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Let's say I have a ket which is a momentum eigenket $| p \rangle$ and then I measure the position and obtain $|x' \rangle$.

$$ | p \rangle = \int | x \rangle \langle x | p \rangle dx \to | x' \rangle $$

My question is what is the minimum number of ancilla qubits required to simulate this transformation unitarily?

Note: Since the cardinality of kets involved here is $\aleph_1$ I am unaware how to implement this

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  • $\begingroup$ do you think this is a measurement that you could actually, physically, do? $\endgroup$
    – DaftWullie
    Dec 6, 2019 at 12:08
  • $\begingroup$ Sorry do mean measure the momentum and then position? $\endgroup$
    – More Anonymous
    Dec 6, 2019 at 12:10
  • $\begingroup$ No, I mean do you think it's possible to exactly perform a position measurement? $\endgroup$
    – DaftWullie
    Dec 6, 2019 at 12:10
  • $\begingroup$ Ah well I am aware in relativistic QM the dirac Delta function is a guassian ... If that's where this discussion is headed? $\endgroup$
    – More Anonymous
    Dec 6, 2019 at 12:12
  • $\begingroup$ Regarding the cardinality of the kets...check Moretti's answer. $\endgroup$
    – Sanchayan Dutta
    Dec 7, 2019 at 1:52

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While talking about knowing the position exactly is a nice theoretical ideal, in practice, you cannot do that. You'll really be asking: "In which 'bin' of width $\delta x$ where $x$ spans from $x_{\min}$ to $x_{\max}$ is the particle confined to?". This means that there's $(x_{\max}-x_{\min})/\delta x$ bins, and so you basically need $$ \log_2\left((x_{\max}-x_{\min})/\delta x\right) $$ qubits to represent that information. Hence, this is the number of ancillas you would need.

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  • $\begingroup$ Does this mean the cardinality of the Hamiltonian (with the ancilla qubits) is aleph 1 ? (Sorry on phone) $\endgroup$
    – More Anonymous
    Dec 6, 2019 at 12:21
  • $\begingroup$ Or is it aleph 2 ? $\endgroup$
    – More Anonymous
    Dec 6, 2019 at 12:22
  • $\begingroup$ I'd add that exactly measuring a particle's position is nonsensical even in theory, as a single $x\in \Bbb R$ has measure zero. Intuitively, that's zero probably for finding a particle at any specific location. $\endgroup$
    – Sanchayan Dutta
    Dec 7, 2019 at 2:08
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    $\begingroup$ @MoreAnonymous: 1. You're discretizing the space and performing the simulation with qubits; infinite cardinalities aren't relevant here (it's finite-dimensional after the discretization!). 2. Even in the infinite-dimensional case, the exact cardinality type is irrelevant to the physics at hand (Motl had written a blog post precisely on this topic; also this). $\endgroup$
    – Sanchayan Dutta
    Dec 7, 2019 at 2:27
  • $\begingroup$ @MoreAnonymous: [cont.] 3. Regardless, if you're still interested in the cardinality from a purely theoretical perspective, it's $\aleph_0$ for $L^2(\Bbb R)$ (cf. this answer). However, in this case, DW is performing the simulation using a finite number of basis states. It should be easy to calculate the dimension of the Hamiltonian matrix given that you know the number of qubits exactly. $\endgroup$
    – Sanchayan Dutta
    Dec 7, 2019 at 2:43

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