Timeline for How is a quantum superposition different from a mixed state?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 4, 2020 at 21:51 | comment | added | Andrew | Actually in the above comment I am intentionally simplifying things by saying $a(A\ {\rm OR}\ B)=a(A)+a(B)$. Really one should ask for the probability that a given observable will have outcomes $A$ or $B$ given that the state is $\psi$. Then $p(A\ {\rm OR}\ B)=|\langle B | \psi \rangle + \langle A | \psi \rangle|^2 = |a(B) + a(A)|^2 = |a(A)|^2 + |a(B)|^2 + 2{\rm Re}[a(A)^\star a(B)]$. The term $2{\rm Re}[a(A)^\star a(B)]$ is the interference term that arises for quantum amplitudes and not probabilities. This term allows $p(A\ {\rm OR}\ B)=0$ even if $p(A)$ and $p(B)$ are not zero. | |
| Oct 4, 2020 at 21:47 | comment | added | Andrew | @anniemarieheart In normal probability theory, to obtain the probability of two distinct outcomes (A OR B), one adds probabilities $p(A\ {\rm OR} B)=p(A)+p(B)$. Since probabilities are always nonnegative, $p(A\ {\rm OR}\ B) \geq 0$, and is only 0 if $p(A)=p(B)=0$. In quantum mechanics, one adds probability amplitudes for distinct final states. So then the amplitude to obtain $A$ or $B$, $a(A\ {\rm OR}\ B) = a(A)+a(B)$. Since amplitudes are in general complex, the amplitude for A OR B can be 0, even if $a(A)$ and $a(B)$ are nonzero. | |
| Oct 4, 2020 at 17:02 | comment | added | ann marie cœur | Can you clarify your: the difference between lies in the ability of quantum amplitudes to interfere, which you can measure by preparing many copies of the same state and then measuring incompatible observables. ---> what do you mean by the ability of quantum amplitudes to interfere say for pure vs mixed states? | |
| Jan 8, 2019 at 11:55 | comment | added | Wang Yun | J.J.Sakurai gave a good explanation of the mixed ensemble and pure ensemble. He said, " A pure ensemble is by definition a collection of physical systems such that every member is characterized by the smae ket $|\alpha \rangle$. In contrast, in a mixed ensemble, a fraction of the members with relative population $\omega_1$ are characterized by $|\alpha^{(1)}\rangle$; some other fraction with relative population $\omega_2$, by $|\alpha^{(2)}\rangle$; and so on." | |
| Oct 3, 2016 at 19:33 | comment | added | Alex | @Andrew How would you present you mixed state and density matrix in your example? | |
| Nov 19, 2015 at 8:46 | comment | added | user929304 | Wonderful answer, thanks. 2 subquestions if I may: i)when you said that the non-commutativity of the two operators leads to one being in a superposition when written in the basis of the other and vice versa, is it because otherwise both would be in eigenstates and thus contradict the compatibility? ii) why was it necessary to choose incompatible operators for the example you wanted to show? By the way why do we still call it a collapse when a mixed state is measured? It is weird because we already know it is in an eigenstate but we just dont know which one, right? | |
| S Jan 20, 2015 at 9:54 | history | suggested | Mehrdad | CC BY-SA 3.0 |
corrected some typos
|
| Jan 20, 2015 at 9:07 | review | Suggested edits | |||
| S Jan 20, 2015 at 9:54 | |||||
| Oct 12, 2013 at 14:41 | comment | added | Andrew | I'm not entirely sure what you mean. Given a state, mixed or pure, you can compute the probability distribution $P(\lambda_n)$ for measuring eigenvalues $\lambda_n$, for any observable you want. The difference is the way you combine probabilities, in a quantum superposition you have complex numbers that can interfere. In a classical probability distribution things only add positively. | |
| Oct 12, 2013 at 12:21 | vote | accept | Ruslan | ||
| Oct 12, 2013 at 8:02 | comment | added | Ruslan | So basically, in measurement of pure state I'd get probability density for sum of eigenstates, while for mixed state I'd just get sum of their probability densities, right? | |
| Oct 12, 2013 at 1:42 | history | answered | Andrew | CC BY-SA 3.0 |