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In quantum mechanics $A$ and $B$ can be simultaneous measured if mathematically $\hat{A}\hat{B}=\hat{B}\hat{A}$. But how do we actually measure thing simultaneously. $\hat{A}\hat{B}$ is not simultaneous measurement because because here we measure $B$ and then $A$. So the above mathematical equation only tells us that $\hat{A}\hat{B}$ ($B$ first, $A$ later) = $\hat{B}\hat{A}$ ( $A$ first, $B$ later) - how does this relate to simultaneous measurement?

Manish Kumar Singh
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    I don't understand your question. You can't measure observables that don't commute "simultaneously", only in succession, and then it's your choice in which order ($XP$ or $PX$) you do that. What are you trying to ask? – ACuriousMind Dec 26 '15 at 14:43
  • say A and B are commutable, then how do I measure it simultaneously, because the only measurement I can do is AB or BA which are not simultaneous but comes one after another. – Manish Kumar Singh Dec 26 '15 at 14:47
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    ...if they commute, $AB = BA$, so it doesn't matter. – ACuriousMind Dec 26 '15 at 15:34
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    First, you say we can't do simultaneous measurement. Then you ask how to do it. Didn't you already give the answer? You can't. – Peter Shor Dec 26 '15 at 19:12
  • @PeterShor my question was if you are able to do simultaneous measurement, how do you do it, if its AB then its not simultaneous as B comes before A. – Manish Kumar Singh Dec 26 '15 at 19:16
  • That one can't measure non-commuting variables simultaneously is simply not true. In experiments one actually can't do anything else. Nobody can buy equipment with infinitely narrow slits for position and infinitely large crystals for momentum measurements. In effect we are always making a mixed measurement... it just doesn't matter within the experimental error that we are trying to achieve! – CuriousOne Dec 26 '15 at 21:38
  • Is your question how to actually measure commuting observables simultaneously? It might help if you reword your question to be more precise, otherwise your question might get closed. – Norbert Schuch Dec 27 '15 at 01:10
  • I can answer your question with one phrase: "the limit exist" ;). I mean, if you measure A and then B, in the limit of zero interval of time is the same as measure B and then A. In other words, the limit of simultaneous measurement if A and B exist because in this limit I do not need to recorded when is measured first! – Nogueira Dec 27 '15 at 08:19
  • @AcuriousMind measure of A with some width is morally another measure because is another preparation, given you different states as outcomes. So, wavepackets is prepared by a different measure then x o p, but something f(x,p). – Nogueira Dec 27 '15 at 08:26

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simultaneous measurement doesn't mean that commuting operators are measured simultaneously (in time, as you asked in question) and not one by one. It means that when the two operators commute, then they are diagonal in the same ket basis, which are referred to as simultaneous eigenkets, and you can now measure them one by one without disturbing the collapsed state vector. The fact that they are diagonal in the basis of same eigenkets allows them to be measured simultaneously without disturbing the state vector. Now coming to your example of two commuting operators, it does not matter in which order you measure the operators, since the two operators commute with each other. You can measure them like $ XY$ or $YX$.

And simultaneous measurement is not prohibited by quantum mechanics. It is prohibited for those operators that don't obey commutation relations and thus are not diagonal in the same ket basis, and therefore the state vector is disturbed each time they are measured. A famous example is that of position and momentum, they have a non-zero commutator and thus cannot be measured simultaneously.

Bruce Lee
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    You can also actually measure commuting observables simultaneously, e.g., by doing a projective measurement in a joint eigenbasis of both operators. – Norbert Schuch Dec 27 '15 at 01:11
  • i was clarifying the apparent misunderstanding raised in the question when i said "simultaneous measurement doesn't mean that commuting operators are measured simultaneously". it seemed that the guy was confusing simultaneity in time for simultaneous measurements as deduced from the comments section. – Bruce Lee Dec 27 '15 at 01:31