# What is the Physical Meaning of Commutation of Two Operators?

I understand the mathematics of commutation relations and anti-commutation relations, but what does it physically mean for an observable (self-adjoint operator) to commute with another observable (self-adjoint operator) in quantum mechanics?

E.g. an operator $A$ with the Hamiltonian $H$?

• When you say you understand the math, do you also understand the relation to representation theory and symmetries? – Marek Apr 27 '11 at 16:09
• It means the commutator is zero, i.e. AH = HA – Mark Eichenlaub Apr 27 '11 at 16:14
• Are you asking about the physical consequences and implications (rather than the mathematical definition, which you already know) when two operators commute? – Qmechanic Apr 27 '11 at 16:41
• Actually if two operators commute, there is a clear physical interpretation - they are both elements of a complete set of commuting observables - i.e. you can make a measurement on the system of both observables in any order and also simultaneously - e.g. $L$ - the total angular momentum and $L_z$ the $z$ component. If two operators do not commute any measurement made of one of these with a certain accuracy will result in an uncertainty of the expectation value of the second operator - e.g. position and momentum. – user346 Apr 27 '11 at 16:43
• @Qmechanic yes. – adustduke Apr 28 '11 at 10:18

Let us first restate the mathematical statement that two operators $\hat A$ and $\hat B$ commute with each other. It means that $$\hat A \hat B - \hat B \hat A = 0,$$ which you can rearrange to $$\hat A \hat B = \hat B \hat A.$$

If you recall that operators act on quantum mechanical states and give you a new state in return, then this means that with $\hat A$ and $\hat B$ commuting, the state you obtain from letting first $\hat A$ and then $\hat B$ act on some initial state is the same as if you let first $\hat B$ and then $\hat A$ act on that state: $$\hat A \hat B | \psi \rangle = \hat B \hat A | \psi \rangle.$$

This is not a trivial statement. Many operations, such as rotations around different axes, do not commute and hence the end-result depends on how you have ordered the operations.

So, what are the important implications? Recall that when you perform a quantum mechanical measurement, you will always measure an eigenvalue of your operator, and after the measurement your state is left in the corresponding eigenstate. The eigenstates to the operator are precisely those states for which there is no uncertainty in the measurement: You will always measure the eigenvalue, with probability $1$. An example are the energy-eigenstates. If you are in a state $|n\rangle$ with eigenenergy $E_n$, you know that $H|n\rangle = E_n |n \rangle$ and you will always measure this energy $E_n$.

Now what if we want to measure two different observables, $\hat A$ and $\hat B$? If we first measure $\hat A$, we know that the system is left in an eigenstate of $\hat A$. This might alter the measurement outcome of $\hat B$, so, in general, the order of your measurements is important. Not so with commuting variables! It is shown in every textbook that if $\hat A$ and $\hat B$ commute, then you can come up with a set of basis states $| a_n b_n\rangle$ that are eigenstates of both $\hat A$ and $\hat B$. If that is the case, then any state can be written as a linear combination of the form $$| \Psi \rangle = \sum_n \alpha_n | a_n b_n \rangle$$ where $|a_n b_n\rangle$ has $\hat A$-eigenvalue $a_n$ and $\hat B$-eigenvalue $b_n$. Now if you measure $\hat A$, you will get result $a_n$ with probability $|\alpha_n|^2$ (assuming no degeneracy; if eigenvalues are degenerate, the argument still remains true but just gets a bit cumbersome to write down). What if we measure $\hat B$ first? Then we get result $b_n$ with probability $|\alpha_n|^2$ and the system is left in the corresponding eigenstate $|a_n b_n \rangle$. If we now measure $\hat A$, we will always get result $a_n$. The overall probability of getting result $a_n$, therefore, is again $|\alpha_n|^2$. So it didn't matter that we measure $\hat B$ before, it did not change the outcome of the measurement for $\hat A$.

EDIT Now let me expand even a bit more. So far, we have talked about some operators $\hat A$ and $\hat B$. We now ask: What does it mean when some observable $\hat A$ commutes with the Hamiltonian $H$? First, we get all the result from above: There is a simultaneous eigenbasis of the energy-eigenstates and the eigenstates of $\hat A$. This can yield a tremendous simplification of the task of diagonalizing $H$. For example, the Hamiltonian of the hydrogen atom commutes with $\hat L$, the angular momentum operator, and with $\hat L_z$, its $z$-component. This tells you that you can classify the eigenstates by an angular- and magnetic quantum number $l$ and $m$, and you can diagonalize $H$ for each set of $l$ and $m$ independently. There are more examples of this.

