# when I take two measure in same time, probability is product of probability?

let |φ› be a state, A the associate operator of "a" and B the associate operator of "b". I want to measure value "a" and then measure value "b", What is the probability? How is this related to the fact that [A, B] commute, if they commute? and if I measured "b" and then "a"? I'm little confused about interference terms and Bayes theorem.

Finally, I want to know if the probability of measuring "a" and then measuring "b" is given by | a | ² + | b | ² or | b || a | or something else

for example: I want to know the probability of a particle being x1

is this related to the fact that there are states in E:=|x›⊗|z› that are not represented by a tensor product?

...and in degenerate case? {i.e. [A,B] = 0, A v = λ1 v A w = λ1 w}

• sorry! I should have said "if I measured b and then a" – user150075 Jun 29 '18 at 13:34
• You can edit it. – Nathaniel Jul 1 '18 at 12:39
• Note that we have an equation editor built into the site which helps for readability of your post. – Kyle Kanos Jul 6 '18 at 10:17

First, let's consider your questions about commutation: If two operators commute then:

$$[\hat A, \hat B] = \hat A \hat B - \hat B \hat A = 0$$

This then means that: $\hat A \hat B = \hat B \hat A$.

Then this means that when we apply the operators to a wave function, the order we apply them does not change the result:

$$\hat A \hat B\ |{\psi} \rangle = \hat B \hat A \ |\psi \rangle$$

This is not a statement that is trivial in QM and does not hold for many operators.

If you do have two operators that commute however, that means that the order in which you measure them does not change the results.

This is because, if you wanted to measure one and then the other, you need to project the original wave function into the eigenspace of the operator.

For example, if I wanted to measure $\hat C$ and then $\hat D$ for arbitrary operators (where they may or may not commute) then I need to perform (ignoring normalization for brevity but you need to renormalize each intermediate state):

$$\hat D ( \hat C \ |\psi\rangle )$$

And similarly, if I want to measure $\hat D$ first I need to perform:

$$\hat C ( \hat D \ |\psi\rangle )$$

From the discussion above, the results are the same iff the operators commute.

• Applying the operator associated with an observable to a state is not the same as making a measurement of that observable on that state. The state $\hat{A}\hat{B}|\psi\rangle$ is not the state you get as a result of measuring $B$ followed by $A$. Hermitian operators do not (generally) represent things we can physically do to a state, they are abstract mathematical operations – By Symmetry Jun 29 '18 at 13:41

If $A$ and $B$ commute, then there exist a common basis of eigenstates for the operators $A$ and $B$ such that $A \, \lvert a_i \rangle=a_i\, \lvert a_i \rangle$ and $B \, \lvert b_i \rangle = b_i \, \lvert b_i \rangle$ and $\lvert a_i \rangle \otimes \lvert b_j \rangle \equiv \lvert a_i,b_j \rangle$ the commum basis. In order to find the probability of measuring $b_n$ and then $a_m$ out of $\lvert \psi\rangle$, you have to express $\lvert \psi \rangle$ as a linear combination of the eigenstates $\lvert a_i,b_j\rangle$. If $A$ and $B$ give discrete eigenvalues, then the most general form of $\lvert \psi \rangle$ is

$$\lvert \psi \rangle = \sum_{i,j} c_{ij}\lvert a_i,b_j \rangle,$$

with $\sum_{i,j} \lvert c_{ij} \lvert^2=1$ from the normalization condition. Since $[A,B] = 0$, it's possible to measure $a$ and $b$ simultaneously. Knowing that, the probability of finding $\lvert b_n \rangle$ then $\lvert a_m \rangle$ is given by the product $$P(b_n,a_m) = P_{\psi}(b_n)P_{b_n}(a_m) =\lvert \langle b_n\lvert \psi \rangle \langle a_m \lvert \psi \rangle \lvert^2 = \lvert\langle a_m,b_n \rvert \psi\rangle\rvert^2 =\, \rvert c_{mn} \rvert^2.$$

Notice that since $[A,B]=0$, the probability of measuring $a_i$ then $b_j$ simultaneously, or vice-versa is the same, $\forall \, i,j \in \Bbb{Z}$.

This happens because of the following: the resulting state vector $\lvert \psi_{ba} \rangle$ that is found calculating the probability of finding $b_n$ firstly, then $a_m$, is $$\lvert \psi_b \rangle = \sum_{i} \langle a_i,b_n \lvert \psi \rangle \lvert a_i,b_n \rangle = \sum_i c_{in} \lvert a_i,b_n \rangle,$$

now we project towards $\lvert a_m \rangle$ to find

$$\lvert \psi_{ba} \rangle = \sum_k \langle a_m,b_k\lvert \psi_b \rangle \lvert a_m,b_k \rangle = c_{mn} \lvert a_m,b_n \rangle.$$

Analogously, if we firstly calculate the projection of $\lvert \psi \rangle$ towards $\lvert a_m \rangle$ (resulting in $\lvert \psi_a \rangle$) and then the projection of $\lvert \psi_a \rangle$ towards $\lvert b_n \rangle$ (resulting in $\lvert \psi_{ab} \rangle$), we find

$$\lvert \psi_{a} \rangle = \sum_i \langle a_m,b_i\lvert \psi \rangle \lvert a_m,b_i \rangle = \sum_ic_{mi}\lvert a_m, b_i \rangle,$$

$$\lvert \psi_{ab} \rangle = \sum_k \langle a_k,b_n\lvert \psi_a \rangle \lvert a_k, b_n \rangle = c_{mn} \lvert a_m, b_n \rangle.$$

It is easily checked that $\lvert \psi_{ab} \rangle = \lvert \psi_{ba} \rangle$ when $[A,B]=0$, and thus the probability does not depend on the order of measurement.

• it is for non degenerate case. [A,B]=0 is not a sufficient hypothesis to ensure that operators have the same eigenstate. – user150075 Jul 1 '18 at 12:31