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Is there a formalism for talking about diagonality/commutativity of operators with respect to an overcomplete basis?

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Consider a density matrix of a free particle in non-relativistic quantum mechanics. Nice, quasi-classical particles will be well-approximated by a wavepacket or a mixture of wavepackets. The coherent superposition of two wavepackets well-separated in phase space is decidedly non-classical.

Is there a formalism I can use to call this density matrix "approximately diagonal in the overcomplete basis of wavepackets"? (For the sake of argument, we can consider a specific class of wavepackets, e.g. of a fixed width $\sigma$ and instantaneously not spreading or contracting.) I am aware of the Wigner phase space representation, but I want something that I can use for other bases, and that I can use for operators that aren't density matrices e.g. observables. For instance: $X$, $P$, and $XP$ are all approximately diagonal in the basis of wavepackets, but $RXR^\dagger$ is not, where $R$ is the unitary operator which maps

$\vert x \rangle \to (\vert x \rangle + \mathrm{sign}(x) \vert - x \rangle) / \sqrt{2}$.

(This operator creates a Schrodinger's cat state by reflecting about $x=0$.)

For two different states $\vert a \rangle$ and $\vert b \rangle$ in the basis, we want to require an approximately diagonal operator $A$ to satisfy $\langle a \vert A \vert b \rangle \approx 0$, but we only want to do this if $\langle a \vert b \rangle \approx 0$. For $\langle a \vert b \rangle \approx 1$, we sensibly expect $\langle a \vert A \vert b \rangle$ to be proportional to a typical eigenvalue.

This post imported from StackExchange Physics at 2014-06-11 21:28 (UCT), posted by SE-user Jess Riedel
asked Nov 9, 2013 in Theoretical Physics by Jess Riedel (220 points) [ no revision ]
See also the related questions physics.stackexchange.com/q/83900 and physics.stackexchange.com/q/117058

This post imported from StackExchange Physics at 2014-06-11 21:28 (UCT), posted by SE-user Jess Riedel

Just wanted to say that I think this question is really important, thanks for asking.

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