# Is there a recognised standard for typesetting quantum mechanical operators?

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Firstly, I wasn't sure exactly where to put this. It's a typesetting query but the scope is greater than $\TeX$; however it's specific also to physics and even more specific to this site.

I've recently been reading a style guide for scientific publications (based on ISO 31-11), however there was no mention of quantum mechanical operators. I've seen them written a few ways and was wondering if there was a decision handed down from "up above" that any particular way is best.

• $H$ -- I see this most commonly but I suspect it's mostly due to (mild) laziness to not distinguish it from a variable.
• $\hat{H}$ -- This is nicer to me because it makes the distinction between operator and variable. From what I understand of the ISO the italic means it's subject to change, which is true of the form of an operator, but not really its meaning? So I'm not totally sure if that's appropriate here.
• $\mathrm{H}$ -- Roman lettering is used for functions e.g. $\sin{x}$, $\mathrm{erf}(x)$, and even the differential operator (as in $\frac{\mathrm{d}}{\mathrm{d}x}$) so this seems to me like the most suitable category to put operators in.
• $\hat{\mathrm{H}}$ -- Probably the least ambiguous but may also be redundant.

Which would be the best to use? Am I being too pedantic?

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asked Mar 6, 2012
Welcome to this site and thanks for participating. Though I gave an answer below, I think this type of question typically suits better in [physics.SE](http://physics.stackexchange.com/), since its scope is rather general. It would suit better to this website if you had asked for guidance to write your, e.g., first paper on some specific field (quantum information, many-body systems, QFT, etc). Please, take some time to read the [faq](http://theoreticalphysics.stackexchange.com/faqhttp://theoreticalphysics.stackexchange.com/faq).

This post has been migrated from (A51.SE)
Just two random comments: (1) to what extent can you say that an operator is not a variable of your problem? This is actually [subjective](http://en.wikipedia.org/wiki/Heisenberg_picture). (2) quantum states can always be seen as [operators](http://en.wikipedia.org/wiki/Density_operator). At the end these things are a matter of taste.

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Yet another option, favourite in my immediate research envronment, is to use $\mathcal{H}$ for the Hamilton operator.

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As you see, there are different notations for quantum mechanical. Typically, even within a journal there is no one typesetting (style guides usually don't touch this topic).

Besides the ones you mentioned, sometimes people use:

• bold font (e.g. ${\mathbf H}$),
• small font for operators acting on subsystems.

Try looking at common notations used by your field. If there is no consensus, just use the one that you like the most.

However, what is important is if you use it consistently and if it is clear to the reader. There aren't many more irritating things than reading a paper when even types of symbols are not clearly stated and one needs to guess.

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answered Mar 7, 2012 by (1,260 points)
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My taste, never overload your notation unless its necessary.

Many people in quantum information try to avoid "hats" or further ornaments for operators that are just linear maps. Simple capital letters are fine to write Hamiltonians, channels, unitaries and measurements (italics are not really important, but its a de-facto standard). When people write many-body hamiltonians in terms of smaller k-body interactions, it is common that they use low-case letters for the latter (example, the Hubbard model). Also, mind that in finite dimensional systems linear maps are in one-to-one correspondence to matrices.

On the other hand, thinking of linear maps as matrices forces you to choose a basis. It might be more clear in some contexts to use a symbol-with-hat to denote an operator without mentioning the basis and the same symbol without hat for a matrix representation. However, I find that this practice can make your notation more complicated without earning much, since typically there is a natural default basis in every problem.

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answered Mar 6, 2012 by (285 points)

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