Undergrad Question about No-Cloning Theorem and State Tomography

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SUMMARY

The no-cloning theorem asserts that it is impossible to create an exact copy of an arbitrary quantum state without prior knowledge of that state. However, quantum state tomography allows for the estimation of an unknown quantum state by preparing multiple identical systems in the same state, provided the preparation method is known. This process does not involve cloning the original state but rather relies on the ability to replicate the preparation method with high fidelity. Quantum teleportation exemplifies a scenario where the original state is destroyed to create a copy, involving classical information transfer.

PREREQUISITES
  • Understanding of the no-cloning theorem in quantum mechanics
  • Familiarity with quantum state tomography techniques
  • Knowledge of quantum teleportation principles
  • Basic concepts of quantum state preparation and measurement
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  • Research quantum state tomography methodologies and their applications
  • Study the implications of the no-cloning theorem on quantum computing
  • Explore quantum teleportation and its role in quantum information theory
  • Investigate methods for achieving high-fidelity quantum state preparation
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Quantum physicists, researchers in quantum information science, and students studying advanced quantum mechanics concepts will benefit from this discussion.

Irishdoug
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Question about No-Cloning Theorem and State Tomography
Hi,

I have a question, or am looking for clarification, about the no-cloning theorem and state tomography. My understanding is that the theorem states one cannot make an exact copy of a quantum state. I was also reading about state state tomography where it was said*

'On the other hand, the no-cloning theorem forbids one to create a perfect copy of the system without already knowing its state in advance. Thus, there is no way out, not even in principle, to infer the quantum state of a single system without having some prior knowledge on it [4]. It is possible to estimate the unknown quantum state of a system when many identical copies are available in the same state, so that a different measurement can be performed on each copy. A procedure of such kind is called quantum tomography.'

How can it be, if one cannot create a perfect copy of a state, one can create many identical copies that are in the same state?

Is it the case that you can (attempt to) prepare each state in the same way, but that each state will be slightly different from last?

*ADVANCES IN IMAGING AND ELECTRON PHYSICS, VOL. 128 Quantum Tomography G. MAURO D'ARIANO, MATTEO G. A. PARIS, and MASSIMILIANO F. SACCHI
 
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Irishdoug said:
How can it be, if one cannot create a perfect copy of a state, one can create many identical copies that are in the same state?
It means that if you have a quantum mechanical system, and you don't know anything about it, you cannot use it to make identical copies of it. But if you know how it was prepared in the first place, you can simply prepare many systems in the same way. You are not using the given system, you are using the same method of preparation.
 
Note also that tomography is in practice very often used to test just how well you are able to control/prepare your system.
That is, the assumption is then that the results of of subsequent measurements would end up being nominally identical if you were able to prepare your system with 100% fidelity.
 
Thanks to both of you.
 
The no-cloning theorem theorem says that an identical copy cannot be made while retaining the original. An identical copy of an arbitary quantum state can be made if the original state is destroyed, e.g., quantum teleportation (also involves passage of some classical information).
 
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Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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