$, so that $\rho$ is a pure state. Next fix a cone $V$ tightly concentrated around $\psi$. As every cone always includes a normalized basis, for every $\epsilon \in (0,1)$, suitably concentrating $V$ around $\psi$ you can find a normalized basis $C:= \{\phi_n\}_{n=1,\ldots, N} \subset V$ with $||\psi -\phi_n||^2 > 1-\epsilon$. Consequently: $$Tr_C(\rho) = \sum_{n}|\langle \psi|\phi_n\rangle|^2 > N(1-\epsilon) > 1 = tr (\rho)\:.$$ This is particularly evident for $N=2$, using $\phi_1:= \psi$ and $\phi_2$ close to $\phi_1$.

  • $\begingroup$ Thanks this looks correct. However for my purpose I am actually trying to understand some properties about my "definition of the trace" (for lack of better way to say it). Namely I ended up with the expression I wrote, and am trying to figure out its properties. But thanks for the insight! $\endgroup$ – zzz Apr 9 '14 at 6:10
  • $\begingroup$ Are the vectors in your basis $C$ normalized? $\endgroup$ – Valter Moretti Apr 9 '14 at 6:20
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    $\begingroup$ Your trace $TR_C(\rho)$ can always be written as: $tr(\rho S_C )$, where $tr$ is the standard trace and $S_C$ is the positive (thus self-adjoint) operator defined as $S_C := \sum_{s\in C}| s\rangle \langle s|$. This allows you to handle your trace using standard formalism. For instance you can easily prove that $0\leq tr(S_C\rho) \leq \sqrt{tr (\rho^2)}\sqrt{tr(S_C^2)}$. If $S$ is strictly positive so that $S^{-1}$ exists, you can get something like $tr \rho \leq \sqrt{tr(\rho S_C^{-1})} \sqrt{tr(\rho S_C)}$... $\endgroup$ – Valter Moretti Apr 9 '14 at 7:03
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    $\begingroup$ I am just using the Cauchy-Schwarz inequality applied to the Hilbert-Schmidt scalar product and something like $tr \rho = tr (S_C^{-1/2}\rho^{1/2}\rho^{1/2} S_C^{1/2})$ arising from positivity of operators and the cyclic property of the trace... $\endgroup$ – Valter Moretti Apr 9 '14 at 7:08
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    $\begingroup$ Yes, that is precisely correct. $\endgroup$ – Emilio Pisanty Apr 9 '14 at 18:42

You can compute the trace of an endomorphism using any basis (including non-orthogonal ones).

In Dirac notation, you show this by inserting the identity expressed in the new basis and re-arranging:

$$\begin{align*} \sum\limits_{|s\rangle \in B} \langle s^*| \rho |s\rangle &= \sum\limits_{|s\rangle \in B} \langle s^*| \left( \sum\limits_{|t\rangle \in C} |t\rangle\langle t^*| \right) \rho |s\rangle \\&= \sum\limits_{|s\rangle \in B} \sum\limits_{|t\rangle \in C} \langle s^*|t\rangle\langle t^*| \rho |s\rangle \\&= \sum\limits_{|s\rangle \in B} \sum\limits_{|t\rangle \in C} \langle t^*| \rho |s\rangle \langle s^*|t\rangle \\&= \sum\limits_{|t\rangle \in C} \langle t^*| \rho \left( \sum\limits_{|s\rangle \in B} |s\rangle \langle s^*| \right) |t\rangle \\&= \sum\limits_{|t\rangle \in C} \langle t^*| \rho |t\rangle \end{align*}$$ Here, I used the non-standard notation $\langle s^*|,\langle t^*|$ to denote the algebraically dual basis, which only consists of the bras $\langle s|,\langle t|$ if we started with an orthonormal one.

  • $\begingroup$ Thanks! I tried to do this as well before posting, but it's not clear to me why the sum of the projection operators in an non-orthogonal basis would still be an resolution of the identity (the identity that you inserted in the first line). In fact for example, if you take C={(1,0),(1,1)}, these have the same span as {(1,0),(0,1)}, but the sum of projectors is not the identity. Am I missing something here? $\endgroup$ – zzz Apr 9 '14 at 1:47
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    $\begingroup$ $\sum\limits_{|t\rangle \in C} |t\rangle\langle t| \neq I$ in general if $C$ is not orthonormal. The trace computed as you wrote is independent from the basis if the basis is orthonormal. Otherwise is not well-defined. $\endgroup$ – Valter Moretti Apr 9 '14 at 5:31

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