# The Physical Meaning behind a Commutator [duplicate]

I've just been introduced to the idea of commutators and I'm aware that it's not a trivial thing if two operators $A$ and $B$ commute, i.e. if two Hermitian operators commute then the eigenvalues of the two operators can be measured with certainty simultaneously.

But what is the physical significance when two operators do not commute such as to give a certain value? For example the position and momentum operator do not commute and give a value of $i\hbar$. What is the significance of the $i\hbar$?

• Possible duplicates: physics.stackexchange.com/q/9194/2451 and links therein. – Qmechanic Apr 7 '14 at 8:09
• I read that post before making this one. My topic is subtly different and not answered in that thread. – Ari Ben Canaan Apr 7 '14 at 14:31
• Tip: If a near-duplicate question exists, it is recommended to link to it in your question, and to spell out for the reader the difference between your question and the near-duplicate. Ideally, the difference should be clear already from the titles. – Qmechanic Apr 7 '14 at 17:25
• I think the difference is clear from both the text and the title. Not a duplicate (and also a better question). +1. – Emilio Pisanty Apr 8 '14 at 0:09

As you said if two operators commute they share eigenvectors. Physically this means that you can have a definite value for both. For example in the hydrogen atom the Hamiltonian $H$, which is the energy, and $J^2$, the magnitude of angular momentum, commute. A hydrogen atom can be in a state of definite energy and definite angular momentum. However, the position operator $x$ does not commute with $H$, so in a state of definite energy the electron doesn't have a well-defined position.
Then conversely the commutator measures the inability for two quantities to have definite values in the same state. More quantitatively, we have the general Heisenberg uncertainty principle $$\Delta A \Delta B \ge \frac{1}{2} |\langle [A,B] \rangle |$$ that is, the product of the uncertainties in $A$ and $B$ is at least half the (absolute value of the) expectation value of their commutator. By uncertainty we mean the usual standard deviation, $$\Delta A = \sqrt{\langle A^2\rangle - \langle A \rangle^2 }.$$
For the position operator $x$ and the momentum operator $p$, the commutator is just a scalar, $i\hbar$; its expectation value is always $i\hbar$. We thus get the most famous instance of the Heisenberg principle $$\Delta x\Delta p \ge \frac{\hbar}{2}.$$
Now you could ask why should we have $[p,x] = i\hbar$ of all things. Well, in the Hamiltonian formulation of classical mechanics there is an operation called the Poisson bracket, $\{F,G\}$. The Poisson bracket has the same algebraic properties as the commutator (they are both brackets in a Lie algebra) and satisfies $$\{ x_i, p_i \} = \begin{cases}1 & i = j \\ 0 & i \neq j\end{cases}.$$ So suppose that you know that whatever quantum mechanics is, quantum states are vectors and observables are operators, and you want to figure out how those operators should be related. Then it would be tempting to just try $$[x,p] \overset{?}{=} 1.$$ The problem is that $x$ and $p$ should be Hermitian (so that expectation values are always real). Then $[x,p]$ must be anti-Hermitian. But that's not a big problem, you can just multiply by $i$: $$[x,p] \overset{?}{=} i.$$ That's okay algebraically, but $x$ has units of length and $p$ has units of momentum, so we need to put a constant there to get the right units too: $$[x,p] \overset{?}{=} i\hbar.$$ I still put a question mark there because this is really just an educated guess, but experiments show that this is the correct commutation relation to use. (Well, you also have to measure $\hbar$ somehow.)
• for completeness. Commutator of Hermitian operators is anti-Hermitian: $[x,p]^\dagger=(xp - px)^\dagger = p^\dagger x^\dagger - x^\dagger p^\dagger = -[x,p]$ – innisfree Apr 7 '14 at 8:51