Scenario: ${\mathcal A}$ and ${\mathcal B}$ are two observables. Mathematically we model them by two Hermitian operators $A\colon H \to H$ and $B\colon H \to H$ on a separable Hilbert space. Physically they correspond to experiments $E_A$ and $E_B$, whose results are values in $Spec(A)$ and $Spec(B)$; repetitions produce value distributions on these spectra, expectation values, variances and higher momenta. The mathematical operator $A+B$ also is Hermitian. So let us look for an experiment which corresponds to this operator and let us study its expectation value in state $\varphi$.

Naive approach: Let us try pair-experiments. Assume we have a black box producing samples of state $\varphi$. Take a sample of the state, do experiment $E_A$ and get result $a$. Sample the state again, do experiment $E_B$ and get result $b$. Call the sum $a+b$ the result of the pair-experiment.

If $Spec(A) = \{ a_1, a_2 \}$ and $Spec(B) = \{b_1, b_2\}$ then the pair experiment has spectrum $\{ a_1 + b_1, a_1 + b_2, a_2 + b_1, a_2 + b_2 \}$. Obviously the pair-experiment has to be described in $H \otimes H$ and with a completely different observable. Details are straight forward, but we have no experiment for $A + B$. :-(

Second attempt: Let us assume that ${\mathcal A}$ and ${\mathcal B}$ are compatible and $A$ and $B$ commute. Then we can do the following: Sample the state once, on that sample do experiments $E_A$ and $E_B$ in whatever sequence, receive sequence independent values $a$ and $b$ and add them. Mathematically all is good. $A$ and $B$ share an eigenbasis, the spectrum of $A + B$ is the sum of the eigenvalues (belonging to the same shared eigenspace). Expectation values work out as expected. :-)

Now my question: $A + B$ still is a Hermitian operator, even if $A$ and $B$ do not commute. So I still am curious to which experiment this operator belongs to.

Note: In case of the product $A \cdot B$, the operator $A\cdot B$ is no longer Hermitian if the operators do not commute, and this makes it impossible for me to ask that question for the product. My question would break the preconditions of the formalism. But in $A + B$ the formalism allows to pose this question...

Update: In consequence of some comments I will try to specify my question more clearly: What is the physical meaning of the sum of two observables?

Obviously the "sum of two observables" is not the "sum of the values of the two observables". Assume that observable $A$ may have the values $2$ or $3$ and assume that observable $B$ may have the values $100$ or $200$ then the observable $A+B$ does not have the values $102$, $103$, $202$ or $203$ as a simple, naive approach might suggest or as an understanding of "sum of the values of the two observables" might suggest.

With this intuition failing, I would like to get an understanding of the physical meaning of $A + B$ starting from an understanding of $A$ and $B$.

Update 2: Adjusted the description of the pair experiment to a less misleading form.

Updaget 3: While I appreciate the hints given and while my "naive approach" and "second attempt" both are miserable, my question still is: When I am proceeding from $A$ and $B$ to $A+B$, what is the physical process or content of this mathematical operation?

  • 8
    $\begingroup$ This is a central part of QM; the core example is where $A$ is a particle's kinetic energy, and $B$ is its potential energy: both are trivial to diagonalize, but their sum rarely is. Basically all hard diagonalization problems boil down to sums of non-commuting operators where each individual term is easy or trivial to diagonalize. $\endgroup$ Aug 29, 2019 at 9:49
  • 1
    $\begingroup$ But that said, I really don't understand what the question really is here. $\endgroup$ Aug 29, 2019 at 9:54
  • $\begingroup$ SE posts are version controlled, so please do not make your post look like a revision table, instead just seamlessly integrate the new material into the post. There is an edit history button at the bottom of the post for those interested in seeing what changed. $\endgroup$
    – Kyle Kanos
    Sep 11, 2019 at 11:31
  • 1
    $\begingroup$ the eigenvalues of $A+B$ can be written as convex combinations of the eigenvalues of $A$ and $B$, with the weights being dependent on the relations between their eigenvectors. So in some sense, $A+B$ is a measurement whose outcomes are something between those of $A$ and $B$, although to say more one would have to specify the relations between the eigenvectors of $A$ and $B$. Would this count as "physical meaning" of measuring $A+B$? $\endgroup$
    – glS
    Sep 13, 2019 at 0:37
  • $\begingroup$ @gIS The hint on a convex combination is a useful mathematical remark, but it does not point us to an understanding of the physical aspect. Given an understanding of how $A$ and $B$ are measured in an experiment, how would I obtain an experiment for measuring $A + B$. So I really want to penetrate the physics aspect here, going beyond the mathematical side, and in particular for non-commuting $A$ and $B$. $\endgroup$ Sep 19, 2019 at 21:05

5 Answers 5


This is a still open issue about the foundation of quantum theories.

