I came across a webpage where they showed $[A,B]=0$ implies that we can measure it's corresponding eigenvalues simultaneously. I don't understand which step of the mathematical proof points to this possibility.

  • $\begingroup$ Should be same set of eigenvectors, not eigenvalues. $\endgroup$ – K_inverse Sep 10 '18 at 7:42

Suppose that $\hat A$ and $\hat B$ have the common eigenstates $\psi_{A_i,B_j}$, i.e. $$\hat A\psi_{A_i,B_j}=A_i\psi_{A_i,B_j}$$ $$\hat B\psi_{A_i,B_j}=B_j\psi_{A_i,B_j},$$ where $A_i$ and $B_j$ are the respective eigenvalues. From the above equations we have $$\hat B\hat A\psi_{A_i,B_j}=A_i\hat B\psi_{A_i,B_j}=A_iB_j\psi_{A_i,B_j}$$ $$\hat A\hat B\psi_{A_i,B_j}=B_j\hat A\psi_{A_i,B_j}=B_jA_i\psi_{A_i,B_j}=A_iB_j\psi_{A_i,B_j},$$ so subtracting these gives: $$[\hat A,\hat B]\psi_{A_i,B_j}=0.$$ This means that two operators with the same set of eigenstates must commute.

The above statement means that you can simultaneously measure the eigenvalues $A_i$ and $B_j$. That is, you can either first measure $\langle\hat A\rangle$ (and find $A_i$) and then measure $\langle \hat B\rangle$ (and find $B_j$) or vice versa. It doesn't matter which physical quantity you measure first.

Due to the crucial comment by @WillO, I will explain the reverse procedure.

Suppose that $[\hat A,\hat B]=0$, we have to show that they have the same eigenstates. Let $$\hat A\psi_{A_i}=A_i\psi_{A_i}\qquad \Rightarrow\qquad \hat B\hat A\psi_{A_i}=\hat B(A_i\psi_{A_i})=A_i\hat B\psi_{A_i}\equiv A_i\phi .$$ Now, due to the vanishing of the commutator we have that $$\hat B\hat A\psi_{A_i}=\hat A\hat B\psi_{A_i}=\hat A\phi$$ From the RHS of the last equations, we have that $$\hat A\phi=A_i\phi,$$ meaning that $\phi$ is also an eigenstate of $\hat A$ with eigenvalue $A_i$. This could happen for the following reasons:

  1. $\phi=c\psi_{A_i}$, with $c$ a constant. Hence, commuting operators have simultaneous eigenstates.
  2. $\phi\neq c\psi_{A_i}$. In this case the operator $\hat A$ must have degenerate eigenstates, namely $\phi$ and $\psi_{A_i}$. Even at this case, the non-degenerate eigenstates of $\hat A$ are simultaneously eigenstates of $\hat B$.
  • $\begingroup$ My query is focused on your statement, "The above statement means that you simultaneously measure the eigen values Ai and Bj". Can you justify this? $\endgroup$ – Sameer Dambal Sep 10 '18 at 8:00
  • $\begingroup$ @SameerDambal It means that you can measure $<\psi_{A_i,B_j}|\hat A|\psi_{A_i,B_j}>=A_i$ and $<\psi_{A_i,B_j}|\hat B|\psi_{A_i,B_j}>=B_j$ in any desired order. The first measurement will never affect the second, since both operators $\hat A,\hat B$ will act on an eigenstate of theirs in the above "sandwiches". $\endgroup$ – G K Sep 10 '18 at 8:05
  • $\begingroup$ Please mind the difference between <\phi|\psi> (wrong) and \langle \phi|\psi\rangle (right). $\endgroup$ – Emilio Pisanty Sep 10 '18 at 12:03
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    $\begingroup$ This appears to be the opposite of what the OP asked. You've shown that operators with the same eigenstates must commute, but the OP asked ((in essence) why operators that commute must have the same set of eigenstates. $\endgroup$ – WillO Sep 10 '18 at 12:46
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    $\begingroup$ @WillO Thanks for the observation! I edited my answer a bit. $\endgroup$ – G K Sep 10 '18 at 13:05

If two operators commute, they have simultaneous eigenfunctions, i.e. the same functions are eigenfunctions of both these functions.

If you additionally invoke the Cauchy-Schwarz inequality, and use it in conjunction with the variance formulation of physical quantities in QM, you can easily establish that the uncertainty product of two operators $\hat A$ and $\hat B$, obeys: $$( \sigma_A \ \sigma_B )^2 \geq \left( \frac{\langle[\hat{A},\hat{B}]\rangle}{2i} \right)^2$$

This is called the generalized Uncertainty Principle (See, e.g. Griffiths, QM 2e/d section 3.5 for a detailed derivation.)

(Or Pg 108, here, in an older version.)

The spirit behind that statement is the following: all non-commuting pairs have respective uncertainty principles defined for them, i.e. are not simultaneously determinable, while the ones which commute do not have any such uncertainty product applicable to them. Hence, the eigenvalues of $\hat A$ and $\hat B$, (with regard to their simultaneous eigenfunctions), can be "simultaneously" determined.


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