Another consequence is that of time dependence. If your observable $\hat A$ has no explicit time dependency introduced in its definition, then if $\hat A$ commutes with $\hat H$, you immediately know that $\hat A$ is a constant of motion. This is due to the Ehrenfest Theorem $$\frac{d}{dt} \langle \hat A \rangle = \frac{-i}{\hbar} \langle [\hat A, \hat H] \rangle + \underbrace{\left\langle \frac{\partial \hat A}{\partial t} \right\rangle}_{=\;0\,\text{by assumption}}$$

• "The order of your measurements is in general important". I'm not sure how that can be the case. In relativity, the order of measurements may in general depend of the frame of reference. – user7348 Dec 20 '14 at 5:43
• @user7348 Not if the measurements are causally connected. – Sebastian Riese Dec 7 '15 at 17:06
• When two operators commute their observables are said to be compatible. Compatible observables do not obey uncertainty relationships, as far as I'm informed. This is equivalent to saying they are (in a way) independent (maybe linked through other inequalities or other boundary conditions of sorts). – Vendetta Sep 26 '17 at 17:34

Physically it means, it does not matter in which temporal order you measure the two commuting observables.

Answer to this question should start from why we want the physical observables to be represented by linear operators.

Theoretical physics is about constructing a mathematical model which we hope describes the phenomena it's being modeled for and hence helps predicting stuff. In classical physics this mathematical model is based simply on the real numbers (at least locally) because of the nice behaviour of things. In quantum mechanics it's not the case. Experiments started giving discrete values as well as continuous values(such as energy of electrons,etc.). So there is a need for some class of mathematical objects that give equal importance to continuous and discrete cases.

We know linear operators have the property that they posses both discrete and continuous spectra which can act as the required class of mathematical objects. Hence we start identifying physical observables with appropriate linear operators.

Postulate 1. To each dynamic variable there exists a linear operator such that possible values are the eigenvalues of the operator.

We need some place where all the physics happens and where these operators act to give us the required results. So we construct a Hilbert space consisting of states of the system which we are observing.

In quantum mechanics the measuring process plays an important role. It will alter the state of the system it's supposed to measure. If we are going to perform two experiments one after other then there is a possibility that some of the information is changed.

The commutator of two observables $A$ and $B$ with operators $\hat{A}$ and $\hat{B}$ is defined to be, $$[\hat{A},\hat{B}]=\hat{A}\hat{B}-\hat{B}\hat{A}$$

A commutator is a mathematical construct that tells us whether two operators commute or not. Suppose $A$ corresponds to a dynamic observable $A$ and $B$ corresponds to the dynamic observable $B$. Then the product $AB$ corresponds to measuring the observable $A$ after measuring $B$. If the measuring process is going to alter (disturb) the result of the next experiment in such a way that, measuring $A$ after measuring $B$ gives different values as measuring $B$ after measuring $A$ then we say they don't commute. Hence it means the commutator is not equal to zero. $$\hat{A}\hat{B}|\Psi\rangle\neq\hat{B}\hat{A}|\Psi\rangle$$ It is written in terms of commutator as, $$[\hat{A},\hat{B}]=\hat{A}\hat{B}-\hat{B}\hat{A}\neq 0$$

Otherwise it's zero. Which means that the two observables can be simultaneously measured. So a commutator tells us if we can measure two physical observables at the same time (which are called compatible observables) or not

If we know the value of the commutator then it tells how the measurements are going to alter things. It gives more information such as the uncertainty.

• Use $\$ $between the codes. – user36790 Nov 6 '15 at 12:52 It means you can (in principle) measure both quantities to arbitrary precision at the same time. If they didnt commute then this would be impossible by the uncertainty principle. • "Precision" depends on the state. There are QM states where commuting variables are still uncertain. – Vladimir Kalitvianski Apr 27 '11 at 16:35 You may consider commutativity of different variables as physical independence, something like separated independent variables:$\frac{\partial}{\partial x} \frac{\partial}{\partial y} = \frac{\partial}{\partial y} \frac{\partial}{\partial x}\$.

commuting operators are any two operators which can be applied to a function in any order without altering the outcome

• This is the mathematical explanation and offers little/no physical meaning that the original poster wanted. – Kyle Kanos Jul 8 '15 at 18:12

When two qm operators do not commute, it means that we are missing stuff in Nature. That is quantum mechanics is a theory of measurement but not of Nature because of non-commutation. Hence this means that the stuff we miss cannot be described by quantum mechanics, and this leads to the conclusion that qm is not a complete description of Nature.

## protected by Emilio PisantyJul 29 '15 at 16:34

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