Usually it is only assumed that for every pair of bounded observables $A,B$ (selfadjoint operators on a complex separable Hilbert space) there is a third bounded observable $C$ whose expectation values are the sums of the expectation values of $A$ and $B$, for every given state $\psi$.

Since $C=A+B$ trivially satisfies this requirement and the states separate the observables, then $A+B$ is the wanted observable.

Here the problem becomes twofold.

From the physical side, a natural issue is

how to measure $A+B$?

In other words,

Q1 What is a measurement instrument for $A+B$ if we know the ones of $A$ and $B$?

From the mathematical side,

Q2 How can we construct the projection valued measure (PVM) of $A+B$ when knowing the ones of $A$ and $B$?

If $A$ and $B$ are compatible, the answers are elementary. Regarding the former issue, we can measure $A$ and $B$ on the same state and the outcome of $C$ is the sum of the outcomes of $A$ and $B$. (If we are instead interested in the post-measurement state, then the situation becomes much more complicated and there is no an definite answer.)

The answer to the second question, when the observables are compatible, it is easily found by taking advantage of the joint PVM of $A$ and $B$.

If $A$ and $B$ are not compatible, there is no definite answer especially to the former question. However, if $A$ and $B$ belong to a Lie algebras of generators of a group of symmetry of the considered physical system, then also $A+B$ is a generator of a symmetry (here we are dealing with unbounded observables defined on a common dense domain of essential self-adjointness). In that case $A+B$ has a concrete physical meaning and a measurement instrument should be suggested by physics.

Regarding the latter issue (to find the PVM of $A+B$ in terms of the PVM of $A$ and $B$ for incompatible observables), in collaboration with two colleagues, we have recently published a paper on it which also includes the case of unbounded non-compatible observables.

There is a procedure to construct the PVM of $aA+bB$, $f(aA+bB)$ (for a suitably interesting class of functions $f$) and also of some other operators constructed out of $A$ and $B$ as their Jordan product $\frac{1}{2}(AB+BA)$.

The final part of the paper furnishes some suggestions regarding the former issue.

N.Drago, S. Mazzucchi, V.Moretti: An operational construction of the sum of two non-commuting observables in quantum theory and related constructions Lett. Math. Phys, 110 (2020) 3197–3242 DOI: 10.1007/s11005-020-01332-7 arxiv.org/abs/1909.10974

The final, in my view quite suggestive, formula, with several hypotheses reads $$f(aA+bB)$$ $$ = \lim_{N\to + \infty} \int_{\mathbb{R}^{2N}} f\left(\frac{1}{N} \sum_{n=1}^N\left(a\lambda_n + b\mu_n\right) \right) dP^{(A)}_{\lambda_1} dP^{(B)}_{\mu_1} \cdots dP^{(A)}_{\lambda_N} dP^{(B)}_{\mu_N} $$ where $P^{(A)}$ and $P^{(B)}$ are the spectral measures (PVM) of $A$ and $B$ respectively.

In the finite dimensional case, the integral becomes a sum on the eigenvalues of $A$ and $B$ and the PVMs are made of projectors on the respective eigenspaces.

$$f(aA+bB)$$ $$ = \lim_{N\to + \infty} \sum_{\lambda_1,\ldots, \lambda_N \in \sigma(A)\:, \mu_1,\ldots, \mu_N \in \sigma(B)}f\left(\frac{1}{N} \sum_{n=1}^N\left(a\lambda_n + b\mu_n\right) \right) P^{(A)}_{\lambda_1} P^{(B)}_{\mu_1} \cdots P^{(A)}_{\lambda_N} P^{(B)}_{\mu_N} $$


Your question lacks focus. All I can do is to give you an example of a sum of non-commuting operators that make physical sense. The kinetic energy $p^2/2m$ and the electromagnetic interaction potential $V$ do not commute and together form the Schrödinger hamiltonian.

  • 1
    $\begingroup$ Isn't this only because energy is additive so it's a special case? Adding any other set of two operators generally isn't. $\endgroup$ May 28, 2022 at 11:54

If $Spec(A) = \{ a_1, a_2 \}$ and $Spec(B) = \{b_1, b_2\}$ then the pair experiment has spectrum $\{ a_1 + b_1, a_1 + b_2, a_2 + b_1, a_2 + b_2 \}$.

I don't think this is even formally correct. Take $A = s_z = \frac\hbar2\left(\begin{array}{}1&0\\0&-1\end{array}\right)$ and $B = s_x = \frac\hbar2\left(\begin{array}{}0&1\\1& 0\end{array}\right)$. Compute the spectrum of $A+B$ and verify that it doesn't satisfy your condition.

