2022
DOI: 10.3390/e24070877
|View full text |Cite
|
Sign up to set email alerts
|

Unitary Evolution and Elements of Reality in Consecutive Quantum Measurements

Abstract: Probabilities of the outcomes of consecutive quantum measurements can be obtained by construction probability amplitudes, thus implying the unitary evolution of the measured system, broken each time a measurement is made. In practice, the experimenter needs to know all past outcomes at the end of the experiment, and that requires the presence of probes carrying the corresponding records. With this in mind, we consider two different ways to extend the description of a quantum system beyond what is actually meas… Show more

Help me understand this report

Search citation statements

Order By: Relevance

Paper Sections

Select...
4

Citation Types

0
4
0

Year Published

2023
2023
2023
2023

Publication Types

Select...
1

Relationship

1
0

Authors

Journals

citations
Cited by 1 publication
(4 citation statements)
references
References 33 publications
0
4
0
Order By: Relevance
“…Or one can simply say that Alice needs to know matrix elements of a unitary operator ÛS+P(t2,t0)$\hat{U}^{S+P}(t_2,t_0)$ between particular states in the Hilbert space of the composite, and not mention the evolution at all. [ 27 ]…”
Section: Bob Wants To Measure Three Quantitiesmentioning
confidence: 99%
See 3 more Smart Citations
“…Or one can simply say that Alice needs to know matrix elements of a unitary operator ÛS+P(t2,t0)$\hat{U}^{S+P}(t_2,t_0)$ between particular states in the Hilbert space of the composite, and not mention the evolution at all. [ 27 ]…”
Section: Bob Wants To Measure Three Quantitiesmentioning
confidence: 99%
“…Furthermore, the probabilities in Equation (4) can be obtained without mentioning the probes explicitly, since the couplings (Equations (2) and (3)) are simple, and their actions are easily taken into account (details can be found, e.g., in ref. [27]). In particular, one has for the probability of measuring the sequence (Bm$B_m$, Cn$C_n$), provided the first detection gave Ak$A_k$, an equally valid form, Pfalse(CnBmAkfalse)badbreak=false|false⟨cnfalse|trueÛSfalse(t2,t1false)π̂false(Bmfalse)trueÛSfalse(t1,t0false)|ak|2$$\begin{eqnarray} P(C_n\leftarrow B_m\leftarrow A_k) =|{\langle }c_n| \hat{U}^S(t_2,t_1)\hat{\pi }(B_m)\hat{U}^S(t_1,t_0)|a_k{\rangle }|^2 \end{eqnarray}$$if the measured eigenvalue of Ĉ$\hat{C}$ is not degenerate.…”
Section: Bob Wants To Measure Three Quantitiesmentioning
confidence: 99%
See 2 more Smart Citations