# Minimum number of ancilla qubits to unitarily simulate a measurement?

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

• do you think this is a measurement that you could actually, physically, do? – DaftWullie Dec 6 '19 at 12:08
• Sorry do mean measure the momentum and then position? – More Anonymous Dec 6 '19 at 12:10
• No, I mean do you think it's possible to exactly perform a position measurement? – DaftWullie Dec 6 '19 at 12:10
• Ah well I am aware in relativistic QM the dirac Delta function is a guassian ... If that's where this discussion is headed? – More Anonymous Dec 6 '19 at 12:12
• Regarding the cardinality of the kets...check Moretti's answer. – Sanchayan Dutta Dec 7 '19 at 1:52

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.
• 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 7 '19 at 2:08
• @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 7 '19 at 2:43