To answer the question in the context of this example: the physical meaning of $A+B$ is the spin measurement along the $\vec{n}_x + \vec{n}_z$ axis (multiplied by $\sqrt{2}$, to be pedantic)

  • $\begingroup$ My description of the pair experiment was misleading, so I adjusted it. The pair experiment does not correspond to the mathematical operator $A + B$. The physical meaning of the mathematical operator $A + B$ is just as you describe and it is not the sum of two values obtained from experiments $A$ and $B$ but rather a completely different experiment. How experiment $A+B$ is obtained in the general case from the individual experiments $A$ and $B$ is exactly my question. $\endgroup$ Sep 9, 2019 at 15:39

What you call "Naive approach" deals with two particules experiments (hence the espace $H\otimes H$) and what you describe is actually a measurement of $A+B$ as a shortcut for operator $A\otimes \mathbb{I}+\mathbb{I}\otimes B$, for the product state $\vert\phi\rangle\otimes \vert\phi\rangle$, where $\mathbb{I}$ is the identity of $H$. Its expectation value is $\langle \phi \vert A\vert\phi\rangle +\langle \phi \vert B\vert\phi\rangle$ ( using $\langle \phi \vert \mathbb{I}\vert\phi\rangle=1$) independently of the comparability or not of $A$ and $B$. As you have a product state, the measurement of $A$ on particle 1 can't influence the measurement of $B$ on particle 2. ($A\otimes \mathbb{I}$ and $\mathbb{I}\otimes B$ commute for any $A$ and $B$).

If you now have to measure the sum $A+B$ for one particle, the prediction will involve conditional probablities like $p(b|a)$ (assuming $b$ is one outcome of $B$ and $a$ the outcome is $A$ previously obtained) and these simplify only when $A$ and $B$ commute. This does not provide an expetimemt to measure $A+B$ but explain why the expected result is different.

Mathematically, the successive measurement of $A$ and $B$ has an expectation value $$ \langle \phi \vert A+B\vert\phi\rangle = \sum_a p(a) \Big(a+ \sum_b p(b|a) b\Big) $$ If $P_a$ denote the projector on eigenspace of $A$ with eigenvalue $a$, $p(a)= \langle \phi \vert P_a\vert\phi\rangle$ and $p(b|a)=\displaystyle\frac{\langle \phi \vert P_a P_b P_a\vert\phi\rangle}{\langle \phi \vert P_a\vert\phi\rangle}$

After expansion, the first term is simply $\langle \phi \vert A\vert\phi\rangle$ and the second, using $\sum_b b P_b =B$ is $$ \sum_{a,b}\langle \phi \vert P_a P_b P_a\vert\phi\rangle b = \langle\phi \vert \tilde B\vert\phi\rangle \quad\text{with}\quad \tilde B = \sum_a P_a B P_a $$ Finally : $$ \langle \phi \vert A+B\vert\phi\rangle = \langle\phi \vert A\vert\phi\rangle + \langle\phi \vert \tilde B\vert\phi\rangle $$ This reduces to the naive value in the special case where $\vert\phi\rangle$ is an eigen state of $A$, or more interesting for any $\vert\phi\rangle$ if $A$ and $B$ are compatible: then $P_a$ and $B$ commute, and $\tilde B = B$.

Nevertheless, in this scheme, the output state will be an eigen vector of $B$ and not of $A+B$...


If $A$ and $B$ commute they share eigenstates, and these are also the eigenstates of $(A+B)$: measuring "A+B" will give the same outcome as measuring "A" then "B" (or "B" then "A"). The operator $(A+B)$ corresponds to the observable you would expect: measuring the sum of "A" and "B" in a single measurement is the same as the sum of individual (back-to-back) measurements of "A" and then "B".

If $A$ and $B$ do not commute then the eigenstates of $(A+B)$ aren't normally shared with $A$ or $B$. The operator $(A+B)$ still corresponds to the observable $(A+B)$ but this is no longer the same thing as measuring $A$ and then $B$. If you have an "A" measuring machine and a "B" measuring machine, you need a brand new "A+B" measuring machine and the surprising fact about quantum mechanics is that the possible outcomes from that machine are not the same as the possible values you can form from adding the possible outcomes of the "A" machine to the possible outcomes of the "B" machine.

To help convince you why $(A+B)$ is the operator for a single measurement of "A+B" think about how $\frac{P^2}{2m} + V$ is the operator for a single measurement of total energy, and the operator is formed from the physical relationship we expect for kinetic and potential energy. But the possible outcomes of a total energy measurement are very different to the sum of a potential energy measurement followed by a kinetic energy measurement: measuring the potential energy (aka the position) will completely "mess with" your subsequent kinetic energy (aka momentum) measurement because they don't commute.

  • $\begingroup$ Yes. I agree and thank for the contribution. My question remains, however, and I shall reflect this in an update to the posting. $\endgroup$ Feb 20, 2022 at 11:02
  • $\begingroup$ @Nobody-Knows-I-am-a-Dog "My question remains" you never had an actual question in the first place $\endgroup$
    – Señor O
    Feb 20, 2022 at 11:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.