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So here's something that's been bothering me. Given the time evolution of the wavefunction can only be unitary or discontinuous as a process of the measurement. So let the observables for our Hamiltonian be position $\hat x$, momentum $\hat p$ and energy $\hat H$. Does this mean the only possible states I can prepare in the lab are:

$$|\phi_1 \rangle = |E_0 \rangle $$

or

$$|\phi_2 \rangle = U|x_0 \rangle $$

or

$$|\phi_3 \rangle = U|p_0 \rangle $$

where $U$ is the unitary operator at arbitrary $t$, $| x_0\rangle$ is an arbitrary position operator, $|p_0 \rangle$ is an arbitrary momentum eigenket and $|E_0 \rangle$ is an arbitrary energy eigenket ? Am I correct in interpreting this means there are only certain quantum computations which can be performed with such a system (in the sense there is limited initial data one might input)?


Cross-posted on quantumcomputing.SE

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  • $\begingroup$ I would like to point out that in a quantum computer in the abstract canonical sense we usually don't have position, momentum, and energy. Typically we would work with qubits that can take two distinct states, and unitary operations as well as measurements. $\endgroup$
    – noah
    Jun 20, 2021 at 19:27
  • $\begingroup$ @noah I think one can make qubits out of these (energy, position and momentum) $\endgroup$ Jun 20, 2021 at 19:41
  • $\begingroup$ How many qubits would you want to make from this? I would be surprised if this resulted in a universal gate set. $\endgroup$
    – noah
    Jun 20, 2021 at 20:30
  • $\begingroup$ could you please link together cross-posts? $\endgroup$
    – glS
    Jun 23, 2021 at 8:33

3 Answers 3

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Energy, momentum and position are not the only things you can measure in a lab. Any other observable is, in principle, measurable (e.g. the spin). These other observables can also be used to set the initial state of the particle.

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  • $\begingroup$ I was asking in particular of "such a system" where the observables are position energy and momentum $\endgroup$ Jun 20, 2021 at 19:42
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The term "observables" in quantum theory usually means anything which could in principle be observed, and this corresponds to physical quantities represented by Hermitian operators. However I think the question has been asked using the term "observables" in a more restricted sense, to mean "I only have the means to measure in either the $x$ basis or the $p$ basis".

Equally, the term "quantum computer" ordinarily means a machine which has enough flexibility that it can be made to simulate any Hamiltonian by stringing together logic gates (and can do this as efficiently, in the computer science sense, as other such machines). In this case the unitary operator $U$ mentioned in the question can be any unitary operator at all, and therefore any state can be prepared by first preparing $| x \rangle$ and then performing the logic gates corresponding to the desired $U$. (Of course we usually think of quantum computers acting on discrete not continuous basis states, but we can suppose some finite degree of approximation etc.)

If the question intends to imply that only one Hamiltonian is available, and therefore only one unitary at any given time, then it becomes the question as to whether this could be sufficient for universal quantum computation. I am not sure but I would guess the answer could be yes if the Hamiltonian describes some sort of universal-Turing-machine-like device.

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I’m not aware of any proposals to use a particle’s position and momentum as the active degrees of freedom in a quantum computer, although I suppose it’s probably theoretically possible. I think that which states could be feasibly initialized would depend entirely in the details of the specific experimental proposal.

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