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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

glS
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More Anonymous
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1 Answers1

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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.

DaftWullie
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  • Does this mean the cardinality of the Hamiltonian (with the ancilla qubits) is aleph 1 ? (Sorry on phone) – More Anonymous Dec 06 '19 at 12:21
  • Or is it aleph 2 ? – More Anonymous Dec 06 '19 at 12:22
  • 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. – Sanchayan Dutta Dec 07 '19 at 02:08
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    @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). – Sanchayan Dutta Dec 07 '19 at 02:27
  • @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. – Sanchayan Dutta Dec 07 '19 at 02:43